Previous set of notes: Notes 1. Next set of notes: Notes 3.

In Exercise 5 (and Lemma 1) of 246A Notes 4 we already observed some links between complex analysis on the disk (or annulus) and Fourier series on the unit circle:

- (i) Functions that are holomorphic on a disk are expressed by a convergent Fourier series (and also Taylor series) for (so in particular ), where conversely, every infinite sequence of coefficients obeying (1) arises from such a function .
- (ii) Functions that are holomorphic on an annulus are expressed by a convergent Fourier series (and also Laurent series) , where conversely, every doubly infinite sequence of coefficients obeying (2) arises from such a function .
- (iii) In the situation of (ii), there is a unique decomposition where extends holomorphically to , and extends holomorphically to and goes to zero at infinity, and are given by the formulae where is any anticlockwise contour in enclosing , and and where is any anticlockwise contour in enclosing but not .

This connection lets us interpret various facts about Fourier series through the lens of complex analysis, at least for some special classes of Fourier series. For instance, the Fourier inversion formula becomes the Cauchy-type formula for the Laurent or Taylor coefficients of , in the event that the coefficients are doubly infinite and obey (2) for some , or singly infinite and obey (1) for some .

It turns out that there are similar links between complex analysis on a half-plane (or strip) and Fourier *integrals* on the real line, which we will explore in these notes.

We first fix a normalisation for the Fourier transform. If is an absolutely integrable function on the real line, we define its Fourier transform by the formula

From the dominated convergence theorem will be a bounded continuous function; from the Riemann-Lebesgue lemma it also decays to zero as . My choice to place the in the exponent is a personal preference (it is slightly more convenient for some harmonic analysis formulae such as the identities (4), (5), (6) below), though in the complex analysis and PDE literature there are also some slight advantages in omitting this factor. In any event it is not difficult to adapt the discussion in this notes for other choices of normalisation. It is of interest to extend the Fourier transform beyond the class into other function spaces, such as or the space of tempered distributions, but we will not pursue this direction here; see for instance these lecture notes of mine for a treatment.

Exercise 1 (Fourier transform of Gaussian)If is a complex number with and is the Gaussian function , show that the Fourier transform is given by the Gaussian , where we use the standard branch for .

The Fourier transform has many remarkable properties. On the one hand, as long as the function is sufficiently “reasonable”, the Fourier transform enjoys a number of very useful identities, such as the Fourier inversion formula

the Plancherel identity and the Poisson summation formula On the other hand, the Fourier transform also intertwines various*qualitative*properties of a function with “dual” qualitative properties of its Fourier transform ; in particular, “decay” properties of tend to be associated with “regularity” properties of , and vice versa. For instance, the Fourier transform of rapidly decreasing functions tend to be smooth. There are complex analysis counterparts of this Fourier dictionary, in which “decay” properties are described in terms of exponentially decaying pointwise bounds, and “regularity” properties are expressed using holomorphicity on various strips, half-planes, or the entire complex plane. The following exercise gives some examples of this:

Exercise 2 (Decay of implies regularity of )Let be an absolutely integrable function.

- (i) If has super-exponential decay in the sense that for all and (that is to say one has for some finite quantity depending only on ), then extends uniquely to an entire function . Furthermore, this function continues to be defined by (3).
- (ii) If is supported on a compact interval then the entire function from (i) obeys the bounds for . In particular, if is supported in then .
- (iii) If obeys the bound for all and some , then extends uniquely to a holomorphic function on the horizontal strip , and obeys the bound in this strip. Furthermore, this function continues to be defined by (3).
- (iv) If is supported on (resp. ), then there is a unique continuous extension of to the lower half-plane (resp. the upper half-plane ) which is holomorphic in the interior of this half-plane, and such that uniformly as (resp. ). Furthermore, this function continues to be defined by (3).
Hint:to establish holomorphicity in each of these cases, use Morera’s theorem and the Fubini-Tonelli theorem. For uniqueness, use analytic continuation, or (for part (iv)) the Schwartz reflection principle.

Later in these notes we will give a partial converse to part (ii) of this exercise, known as the Paley-Wiener theorem; there are also partial converses to the other parts of this exercise.

From (3) we observe the following intertwining property between multiplication by an exponential and complex translation: if is a complex number and is an absolutely integrable function such that the modulated function is also absolutely integrable, then we have the identity

whenever is a complex number such that at least one of the two sides of the equation in (7) is well defined. Thus, multiplication of a function by an exponential weight corresponds (formally, at least) to translation of its Fourier transform. By using contour shifting, we will also obtain a dual relationship: under suitable holomorphicity and decay conditions on , translation by a complex shift will correspond to multiplication of the Fourier transform by an exponential weight. It turns out to be possible to exploit this property to derive many Fourier-analytic identities, such as the inversion formula (4) and the Poisson summation formula (6), which we do later in these notes. (The Plancherel theorem can also be established by complex analytic methods, but this requires a little more effort; see Exercise 8.)The material in these notes is loosely adapted from Chapter 4 of Stein-Shakarchi’s “Complex Analysis”.

** — 1. The inversion and Poisson summation formulae — **

We now explore how the Fourier transform of a function behaves when extends holomorphically to a strip. For technical reasons we will also impose a fairly mild decay condition on at infinity to ensure integrability. As we shall shortly see, the method of contour shifting then allows us to insert various exponentially decaying factors into Fourier integrals that make the justification of identities such as the Fourier inversion formula straightforward.

Proposition 3 (Fourier transform of functions holomorphic in a strip)Let , and suppose that is a holomorphic function on the strip which obeys a decay bound of the form for all , , , and some and (or in asymptotic notation, one has whenever and ).

- (i) (Translation intertwines with modulation) For any in the strip , the Fourier transform of the function is .
- (ii) (Exponential decay of Fourier transform) For any , there is a quantity such that for all (or in asymptotic notation, one has for and ).
- (iii) (Partial Fourier inversion) For any and , one has and
- (iv) (Full Fourier inversion) For any , the identity (4) holds for this function .
- (v) (Poisson summation formula) The identity (6) holds for this function .

*Proof:* We begin with (i), which is a standard application of contour shifting. Applying the definition (3) of the Fourier transform, our task is to show that

For (ii), we apply (i) with to observe that the Fourier transform of is . Applying (8) and the triangle inequality we conclude that

for both choices of sign and all , giving the claim.For the first part of (iii), we write . By part (i), we have , so we can rewrite the desired identity as

By (3) and Fubini’s theorem (taking advantage of (8) and the exponential decay of as ) the left-hand side can be written as But a routine calculation shows that giving the claim. The second part of (iii) is proven similarly.To prove (iv), it suffices in light of (iii) to show that

for any . The left-hand side can be written after a change of variables as On the other hand, from dominated convergence as in the proof of (i) we have while from the Cauchy integral formula one has giving the claim.Now we prove (v). Let . From (i) we have

and for any . If we sum the first identity for we see from the geometric series formula and Fubini’s theorem that and similarly if we sum the second identity for we have Adding these two identities and changing variables, we conclude that We would like to use the residue theorem to evaluate the right-hand side, but we need to take a little care to avoid the poles of the integrand , which are at the integers. Hence we shall restrict to be a half-integer . In this case, a routine application of the residue theorem shows that Noting that stays bounded for in or when is a half-integer, we also see from dominated convergence as before that The claim follows.

Exercise 4 (Hilbert transform and Plemelj formula)Let be as in Proposition 3. Define the Cauchy-Stieltjes transform by the formula

- (i) Show that is holomorphic on and has the Fourier representation in the upper half-plane and in the lower half-plane .
- (ii) Establish the Plemelj formulae and uniformly for any , where the Hilbert transform of is defined by the principal value integral
- (iii) Show that is the unique holomorphic function on that obeys the decay bound and solves the (very simple special case of the) Riemann-Hilbert problem uniformly for all , with both limits existing uniformly in .
- (iv) Establish the identity where the signum function is defined to equal for , for , and for .
- (v) Assume now that has mean zero (i.e., ). Show that extends holomorphically to the strip and obeys the bound (8) (but possibly with a different constant , and with replaced by a slightly smaller quantity ), with the identity holding for . ({
Hint:To exploit the mean value hypothesis to get good decay bounds on and hence , write as the sum of and and use the mean value hypothesis to manage the first term. For the contribution of the second term, take advantage of contour shifting to avoid the singularity at . One may have to divide the integrals one encounters into a couple of pieces and estimate each piece separately.)- (vi) Continue to assume that has mean zero. Establish the identities and (
Hint:for the latter inequality, square both sides of (9) and use (iii).)

Exercise 5 (Kramers-Kronig relations)Let be a continuous function on the upper half-plane which is holomorphic on the interior of this half-plane, and obeys the bound for all non-zero in this half-plane and some . Establish the Kramers-Kronig relations and relating the real and imaginary parts of to each other.

Exercise 6

- (i) By applying the Poisson summation formula to the function , establish the identity for any positive real number . Explain why this is consistent with Exercise 24 from Notes 1.
- (ii) By carefully taking limits of (i) as , establish yet another alternate proof of Euler’s identity

Exercise 7For in the upper half-plane , define the theta function . Use Exercise 1 and the Poisson summation formula to establish the modular identity for such , where one takes the standard branch of the square root.

Exercise 8 (Fourier proof of Plancherel identity)Let be smooth and compactly supported. For any with , define the quantityRemarkably, this proof of the Plancherel identity generalises to a nonlinear version involving a trace formula for the scattering transform for either Schrödinger or Dirac operators. For Schrödinger operators this was first obtained (implicitly) by Buslaev and Faddeev, and later more explicitly by Deift and Killip. The version for Dirac operators more closely resembles the linear Plancherel identity; see for instance the appendix to this paper of Muscalu, Thiele, and myself. The quantity is a component of a nonlinear quantity known as the

- (i) When is real, show that . (
Hint:find a way to rearrange the expression .)- (ii) For non-zero, show that , where the implied constant in the notation can depend on . (
Hint:integrate by parts several times.)- (iii) Establish the Plancherel identity (5).
transmission coefficientof a Dirac operator with potential and spectral parameter (or , depending on normalisations).

The Fourier inversion formula was only established in Proposition 3 for functions that had a suitable holomorphic extension to a strip, but one can relax the hypotheses by a limiting argument. Here is one such example of this:

Exercise 9 (More general Fourier inversion formula)Let be continuous and obey the bound for all and some . Suppose that the Fourier transform is absolutely integrable.

Exercise 10 (Laplace inversion formula)Let be a continuously twice differentiable function, obeying the bounds for all and some .The Laplace-Mellin inversion formula in fact holds under more relaxed decay and regularity hypotheses than the ones given in this exercise, but we will not pursue these generalisations here. The limiting integral in (10) is also known as the

- (i) Show that the Fourier transform obeys the asymptotic for any non-zero .
- (ii) Establish the principal value inversion formula for any positive real . (
Hint:modify the proof of Exercise 9(ii).) What happens when is negative? zero?- (iii) Define the Laplace transform of for by the formula Show that is continuous on the half-plane , holomorphic on the interior of this half-plane, and obeys the
Laplace-Mellin inversion formulafor any and , where is the line segment contour from to . Conclude in particular that the Laplace transform is injective on this class of functions .Bromwich integral, and often written (with a slight abuse of notation) as . The Laplace transform is a close cousin of the Fourier transform that has many uses; for instance, it is a popular tool for analysing ordinary differential equations on half-lines such as .

Exercise 11 (Mellin inversion formula)Let be a continuous function that is compactly supported in . Define theMellin transformby the formula Show that is entire and one has theMellin inversion formulafor any and . The regularity and support hypotheses on can be relaxed significantly, but we will not pursue this direction here.

Exercise 12 (Perron’s formula)Let be a function which is of subpolynomial growth in the sense that for all and , where depends on (and ). For in the half-plane , form the Dirichlet series For any non-integer and any , establish Perron’s formula What happens when is an integer? (The Perron formula and its many variants are of great utility in analytic number theory; see these previous lecture notes for further discussion.) Note that one can use the special case of this exercise from Exercise 27 of 246A Notes 4 to help out here.

Exercise 13 (Solution to Schrödinger equation)Let be as in Proposition 3. Define the function by the formula \{ u(t,x) := \int_R\hat f(\xi) e^{2\pi i x \xi – 4 \pi^2 i \xi^2 t} d\xi.\}

- (i) Show that is a smooth function of that obeys the Schrödinger equation with initial condition for .
- (ii) Establish the formula for and , where we use the standard branch of the square root.

** — 2. Phragmén-Lindelöf and Paley-Wiener — **

The maximum modulus principle (Exercise 26 from 246A Notes 1) for holomorphic functions asserts that if a function continuous on a compact subset of the plane and holomorphic on the interior of that set is bounded in magnitude by a bound on the boundary , then it is also bounded by on the interior. This principle does not directly apply for noncompact domains : for instance, on the entire complex plane , there is no boundary whatsoever and the bound is clearly vacuous. On the half-plane , the holomorphic function (for instance) is bounded in magnitude by on the boundary of the half-plane, but grows exponentially in the interior. Similarly, in the strip , the holomorphic function is bounded in magnitude by on the boundary of the strip, but is grows double-exponentially in the interior of the strip. However, if one does not have such absurdly high growth, one can recover a form of the maximum principle, known as the Phragmén-Lindelöf principle. Here is one formulation of this principle:

Theorem 14 (Lindelöf’s theorem)Let be a continuous function on a strip for some , which is holomorphic in the interior of the strip and obeys the bound for all and some constants . Suppose also that and for all and some . Then we have for all and .

Remark 15The hypothesis (12) is a qualitative hypothesis rather than a quantitative one, since the exact values of do not show up in the conclusion. It is quite a mild condition; any function of exponential growth in , or even with such super-exponential growth as or , will obey (12). The principle however fails without this hypothesis, as discussed previously.

*Proof:* By shifting and dilating (adjusting as necessary) we can reduce to the case , , and by multiplying by a constant we can also normalise .

Suppose we temporarily assume that as . Then on a sufficiently large rectangle , we have on the boundary of the rectangle, hence on the interior by the maximum modulus principle. Sending , we obtain the claim.

To remove the assumption that goes to zero at infinity, we use the trick of giving ourselves an epsilon of room. Namely, we multiply by the holomorphic function for some . A little complex arithmetic shows that the function goes to zero at infinity in . Applying the previous case to this function, then taking limits as , we obtain the claim.

Corollary 16 (Phragmén-Lindelöf principle in a sector)Let be a continuous function on a sector for some , which is holomorphic on the interior of the sector and obeys the bound for some and . Suppose also that on the boundary of the sector for some . Then one also has in the interior.

*Proof:* Apply Theorem 14 to the function on the strip .

Exercise 17With the notation and hypotheses of Theorem 14, show that the function is log-convex on .

Exercise 18 (Hadamard three-circles theorem)Let be a holomorphic function on an annulus . Show that the function is log-convex on .

Exercise 19 (Phragmén-Lindelöf principle)Let be as in Theorem 14 with , but with the hypotheses after “Suppose also” in that theorem replaced instead by the bounds and for all and some exponents and a constant . Show that one has for all and some constant (which is allowed to depend on the constants in (12), as well as ). (Hint: it is convenient to work first in a half-strip such as for some large . Then multiply by something like for some suitable branch of the logarithm and apply a variant of Theorem 14 for the half-strip. A more refined estimate in this regard is due to Rademacher.) This particular version of the principle gives theconvexity boundfor Dirichlet series such as the Riemann zeta function. Bounds which exploit the deeper properties of these functions to improve upon the convexity bound are known assubconvexity boundsand are of major importance in analytic number theory, which is of course well outside the scope of this course.

Now we can establish a remarkable converse of sorts to Exercise 2(ii) known as the Paley-Wiener theorem, that links the exponential growth of (the analytic continuation) of a function with the support of its Fourier transform:

Theorem 20 (Paley-Wiener theorem)Let be a continuous function obeying the decay condition for all and some . Let . Then the following are equivalent:

- (i) The Fourier transform is supported on .
- (ii) extends analytically to an entire function that obeys the bound for some .
- (iii) extends analytically to an entire function that obeys the bound for some .

The continuity and decay hypotheses on can be relaxed, but we will not explore such generalisations here.

*Proof:* If (i) holds, then by Exercise 9, we have the inversion formula (4), and the claim (iii) then holds by a slight modification of Exercise 2(ii). Also, the claim (iii) clearly implies (ii).

Now we see why (iii) implies (i). We first assume that we have the stronger bound

for . Then we can apply Proposition 3 for any , and conclude in particular that for any and . Applying (14) and the triangle inequality, we see that If , we can then send and conclude that ; similarly for we can send and again conclude . This establishes (i) in this case.Now suppose we only have the weaker bound on assumed in (iii). We again use the epsilon of room trick. For any , we consider the modified function . This is still holomorphic on the lower half-plane and obeys a bound of the form (14) on this half-plane. An inspection of the previous arguments shows that we can still show that for despite no longer having holomorphicity on the entire upper half-plane; sending using dominated convergence we conclude that for . A similar argument (now using in place of shows that for . This proves (i).

Finally, we show that (ii) implies (iii). The function is entire, bounded on the real axis by (13), bounded on the upper imaginary axis by (iii), and has exponential growth. By Corollary 16, it is also bounded on the upper half-plane, which gives (iii) in the upper half-plane as well. A similar argument (using in place of ) also yields (iii) in the lower half-plane.

** — 3. The Hardy uncertainty principle — **

Informally speaking, the uncertainty principle for the Fourier transform asserts that a function and its Fourier transform cannot simultaneously be strongly localised, except in the degenerate case when is identically zero. There are many rigorous formulations of this principle. Perhaps the best known is the *Heisenberg uncertainty principle*

Another manifestation of the uncertainty principle is the following simple fact:

Lemma 21

- (i) If is an integrable function that has exponential decay in the sense that one has for all and some , then the Fourier transform is either identically zero, or only has isolated zeroes (that is to say, the set is discrete).
- (ii) If is a compactly supported continuous function such that is also compactly supported, then is identically zero.

*Proof:* For (i), we observe from Exercise 2(iii) that extends holomorphically to a strip around the real axis, and the claim follows since non-zero holomorphic functions have isolated zeroes. For (ii), we observe from (i) that must be identically zero, and the claim now follows from the Fourier inversion formula (Exercise 9).

Lemma 21(ii) rules out the existence of a bump function whose Fourier transform is also a bump function, which would have been a rather useful tool to have in harmonic analysis over the reals. (Such functions do exist however in some non-archimedean domains, such as the -adics.) On the other hand, from Exercise 1 we see that we do at least have gaussian functions whose Fourier transform also decays as a gaussian. Unfortunately this is basically the best one can do:

Theorem 22 (Hardy uncertainty principle)Let be a continuous function which obeys the bound for all and some . Suppose also that for all and some . Then is a scalar multiple of the gaussian , that is to say one has for some .

*Proof:* By replacing with the rescaled version , which replaces with the rescaled version , we may normalise . By multiplying by a small constant we may also normalise .

From Exercise 2(i), extends to an entire function. By the triangle inequality, we can bound

for any . Completing the square and using , we conclude the bound In particular, if we introduce the normalised function then In particular, is bounded by on the imaginary axis. On the other hand, from hypothesis is also bounded by on the real axis. We can now*almost*invoke the Phragmén-Lindelöf principle (Corollary 16) to conclude that is bounded on all four quadrants, but the growth bound we have (15) is just barely too weak. To get around this we use the epsilon of room trick. For any , the function is still entire, and is still bounded by in magnitude on the real line. From (15) we have so in particular it is bounded by on the slightly tilted imaginary axis . We can now apply Corollary 16 in the two acute-angle sectors between and to conclude that in those two sectors; letting , we conclude that in the first and third quadrants. A similar argument (using negative values of ) shows that in the second and fourth quadrants. By Liouville’s theorem, we conclude that is constant, thus we have for some complex number . The claim now follows from the Fourier inversion formula (Proposition 3(iv)) and Exercise 1.

One corollary of this theorem is that if is continuous and decays like or better, then cannot decay any faster than without vanishing identically. This is a stronger version of Lemma 21(ii). There is a more general tradeoff known as the Gel’fand-Shilov uncertainty principle, which roughly speaking asserts that if decays like then cannot decay faster than without vanishing identically, whenever are dual exponents in the sense that , and is large enough (the precise threshold was established in work of Morgan). See for instance this article of Nazarov for further discussion of these variants.

Exercise 23If is continuous and obeys the bound for some and and all , and obeys the bound for some and all , show that is of the form for some polynomial of degree at most .

Exercise 24In this exercise we develop an alternate proof of (a special case of) the Hardy uncertainty principle, which can be found in the original paper of Hardy. Let the hypotheses be as in Theorem 22.It is possible to adapt this argument to also cover the case of general that are not required to be even; see the paper of Hardy for details.

- (i) Show that the function is holomorphic on the region and obeys the bound in this region, where we use the standard branch of the square root.
- (ii) Show that the function is holomorphic on the region and obeys the bound in this region.
- (iii) Show that and agree on their common domain of definition.
- (iv) Show that the functions are constant. (You may find Exercise 13 from 246A Notes 4 to be useful.)
- (v) Use the above to give an alternate proof of Theorem 22 in the case when is even. (
Hint:subtract a constant multiple of a gaussian from to make vanish, and conclude on Taylor expansion around the origin that all the even moments vanish. Conclude that the Taylor series coefficients of around the origin all vanish.)

Exercise 25(This problem is due to Tom Liggett; see this previous post.) Let be a sequence of complex numbers bounded in magnitude by some bound , and suppose that the power series obeys the bound for all and some .

- (i) Show that the Laplace transform extends holomorphically to the region and obeys the bound in this region.
- (ii) Show that the function is holomorphic in the region , obeys the bound in this region, and agrees with on the common domain of definition.
- (iii) Show that is a constant multiple of .
- (iv) Show that the sequence is a constant multiple of the sequence .

Remark 26There are many further variants of the Hardy uncertainty principle. For instance we have the following uncertainty principle of Beurling, which we state in a strengthened form due to Bonami, Demange, and Jaming: if is a square-integrable function such that , then is equal (almost everywhere) to a polynomial times a gaussian; it is not difficult to show that this implies Theorem 22 and Exercise 23, as well as the Gel’fand-Shilov uncertainty principle. In recent years, PDE-based proofs of the Hardy uncertainty principle have been established, which have then been generalised to establish uncertainty principles for various Schrödinger type equations; see for instance this review article of Kenig. I also have some older notes on the Hardy uncertainty principle in this blog post. Finally, we mention theBeurling-Malliavin theorem, which provides a precise description of the possible decay rates of a function whose Fourier transform is compactly supported; see for instance this paper of Mashregi, Nazarov, and Khavin for a modern treatment.

## 37 comments

Comments feed for this article

23 January, 2021 at 1:13 pm

Michele Nardelli, Antonio NardelliDear Terence,

The topic of this very interesting post, can be applied also to the some sectors of theoretical physics? Regards

26 January, 2021 at 5:41 am

DiscoverReality CloverYes. I think that it is possible. It seems possible to conjecture that, based on this bound, there is a minimum pressure for a de-pressurising thermal system.

For example, based on this, one can show that if you pop a pressurised balloon (thermal system) in a vacuum then it obeys Newton’s first law – a body stays in motion… – by showing that the pressure inside the balloon does not go to zero. It approaches a minimum, hence the balloon stays in motion.

26 January, 2021 at 5:59 am

AnonymousThank you very much! Regards

23 January, 2021 at 2:50 pm

AnonymousTypos in third bullet point: the integral sign in the wrong place.

[Corrected, thanks – T.]23 January, 2021 at 2:51 pm

AnonymousAt the beginning for and .

24 January, 2021 at 3:40 pm

AnonymousTypo in

[Corrected – T.]23 January, 2021 at 5:47 pm

AnonymousIn the introduction of this notes, I am not sure about the proof of the “conversely” direction in (i),(ii),(iii). It is not intuitive for a sequence to satisfy certain property that it has to arise from a series of a particular form. How do we prove such statement? Say the converse direction in (i)? (Sorry, should have asked about Exe.5 from notes 4 last quarter).

[These are immediate from Proposition 7 and Theorem 15 from Notes 1 of the previous quarter’s notes. -T]23 January, 2021 at 9:56 pm

AnonymousHello,

In the point (iii) there is a typo for defn of f_1 and f_2. 2πi should replace by 1/2πi and unfortunately integral signs are not in right pisition.

24 January, 2021 at 12:20 pm

AnonymousDear Prof. Tao

I’ve read this your very interesting Note. What other note or papers of yours can you recommend me to consult in order to delve into these interesting topics? I’m grateful if You can show/suggest me other Your papers regarding this subject, All my best wishes for your works and all my esteem and gratitude,

All the best

25 January, 2021 at 1:20 pm

AnonymousA simple heuristic interpretation of Poisson summation formula (6) is by observing that it can be written as

(6′)

where is a nonnegative (singular) measure supported on the integers and has a unit measure on each integer.

It is not difficult to verify that is invariant under the Fourier transform, so (6′) is equivalent to

(6”)

Which (at least heuristically) follows from Plancherel’s theorem on the invariance of inner products in under Fourier transform.

5 February, 2021 at 8:47 am

N is a numberThis is nice! Thank you.

25 January, 2021 at 5:27 pm

AnonymousTitle of this notes: “factorisation” should be factorization.

26 January, 2021 at 12:22 pm

AnonymousThis is for 246B, Notes 1: —

26 January, 2021 at 8:50 am

DiscoverReality CloverDear All,

I apologise for accidentally rating my own comment – an action which l couldn’t undo.

28 January, 2021 at 12:47 pm

AnonymousIt is interesting to observe that both the Fourier and Hilbert transforms, as linear operators on , have the property that their fourth power is the identity operator.

28 January, 2021 at 1:48 pm

AnonymousHi Terence Tao. I am only 14, but I understand the fundamentals of this maths work. Please can you teach me more about this. Thank you

2 February, 2021 at 4:36 am

Jim HefferonNot about this article. The link to “Cult of Genius” in the sidebar has changed. A litle googling found that it is now https://www.discovermagazine.com/mind/the-cult-of-genius. Regards.

[Corrected, thanks – T.]5 February, 2021 at 8:46 am

N is a numberJust some nitpicking: in point ii towards the beginning of the post, one should have .

[Corrected, thanks – T.]5 February, 2021 at 6:34 pm

AnonymousAt the end of today’s lecture, you mentioned that Fourier transform is the rotation of the function by 90 degrees in phase space, which gives a good interpretation that take Fourier transform 4 times gets back the original function. Detailed proof involves micro local analysis and something about symplectic structure. Could you provide some reference on this topic? It is a very interesting subject.

6 February, 2021 at 10:56 am

Terence TaoIn the physics literature this is known as the phase space formulation of quantum mechanics, and the Wigner quasiprobability distribution can be used as a candidate for the “phase space transform” of a function, though it is not the only such choice. A more mathematical treatment of some of these ideas can be found in Folland’s “Harmonic analysis in phase space”.

9 February, 2021 at 9:28 am

N is a numberDoesn’t the proof of Exercise 2 part i follow directly by putting to be a complex number and then utilising the super-exponential decay of to obtain a bound for the integral displayed on the RHS of equation 3 (for instance by using the dominated convergence theorem), and then conclude that defined in this way will be entire by observing that what we did actually implies that has no poles or other types of singularities.

9 February, 2021 at 9:41 am

N is a numberOr once we define in this way, we can also use the second funadamental theorem of calculus to obtain the claim.

9 February, 2021 at 8:37 pm

Aditya Guha RoyI think what you wrote above is correct, and then we can use the second fundamental theorem of calculus to conclude that defined in this way will be holomorphic.

9 February, 2021 at 11:39 am

AnonymousYou mentioned in class yesterday that we use Harr measure for the Mellin transform on the half line because Harr measure is scaling invariant, and Lebesgue measure is translation invariant. Translation invariant property is more useful on the whole real line, and not very useful on the half line. Please elaborate on this. Don’t we define the Laplace transform on the half line using Lebesgure measure?

9 February, 2021 at 6:08 pm

Terence TaoOne can still try to exploit translation invariance on the half-line, but one picks up additional terms at the boundary which complicate things a little. Compare for instance the effects of differentiation or translation on the Laplace transform displayed in the table in https://en.wikipedia.org/wiki/Laplace_transform#Properties_and_theorems with the corresponding properties for the Fourier transform in https://en.wikipedia.org/wiki/Fourier_transform#Basic_properties .

10 February, 2021 at 8:26 am

Aditya Guha RoyIn ii of Exercise 2 shouldn’t we have instead of ?

[As it turns out, one does not need a dependence on here. -T]10 February, 2021 at 8:36 am

Aditya Guha RoyIn ii of Exercise 2 I think we should have ….. .

[Here we use to denote the estimate (that is, the absolute values are implicit in the notation) -T.]12 February, 2021 at 9:39 am

246B, Notes 4: The Riemann zeta function and the prime number theorem | What's new[…] function for imaginary values of the argument in the upper half-plane (so ); from Exercise 7 of Notes 2 we have the modularity […]

13 February, 2021 at 6:01 am

Aditya Guha RoyIn the 4th line of the proof of part (iv.) in Proposition 3, the second integral should be over .

[Corrected, thanks – T.]13 February, 2021 at 6:05 am

N is a numberHere we see a proof of the Fourier inversion theorem which becomes easy if one uses the tools from complex analysis as demonstrated above in the proof of Proposition 3.

We also know that for the discrete setting we can give a proof of the Fourier inversion formula which uses some algebraic manipulations and orthogonality properties.

In general is there a unified approach to prove Fourier inversion type results?

14 February, 2021 at 10:28 am

Terence TaoThere are various ways to establish Plancherel’s theorem (which includes the Fourier inversion formula as a component) on arbitrary locally compact abelian (LCA) groups; see Section 4 of these notes for a brief discussion of two approaches, one using Gelfand theory, and another via Bochner’s theorem. For special groups such as the real line or the circle there are shortcuts available that utilise special features of those groups, such as the presence of gaussians or holomorphic extensions to various regions of the complex plane, but these do not seem to generalise well to the LCA setting, which seems to be the most natural framework in which to present Fourier analysis in a unified manner.

18 February, 2021 at 5:32 pm

AnonymousAbout the Beurling uncertainty principle [Remark 26], how do you get the hyperbola geometric interpretation phase portrait? Does it follow directly from the square integrability condition? I know and are kind of “inverse” to each other.

[It arises from consideration of where the weight is exponentially large. -T.]27 February, 2021 at 9:14 am

Course announcement: 246B, complex analysis | What's new[…] Connections with the Fourier transform on the real line; […]

27 February, 2021 at 9:36 am

246B, Notes 1: Zeroes, poles, and factorisation of meromorphic functions | What's new[…] Previous set of notes: 246A Notes 5. Next set of notes: Notes 2. […]

27 February, 2021 at 9:38 am

246B, Notes 3: Elliptic functions and modular forms | What's new[…] set of notes: Notes 2. Next set of notes: Notes […]

18 January, 2022 at 1:38 am

QuanDear Professor Tao,

In the Laplace transform problem, one comes up with a different restriction for {f} that {f} is continuously differentiable and {\vert f(x)e^{-xt}\vert \leq \frac{C_t}{1+x^2}, \forall t \in {\bf R} (1)}. Then, we construct a compactly supported function {h(x)} that agrees with {2\pi f(2\pi x)e^{-2\pi xt}} when {x\geq 0} for the purpose of applying Fourier transform (modify the proof of excercise 9 for function with contable discontinuities), and one can obtain the Laplace transform for the class of function satisfies (1). Is this approach valid? Thank you for your time, professor.

[Large parts of Exercise 10 would still be valid under these more general hypotheses on , though perhaps not all. It would be a instructive exercise for you to work out the details. -T]24 January, 2022 at 1:38 pm

Quân NguyễnI think the proof that Exercise 9 can still be applied for the Laplace inverse transform (exercise 10) if we withdraw the condition of $f$ being differentiable.