In these notes we lay out the basic theory of the Fourier transform, which is of course the most fundamental tool in harmonic analysis and also of major importance in related fields (functional analysis, complex analysis, PDE, number theory, additive combinatorics, representation theory, signal processing, etc.). The Fourier transform, in conjunction with the Fourier inversion formula, allows one to take essentially arbitrary (complex-valued) functions on a group (or more generally, a space that acts on, e.g. a homogeneous space ), and decompose them as a (discrete or continuous) superposition of much more symmetric functions on the domain, such as characters ; the precise superposition is given by *Fourier coefficients* , which take values in some dual object such as the Pontryagin dual of . Characters behave in a very simple manner with respect to translation (indeed, they are eigenfunctions of the translation action), and so the Fourier transform tends to simplify any mathematical problem which enjoys a translation invariance symmetry (or an approximation to such a symmetry), and is somehow “linear” (i.e. it interacts nicely with superpositions). In particular, Fourier analytic methods are particularly useful for studying operations such as convolution and set-theoretic addition , or the closely related problem of counting solutions to additive problems such as or , where are constrained to lie in specific sets . The Fourier transform is also a particularly powerful tool for solving constant-coefficient linear ODE and PDE (because of the translation invariance), and can also approximately solve some variable-coefficient (or slightly non-linear) equations if the coefficients vary smoothly enough and the nonlinear terms are sufficiently tame.

The Fourier transform also provides an important new way of looking at a function , as it highlights the distribution of in *frequency space* (the domain of the frequency variable ) rather than *physical space* (the domain of the physical variable ). A given property of in the physical domain may be transformed to a rather different-looking property of in the frequency domain. For instance:

- Smoothness of in the physical domain corresponds to decay of in the Fourier domain, and conversely. (More generally, fine scale properties of tend to manifest themselves as coarse scale properties of , and conversely.)
- Convolution in the physical domain corresponds to pointwise multiplication in the Fourier domain, and conversely.
- Constant coefficient differential operators such as in the physical domain corresponds to multiplication by polynomials such as in the Fourier domain, and conversely.
- More generally, translation invariant operators in the physical domain correspond to multiplication by symbols in the Fourier domain, and conversely.
- Rescaling in the physical domain by an invertible linear transformation corresponds to an inverse (adjoint) rescaling in the Fourier domain.
- Restriction to a subspace (or subgroup) in the physical domain corresponds to projection to the dual quotient space (or quotient group) in the Fourier domain, and conversely.
- Frequency modulation in the physical domain corresponds to translation in the frequency domain, and conversely.

(We will make these statements more precise below.)

On the other hand, some operations in the physical domain remain essentially unchanged in the Fourier domain. Most importantly, the norm (or *energy*) of a function is the same as that of its Fourier transform, and more generally the inner product of two functions is the same as that of their Fourier transforms. Indeed, the Fourier transform is a unitary operator on (a fact which is variously known as the *Plancherel theorem* or the *Parseval identity*). This makes it easier to pass back and forth between the physical domain and frequency domain, so that one can combine techniques that are easy to execute in the physical domain with other techniques that are easy to execute in the frequency domain. (In fact, one can combine the physical and frequency domains together into a product domain known as phase space, and there are entire fields of mathematics (e.g. microlocal analysis, geometric quantisation, time-frequency analysis) devoted to performing analysis on these sorts of spaces directly, but this is beyond the scope of this course.)

In these notes, we briefly discuss the general theory of the Fourier transform, but will mainly focus on the two classical domains for Fourier analysis: the torus , and the Euclidean space . For these domains one has the advantage of being able to perform very explicit algebraic calculations, involving concrete functions such as plane waves or Gaussians .

** — 1. Generalities — **

Let us begin with some generalities. An abelian topological group is an abelian group with a topological structure, such that the group operations of addition and negation are continuous. (One can of course also consider abelian multiplicative groups , but to fix the notation we shall restrict attention to additive groups.) For technical reasons (and in particular, in order to apply many of the results from the previous quarter) it is convenient to restrict attention to abelian topological groups which are locally compact Hausdorff (LCH); these are known as locally compact abelian (LCA) groups.

Some basic examples of locally compact abelian groups are:

- Finite additive groups (with the discrete topology), such as cyclic groups .
- Finitely generated additive groups (with the discrete topology), such as the standard lattice .
- Tori, such as the standard -dimensional torus with the standard topology.
- Euclidean spaces, such the standard -dimensional Euclidean space (with the standard topology, of course).
- The rationals are
*not*locally compact with the usual topology; but if one uses the discrete topology instead, one recovers an LCA group. - Another example of an LCA group, of importance in number theory, is the adele ring , discussed in this earlier blog post.

Thus we see that locally compact abelian groups can be either discrete or continuous, and either compact or non-compact; all four combinations of these cases are of importance. The topology of course generates a Borel -algebra in the usual fashion, as well as a space of continuous, compactly supported complex-valued functions. There is a translation action of on , where for every , is the translation operation

LCA groups need not be -compact (think of the free abelian group on uncountably many generators, with the discrete topology), but one has the following useful substitute:

Exercise 1Show that every LCA group contains a -compact open subgroup , and in particular is the disjoint union of -compact sets. (Hint:Take a compact symmetric neighbourhood of the identity, and consider the group generated by this neighbourhood.)

An important notion for us will be that of a *Haar measure*: a Radon measure on which is *translation-invariant* (i.e. for all Borel sets and all , where is the translation of by ). From this and the definition of integration we see that integration against a Haar measure (an operation known as the *Haar integral*) is also translation-invariant, thus

for all and . The trivial measure is of course a Haar measure; all other Haar measures are called *non-trivial*.

Let us note some non-trivial Haar measures in the four basic examples of locally compact abelian groups:

- For a finite additive group , one can take either counting measure or normalised counting measure as a Haar measure. (The former measure emphasises the discrete nature of ; the latter measure emphasises the compact nature of .)
- For finitely generated additive groups such as , counting measure is a Haar measure.
- For the standard torus , one can obtain a Haar measure by identifying this torus with in the usual manner and then taking Lebesgue measure on the latter space. This Haar measure is a probability measure.
- For the standard Euclidean space , Lebesgue measure is a Haar measure.

Of course, any non-negative constant multiple of a Haar measure is again a Haar measure. The converse is also true:

Exercise 2 (Uniqueness of Haar measure up to scalars)Let be two non-trivial Haar measures on a locally compact abelian group . Show that are scalar multiples of each other, i.e. there exists a constant such that . (Hint: for any , compute the quantity in two different ways.)

The above argument also implies a useful symmetry property of Haar measures:

Exercise 3 (Haar measures are symmetric)Let be a Haar measure on a locally compact abelian group . Show that for all . (Hint: expand in two different ways.) Conclude that Haar measures on LCA groups aresymmetricin the sense that for all measurable , where is the reflection of .

Exercise 4 (Open sets have positive measure)Let be a non-trivial Haar measure on a locally compact abelian group . Show that for any non-empty open set . Conclude that if is non-negative and not identically zero, then .

Exercise 5Let be an LCA group with non-trivial Haar measure . Let be the space of equivalence classes of functions such that for every set of finite measure, the restriction of to lies in with a norm bounded uniformly in , with two functions in in the same equivalence class if they agree almost everywhere on every set of finite measure, and with the norm of equal to the supremum of the norms of the restrictions. (For -finite groups , is identical to , but the two spaces differ slightly in general.) Show that is identifiable with . (Unfortunately, is not always -finite, and so the standard duality theorem from Notes 3 of 245B does not directly apply. However, one can get around this using Exercise 1.)

It is a (not entirely trivial) theorem, due to André Weil, that all LCA groups have a non-trivial Haar measure. For discrete groups, one can of course take counting measure as a Haar measure. For compact groups, the result is due to Haar, and one can argue as follows:

Exercise 6 (Existence of Haar measure, compact case)Let be a compact metrisable abelian group. For any real-valued , and any Borel probability measure on , define theoscillationof with respect to to be the quantity .

- (a) Show that a Borel probability measure is a Haar measure if and only if for all .
- (b) If a sequence of Borel probability measures converges in the vague topology to another Borel probability measure , show that for all .
- (c) If is a Borel probability measure and is such that , show that there exists a Borel probability measure such that and for all . (
Hint:take to be the an average of certain translations of .)- (d) Given any finite number of functions , show that there exists a Borel probability measure such that for all . (
Hint:Use Prokhorov’s theorem, see Corollary 13 of 245B Notes 12. Try the case first.)- (e) Show that there exists a unique Haar probability measure on . (
Hint:One can identify each probability measure with the element of the product space , which is compact by Tychonoff’s theorem. Now use (d) and the finite intersection property.)(The argument can be adapted to the case when is not metrisable, but one has to replace the sequential compactness given by Prokhorov’s theorem by the topological compactness given by the Banach-Alaoglu theorem.)

For general LCA groups, the proof is more complicated:

Exercise 7 (Existence of Haar measure, general case)Let be an LCA group. Let denote the space of non-negative functions that are not identically zero. Given two , define a-coverof to be an expression of the form that pointwise dominates , where are non-negative numbers and . Let denote the infimum of the quantity for all -covers of .

- (a) (Finiteness) Show that for all .
- (b) Let is a Haar measure on . Show that for all . Conversely, for every and , and any neighbourhood of the identity, show that there exists supported in such that . (Hint: is uniformly continuous. Take to be an approximation to the identity.) Thus Haar integrals are related to certain renormalised versions of the functionals ; this observation underlies the strategy for construction of Haar measure in the rest of this exercise.
- (c) (Transitivity) Show that for all .
- (d) (Translation invariance) Show that for all and .
- (e) (Sublinearity) Show that and for all and .
- (f) (Approximate superadditivity) If and , show that there exists a neighbourhood of the identity such that whenever is supported in . (
Hint:are all uniformly continuous. Take a -cover of and multiply the weight at by weights such as and .)Next, fix a reference function , and define the functional for all by the formula .

- (g) Show that for any fixed , ranges in the compact interval ; thus can be viewed as an element of the product space , which is compact by Tychonoff’s theorem.
- (h) From (d), (e) we have the translation-invariance property , the homogeneity property , and the sub-additivity property for all , , and ; we also have the normalisation . Now show that for all and , there exists such that for all .
- (i) Show that there exists a unique Haar measure on with . (
Hint:Use (h) and the finite intersection property to obtain a translation-invariant positive linear functional on , then use the Riesz representation theorem.)

Now we come to a fundamental notion, that of a character.

Definition 8 (Characters)Let be a LCA group. Amultiplicative characteris a continuous function to the unit circle which is a homomorphism, i.e. for all . Anadditive characterorfrequencyis a continuous function which is a homomorphism, thus for all . The set of all frequencies is called thePontryagin dualof and is denoted ; it is clearly an abelian group. A multiplicative character is callednon-trivialif it is not the constant function ; an additive character is callednon-trivialif it is not the constant function .

Multiplicative characters and additive characters are clearly related: if is an additive character, then the function is a multiplicative character, and conversely every multiplicative character arises uniquely from an additive character in this fashion.

Exercise 9Let be an LCA group. We give the topology of local uniform convergence on compact sets, thus the topology on are generated by sets of the form . Show that this turns into an LCA group. (Hint:Show that for any neighbourhood of the identity in , the sets for (say) are compact.) Furthermore, if is discrete, show that is compact.

The Pontryagin dual can be computed easily for various classical LCA groups:

Exercise 10Let be an integer.

- (a) Show that the Pontryagin dual of is identifiable as an LCA group with , by identifying each with the frequency given by the dot product.
- (b) Show that the Pontryagin dual of is identifiable as an LCA group with , by identifying each with the frequency given by the dot product.
- (c) Show that the Pontryagin dual of is identifiable as an LCA group with , by identifying each with the frequency given by the dot product.
- (d) (Contravariant functoriality) If is a continuous homomorphism between LCA groups, show that there is a continuous homomorphism between their Pontryagin duals, defined by for and .
- (e) If is a closed subgroup of an LCA group (and is thus also LCA), show that is identifiable with , where is the space of all frequencies which annihilate (i.e. for all ).
- (f) If are LCA groups, show that is identifiable as an LCA group with .
- (g) Show that the Pontryagin dual of a finite abelian group is identifiable with itself. (
Hint:first do this for cyclic groups , identifying with the additive character ), then use the classification of finite abelian groups.) Note that this identification is not unique.

Exercise 11Let be an LCA group with non-trivial Haar measure , and let be a measurable function such that for almost every . Show that is equal almost everywhere to a multiplicative character of . (Hint:on the one hand, a.e. for almost every . On the other hand, depends continuously on in, say, the local topology.)

In the remainder of this section, is a fixed LCA group with a non-trivial Haar measure .

Given an absolutely integrable function , we define the *Fourier transform* by the formula

This is clearly a linear transformation, with the obvious bound

It converts translations into frequency modulations: indeed, one easily verifies that

for any , , and . Conversely, it converts frequency modulations to translations: one has

for any and , where is the multiplicative character .

Exercise 12 (Riemann-Lebesgue lemma)If , show that is continuous. Furthermore, show that goes to zero at infinity in the sense that for every there exists a compact subset of such that for . (Hint:First show that there exists a neighbourhood of the identity in such that (say) for all . Now take the Fourier transform of this fact.) Thus the Fourier transform maps continuously to , the space of continuous functions on which go to zero at infinity; the decay at infinity is known as the Riemann-Lebesgue lemma.

Exercise 13Let be an LCA group with non-trivial Haar measure . Show that the topology of is the weakest topology such that is continuous for every .

Given two , recall that the convolution is defined as

From Young’s inequality (Exercise 25 of Notes 1) we know that is defined a.e., and lies in ; indeed, we have

Exercise 14Show that the operation is a bilinear, continuous, commutative, and associative operation on . As a consequence, the Banach space with the convolution operation as “multiplication” operation becomes a commutative Banach algebra. If we also define for all , this turns into a Banach -algebra.for all ; thus the Fourier transform converts convolution to pointwise product.

Exercise 15Let be LCA groups with non-trivial Haar measures respectively, and let , . Show that the tensor product (with product Haar measure ) has a Fourier transform of , where we identify with as per Exercise 10(f). Informally, this exercise asserts that the Fourier transform commutes with tensor products. (Because of this fact, the tensor power trick is often available when proving results about the Fourier transform on general groups.)

Exercise 16 (Convolution and Fourier transform of measures)If is a finite Radon measure on an LCA group with non-trivial Haar measure , define theFourier-Stieltjes transformby the formula (thus for instance for any ). Show that is a bounded continuous function on . Given any , define the convolution to be the functionand given any finite Radon measure , let be the measure

Show that and for all , and similarly that is a finite measure and for all . Thus the convolution and Fourier structure on can be extended to the larger space of finite Radon measures.

** — 2. The Fourier transform on compact abelian groups — **

In this section we specialise the Fourier transform to the case when the locally compact group is in fact compact, thus we now have a compact abelian group with non-trivial Haar measure . This case includes that of finite groups, together with that of the tori .

As is a Radon measure, compact groups have finite measure. It is then convenient to normalise the Haar measure so that , thus is now a probability measure. For the remainder of this section, we will assume that is a compact abelian group and is its (unique) Haar probability measure, as given by Exercise 6.

A key advantage of working in the compact setting is that multiplicative characters now lie in and . In particular, they can be integrated:

Lemma 17Let be a multiplicative character. Then equals when is trivial and when is non-trivial. Equivalently, for , we have , where is the Kronecker delta function at .

*Proof:* The claim is clear when is trivial. When is non-trivial, there exists such that . If one then integrates the identity using (2) one obtains the claim.

Exercise 18Show that the Pontryagin dual of a compact abelian group is discrete (compare with Exercise 9).

Exercise 19Show that the Fourier transform of the constant function is the Kronecker delta function at . More generally, for any , show that the Fourier transform of the multiplicative character is the Kronecker delta function at .

Since the pointwise product of two multiplicative characters is again a multiplicative character, and the conjugate of a multiplicative character is also a multiplicative character, we obtain

Corollary 20The space of multiplicative chararacters is an orthonormal set in the complex Hilbert space .

Actually, one can say more:

Theorem 21 (Plancherel theorem for compact abelian groups)Let be a compact abelian group with probability Haar measure . Then the space of multiplicative characters is an orthonormalbasisfor the complex Hilbert space .

The full proof of this theorem will have to wait until later notes, once we have developed the spectral theorem, though see Exercise 56 below. However, we can work out some important special cases here.

- When is a torus , the multiplicative characters separate points (given any two , there exists a character which takes different values at and at ). The space of finite linear combinations of multiplicative characters (i.e. the space of trigonometric polynomials) is then an algebra closed under conjugation that separates points and contains the unit , and thus by the Stone-Weierstrass theorem, is dense in in the uniform (and hence in ) topology, and is thus dense in (in the topology) also.
- The same argument works when is a cyclic group , using the multiplicative characters for . As every finite abelian group is isomorphic to the product of cyclic groups, we also obtain the claim for finite abelian groups.
- Alternatively, when is finite, one can argue by viewing the linear operators as unitary matrices (in fact, they are permutation matrices) for each . The spectral theorem for unitary matrices allows each of these matrices to be diagonalised; as is abelian, the matrices commute and so one can
*simultaneously*diagonalise these matrices. It is not hard to see that each simultaneous eigenvector of these matrices is a multiple of a character, and so the characters span , yielding the claim. (The same argument will in fact work for arbitrary compact abelian groups, once we obtain the spectral theorem for unitary operators.)

If , the inner product of with any multiplicative character is just the Fourier coefficient of at the corresponding frequency. Applying the general theory of orthonormal bases (see Notes 5), we obtain the following consequences:

Corollary 22 (Plancherel theorem for compact abelian groups, again)Let be a compact abelian group with probability Haar measure .

- (Parseval identity) For any , we have .
- (Parseval identity, II) For any , we have .
- (Unitarity) Thus the Fourier transform is a unitary transformation from to .
- (Inversion formula) For any , the series converges unconditionally in to .
- (Inversion formula, II) For any sequence in , the series converges unconditionally in to a function with as its Fourier coefficients.

We can record here a textbook application of the Riesz-Thorin interpolation theorem from the previous notes. Observe that the Fourier transform map maps to with norm , and also trivially maps to with norm . Applying the interpolation theorem, we conclude the *Hausdorff-Young inequality*

for all and all ; in particular, the Fourier transform maps to , where is the dual exponent of , thus . It is remarkably difficult (though not impossible) to establish the inequality (6) without the aid of the Riesz-Thorin theorem. (For instance, one could use the Marcinkiewicz interpolation theorem combined with the tensor power trick.) The constant cannot be improved, as can be seen by testing (6) with the function and using Exercise 19. By combining (6) with Hölder’s inequality, one concludes that

whenever and . These are the optimal hypotheses on for which (14) holds, though we will not establish this fact here.

Exercise 23If , show that the Fourier transform of is given by the formulaThus multiplication is converted via the Fourier transform to convolution; compare this with (5).

Exercise 24 (Hardy-Littlewood majorant property)Let be an even integer. If are such that for all (in particular, is non-negative), show that . (Hint: use Exercise 23 and the Plancherel identity.) The claim fails for all other values of , a result of Fournier.

Exercise 25In this exercise and the next two, we will work on the torus with the probability Haar measure . The Pontryagin dual is identified with in the usual manner, thus for all . For every integer and , define thepartial Fourier seriesto be the expression

- Show that , where is the Dirichlet kernel .
- Show that for some absolute constant . Conclude that the operator norm of on (with the uniform norm) is at least .
- Conclude that there exists a continuous function such that the partial Fourier series does not converge uniformly. (
Hint:use the uniform boundedness principle.) This is despite the fact that must converge to in norm, by the Plancherel theorem. (Another example of non-uniform convergence of is given by the Gibbs phenomenon.)

Exercise 26We continue the notational conventions of the preceding exercise. For every integer and , define theCésaro-summed partial Fourier seriesto be the expression

- Show that , where is the Fejér kernel .
- Show that . (Hint: what is the Fourier coefficient of at zero?)
- Show that converges uniformly to for every . (Thus we see that Césaro averaging improves the convergence properties of Fourier series.)

Exercise 27Carleson’s inequalityasserts that for any , one has the weak-type inequalityfor some absolute constant . Assuming this (deep) inequality, establish Carleson’s theorem that for any , the partial Fourier series converge for almost every to . (Conversely, a general principle of Stein, analogous to the uniform boundedness principle, allows one to deduce Carleson’s inequality from Carleson’s theorem. A later result of Hunt extends Carleson’s theorem to for any , but a famous example of Kolmogorov shows that almost everywhere convergence can fail for functions; in fact the series may diverge pointwise everywhere.)

** — 3. The Fourier transform on Euclidean spaces — **

We now turn to the Fourier transform on the Euclidean space , where is a fixed integer. From Exercise 10 we can identify the Pontryagin dual of with itself, and then the Fourier transform of a function is given by the formula

Remark 28One needs the Euclidean inner product structure on in order to identify with . Without this structure, it is more natural to identify with the dual space of . (In the language of physics, one should interpret frequency as a covector rather than a vector.) However, we will not need to consider such subtleties here. In other areas of mathematics than harmonic analysis, the normalisation of the Fourier transform (particularly with regard to the positioning of the sign and the factor ) is sometimes slightly different from that presented here. For instance, in PDE, the factor of is often omitted from the exponent in order to slightly simplify the behaviour of differential operators under the Fourier transform (at the cost of introducing factors of in various identities, such as the Plancherel formula or inversion formula).

In Exercise 12 we saw that if was in , then was continuous and decayed to zero at infinity. One can improve both the regularity and decay on by strengthening the hypotheses on . We need two basic facts:

Exercise 29 (Decay transforms to regularity)Let , and suppose that both lie in , where is the coordinate function. Show that is continuously differentiable in the variable, with(

Hint:The main difficulty is to justify differentiation under the integral sign. Use the fact that the function has a derivative of magnitude , and is hence Lipschitz by the fundamental theorem of calculus. Alternatively, one can show first that is the indefinite integral of and then use the fundamental theorem of calculus.)

Exercise 30 (Regularity transforms to decay)Let , and suppose that has a derivative in , for which one has the fundamental theorem of calculusfor almost every . (This is equivalent to being absolutely continuous in for almost every .) Show that

In particular, conclude that goes to zero as .

Remark 31Exercise 30 shows thatFourier transforms diagonalise differentiation: (constant-coefficient) differential operators such as , when viewed in frequency space, become nothing more than multiplication operators . (Multiplication operators are the continuous analogue of diagonal matrices.) It is because of this fact that the Fourier transform is extremely useful in PDE, particularly in constant-coefficient linear PDE, or perturbations thereof.

It is now convenient to work with a class of functions which has an infinite amount of both regularity and decay.

Definition 32 (Schwartz class)Arapidly decreasing functionis a measurable function such that is bounded for every non-negative integer . A Schwartz function is a smooth function such that all derivatives are rapidly decreasing. The space of all Schwartz functions is denoted .

Example 33Any smooth, compactly supported function is a Schwartz function. The gaussian functions

Exercise 34Show that the seminormsfor , where we think of as a -dimensional vector (or, if one wishes, a rank -dimensional tensor), give the structure of a Fréchet space. In particular, is a topological vector space.

Clearly, every Schwartz function is both smooth and rapidly decreasing. The following exercise explores the converse:

Exercise 35

- Give an example to show that not all smooth, rapidly decreasing functions are Schwartz.
- Show that if is a smooth, rapidly decreasing function, and all derivatives of are bounded, then is Schwartz. (Hint: use Taylor’s theorem with remainder.)

One of the reasons why the Schwartz space is convenient to work with is that it is closed under a wide variety of operations. For instance, the derivative of a Schwartz function is again a Schwartz function, and that the product of a Schwartz function with a polynomial is again a Schwartz function. Here are some further such closure properties:

Exercise 36Show that the product of two Schwartz functions is again a Schwartz function. Moreover, show that the product map is continuous from to .

Exercise 37Show that the convolution of two Schwartz functions is again a Schwartz function. Moreover, show that the convolution map is continuous from to .

Exercise 38Show that the Fourier transform of a Schwartz function is again a Schwartz function. Moreover, show that the Fourier transform map is continuous from to .

The other important property of the Schwartz class is that it is dense in many other spaces:

Exercise 39Show that is dense in for every , and is also dense in (with the uniform topology). (Hint: one can either use the Stone-Weierstrass theorem, or convolutions with approximations to the identity.)

Because of this density property, it becomes possible to establish various estimates and identities in spaces of rough functions (e.g. functions) by first establishing these estimates on Schwartz functions (where it is easy to justify operations such as differentiation under the integral sign) and then taking limits.

Having defined the Fourier transform , we now introduce the *adjoint Fourier transform* by the formula

(note the sign change from (8)). We will shortly demonstrate that the adjoint Fourier transform is also the inverse Fourier transform: .

we see that obeys much the same propeties as ; for instance, it is also continuous from to . It is also the adjoint to in the sense that

for all .

Now we show that inverts . We begin with an easy preliminary result:

Next, we perform a computation:

Exercise 41 (Fourier transform of Gaussians)Let . Show that the Fourier transform of the gaussian function is . (Hint:Reduce to the case and , then complete the square and use contour integration and the classical identity .) Conclude that .

Exercise 42With as in the previous exercise, show that converges in the Schwartz space topology to as for all . (Hint:First show convergence in the uniform topology, then use the identities and for .)

From Exercises 40, 41 we see that

for all and . Taking limits as using Exercises 38, 42 we conclude that

for all , or in other words we have the Fourier inversion formula

for all . From (10) we also have

Taking inner products with another Schwartz function , we obtain *Parseval’s identity*

for all , and similarly for . In particular, we obtain *Plancherel’s identity*

for all . We conclude that

Theorem 43 (Plancherel’s theorem for )The Fourier transform operator can be uniquely extended to a unitary transformation .

Exercise 44Show that the Fourier transform on given by Plancherel’s theorem agrees with the Fourier transform on given by (8) on the common domain . Thus we may define for or (or even without any ambiguity (other than the usual identification of any two functions that agree almost everywhere).

Note that it is certainly possible for a function to lie in but not in (e.g. the function ). In such cases, the integrand in (8) is not absolutely integrable, and so this formula does not define the Fourier transform of directly. Nevertheless, one can recover the Fourier transform via a limiting version of (8):

Exercise 45Let . Show that the partial Fourier integrals converge in to as .

Remark 46It is a famous open question whether the partial Fourier integrals of an function also converge pointwise almost everywhere for . For , this is essentially the celebrated theorem of Carleson mentioned in Exercise 27.

Exercise 47 (Heisenberg uncertainty principle)Let . Define theposition operatorandmomentum operatorby the formulaeand the formal self-adjointness relationships

and then establish the inequality

(

Hint:start with the obvious inequality for real numbers , and optimise in and .) If , deduce theHeisenberg uncertainty principlefor any . (

Hint:one can use the translation and modulation symmetries (3), (4) of the Fourier transform to reduce to the case .) Classify precisely the for which equality occurs.

Remark 48For and , define thegaussian wave packetby the formulaThese wave packets are normalised to have norm one, and their Fourier transform is given by

Informally, is localised to the region in physical space, and to the region in frequency space; observe that this is consistent with the uncertainty principle. These packets “almost diagonalise” the position and momentum operators in the sense that (taking for simplicity)

(where the errors terms are morally of the form and respectively). Of course, the non-commutativity of and as evidenced by the last equation in (12) shows that exact diagonalisation is impossible. Nevertheless it is useful, at an intuitive level at least, to view these wave-packets as a sort of (overdetermined) basis for that approximately diagonalises and (as well as other formal combinations of these operators, such as differential operators or pseudodifferential operators). Meanwhile, the Fourier transform morally maps the point in phase space to , as evidenced by (13) or (12); it is the model example of the more general class of Fourier integral operators, which morally move points in phase space around by canonical transformations. The study of these types of objects (which are of importance in linear PDE) is known as

microlocal analysis, and is beyond the scope of this course.

The proof of the Hausdorff-Young inequality (6) carries over to the Euclidean space setting, and gives

for all and all ; in particular the Fourier transform is bounded from to . The constant of on the right-hand side of (14) turns out to not be optimal in the Euclidean setting, in contrast to the compact setting; the sharp constant is in fact , a result of Beckner. (The fact that this constant cannot be improved can be seen by using the gaussians from Exercise 41.)

Exercise 49 (Entropy uncertainty principle)For any with , show that(

Hint:differentiate (!) (14) in at , where one has equality in (14).) Using Beckner’s improvement to (6), improve the right-hand side to the optimal value of .

Exercise 50 (Fourier transform under linear changes of variable)Let be an invertible linear transformation. If and , show that the Fourier transform of is given by the formulawhere is the adjoint operator to . Verify that this transformation is consistent with (14), and indeed shows that the exponent on the left-hand side cannot be replaced by any other exponent. (One can also establish this latter claim by dimensional analysis.)

Remark 51As a corollary of Exercise 50, observe that if is spherically symmetric (thus for all rotation matrices ) then is spherically symmetric also.

Exercise 52 (Fourier transform intertwines restriction and projection)Let , and let . We express as in the obvious manner.

- (Restriction becomes projection) If is the restriction of to , show that for all .
- (Projection becomes restriction) If is the projection of to , show that for all .

Exercise 53 (Fourier transform on large tori)Let , and let be the torus of length with Lebesgue measure (thus the total measure of this torus is . We identify the Pontryagin dual of this torus with in the usual manner, thus we have the Fourier coefficientsfor all and .

- Show that for any , the Fourier series converges unconditionally in .
- Use this to give an alternate proof of the Fourier inversion formula (11) in the case where is smooth and compactly supported.

Exercise 54 (Poisson summation formula)Let . Show that the function defined by has Fourier transform for all (note the two different Fourier transforms in play here). Conclude the Poisson summation formula

Exercise 55Let be a compactly supported, absolutely integrable function. Show that the function is real-analytic. Conclude that it is not possible to find a non-trivial such that and are both compactly supported.

** — 4. The Fourier transform on general groups (optional) — **

The field of *abstract harmonic analysis* is concerned, among other things, with extensions of the above theory to more general groups, for instance arbitrary LCA groups. One of the ways to proceed is via Gelfand theory, which for instance can be used to show that the Fourier transform is at least injective:

Exercise 56 (Fourier analysis via Gelfand theory)(Optional) In this exercise we use the Gelfand theory of commutative Banach *-algebras (see 245B Notes 12) to establish some basic facts of Fourier analysis in general groups. Let be an LCA group. We view as a commutative Banach *-algebra (see Exercise 14).

- (a) If is such that , where is the convolution of copies of , show that there exists a non-zero complex number such that the map is not invertible on . (
Hint:If contains a unit, one can use Exercise 35 of 245B Notes 12; otherwise, adjoin a unit.)- (b) If and are as in (a), show that there exists a character (in the sense of Banach *-algebras, see Definition 16 of 245B Notes 12) such that lies in the kernel of for all . Conclude in particular that is non-zero.
- (c) If is a character, show that there exists a multiplicative character such that for all . (You will need Exercise 5 and Exercise 11.)
- (d) For any and , show that , where is the group identity and is the conjugate of . (
Hint:the inner product is positive semi-definite.)- (e) Show that if is not identically zero, then there exists such that . (
Hint:first find such that and , and conclude using (d) repeatedly that . Then use (a), (b), (c).) Conclude that the Fourier transform is injective on . (The image of under the Fourier transform is then a Banach *-algebra known as theWiener algebra, and is denoted .)- (f) Prove Theorem 21.

It is possible to use arguments similar to those in Exercise 56 to characterise positive measures on in terms of continuous functions on , leading to Bochner’s theorem:

Theorem 57 (Bochner’s theorem)Let be a continuous function on an LCA group . Then the following are equivalent:

- (a) for all and .
- (b) There exists a non-negative finite Radon measure on such that .

Functions obeying either (a) or (b) are known as *positive-definite functions*. The space of such functions is denoted .

Exercise 58Show that (b) implies (a) in Bochner’s theorem. (The converse implication is significantly harder, reprising much of the machinery in Exercise 56, but with taking the place of : see Rudin’s book for details.)

Using Bochner’s theorem, it is possible to show

Theorem 59 (Plancherel’s theorem for LCA groups)Let be an LCA group with non-trivial Haar measure . Then there exists a non-trivial Haar measure on such that the Fourier transform on can be extended continuously to a unitary transformation from to . In particular we have the Plancherel identityfor all , and the Parseval identity

for all . Furthermore, the inversion formula

is valid for in a dense subclass of (in particular, it is valid for ).

Again, see Rudin’s book for details. A related result is that of Pontryagin duality: if is the Pontryagin dual of an LCA group , then is the Pontryagin dual of . (Certainly, every element defines a character on , thus embedding into via the Gelfand transform; the non-trivial fact is that this embedding is in fact surjective.) One can use Pontryagin duality to convert various properties of LCA groups into other properties on LCA groups. For instance, we have already seen that is compact (resp. discrete) if is discrete (resp. compact); with Pontryagin duality, the implications can now also be reversed. As another example, one can show that is connected (resp. torsion-free) if and only if is torsion-free (resp. connected). We will not prove these assertions here.

It is natural to ask what happens for non-abelian locally compact groups . One can still build non-trivial Haar measures (the proof sketched out in Exercise 7 extends without difficulty to the non-abelian setting), though one must now distinguish between left-invariant and right-invariant Haar measures. (The two notions are equivalent for some classes of groups, notably compact groups, but not in general. Groups for which the two notions of Haar measures coincide are called *unimodular*.) However, when is non-abelian then there are not enough multiplicative characters to have a satisfactory Fourier analysis. (Indeed, such characters must annihilate the commutator group , and it is entirely possible for this commutator group to be all of , e.g. if is simple and non-abelian.) Instead, one must generalise the notion of a multiplicative character to that of a *unitary representation* from to the group of unitary transformations on a complex Hilbert space ; thus the Fourier coefficients of a function will now be operators on thisl Hilbert space , rather than complex numbers. When is a compact group, it turns out to be possible to restrict attention to finite-dimensional representations (thus one can replace by the matrix group for some ). The analogue of the Pontryagin dual is then the collection of (irreducible) finite-dimensional unitary representations of , up to isomorphism. There is an analogue of the Plancherel theorem in this setting, closely related to the Peter-Weyl theorem in representation theory. We will not discuss these topics here, but refer the reader instead to any representation theory text.

The situation for non-compact non-abelian groups (e.g. ) is significantly more subtle, as one must now consider infinite-dimensional representations as well as finite-dimensional ones, and the inversion formula can become quite non-trivial (one has to decide what “weight” each representation should be assigned in that formula). At this point it seems unprofitable to work in the category of locally compact groups, and specialise to a more structured class of groups, e.g. algebraic groups. The representation theory of such groups is a massive subject and well beyond the scope of this course.

** — 5. Relatives of the Fourier transform (optional) — **

There are a number of other Fourier-like transforms used in mathematics, which we will briefly survey here. Firstly, there are some rather trivial modifications one can make to the definition of Fourier transform, for instance by replacing the complex exponential by trigonometric functions such as and , or moving around the various factors of , , , etc. in the definition. In this spirit, we have the Laplace transform

of a measurable function with some reasonable growth at infinity, where . Roughly speaking, the Laplace transform is “the Fourier transform without the ” (cf. Wick rotation), and so has the (mild) advantage of being definable in the realm of real-valued functions rather than complex-valued functions. It is particularly well suited for studying ODE on the half-line (e.g. initial value problems for a finite-dimensional system). The Laplace transform and Fourier transform can be unified by allowing the parameter in (15) to vary in the right-half plane .

When the Fourier transform is applied to a spherically symmetric function on , then the Fourier transform is also spherically symmetric, given by the formula where is the Fourier-Bessel transform (or *Hankel transform*)

where is the Bessel function of the first kind with index . In practice, one can then analyse the Fourier-analytic behaviour of spherically symmetric functions in terms of one-dimensional Fourier-like integrals by using various asymptotic expansions of the Bessel function.

There is a relationship between the -dimensional Fourier transform and the one-dimensional Fourier transform, provided by the Radon transform, defined for (say) by the formula

where , , and the integration is with respect to -dimensional measure. Indeed one checks that the -dimensional Fourier transform of at for some and is nothing more than the one-dimensional Fourier coefficient of the function at . The Radon transform is often used in scattering theory and related areas of analysis, geometry, and physics.

In analytic number theory, a multiplicative version of the Fourier-Laplace transform is often used, namely the Mellin transform

(Note that is a Haar measure for the *multiplicative* group .) To see the relation with the Fourier-Laplace transform, write , then the Mellin transform becomes

Many functions of importance in analytic number theory, such as the Gamma function or the zeta function, can be expressed neatly in terms of Mellin transforms.

In electrical engineering and signal processing, the z-transform is often used, transforming a sequence of complex numbers to a formal Laurent series

(some authors use instead of here). If one makes the substitution then this becomes a (formal) Fourier series expansion on the unit circle. If the sequence is restricted to only be non-zero for non-negative , and does not grow too quickly as , then the -transform becomes holomorphic on the unit disk, thus providing a link between Fourier analysis and complex analysis. For instance, the standard formula

for the Taylor coefficients of a holomorphic function at the origin can be viewed as a version of the Fourier inversion formula for the torus . Just as the Fourier or Laplace transforms are useful for analysing differential equations in continuous settings, the -transform is useful for analysing difference equations in discrete settings. The -transform is of course also very similar to the method of generating functions in combinatorics and probability.

In probability theory one also considers the characteristic function of a real-valued random variable ; this is essentially the Fourier transform of the probability distribution of . Just as the Fourier transform is useful for understanding convolutions , the characteristic function is useful for understanding sums of independent random variables.

We have briefly touched upon the role of Gelfand theory in the general theory of the Fourier transform. Indeed, one can view the Fourier transform as the special case of the Gelfand transform for Banach *-algebras, which we already discussed in 245B Notes 12.

The Fast Fourier Transform (FFT) is not, strictly speaking, a variant of the Fourier transform, but rather is an efficient algorithm for *computing* the Fourier transform

on a cyclic group , when is large but composite. Note that a brute force computation of this transform for all values of would require about addition and multiplication operations. The FFT algorithm, in contrast, takes only operations, and is based on reducing the FFT for a large to FFT for smaller . For instance, suppose is even, say , then observe that

where are the functions . Thus one can obtain the Fourier transform of the length vector from the Fourier transforms of the two length vectors after about operations. Iterating this we see that we can indeed compute in operations, at least in the model case when is a power of two; the general case has a similar but more complicated analysis.

In many situations (particularly in ergodic theory), it is desirable not to perform Fourier analysis on a group directly, but instead on another space that acts on. Suppose for instance that is a compact abelian group, with probability Haar measure , which acts in a measure-preserving (and measurable) fashion on a probability space . Then one can decompose any into Fourier components , where , where the series is unconditionally convergent in . The reason for doing this is that each of the behaves in a simple way with respect to the group action, indeed one has for (almost) all . This decomposition is closely related to the decomposition in representation theory of a given representation into irreducible components. Perhaps the most basic example of this type of operation is the decomposition of a function into even and odd components , ; here the underlying group is , which acts on by reflections, .

The operation of converting a square matrix of numbers into eigenvalues or singular values can be viewed as a sort of non-commutative generalisation of the Fourier transform. (Note that the eigenvalues of a circulant matrix are essentially the Fourier coefficients of the first row of that matrix.) For instance, the identity can be viewed as a variant of the Plancherel identity. More generally, there are close relationships between spectral theory and Fourier analysis (as one can already see from the connection to Gelfand theory). For instance, in and , one can view Fourier analysis as the spectral theory of the gradient operator (note that the characters are joint eigenfunctions of ). As the gradient operator is closely related to the Laplacian , it is not surprising that Fourier analysis is also closely related to the spectral theory of the Laplacian, and in particular to various operators built using the Laplacian (e.g. resolvents, heat kernels, wave operators, Schrödinger operators, Littlewood-Paley projections, etc.). Indeed, the spectral theory of the Laplacian can serve as a partial substitute for the Fourier transform in situations in which there is not enough symmetry to exploit Fourier-analytic techniques (e.g. on a manifold with no translation symmetries).

Finally, there is an analogue of the Fourier duality relationship between an LCA group and its Pontryagin dual in algebraic geometry, known as the Fourier-Mukai transform, which relates an abelian variety to its dual , and transforms coherent sheaves on the former to coherent sheaves on the latter. This transform obeys many of the algebraic identities that the Fourier transform does, although it does not seem to have much of the analytic structure.

## 160 comments

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6 April, 2009 at 4:29 pm

AnonymousDear Prof. Tao,

In exercise 30, do you mean that Schwartz class is dense in continuous compactly supported functions? I think this is not true.

thanks

6 April, 2009 at 4:32 pm

Anonymousin equation (10), on the right hand side, is it f or F? I think it should be F.

6 April, 2009 at 4:39 pm

Anonymousin (14), the left hand side is it ?

6 April, 2009 at 6:30 pm

Terence TaoThanks for the corrections! The Schwartz class is dense both in (the space of continuous functions going to zero at infinity), as well as in the subspace (compactly supported continuous functions) in the uniform topology, although in the latter case one of course has to intersect the Schwartz space with (which gives , the space of smooth compactly supported functions).

6 April, 2009 at 10:17 pm

timurThanks for the notes!

In the FFT paragraph, probably you meant “operations” in “… takes only algorithms”.

[Corrected, thanks – T.]7 April, 2009 at 12:14 am

Anonymouswonderful notes. Just a small remark: the Latin plural form of “torus” is “tori”, not “torii” :)

[Corrected, thanks – T.]7 April, 2009 at 6:56 am

HunterThese notes are a pleasure to read. The characterization of the topology on G hat in exercise 8 as generated by certain sets seems to have a few missing characters? Also, the second G in exercise 11 should have a hat.

[Corrected, thanks – T.]7 April, 2009 at 7:55 am

mfrascaIn exercise 8 and toward the end of the notes there seem to be a couple of formulas that do not parse.

Marco

[Corrected, thanks – T.]8 April, 2009 at 6:07 pm

Qiaochu YuanThere’s a strong superficial similarity between the Fourier transform and its inverse and similar inverse pairs in combinatorics, e.g. the binomial transforms, the Stirling transforms, and Mobius inversion. If the Fourier transform is essentially group-theoretic and Mobius inversion is essentially poset-theoretic, how do we unify the two approaches? In more concrete terms:

– In what sense is the Fourier transform a use of the principle of inclusion-exclusion, or

– In what sense is the principle of inclusion-exclusion a Fourier transform?

9 April, 2009 at 7:36 am

Terence TaoDear Qiaochu,

I think they are not quite the same thing; the Fourier transform of a function of a spatial variable becomes a function in a dual frequency variable, whereas partial summation and Mobius inversion (or related concepts, such as inclusion-exclusion) stay exclusively in the spatial domain. The poset-theoretic concepts seem to be more related to the operations of integration and differentiation, or convolution and deconvolution, than to the Fourier transform – thus they are more like Fourier

multipliersthan Fouriertransforms.I don’t see a way to properly formalise this analogy, though – Fourier analysis really requires some sort of underlying translation symmetry to be effective, and I don’t see that sort of symmetry in the poset world.

12 April, 2009 at 7:45 pm

AnonymousDear Prof. Tao,

in the paragraph after corollary 5, G in little l2 and l(infinity) should be G^.

in your notes is p’ always conjugate of p?

thanks

[Corrected and clarified, thanks, – T.]17 April, 2009 at 12:09 am

josephHi, I’m sorry to bother you. My major is math and I ‘m also interested in physics. But my time is not adequate for learning two subjects. So what physics modules are useful to be a mathematician? Esp. studying analysis or geometry? Thank you.

17 April, 2009 at 10:48 am

maxbaroiI believe there’s a typo in 9a. I think you wanted to say to identiy each

[Corrected, thanks – T]17 April, 2009 at 6:07 pm

AnonymousAlso, I just went on your UCLA site, and you listed in your previous teaching section the link text says you taught 254B not 245B in Winter 09.

[Corrected, thanks – T]18 April, 2009 at 1:10 pm

anonymous joeNice article! Could you also post something on “Eigen Vectors” in a future article perhaps?

18 April, 2009 at 1:48 pm

Terence TaoDear joe,

I am not planning to teach linear algebra in the near future, but you can have a look at my lecture notes on the topic from 2002 at

http://www.math.ucla.edu/~tao/resource/general/115a.3.02f/

18 April, 2009 at 2:38 pm

AnonymousHi Prof. Tao,

What are you going to teach just after that course? If you announce it before, we can read some books and prepare ourself.

thanks

18 April, 2009 at 6:30 pm

Terence TaoMy next class is going to be in Winter of 2010. It is a topics course in analysis; I have not yet decided what I will lecture on, but I am thinking to do something involving random matrices.

18 April, 2009 at 7:29 pm

AnonymousThank you very much Prof. Tao.

Would you please recommend any book or lecture notes on it ?

19 April, 2009 at 7:31 am

josephHi,

I am curious that can two cont. functions have the same divergent Fourier series? Thank you.

19 April, 2009 at 10:23 am

anonymousJoseph,

By subtracting the two Fourier series, one can reformulate the question as; can a continuous (or more generally L^1) function have an identically vanishing Fourier series? While it’s true that the partial sums of the Fourier expansion of a L^1 function can behave badly (for example, diverge everywhere), we can “fix” this in several ways. For example, if F_n is the Fejer kernel (see http://en.wikipedia.org/wiki/Fejer_kernel), then F_n * f converges to f in L^1 norm (as well as almost everywhere). It is easy to see from this that if the Fourier series of f is identically zero, f must be identically zero.

19 April, 2009 at 12:40 pm

anonymousThe previous comments, and exercise 22, made me think of the following question. In several of your papers you briefly mention the problem of establishing pointwise convergence for , (i.e. Carleson’s theorem in 2 and more dimensions). Of course, this is an endpoint case of the maximal Bochner-Riesz maximal conjecture. In fact, you touch on this problem in both your work on the previously mentioned conjecture, as well as your work on multilinear operators. I’ve often been curious, why does this problem lie out of the reach of current technology? It seems the issues here will be different than, say, that in the tri-linear Hilbert transform, since there is no quadratic modulation invariance.

19 April, 2009 at 2:03 pm

Terence TaoDear anonymous @ 12:40pm,

Perhaps the best place where this issue is discussed is Lacey’s survey paper at http://arxiv.org/abs/math/0307008 . The time-frequency technology that is used to (say) reduce the bilinear Hilbert transform to that of bounding certain paraproducts associated to “trees”, seems to similarly reduce the problem of two-dimensional pointwise convergence to that of understanding (more complicated variants of) the lacunary maximal operator

,

i.e. the supremum of a lacunary family of disk multipliers. In order to control this, it is not enough to know the size (e.g. in $L^p$) of each individual disk multiplier; one must also know how each multiplier is distributed (the enemy is if they are each large in different regions of space, thus causing the sup to be enormous). Unfortunately our Bochner-Riesz technology is still quite poor at controlling the spatial distribution of disk multipliers (there are some conjectures in this direction, most notably a weighted conjecture of Stein, but progress is still very partial here, and also there are a variety of logarithmic losses which could be quite fatal when applying these conjectured estimates to the Carleson problem). But the problem may not be totally out of reach, and it is conceivable that finer endpoint control on the size and distribution of Bochner-Riesz and Kakeya type operators will help.

Separately to this, there is also a technical combinatorial problem, which is that the combinatorics of organising time-frequency tiles into trees is not well understood in higher dimensions, in which tiles now have some eccentricity in both physical space and frequency space. There are however some advances in understanding how to organise eccentric tiles, in particular in the recent work of Lie on the quadratic one-dimensional Carleson operator, and of Lacey and Li on the vector fields differentiation problem. So this may be more of a technical issue than a fundamental one.

19 April, 2009 at 3:56 pm

245C, Notes 3: Distributions « What’s new[…] is not a test function, but is instead just a Schwartz function. (Indeed, by Exercise 46 of Notes 2, it is not possible to find a non-trivial test function whose Fourier transform is again a test […]

20 April, 2009 at 7:41 am

ObsessiveMathsFreakDespite its great usefulness, the Fast Fourier transform is not without cost. The terrible unspoken secret of the FFT is that it does not in fact compute or approximate the FT of the function. Rather, it computes the Fourier transform of a quite different, albeit closely related, function.

The nature of this relation is a quite complicated series of delta function multiplications, truncations, cutoffs and convolutions. The book “The fast Fourier transform and its applications” by E. Brigham, has a very good description of the relationship between the discrete and continuous Fourier transforms, with pictures!

While this may not seem very serious, it becomes so when taking the Fourier transforms of multivariate functions. When moving to two dimensions, the implicit truncations, convolutions and multiplications in the FFT process will introduce fairly serious artefact’s in the computed FFT that can make it fairly unusable for analytical purposes. You’re often a lot better off computing the continuous Fourier transform numerically in some case(This is in fact feasible on modern desktops).

Speaking for myself, I can say that discovery of this fact was a cause of some dismay.

This is all aside form the fact that the standard FFT is rather awkward to use when it comes to analysis, as it ignores the normal and frequency domains of the problem, treating the transform as a transformation of one flat array to another, making it quite difficult to translate the produced transforms into something with analytical meaning.

So the FFT, while undoubtedly necessary in many areas, is not quite sufficient in others.

21 April, 2009 at 3:52 pm

Manuel MoreiraDear Prof, Tao,

The Fourier and generalized functions, were a really nice.

Cheer and thank,

Manuel

1 May, 2009 at 8:48 am

Takis KonstantopoulosDear Terry,

Thanks for the excellent notes. Can I ask a question? How do you write formulas on this blog (apparently you use LaTeX directly?) Several months ago I tried to post something on Tim Gowers’ blog but failed miserably and then had to link a pdf file. I am wondering whether there is something special about this type of blog you are writing on or whether it can be done on other blogs (e.g. blogspot).

Thanks!

1 May, 2009 at 8:55 am

Terence TaoDear Takis,

Please see my “about” page

https://terrytao.wordpress.com/about/

for details.

18 June, 2009 at 6:29 pm

JeffDear Prof Tao,

I have a question regarding probability theory but it has relation with Fourier transfrom.

Let X be a random variable and be its characteristic function, i.e Fourier transform of its distribution.

if is differentiable, is again a characteristic function of a random variable? Which one?

thanks

18 June, 2009 at 9:10 pm

PDEbeginnerDear Jeff,

I think may not be a characteristic function of some random variable, because $\phi^{‘}(0)=E[X]$ (assuming that is the characteristic function of ) is not necessarily equal to 1.

18 June, 2009 at 9:56 pm

jeffDear PDE beginner,

excuse me I did not understand your argument,

could you please explain more?

thanks

18 June, 2009 at 10:34 pm

PDEbeginnerI think each characteristic funtion of a random variable has the property . Because , which can be some value other than $1$, is not a characteristic function.

19 June, 2009 at 10:06 am

jeffOne of my friend said that ”yes it is, think about Radon- Nykodm theorem” but now your argument seems reasonable, I am really confused.

I hope that Prof Tao explain and clarify that question.

3 July, 2009 at 12:37 am

liuxiaochuanDear Professor Tao:

In Exercise 7, (b) I guess what you mean is .

In exercise 7 (f). I think the weights in the end are something like for some .

don’t know if I’m right.

3 July, 2009 at 8:06 am

Terence TaoDear Liuxiaochuan,

Thanks for the corrections! But compactness is not needed in exercise 8. (Note that the larger U gets, the smaller the dual set gets.)

For Exercise 10, Lusin’s theorem does not work directly, because the continuity is only obtained on a restriction of the character to a set of strictly smaller measure. (Note that there certainly exist measurable functions which are discontinuous at every point, e.g. the indicator function of the rationals.) The hint asks for you to exploit the continuity of the map from to , which can be done by approximating locally by a continuous function in .

(An alternate proof (which is essentially equivalent) proceeds by using the Steinhaus theorem.)

3 July, 2009 at 1:13 am

liuxiaochuanDear Professor Tao:

In exercise 8, does U in the hint has to be a compact neighborhood of the identity to make it work?

3 July, 2009 at 6:15 am

liuxiaochuanDear professor Tao:

About exercise 10,I didn’t quite get the hint. I want to know if what we want to do is just prove the continuity of . If so, then will the following be correct?

Use the Lusin Theorem to show there are at least one point x where f is continuous, then we prove from the condition that if f is continuous at one point, then it is continuous everywhere.

3 July, 2009 at 9:12 am

liuxiaochuanDear Professor Tao ：

there is a small correction:

in (4), the left-hand side should be

Again about exercise 8, although I noticed that the larger U gets, the smaller the dual set gets, I still tried using countable many of sets of the form to cover the set, which cann't be reduced to finite many ones.

3 July, 2009 at 9:35 am

Terence TaoDear Liuxiaochuan,

Actually, I think (4) is correct as it stands.

In a Hausdorff space, the intersection of uncountably many compact sets is still compact.

3 July, 2009 at 10:35 am

IqbalHello

I am glad I found this website. I am trying to understand your paper titled “Near Optimal Signal Recovery…..Strategies?”. On page 7, you present a geometric picture of the problem which I don’t understand. Forgive me for my lack of background knowledge but if you can suggest a good reference to strengthen my background for my understanding,I will be very grateful. I have just started my research on Compressed Sensing and am having trouble in the understanding.

Thanks

Iqbal

4 July, 2009 at 8:41 am

Terence TaoDear Iqbal,

Several tutorials, background references, etc. can be found at

http://dsp.rice.edu/cs

3 July, 2009 at 8:11 pm

Benford’s law, Zipf’s law, and the Pareto distribution « What’s new[…] for all integers , where the left-hand side denotes the proportion of data for which lies between and for some integer . Suppose now that we generalise Benford’s law to the continuous Benford’s law, which asserts that (1) is true for all real numbers . Then it is not hard to show that a statistic obeys the continuous Benford’s law if and only if its dilate does, and similarly with replaced by any other constant growth factor. (This is easiest seen by observing that (1) is equivalent to asserting that the fractional part of is uniformly distributed.) In fact, the continuous Benford law is the only distribution for the quantities on the left-hand side of (1) with this scale-invariance property; this fact is a special case of the general fact that Haar measures are unique (see e.g. these lecture notes). […]

4 July, 2009 at 9:17 pm

AnonymousDear Professor Tao:

Is it correct in understanding that multiplier and symbol are same thing and that the order of symbol can be negative?

6 July, 2009 at 12:58 pm

Terence TaoThe two terms are sometimes used synonymously, though I prefer to use the term “multiplier” (or more precisely, “Fourier multiplier operator”) to refer to the operator , and “symbol” to refer to the function . The two concepts are then related by the formula .

10 July, 2009 at 6:13 pm

Some examples of graded algebras « Annoying Precision[…] the direct sum is replaced by an integral.) For more details, see Terence Tao’s notes on the Fourier transform. With (the circle group), (the circle), and we recover the usual gradation on the space of […]

7 August, 2009 at 5:39 am

GabrielDear Prof. Tao,

Thank you very much for posting this notes. They were a very nice resume. Reading them the following question came to my mind:

Given a function in or more generally in consider its Fourier Series. Let and let us define

. We are interested to know if the following series converges:

The case it is trivial and is essentially Parseval's identity. What can be said about ? Does this series converges for all ?

Thanks!

7 August, 2009 at 6:24 am

Terence TaoThis series diverges without additional regularity assumptions on f. Consider for instance a random sign function at scale 1/N for some N, i.e. a function of the form for some iid signs . This function is bounded in every space, but Khintchine’s inequality tells us that the first N Fourier coefficients of this function are of size on the average, which will lead to divergence of the above sum for in the limit . (One can formalise this divergence by creating a suitable linear combination of the above examples over all N, or appealing to the uniform boundedness principle.)

More generally, random sign patterns are one of two the fundamental counterexamples in Fourier analysis (the others being rescaled bump functions). Most of the more sophisticated counterexamples are some combination of the two (e.g. the unboundedness of the disc multiplier stems from random combinations of bump functions rescaled to tubes forming a Besicovitch set).

Note however that if the absolute values of the Fourier transform are on the outside, then is essentially the k^th moment of f, which is finite. Thus we see that there is significant cancellation between the Fourier coefficients of f that is lost when one only considers magnitude information.

Incidentally, this type of divergence is the main reason why the trilinear Hilbert transform cannot be controlled by existing time-frequency analysis techniques (which are based on magnitudes of various wave packet coefficients, ignoring phase information), whereas the bilinear Hilbert transform can. (See my earlier blog post on this.)

21 October, 2009 at 9:38 pm

Integral Transforms and Partial Differential Equations « Stevethemathsman's Blog[…] at the above links. An interesting discussion of the Fourier transform is on Terry Tao’s blog here esp is you have seen LCA groups […]

29 October, 2009 at 11:51 pm

PDEbeginnerDear Prof. Tao,

I have some some (possibly) trivilal questions about the section 1, because I didn’t systematically learn group theory before and no people around me are working on group theory.

1. Excerse 1: is a LCA group. According to the hint, we can take , which is a compact neighbourhood of identity . But generates .

2. Exercise 8: I don’t know how to prove that ‘If is discrete, then is compact’. I also don’t know how to prove Exercise 16, which is vice versa of Exerecise 16.

By the way, there seem some typos:

1. Exercise 7 (f), should be .

2. Exercise 8, The in the should be .

30 October, 2009 at 10:55 am

Terence TaoThanks for the corrections!

Regarding : this set is compact but not a neighbourhood (if one endows with the product topology), or a neighbourhood but not compact (if one instead uses the box topology). Also, it doesn’t generate all the functions in , but only those functions that are bounded.

To show that the dual of a discrete group G is compact, note that the topology on is compatible with the product topology on .

The main tool used to prove Exercise 16 is Lemma 2.

19 December, 2009 at 12:09 am

Functoriality « Annoying Precision[…] A basic reason to be interested in locally compact Hausdorff spaces is that, if we in addition allow the algebras to be over instead of over , we recover Pontryagin duality. One constructs a C*-algebra which generalizes the group algebra of a locally compact Hausdorff abelian group , and the spectrum of this algebra is the dual group . However, I’m not sure of the technical details, so I’ll just refer to this post of Terence Tao. […]

6 January, 2010 at 11:59 am

254A, Notes 2: The central limit theorem « What’s new[…] make the characteristic function defined on the dual space instead of on itself; see for instance my notes on the Fourier transform in general locally compact abelian groups. But we will not need to care about this subtlety in our […]

25 June, 2010 at 6:43 pm

The uncertainty principle « What’s new[…] Pontryagin group duality A (locally compact Hausdorff) abelian group can be described either by listing its elements , or by listing the characters (i.e. continuous homomorphisms from to the unit circle, or equivalently elements of ). The connection between the two is the focus of abstract harmonic analysis. […]

21 January, 2011 at 2:19 pm

a.sJust a minor typo:

Exercise 3, dx should be replaced by $\mu(dx)$

[Corrected, thanks – T.]9 May, 2011 at 7:11 pm

Wenying GanDear Prof. Tao:

I think there is a typo in Ex 20. For the Dirichlet kernel in Ex20 part 1, it should be

Wenying

[Corrected, thanks (and a related error for the Fejer kernel on the next exercise has also been corrected – T.]2 June, 2011 at 12:19 am

局部紧群上的Fourier分析 « Fight with Infinity[…] Tao The Fourier transform […]

12 July, 2011 at 5:53 pm

Gleason’s lemma « What’s new[…] be a non-trivial left-invariant Haar measure on (see for instance this previous blog post for a construction of Haar measure on locally compact groups). We then form the convolution , with […]

30 September, 2011 at 8:11 am

254A, Notes 3: Haar measure and the Peter-Weyl theorem « What’s new[…] more precisely the unique constant function that is an average of translates of . See Exercise 6 of these notes for further discussion (the post there focuses on the abelian case, but the argument extends to the […]

6 December, 2011 at 10:59 am

254B, Notes 2: Cayley graphs and Kazhdan’s property (T) « What’s new[…] Now we turn to the higher-dimensional cases . The idea is to first use Fourier analysis to understand the action of various simpler subgroups of acting on a space with approximately invariant vectors, and obtain non-trivial vectors that are invariant with respect to those simpler subgroups. Then, we will use an asymptotic conjugation trick of Mautner to boost this invariance up to increasingly larger groups, until we obtain a non-trivial vector invariant under the whole group . (As such, this section will presuppose some familiarity with Fourier analysis, as reviewed for instance in this blog post.) […]

11 February, 2012 at 2:24 pm

RexFor Exercise 4, I can see how the assumptions imply that every compact set has zero measure. If G is sigma-compact, then we can cover G with countably many compact sets and come to the desired conclusion.

However, if G is not assumed to be sigma-compact, I don’t see why the fact that every compact set has zero measure would imply that the Haar measure in question is 0.

In other words, why can’t there exist a locally compact group with nontrivial Haar measures vanishing on every compact set?

11 February, 2012 at 2:43 pm

Terence TaoHaar measures are (by definition) Radon measures, and thus inner regular. In particular, if they vanish on compact sets, they vanish everywhere.

12 February, 2012 at 3:56 am

RexFor Exercise 8, I see how to prove local compactness when G is sigma-compact (invoking the Arzela-Ascoli theorem to deduce sequential compactness), but how does one handle the case when G is not sigma-compact? It seems that there is no Arzela-Ascoli to appeal to here.

12 February, 2012 at 8:33 am

Terence TaoIn the non-sigma-compact case one can use Tychonoff’s theorem instead of Arzela-Ascoli to obtain the desired compactness. The point is that the equicontinuity allows one to upgrade the product topology to the locally uniform topology.

12 February, 2012 at 4:47 pm

RexIn the definition of the Fourier transform, you write:

“In the remainder of this section, is a fixed LCA group with a non-trivial Haar measure .

Given an absolutely integrable function , we define the Fourier transform by the formula

”

That is the Fourier transform formula for R^n, but I think here you mean to write the general formula for a LCA group i.e.

.

12 February, 2012 at 4:53 pm

Terence TaoThe action of on is, by convention, denoted instead of (see Definition 1), precisely in order to align the notation in the general LCA case with the Euclidean cases.

12 February, 2012 at 5:02 pm

RexIn this case, wouldn’t you want the formula to be

rather than

For example, if $\latex G = \mathbb{R}$ and we fix a character $\latex \xi_t(x) = e^{2 \pi i tx}$ with the notation $\latex \xi \cdot x = \xi(x)$, then the formula that you have currently written would give

12 February, 2012 at 5:07 pm

RexOh, I see. You write to denote an additive character, rather than a multiplicative one. I think this makes more sense now.

14 February, 2012 at 7:25 am

RexTypo: “Corollary 3 The space of multiplicative chararacters…” should be “characters”.

14 February, 2012 at 1:32 pm

RexIt seems that, lurking behind the formulation of the Plancherel theorem for compact groups, is the claim that a compact group has at most countably many characters. How does one see this?

14 February, 2012 at 2:51 pm

Terence TaoCompact metrisable groups have countably many characters, but in general a compact group may have uncountably many characters (e.g. an uncountable power of the unit circle). But this does not really cause any difficulty (one has to sum over uncountable index sets rather than countable ones in the Plancherel theorem, but in practice at most countably many of the summands are non-zero).

8 May, 2012 at 6:59 pm

alabairDear Professor,

My sincere thanks for this post which I qualified it as a true jewel.

my question is if it’s possible to define white noise probability space and white noise analysis by using the theorem 8(Bochner theorem) in LCA group G or more simply in a Lie group as is done by Bernt Øksendal in the book:” Malliavin Calculus for Lévy Processes with Applications for Finance”. I pain to choose a positive-definite functions to have a white noise probability measure. I think we must have something like:

$\latex e^{-\frac{1}{2}||\phi||^2}$.

My deep apologies for the inconvenience caused if my question seems too silly.

With respect.

9 May, 2012 at 3:02 am

alabairReblogged this on Alabair's Weblog.

23 June, 2012 at 3:25 pm

JuritoDear prof. Tao,

I need a question regarding Exercise 11 (Riemann-Lebesgue lemma). I see how to finish the proof once the statement from your hint is prove. To prove the statement it is enough to prove it in the case where f is continuous (because of density of continuous functions in L1). I have to prove that

\int | f(x-y) – f(x) | dx < epsilon, for all y in some open neighbourhood U of identity

This sounds trivial, but I don't see a correct argument. Of course, it seems that dominated convergence theorem should do the job, but one needs to have a sequence of ys converging to identitiy in order to properly use it. However, I don't see how to get desired set U from this argument.

23 June, 2012 at 3:33 pm

Terence TaoUse the fact that continuous, compactly supported functions are uniformly continuous.

23 June, 2012 at 4:33 pm

JuritoThat was my second idea. The problem with that one was that G is not necessarily metrisable. Of course, I see how I would define uniform convergence for functions on general topological groups, but I don’t see how to adapt proof that continuous function with compact support implies uniformity. Could you recommend some reference for this?

23 June, 2012 at 6:48 pm

Terence TaoWrite out the proof in full in the metrisable case (including the proof that continuous functions on compact sets are uniformly continuous). Try not to refer to the metric directly, but rely instead on balls. Then, replace the balls by more general open neighbourhoods.

23 June, 2012 at 3:31 pm

JuritoOh, and I also need a bit of help with proving that Fourier transform of every L1 function is continuous (first part of the same Exercise). It seems to rely on fact that most of the L1 mass of f comes from a compact set. Is this true and how to prove it?

4 July, 2012 at 1:57 pm

AnonymousDear Professor Tao: You mentioned ” The poset-theoretic concepts seem to be more related to the operations of integration and differentiation, or convolution and deconvolution, than to the Fourier transform – thus they are more like Fourier multipliers than Fourier transforms.” Is there a good reference to understand what you are informing here? I am not a professional mathematician but an engineer. Is there a reference that I could access??

10 June, 2013 at 5:28 am

alabairDear Professor,

Can you give a probabilistic version of Bochner-(Milnos) theorem’s, in order to define the white noise process taking values in Lie group for exemple. As it was done in B. øksendal, white noise process is defined in the case.

Thank’s

17 June, 2015 at 5:17 am

rajeshd007http://mathoverflow.net/q/208867/14414

17 June, 2015 at 5:24 am

rajeshd007Dear Prof. Tao,

I wish you had mentioned about the jump discontinuties and Fourier transform, in this article. Does this deviate from the article. Do Jump discontinuties are being studied by any one in the generalized setup of Fourier transform on groups/harmonic analasys. Appreciate any references.

15 January, 2016 at 4:04 pm

杜鹏Dear Prof. Tao,

Do you have some opinion about this problem http://mathoverflow.net/questions/228379/interpret-fourier-transform-as-limit-of-fourier-series

on mathoverflow?

4 February, 2016 at 5:14 pm

AnonymousDear Prof. Tao,

In Exercise 7. (b) (existence of Haar measure),

can one just take g as the given f for the second claim?

Then the number (f:g) is surely greater or equal to 1.

Even we don`t need any epsilon concession.

Something`s strange.

7 February, 2016 at 3:46 am

AnonymousDear Prof. Tao,

In Exercise 7 (the existence of Haar measure, general case),

I am trying to prove the second statement of (b), with the idea that

we cover the support of f by finitely many translations U_{i} (i=1, …, n) of U the open neighborhood of the identity. We may assume that this U possesses the nice property about the uniform continuity of f.

Up to now all are good but I have some troubles in dealing with inequalities.

I guess that we have to cover K by U_{i}`s nicely (assuming the minimality of n, for instance). Am I right? Or, could you give me more hints?

Best

7 February, 2016 at 4:49 pm

Terence TaoFirstly, there is an erratum to this exercise that I forgot to update here, which is that the function should be supported in a small neighbourhood of the identity, but you have already chosen in the spirit intended. Actually this part (b) turns out to be a bit trickier than the subsequent parts of the exercise (which do not directly depend on (b), but for which (b) serves as motivation). I may separate it off into a separate exercise in a subsequent edition of the textbook that these notes are based off of.

We can normalise . The point is that one can approximate the function (using the uniform continuity of ) by the convolution integral , and this integral can in turn be approximated (using the uniform continuity of both and ) by a “Riemann sum” for some partition of into sets , where is a point in $U_j$. This uniformly approximates by a suitable combination of shifts of , but isn’t quite a majorisation. To achieve the majorisation, we can apply a similar uniform approximation to some bump function that equals (say) on the support of , and add a small multiple of that approximation to the previous approximation.

12 February, 2016 at 12:49 pm

AnonymousIf is an additive character, then the function is a multiplicative character. This can be checked directly. Why

conversely every multiplicative character arises uniquely from an additive character in this fashion?12 February, 2016 at 9:21 pm

Terence TaoThe map that sends to is a homeomorphism.

12 February, 2016 at 12:58 pm

AnonymousDo you have a hint for Exercise 9a (e.g. d=1)? I have no idea how to approach it by using the definition of the Pontryagin dual.

12 February, 2016 at 9:24 pm

Terence TaoA homomorphism from to any group is completely determined by its value at the generator . (This case of Pontryagin duality corresponds to the operation of forming a Fourier series out of a discrete sequence of coefficients, also known as the z-transform in electrical engineering.)

12 February, 2016 at 1:38 pm

AnonymousWhen doing Fourier transform to functions on , one gets functions on . This corresponds to Exercise 9 (b). When doing Fourier transform to functions on , (or functions on bounded intervals), one gets the Fourier coefficients which are functions on . This corresponds to Exercise 9(c). These two cases appear in the undergraduate Fourier Analysis course. Would you explain what is the case in Ex 9(a) (probably also in the context of an undergraduate Fourier Analysis course)?

15 June, 2016 at 8:12 am

AnonymousAssuming the Fourier inversion theorem ( for ) is true in the one-dimensional case, can one directly show that it is true for the general dimensional case?

15 June, 2016 at 8:47 am

Terence TaoYes; the -dimensional Fourier transform is the composition of one-dimensional Fourier transforms (applied in each variable separately, and composed in an arbitrary order), and similarly for the adjoint Fourier transform.

27 June, 2016 at 7:20 pm

AnonymousI’m confused about the topology on . On the one hand, it is not a normed vector space but a seminormed space, though it has a metrizable topology. One the other hand, in the setting of Theorem 7, is a normed vector space with the L^2 norm . How should one really understand the structure of ?

28 June, 2016 at 4:31 pm

AnonymousIn Exercise 30, what is ? Do you mean continuous functions with compact support (or decaying to at infinity)? I don’t find the definition of it in this note.

[ is defined in Exercise 11. -T.]15 July, 2016 at 4:17 pm

NickDear Prof. Tao,

is there an elegant solution to exercise 9 (e)? One needs to show, that every character of the closed subgroup can be extended to , but I don’t see how this can be done without further methods.

17 July, 2016 at 8:06 am

Terence TaoYes, you’re right, this part of the exercise is non-trivial and requires Pontryagin duality.

21 October, 2016 at 4:07 pm

AnonymousDear Prof Tao,

actual computation of fourier transforms can be tricky, for example, is there a way, or possible approach to compute the fourier transform of: exp(-icosx)?

21 October, 2016 at 9:54 pm

Terence TaoWell, an exact formula for the Fourier transform is usually too unrealistic of an expectation (for much the same reason as most indefinite integrals do not have any explicit closed form). However, the method of stationary phase (and its older sibling, the method of steepest descent) can be used to obtain asymptotics for many Fourier transforms, and this suffices for many applications in analysis and PDE.

23 October, 2016 at 1:17 pm

Jhon Manugalhttps://en.wikipedia.org/wiki/Bessel_function#Properties

23 February, 2017 at 2:05 am

AnonymousIn the LHS (according to the above Wikipedia article) should be .

23 February, 2017 at 2:19 am

AnonymousCorrection: according to the Wikipedia article on the Jacobi-Anger expansion, if in the LHS is replaced by then the coefficients in the RHS should be multiplied by .

18 February, 2017 at 6:31 am

AnonymousTo anyone who reads this,

In Exercise 43 of the proof of Plancherel theorem for compact case via Gelfand theory, I don`t see where the compactness is critically used. Can anybody tell me this?

25 February, 2017 at 10:11 am

Terence TaoCorollary 3 breaks down in the non-compact case (the characters do not even lie in , in fact, as the Haar measure now has infinite measure.) Also, no longer embeds into . Theorem 4 is basically Corollary 3 combined with the injectivity of the Fourier transform on ; the latter is indeed true for all LCA groups, which is the main conclusion of Exercise 43.

22 February, 2017 at 6:59 pm

AnonymousHow can we see Fourier analysis as Laplacian in spectral theory, in the last paragraph? Is there any reference?

[See e.g. http://www.sciencedirect.com/science/article/pii/0022123689900049 -T.]27 February, 2017 at 6:24 pm

Anonymous“For the standard torus , one can obtain a Haar measure by identifying this torus with in the usual manner and then taking Lebesgue measure on the latter space. This Haar measure is a probability measure.”

Would you elaborate what “identifying” means here? And what is the “usual manner” you are referring to?

28 February, 2017 at 9:43 am

Terence Taois a fundamental domain for the torus , so in particular there is a bijection that can be used to perform the identification. Thus, for instance, the Haar measure of a set can be defined as the Lebesgue measure of . (One has to check that this is indeed a Haar measure, but this is not too difficult.)

Alternatively, one can identify (= give an explicit one-to-one correspondence between) functions from to (say) the complex numbers with functions from to that are -periodic (thus whenever and ). The integral of such a function against Haar measure is then the same as (many other formulae for this integral are also available, e.g. ).

28 February, 2017 at 2:35 pm

Romain Viguieri see what you mean by Haar measure, but there is too many dimensions that are in each other and find

6 March, 2017 at 6:11 pm

AnonymousLet be the Lebesgue measure on . How should one should the inner regularity for ?

6 March, 2017 at 10:34 pm

Terence TaoThis follows easily from the inner regularity of Lebesgue measure (it may help to divide into two cases, e.g. , and , and then glue the two cases together). Alternatively one can apply the Riesz representation theorem starting with the functional from to that takes a function to its mean .

7 March, 2017 at 8:21 am

AnonymousThanks for the comment! I would be interested in seeing a little bit more details. Now I don’t see how to build up a clear connection between compact sets in and those in the fundamental domain (which is not homeomorphic to ), which should give the desired inner regularity. For instance, let . Then clearly by the inner regularity of Lebesgue measure, one has

$displaystyle m(\iota(E))=\sup\{m(K)\:K\subset \iota(E), K\textrm{ compact in } [0,1)\}.$

However, how can one show that

$displaystyle m(\iota(E))=\sup\{m(\iota(K))\:K\subset E, K\textrm{ compact in } {\bf R}/{\bf Z}\}?$

28 February, 2017 at 11:34 am

Romain ViguierLet A and B be 2 set such as the cartesian product A x B is in the form : f(A) f(B) and f(A) \bigcup f(B).

We can decompose f(a) as following : f(A) = f(a1) \times f(ai) \times … \times \infty. These functions are local horizontal asymptotes.

In the same way, one has f(b1) \times f(bi) \times … \times \infty. These functions are local vertical asymptotes.

For all a and all b, we have a space whish is defined as this :

lim \stackrel{\rightarrow}{0} f(a) \times lim \stackrel{\rightarrow}{0} f(b) \cong

\phi with \phi a coefficient of proportionality wish is the relation between the local to the global.

Thus, we have N \Rightarrow C \Rightarrow N an infinity of time.

For exemple : if E = [ \sum \phi ], then E \times E = C.

28 February, 2017 at 11:39 am

Romain Viguier* f(A) \propto f(B) at the beginning

9 March, 2017 at 8:39 am

AnonymousIn Definition 1, and have the same topological structure as well as the group structure. Does this imply that the multiplicative character and the additive character are actually the same thing in some sense?

[They are related to each other by an isomorphism of topological groups, but strictly speaking they are still distinct characters. -T.]9 March, 2017 at 10:32 am

AnonymousUnder the Definition 1, it is said that the connection is given by the map , the notation is a little confusing. Isn’t an element of ? Then what does mean?

[The function is -periodic, so it descends from a function from to , to a function from to , and it is a common “abuse of notation” to identify the two functions. -T]9 March, 2017 at 10:23 am

AnonymousA topological space is compact if and only if any collection of its closed sets having the finite intersection property has non-empty intersection. In Exercise 6e, since one can identify each probability measure wit an element of the product space, I guess the nonempty-ness of the intersection some collection of closed sets implies the desired Haar measure. However, I don’t understand what collection of closed sets to consider. How does it have anything to do with (d)?

[Part (a) describes the Haar measure elements of the product space as the intersection of a family of closed sets, each of which will be nonempty thanks to (d). -T]9 March, 2017 at 10:54 am

AnonymousFor Exercise 9, I’m not quite sure how the identification of LCA groups should be shown. For instance, in 9a, the (set) bijection is given already. Then one shows (i) the bijection is a (topology) homeomorphism (ii) the bijection is a (group) isomorphism. Would (i) and (ii) together be enough? (Does one need to prove further that the topological structure and group structure are “compatible” (the group operations are continuous) or it just follows from (i) and (ii)? )

[Yes, this is enough (assuming that one has already demonstrated that at least one of the spaces is a topological group) -T.]9 March, 2017 at 11:23 am

AnonymousIs the topology in Exercise 8 for the Pontryagin dual a “canonical” one (does one have other choice?) so that all the exercise (e.g. Exercise 11) in this post use this one?

[Yes – T.]28 June, 2017 at 5:20 am

Legendre transform – foreXiv[…] Fourier transform for understanding translationally invariant systems?” can be answered by something like “Translationally invariant operations in the spatial domain correspond to multiplication in […]

9 May, 2018 at 8:24 am

AnonymousTo prove Theorem 7 using Exercise 30, I believe the bounded linear transformation theorem is needed (https://en.wikipedia.org/wiki/Continuous_linear_extension) or one needs to repeat essentially the argument of the whole proof. (To my knowledge, I don’t know if there is another way to do it.) Was it proved or mentioned somewhere in the course notes or do you assume it as a trivial fact?

10 May, 2018 at 5:45 am

AnonymousIn Exercise 30, suppose is in for two different ‘s, say . Can one find a sequence in so that the sequence converges to in

boththe norm and the norm?10 May, 2018 at 5:02 pm

Terence TaoYes; this is a good exercise for you to work out.

11 May, 2018 at 6:11 am

AnonymousIf one uses an approximation to the identity, , then

in both and for . This seems to be quite close to what one wants here. But there is no guarantee that

:-(

On the other hand, since , there exists so that and . So one can approximate by

where .

Since as well, similarly one has

where . But then one has two different sequences, and . :-(

Is there a way to "combine" them together?

10 May, 2018 at 6:07 am

AnonymousDenote the Fourier transform definition as and the version by Plancherel as . If then these two versions agree.

In Exercise 34, suppose I can find a sequence in the Schwartz space that in both the and norms. Then I can conclude that uniformly and in by continuity. But these two modes of convergence are different… How can one conclude that

?

10 May, 2018 at 5:03 pm

Terence TaoThere are several ways to proceed here. One is to observe that both modes of convergence are stronger than distributional convergence, which has unique limits. Another is to upgrade the L^2 convergence to pointwise almost everywhere convergence by passing to a subsequence.

10 May, 2018 at 8:10 am

AnonymousWolff made an argument in his notes on harmonic analysis that if one can show that

for and , then one can conclude that this is also true for and by Plancherel theorem. Can anyone elaborate the implication?

10 May, 2018 at 5:18 pm

Terence TaoHint: is dense in , and inner products with or are continuous in the topology.

6 August, 2018 at 6:07 am

Paul RamondDear Pr Tao

Thank you for this great note. I am concerned about the Heisenberg uncertainty principle, Ex 36, the very last equation. Shouldn’t the whole integrand be squared, as it is the L_2 norm ?

Besides, can you give a reference as to how to optimize with respect to a and b in this exercise ? I don’t know how to start, and not sure what it means actually.

Thank you again.

– Paul R.

6 August, 2018 at 7:59 am

Terence TaoThanks for the correction! Examples of optimisation can be found in the previous blog post https://terrytao.wordpress.com/2007/09/05/amplification-arbitrage-and-the-tensor-power-trick/ .

27 August, 2018 at 1:20 pm

AnonymousIn (8), is the minus sign purely from the convention? Remark1 says something about the constant . I’m wondering if there is any particular reason one puts a minus sign in the exponent in the definition (8).

27 August, 2018 at 8:48 pm

Terence TaoThe sign is mostly a convention. I prefer to have the minus sign in the Fourier transform so that it does not appear in the inverse Fourier transform, so that the Fourier inversion formula becomes a device for representing a function as a superposition of plane waves without the minus sign present.

15 November, 2018 at 7:18 am

Anonymous(I’m not good at English, so there may be many English mistakes.)

Dear Professor Tao,

In Exercise 5, how can I prove that the dual of L^1 is L^(infinity)?

I used exercise 1 and took coset decomposition of G, and used duality theorem, I could get elements of the L^(infinity) space over each coset of G, but couldn’t prove that they are glued together to the element of the L^(infinity) space over G.

21 November, 2018 at 2:49 pm

Terence TaoAh, this is a subtlety I had not previously realised… the question was not correct as stated, one needs to replace by a “local” version in which one works with functions (not necessarily globally measurable!) whose restriction to every finite measure set lies in (with uniform bounds on the norm), and with two such functions declared equivalent if they agree locally almost everywhere (a.e. on every finite measure subset). I’ve reworded the exercise accordingly.

19 December, 2018 at 9:17 am

AnonymousIn Wolff’s Lecture notes on Harmonic Analysis (Chapter 4), when discussing

the Fourier transform of the function

the author claims without proof the following:

for , where denotes the Schwartz space, the map

is analytic in the “indicated regime” (It is unclear in Wolff’s notes where the “indicated regime” is.), which may be done by using the dominated convergence theorem to justify complex differentiation under the integral sign.

How would you apply the dominated convergence theorem here? Can one find a dominated function to bound

19 December, 2018 at 2:01 pm

Terence TaoSure (as long as for some ), one can just use as the dominating function. Actually it is slightly easier to use Morera's theorem to establish analyticity, as one can then appeal to Fubini's theorem rather than trying to justify differentiation under the integral sign.

22 December, 2018 at 7:22 pm

AnonymousIn your 247A notes2(http://www.math.ucla.edu/~tao/247a.1.06f/notes2.pdf), in the discussion of the Hausdorff-Young inequality (page 21), the following function is considered:

where is the standard Gaussian and , . (I consider only one term in the definition of and take .)

I got that the Fourier transform is given by

which is different from your calculation:

I tried several times but cannot find my mistakes. Where did I go wrong?

23 December, 2018 at 10:07 am

Terence TaoAh, there is a phase correction that needs to be inserted. If one inserts a factor of in the summand for (and for ) then the rest of the discussion can proceed without difficulty.

24 December, 2018 at 3:21 pm

AnonymousI think you mean a factor of ? The calculation in the previous comment shows that

if

then

So the summand for and its Fourier transform still have the desired properties.

[Ah, you are right; I had made a sign error. -T]24 December, 2018 at 3:31 pm

AnonymousBy translation invariance and the triangle inequality, one can show that

for . In order to establish the estimate for all one needs the case for as well. But applying triangle inequality again one gets

which is different from the desired bound

Intuitively, since can not be zero for all , one should expect that

would be an overkill.

24 December, 2018 at 4:36 pm

Terence TaoFor any given , the arithmetic progression is only close to zero for at most one value of ; indeed, if is the integer that minimises , then we have for all the other values of , by the triangle inequality. So the sum is highly convergent (thanks the rapid decay of the Gaussian) giving the required bound of O(1).

(As a general rule of thumb, one expects rapidly decreasing functions such as the gaussian to behave in the same fashion as compactly supported functions, in that estimates that are true for the latter tend to also be true for the former. The intuition to have is not that “ is non-zero everywhere”, but rather “ is negligible unless “.)

24 December, 2018 at 8:25 pm

AnonymousIn this argument (and the whole proof for showing ), it seems that one only needs . Where does one need be large?

24 December, 2018 at 11:30 pm

Terence TaoIf is too small, one runs into the danger that the various summands in may start interfering with each other, leading to an norm that is unexpectedly small due to cancellation.

25 December, 2018 at 10:25 am

AnonymousHow can one calculate by hand and thus establish the lower bound?

where

Without $J$ (but unfortunately it is there), one could immediately conclude by translation invariance that

and hence the desired lower bound:

How could the magnitude of $v$ cause cancellation in $I+J$?

25 December, 2018 at 10:46 am

Terence TaoIf , then as .

I recommend you draw some pictures to sketch what all the functions look like here; it should then become visually obvious why the norms should be what they are predicted to be, and why having large would be advantageous.

25 December, 2018 at 1:10 pm

AnonymousTrue that when , as . But this is also true for as well?

I guess you meant that

where

and

as .

25 December, 2018 at 6:18 pm

AnonymousBy the triangle inequality,

How did you get rid of the dependent of on ? Or the triangle inequality is used differently?

25 December, 2018 at 8:58 pm

Terence Tao.

Seriously, I recommend drawing pictures.

27 December, 2018 at 10:27 am

AnonymousThanks for the suggestion!

One can use Mathematica to do the job of sketching. I plotted the function where

for various values of and .

For fixed (being large), one can see that the graph of the function looks very much like a summation of functions that have () disjoint supports. The bigger is, the larger distance between the “supports” (the place where the function takes non-negligible values) of the functions in the summand, which matches intuitively that

While one can indeed observe that those “supports” interfered with each other when is small, I don’t see how exactly small could “fail” the proof of

if one calculates directly the norm.

Could you give an example that has “unexpectedly small L^2 norm” you mentioned in the previous comment (https://terrytao.wordpress.com/2009/04/06/the-fourier-transform/#comment-509369)?

(It seems that as and I don’t observe any “cancellation”.)

27 December, 2018 at 2:20 pm

Terence TaoThe type of cancellation that could potentially happen would for instance occur in the sum , whose norm becomes very small as is nonzero. Now, the function might not actually exhibit this sort of cancellation, but actually proving this when is small is non-trivial, whereas there is no difficulty in doing so when is large due to the essentially disjoint supports.

25 December, 2018 at 7:10 am

AnonymousWolff gave three different arguments in his notes demonstrating the failure of Hausdorff-Young for . The first one is the following “most elementary” one (which appears as an “exercise” in Wolff’s notes):

There exists a sequence of Schwartz functions so that

(1). Each has the same norm.

(2). Each has the same norm.

(3). The supports of the are disjoint.

(4). The supports of the are “essentially disjoint”:

uniformly in (which, I guess, means that is and does not depends on ).

Can one adapt the example in the previous comment (denoted as ) to get one that satisfies (1)–(4)?

The difficulty is in (3): do not have disjoint supports; on the other hand, if one uses a function instead of the standard Gaussian in the construction of , then (1)–(3) could be resolved but one has the difficulty in calculating the inverse Fourier transform of and getting (4).

24 December, 2018 at 7:12 am

AnonymousIn a discussion of the Khinchin’s inequality in your notes on the restriction conjecture (https://arxiv.org/pdf/math/0311181.pdf), things boil down to proving

You first show by direct calculation that for ,

which implies by Holder that for ,

You then show the upper bound for :

How do you use the log-convexity of the norm to show the lower bound

(Have you ever published this set of notes? Something is missing in the arxiv version. For instance, part of the picture on page 8 is gone.)

24 December, 2018 at 9:29 am

Terence TaoThe log convexity lower bounds the norm by a combination of an norm for some and an norm for . Since one has an exact formula (actually just a lower bound is needed) for the norm and an upper bound for the norm, one deduces a lower bound for the norm. (If one prefers, one can think in the contrapositive: if the norm were small, then the norm would also have to be small by interpolation with the norm, contradicting the formula one has for that norm.)

The notes were supposed to be published in the Park City Proceedings series, but unfortunately this particular issue had technical difficulties getting published.

30 December, 2018 at 6:56 pm

AnonymousIn Wolff’s notes on the stationary phase method (based on which are the 247AB notes), in the discussion of the function

where the author discussed the nondegenerate critical point case, he made a quantitative argument that

“uniformly” in and and accordingly

How can one make sense of the “uniform” big-O in and ? One can see that the remainder term is dominated by

when is small (and being big). But how could it be uniformly true for

all(and thus justify the later integral)?30 December, 2018 at 8:02 pm

Terence TaoI do not have these notes handy, but presumably Wolff is implicitly restricting to the range of interest, namely in the support of and large.

More generally, one does not need to halt one’s reading comprehension every time there is a line that you can’t quite justify. I wrote about these sorts of “compilation errors” and how to avoid them at https://plus.google.com/u/0/114134834346472219368/posts/TGjjJPUdJjk

30 December, 2018 at 8:31 pm

AnonymousIncidentally, I read the linked post on Google Buzz before posting the comment and I was searching on this list on Tricki: http://www.tricki.org/article/Estimating_integrals

(Wolff’s has been available online: http://www.math.ubc.ca/~ilaba/wolff/notes_march2002.pdf)

I was trying to give an argument directly showing that

which is “what the author is trying to do”. But I somehow ended up with needing the argument in Wolff’s notes.

If has compact support, then this is fine. But (as mentioned in the comment) only implies that is Schwartz. On the other hand, the tail

contains with

arbitrarily large. So I was not convinced thatwhich would give the desired order in using the Schwartz norm in .

30 December, 2018 at 8:58 pm

Terence TaoI think here one should use Taylor’s theorem with remainder, rather than the infinite series version of Taylor’s theorem. (It is here that the “i” in the exponential becomes important, as it keeps all the derivatives of bounded, in contrast to ; this is an important distinction that is difficult to see purely at the Taylor series level.)

9 January, 2019 at 8:14 pm

AnonymousThe proof of the Tomas-Stein restriction theorem in Wolff’s notes (http://www.math.ubc.ca/~ilaba/wolff/notes_march2002.pdf) uses the convolution of a Schwartz function and a measure that has compact support:

But the group as in Exercise 16 is not written out explicitly and later the following integral is used:

where is the surface measure for the sphere in .

It seems by the context that should be “a” unit sphere. But where should be the center of the sphere?

10 January, 2019 at 9:38 am

Terence TaoThe group here is ; the sphere is not a group. (But of course, the surface measure on the sphere would assign a zero measure to the portion of outside of the sphere.)

10 January, 2019 at 1:02 pm

AnonymousThanks for the comment! Then one should understand the integral in the previous comment as

where $S\subset \mathbf{R}^n$ is the unit sphere centered

at the origin. But I can’t make sense of the latter estimates:Reading around the context of the notes one can see that

It seems that should be a ball in .

How can one make sense of the canonical measure , which is originally defined for the sphere , “evaluating” on the ball in , and thus make sense of the second inequality above?

10 January, 2019 at 7:34 pm

Terence TaoI would imagine would be a ball in , which contains the sphere . If one wishes one could replace with since assigns zero mass to the complement of in .

10 March, 2019 at 7:22 pm

EricDear Prof Tao,

Thanks a lot for your notes. I think that there might be a typo in Exercise 42. The formula in the hint should read x_j (f*g)= (x_j f)*g+f*(x_j g). This is the only way that I can get to show that the left hand side converges uniformly to x_j f, when g=g_r, as r tends to zero.

[Corrected, thanks – T.]