In set theory, a function is defined as an object that evaluates every input
to exactly one output
. However, in various branches of mathematics, it has become convenient to generalise this classical concept of a function to a more abstract one. For instance, in operator algebras, quantum mechanics, or non-commutative geometry, one often replaces commutative algebras of (real or complex-valued) functions on some space
, such as
or
, with a more general – and possibly non-commutative – algebra (e.g. a
-algebra or a von Neumann algebra). Elements in this more abstract algebra are no longer definable as functions in the classical sense of assigning a single value
to every point
, but one can still define other operations on these “generalised functions” (e.g. one can multiply or take inner products between two such objects).
Generalisations of functions are also very useful in analysis. In our study of spaces, we have already seen one such generalisation, namely the concept of a function defined up to almost everywhere equivalence. Such a function
(or more precisely, an equivalence class of classical functions) cannot be evaluated at any given point
, if that point has measure zero. However, it is still possible to perform algebraic operations on such functions (e.g. multiplying or adding two functions together), and one can also integrate such functions on measurable sets (provided, of course, that the function has some suitable integrability condition). We also know that the
spaces can usually be described via duality, as the dual space of
(except in some endpoint cases, namely when
, or when
and the underlying space is not
-finite).
We have also seen (via the Lebesgue-Radon-Nikodym theorem) that locally integrable functions on, say, the real line
, can be identified with locally finite absolutely continuous measures
on the line, by multiplying Lebesgue measure
by the function
. So another way to generalise the concept of a function is to consider arbitrary locally finite Radon measures
(not necessarily absolutely continuous), such as the Dirac measure
. With this concept of “generalised function”, one can still add and subtract two measures
, and integrate any measure
against a (bounded) measurable set
to obtain a number
, but one cannot evaluate a measure
(or more precisely, the Radon-Nikodym derivative
of that measure) at a single point
, and one also cannot multiply two measures together to obtain another measure. From the Riesz representation theorem, we also know that the space of (finite) Radon measures can be described via duality, as linear functionals on
.
There is an even larger class of generalised functions that is very useful, particularly in linear PDE, namely the space of distributions, say on a Euclidean space . In contrast to Radon measures
, which can be defined by how they “pair up” against continuous, compactly supported test functions
to create numbers
, a distribution
is defined by how it pairs up against a smooth compactly supported function
to create a number
. As the space
of smooth compactly supported functions is smaller than (but dense in) the space
of continuous compactly supported functions (and has a stronger topology), the space of distributions is larger than that of measures. But the space
is closed under more operations than
, and in particular is closed under differential operators (with smooth coefficients). Because of this, the space of distributions is similarly closed under such operations; in particular, one can differentiate a distribution and get another distribution, which is something that is not always possible with measures or
functions. But as measures or functions can be interpreted as distributions, this leads to the notion of a weak derivative for such objects, which makes sense (but only as a distribution) even for functions that are not classically differentiable. Thus the theory of distributions can allow one to rigorously manipulate rough functions “as if” they were smooth, although one must still be careful as some operations on distributions are not well-defined, most notably the operation of multiplying two distributions together. Nevertheless one can use this theory to justify many formal computations involving derivatives, integrals, etc. (including several computations used routinely in physics) that would be difficult to formalise rigorously in a purely classical framework.
If one shrinks the space of distributions slightly, to the space of tempered distributions (which is formed by enlarging dual class to the Schwartz class
), then one obtains closure under another important operation, namely the Fourier transform. This allows one to define various Fourier-analytic operations (e.g. pseudodifferential operators) on such distributions.
Of course, at the end of the day, one is usually not all that interested in distributions in their own right, but would like to be able to use them as a tool to study more classical objects, such as smooth functions. Fortunately, one can recover facts about smooth functions from facts about the (far rougher) space of distributions in a number of ways. For instance, if one convolves a distribution with a smooth, compactly supported function, one gets back a smooth function. This is a particularly useful fact in the theory of constant-coefficient linear partial differential equations such as , as it allows one to recover a smooth solution
from smooth, compactly supported data
by convolving
with a specific distribution
, known as the fundamental solution of
. We will give some examples of this later in these notes.
It is this unusual and useful combination of both being able to pass from classical functions to generalised functions (e.g. by differentiation) and then back from generalised functions to classical functions (e.g. by convolution) that sets the theory of distributions apart from other competing theories of generalised functions, in particular allowing one to justify many formal calculations in PDE and Fourier analysis rigorously with relatively little additional effort. On the other hand, being defined by linear duality, the theory of distributions becomes somewhat less useful when one moves to more nonlinear problems, such as nonlinear PDE. However, they still serve an important supporting role in such problems as a “ambient space” of functions, inside of which one carves out more useful function spaces, such as Sobolev spaces, which we will discuss in the next set of notes.
— 1. Smooth functions with compact support —
In the rest of the notes we will work on a fixed Euclidean space . (One can also define distributions on other domains related to
, such as open subsets of
, or
-dimensional manifolds, but for simplicity we shall restrict attention to Euclidean spaces in these notes.)
A test function is any smooth, compactly supported function ; the space of such functions is denoted
. (In some texts, this space is denoted
instead.)
From analytic continuation one sees that there are no real-analytic test functions other than the zero function. Despite this negative result, test functions actually exist in abundance:
- (i) Show that there exists at least one test function that is not identically zero. (Hint: it suffices to do this for
. One starting point is to use the fact that the function
defined by
for
and
otherwise is smooth, even at the origin
.)
- (ii) Show that if
and
is absolutely integrable and compactly supported, then the convolution
is also in
. (Hint: first show that
is continuously differentiable with
.)
- (iii) (
Urysohn lemma) Let
be a compact subset of
, and let
be an open neighbourhood of
. Show that there exists a function
supported in
which equals
on
. (Hint: use the ordinary Urysohn lemma to find a function in
that equals
on a neighbourhood of
and is supported in a compact subset of
, then convolve this function by a suitable test function.)
- (iv) Show that
is dense in
(in the uniform topology), and dense in
(with the
topology) for all
.
The space is clearly a vector space. Now we place a (very strong!) topology on it. We first observe that
, where
ranges over all compact subsets of
and
consists of those functions
which are supported in
. Each
will be given a topology (called the smooth topology) generated by the norms
for , where we view
as a
-dimensional vector (or, if one wishes, a
-dimensional rank
tensor); thus a sequence
converges to a limit
if and only if
converges uniformly to
for all
. (This gives
the structure of a Fréchet space, though we will not use this fact here.)
We make the trivial remark that if are compact sets, then
is a subspace of
, and the topology on the former space is the restriction of the topology of the latter space. Because of this, we are able to give
a (very strong) topology as follows. Call a seminorm
on
good if it is continuous function on
for each compact
(or equivalently, the ball
is open in
for each compact
). We then give
the topology defined by all good seminorms. Clearly, this makes
a (locally convex) topological vector space.
Exercise 2 Let
be a sequence in
, and let
be another function in
. Show that
converges in the topology of
to
if and only if there exists a compact set
such that
are all supported in
, and
converges to
in the smooth topology of
.
Exercise 3
- (i) Show that the topology of
is first countable for every compact
.
- (ii) Show that the topology of
is not first countable. (Hint: given any countable sequence of open neighbourhoods of
, build a new open neighbourhood that does not contain any of the previous ones, using the
-compact nature of
.)
- (iii) As an additional challenge, construct a set
such that
is an adherent point of
, but
is not as the limit of any sequence in
.
There are plenty of continuous operations on :
- (i) Let
be a compact set. Show that a linear map
into a normed vector space
is continuous if and only if there exists
and
such that
for all
.
- (ii) Let
be compact sets. Show that a linear map
is continuous if and only if for every
there exists
and a constant
such that
for all
.
- (iii) Show that a linear map
from the space of test functions into a topological vector space generated by some family of seminorms (i.e., a locally convex topological vector space) is continuous if and only if it is sequentially continuous (i.e. whenever
converges to
in
,
converges to
in
), and if and only if
is continuous for each compact
. Thus while first countability fails for
, we have a serviceable substitute for this property.
- (iv) Show that the inclusion map from
to
is continuous for every
.
- (v) Show that a map
is continuous if and only if for every compact set
there exists a compact set
such that
maps
continuously to
.
- (vi) Show that every linear differential operator with smooth coefficients is a continuous operation on
.
- (vii) Show that convolution with any absolutely integrable, compactly supported function is a continuous operation on
.
- (viii) Show that the product operation
is continuous from
to
.
A sequence of continuous, compactly supported functions is said to be an approximation to the identity if the
are non-negative, have total mass
equal to
, and whose supports shrink to the origin, thus for any fixed
,
is supported on the ball
for
sufficiently large. One can generate such a sequence by starting with a single non-negative continuous compactly supported function
of total mass
, and then setting
; many other constructions are possible also.
One has the following useful fact:
Exercise 5 Let
be a sequence of approximations to the identity.
- (i) If
is continuous, show that
converges uniformly on compact sets to
.
- (ii) If
for some
, show that
converges in
to
. (Hint: use (i), the density of
in
, and Young’s inequality.)
- (iii) If
, show that
converges in
to
. (Hint: use the identity
, cf. Exercise 1(ii).)
Exercise 6 Show that
is separable. (Hint: it suffices to show that
is separable for each compact
. There are several ways to accomplish this. One is to begin with the Stone-Weierstrass theorem, which will give a countable set which is dense in the uniform topology, then use the fundamental theorem of calculus to strengthen the topology. Another is to use Exercise 5 and then discretise the convolution. Another is to embed
into a torus and use Fourier series, noting that the Fourier coefficients
of a smooth function
decay faster than any power of
.)
— 2. Distributions —
Now we can define the concept of a distribution.
Definition 1 (Distribution) A distribution on
is a continuous linear functional
from
to
. The space of such distributions is denoted
, and is given the weak-* topology. In particular, a sequence of distributions
converges (in the sense of distributions) to a limit
if one has
for all
.
A technical point: we endow the space
with the conjugate complex structure. Thus, if
, and
is a complex number, then
is the distribution that maps a test function
to
rather than
; thus
. This is to keep the analogy between the evaluation of a distribution against a function, and the usual Hermitian inner product
of two test functions.
From Exercise 4, we see that a linear functional is a distribution if, for every compact set
, there exists
and
such that
Exercise 7 Show that
is a Hausdorff topological vector space.
We note two basic examples of distributions:
- Any locally integrable function
can be viewed as a distribution, by writing
for all test functions
.
- Any complex Radon measure
can be viewed as a distribution, by writing
, where
is the complex conjugate of
(thus
). (Note that this example generalises the preceding one, which corresponds to the case when
is absolutely continuous with respect to Lebesgue measure.) Thus, for instance, the Dirac measure
at the origin is a distribution, with
for all test functions
.
Exercise 8 Show that the above identifications of locally integrable functions or complex Radon measures with distributions are injective. (Hint: use Exercise 1(iv).)
From the above exercise, we may view locally integrable functions and locally finite measures as a special type of distribution. In particular, and
are now contained in
for all
.
Exercise 9 Show that if a sequence of locally integrable functions converge in
to a limit, then they also converge in the sense of distributions; similarly, if a sequence of complex Radon measures converge in the vague topology to a limit, then they also converge in the sense of distributions.
Thus we see that convergence in the sense of distributions is among the weakest of the notions of convergence used in analysis; however, from the Hausdorff property, distributional limits are still unique.
Exercise 10 If
is a sequence of approximations to the identity, show that
converges in the sense of distributions to the Dirac distribution
.
More exotic examples of distributions can be given:
Exercise 11 (Derivative of the delta function) Let
. Show that the functional
for all test functions
is a distribution which does not arise from either a locally integrable function or a Radon measure. (Note how it is important here that
is smooth (and in particular differentiable, and not merely continuous.) The presence of the minus sign will be explained shortly.
Exercise 12 (Principal value of
) Let
. Show that the functional
defined by the formula
is a distribution which does not arise from either a locally integrable function or a Radon measure. (Note that
is not a locally integrable function!)
Exercise 13 (Distributional interpretations of
) Let
. For any
, show that the functional
defined by the formula
is a distribution that does not arise from either a locally integrable function or a Radon measure. Note that any two such functionals
differ by a constant multiple of the Dirac delta distribution.
Exercise 14 A distribution
is said to be real if
is real for every real-valued test function
. Show that every distribution
can be uniquely expressed as
for some real distributions
.
Exercise 15 A distribution
is said to be non-negative if
is non-negative for every non-negative test function
. Show that a distribution is non-negative if and only if it is a non-negative Radon measure. (Hint: use the Riesz representation theorem and Exercise 1(iv).) Note that this implies that the analogue of the Jordan decomposition fails for distributions; any distribution which is not a Radon measure will not be the difference of non-negative distributions.
We will now extend various operations on locally integrable functions or Radon measures to distributions by arguing by analogy. (Shortly we will give a more formal approach, based on density.)
We begin with the operation of multiplying a distribution by a smooth function
. Observe that
for all test functions . Inspired by this formula, we define the product
of a distribution with a smooth function by setting
for all test functions . It is easy to see (e.g. using Exercise 4(vi)) that this defines a distribution
, and that this operation is compatible with existing definitions of products between a locally integrable function (or Radon measure) with a smooth function. It is important that
is smooth (and not merely, say, continuous) because one needs the product of a test function
with
to still be a test function.
Exercise 16 Let
. Establish the identity
for any smooth function
. In particular,
where we abuse notation slightly and write
for the identity function
. Conversely, if
is a distribution such that
show that
is a constant multiple of
. (Hint: Use the identity
to write
as the sum of
and
times a test function for any test function
, where
is a fixed test function equalling
at the origin.)
Remark 1 Even though distributions are not, strictly speaking, functions, it is often useful heuristically to view them as such, thus for instance one might write a distributional identity such as
suggestively as
. Another useful (and rigorous) way to view such identities is to write distributions such as
as a limit of approximations to the identity
, and show that the relevant identity becomes true in the limit; thus, for instance, to show that
, one can show that
in the sense of distributions as
. (In fact,
converges to zero in the
norm.)
Exercise 17 Let
. With the distribution
from Exercise 12, show that
is equal to
. With the distributions
from Exercise 13, show that
, where
is the signum function.
A distribution is said to be supported in a closed set
in
for all
that vanish on an open neighbourhood of
. The intersection of all
that
is supported on is denoted
and is referred to as the support of the distribution; this is the smallest closed set that
is supported on. Thus, for instance, the Dirac delta function is supported on
, as are all derivatives of that function. (Note here that it is important that
vanish on a neighbourhood of
, rather than merely vanishing on
itself; for instance, in one dimension, there certainly exist test functions
that vanish at
but nevertheless have a non-zero inner product with
.)
Exercise 18 Show that every distribution is the limit of a sequence of compactly supported distributions (using the weak-* topology, of course). (Hint: Approximate a distribution
by the truncated distributions
for some smooth cutoff functions
constructed using Exercise 1(iii).)
In a similar spirit, we can convolve a distribution by an absolutely integrable, compactly supported function
. From Fubini’s theorem we observe the formula
for all test functions , where
. Inspired by this formula, we define the convolution
of a distribution with an absolutely integrable, compactly supported function by the formula
for all test functions . This gives a well-defined distribution
(thanks to Exercise 4(vii)) which is compatible with previous notions of convolution.
Example 1 One has
for all test functions
. In one dimension, we have
(why?), thus differentiation can be viewed as convolution with a distribution.
A remarkable fact about convolutions of two functions is that they inherit the regularity of the smoother of the two factors
(in contrast to products
, which tend to inherit the regularity of the rougher of the two factors). (This disparity can be also be seen by contrasting the identity
with the identity
.) In the case of convolving distributions with test functions, this phenomenon is manifested as follows:
Lemma 2 Let
be a distribution, and let
be a test function. Then
is equal to a smooth function.
Proof: If were itself a smooth function, then one could easily verify the identity
where . As
is a test function, it is easy to see that
varies smoothly in
in any
norm (indeed, it has Taylor expansions to any order in such norms) and so the right-hand side is a smooth function of
. So it suffices to verify the identity (3). As distributions are defined against test functions
, it suffices to show that
On the other hand, we have from (2) that
So the only issue is to justify the interchange of integral and inner product:
Certainly, (from the compact support of ) any Riemann sum can be interchanged with the inner product:
where ranges over some lattice and
is the volume of the fundamental domain. A modification of the argument that shows convergence of the Riemann integral for smooth, compactly supported functions then works here and allows one to take limits; we omit the details.
This has an important corollary:
Lemma 3 Every distribution is the limit of a sequence of test functions. In particular,
is dense in
.
Proof: By Exercise 18, it suffices to verify this for compactly supported distributions . We let
be a sequence of approximations to the identity. By Exercise 5(iii) and (2), we see that
converges in the sense of distributions to
. By Lemma 2,
is a smooth function; as
and
are both compactly supported,
is compactly supported also. The claim follows.
Because of this lemma, we can formalise the previous procedure of extending operations that were previously defined on test functions, to distributions, provided that these operations were continuous in distributional topologies. However, we shall continue to proceed by analogy as it requires fewer verifications in order to motivate the definition.
Exercise 19 Another consequence of Lemma 2 is that it allows one to extend the definition (2) of convolution to the case when
is not an integrable function of compact support, but is instead merely a distribution of compact support. Adopting this convention, show that convolution of distributions of compact support is both commutative and associative. (Hint: this can either be done directly, or by carefully taking limits using Lemma 3.)
The next operation we will introduce is that of differentiation. An integration by parts reveals the identity
for any test functions and
. Inspired by this, we define the (distributional) partial derivative
of a distribution
by the formula
This can be verified to still be a distribution, and by Exercise 4(vi), the operation of differentiation is a continuous one on distributions. More generally, given any linear differential operator with smooth coefficients, one can define
for a distribution
by the formula
where is the adjoint differential operator
, which can be defined implicitly by the formula
for test functions , or more explicitly by replacing all coefficients with complex conjugates, replacing each partial derivative
with its negative, and reversing the order of operations (thus for instance the adjoint of the first-order operator
would be
).
Example 2 The distribution
defined in Exercise 11 is the derivative
of
, as defined by the above formula.
Many of the identities one is used to in classical calculus extend to the distributional setting (as one would already expect from Lemma 3). For instance:
Exercise 20 (Product rule) Let
be a distribution, and let
be smooth. Show that
for all
.
Exercise 21 Let
. Show that
in three different ways:
- Directly from the definitions;
- using the product rule;
- Writing
as the limit of approximations
to the identity.
- (i) Show that if
is a distribution and
is an integer, then
if and only if is a linear combination of
and its first
derivatives
.
- (ii) Show that a distribution
is supported on
if and only if it is a linear combination of
and finitely many of its derivatives.
- (iii) Generalise (ii) to the case of general dimension
(where of course one now uses partial derivatives instead of derivatives).
Exercise 23 Let
.
- Show that the derivative of the Heaviside function
is equal to
.
- Show that the derivative of the signum function
is equal to
.
- Show that the derivative of the locally integrable function
is equal to
.
- Show that the derivative of the locally integrable function
is equal to the distribution
from Exercise 13.
- Show that the derivative of the locally integrable function
is the locally integrable function
.
If a locally integrable function has a distributional derivative which is also a locally integrable function, we refer to the latter as the weak derivative of the former. Thus, for instance, the weak derivative of is
(as one would expect), but
does not have a weak derivative (despite being (classically) differentiable almost everywhere), because the distributional derivative
of this function is not itself a locally integrable function. Thus weak derivatives differ in some respects from their classical counterparts, though of course the two concepts agree for smooth functions.
Exercise 24 Let
. Show that for any
, and any distribution
, we have
, thus weak derivatives commute with each other. (This is in contrast to classical derivatives, which can fail to commute for non-smooth functions; for instance,
at the origin
, despite both derivatives being defined. More generally, weak derivatives tend to be less pathological than classical derivatives, but of course the downside is that weak derivatives do not always have a classical interpretation as a limit of a Newton quotient.)
Exercise 25 Let
, and let
be an integer. Let us say that a compactly supported distribution
has of order at most
if the functional
is continuous in the
norm. Thus, for instance,
has order at most
, and
has order at most
, and every compactly supported distribution is of order at most
for some sufficiently large
.
- Show that if
is a compactly supported distribution of order at most
, then it is a compactly supported Radon measure.
- Show that if
is a compactly supported distribution of order at most
, then
has order at most
.
- Conversely, if
is a compactly supported distribution of order
, then we can write
for some compactly supported distributions of order
. (Hint: one has to “dualise” the fundamental theorem of calculus, and then apply smooth cutoffs to recover compact support.)
- Show that every compactly supported distribution can be expressed as a finite linear combination of (distributional) derivatives of compactly supported Radon measures.
- Show that every compactly supported distribution can be expressed as a finite linear combination of (distributional) derivatives of functions in
, for any fixed
.
We now set out some other operations on distributions. If we define the translation of a test function
by a shift
by the formula
, then we have
for all test functions , so it is natural to define the translation
of a distribution
by the formula
Next, we consider linear changes of variable.
Exercise 26 (Linear changes of variable) Let
, and let
be a linear transformation. Given a distribution
, let
be the distribution given by the formula
for all test functions
. (How would one motivate this formula?)
- Show that
for all linear transformations
.
- If
, show that
for all linear transformations
.
- Conversely, if
and
is a distribution such that
for all linear transformations
. (Hint: first show that there exists a constant
such that
whenever
is a bump function supported in
. To show this, approximate
by the function
for
an approximation to the identity.)
Remark 2 One can also compose distributions with diffeomorphisms. However, things become much more delicate if the map one is composing with contains stationary points; for instance, in one dimension, one cannot meaningfully make sense of
(the composition of the Dirac delta distribution with
); this can be seen by first noting that for an approximation
to the identity,
does not converge to a limit in the distributional sense.
Exercise 27 (Tensor product of distributions) Let
be integers. If
and
are distributions, show that there is a unique distribution
with the property that
for all test functions
,
, where
is the tensor product
of
and
. (Hint: like many other constructions of tensor products, this is rather intricate. One way is to start by fixing two cutoff functions
on
respectively, and define
on modulated test functions
for various frequencies
, and then use Fourier series to define
on
for smooth
. Then show that these definitions of
are compatible for different choices of
and can be glued together to form a distribution; finally, go back and verify (4).)
We close this section with one caveat. Despite the many operations that one can perform on distributions, there are two types of operations which cannot, in general, be defined on arbitrary distributions (at least while remaining in the class of distributions):
- Nonlinear operations (e.g. taking the absolute value of a distribution); or
- Multiplying a distribution by anything rougher than a smooth function.
Thus, for instance, there is no meaningful way to interpret the square of the Dirac delta function as a distribution. This is perhaps easiest to see using an approximation
to the identity:
converges to
in the sense of distributions, but
does not converge to anything (the integral against a test function that does not vanish at the origin will go to infinity as
). For similar reasons, one cannot meaningfully interpret the absolute value
of the derivative of the delta function. (One also cannot multiply
by
– why?)
Exercise 28 Let
be a normed vector space which contains
as a dense subspace (and such that the inclusion of
to
is continuous). The adjoint (or transpose) of this inclusion map is then an injection from
to the space of distributions
; thus
can be viewed as a subspace of the space of distributions.
- Show that the closed unit ball in
is also closed in the space of distributions.
- Conclude that any distributional limit of a bounded sequence in
for
, is still in
.
- Show that the previous claim fails for
, but holds for the space
of finite measures.
— 3. Tempered distributions —
The list of operations one can define on distributions has one major omission – the Fourier transform . Unfortunately, one cannot easily define the Fourier transform for all distributions. One can see this as follows. From Plancherel’s theorem one has the identity
for test functions , so one would like to define the Fourier transform
of a distribution
by the formula
Unfortunately this does not quite work, because the adjoint Fourier transform of a test function is not a test function, but is instead just a Schwartz function. (Indeed, by Exercise 55 of Notes 2, it is not possible to find a non-trivial test function whose Fourier transform is again a test function.) To address this, we need to work with a slightly smaller space than that of all distributions, namely those of tempered distributions:
Definition 4 (Tempered distributions) A tempered distribution is a continuous linear functional
on the Schwartz space
(with the topology given by Exercise 25 of Notes 2), i.e. an element of
.
Since embeds continuously into
(with a dense image), we see that the space of tempered distributions can be embedded into the space of distributions. However, not every distribution is tempered:
Example 3 The distribution
is not tempered. Indeed, if
is a bump function, observe that the sequence of functions
converges to zero in the Schwartz space topology, but
does not go to zero, and so this distribution does not correspond to a tempered distribution.
On the other hand, distributions which avoid this sort of exponential growth, and instead only grow polynomially, tend to be tempered:
Exercise 29 Show that any Radon measure
which is of polynomial growth in the sense that
for all
and some constants
, where
is the ball of radius
centred at the origin in
, is tempered.
Remark 3 As a zeroth approximation, one can roughly think of “tempered” as being synonymous with “polynomial growth”. However, this is not strictly true: for instance, the (weak) derivative of a function of polynomial growth will still be tempered, but need not be of polynomial growth (for instance, the derivative
of
is a tempered distribution, despite having exponential growth). While one can eventually describe which distributions are tempered by measuring their “growth” in both physical space and in frequency space, we will not do so here.
Most of the operations that preserve the space of distributions, also preserve the space of tempered distributions. For instance:
Exercise 30
- Show that any derivative of a tempered distribution is again a tempered distribution.
- Show that and any convolution of a tempered distribution with a compactly supported distribution is again a tempered distribution.
- Show that if
is a measurable function which is rapidly decreasing in the sense that
is an
function for each
, then a convolution of a tempered distribution with
can be defined, and is again a tempered distribution.
- Show that if
is a smooth function such that
and all its derivatives have at most polynomial growth (thus for each
there exists
such that
for all
) then the product of a tempered distribution with
is again a tempered distribution. Give a counterexample to show that this statement fails if the polynomial growth hypotheses are dropped.
- Show that the translate of a tempered distribution is again a tempered distribution.
But we can now add a new operation to this list using (5): as the Fourier transform maps Schwartz functions continuously to Schwartz functions, it also continuously maps the space of tempered distributions to itself. One can also define the inverse Fourier transform
on tempered distributions in a similar manner.
It is not difficult to extend many of the properties of the Fourier transform from Schwartz functions to distributions. For instance:
Exercise 31 Let
be a tempered distribution, and let
be a Schwartz function.
- (Inversion formula) Show that
.
- (Multiplication intertwines with convolution) Show that
and
.
- (Translation intertwines with modulation) For any
, show that
, where
. Similarly, show that for any
, one has
.
- (Linear transformations) For any invertible linear transformation
, show that
.
- (Differentiation intertwines with polynomial multiplication) For any
, show that
, where
and
is the
coordinate function in physical space and frequency space respectively, and similarly
.
Exercise 32 Let
.
- (Inversion formula) Show that
and
.
- (Orthogonality) Let
be a subspace of
, and let
be Lebesgue measure on
. Show that
is Lebesgue measure on the orthogonal complement
of
. (Note that this generalises the previous exercise.)
- (Poisson summation formula) Let
be the distribution
Show that this is a tempered distribution which is equal to its own Fourier transform.
One can use these properties of tempered distributions to start solving constant-coefficient PDE. We first illustrate this by an ODE example, showing how the formal symbolic calculus for solving such ODE that you may have seen as an undergraduate, can now be (sometimes) justified using tempered distributions.
Exercise 33 Let
, let
be real numbers, and let
be the operator
.
- If
, use the Fourier transform to show that all tempered distribution solutions to the ODE
are of the form
for some constants
.
- If
, show that all tempered distribution solutions to the ODE
are of the form
for some constants
.
Remark 4 More generally, one can solve any homogeneous constant-coefficient ODE using tempered distributions and the Fourier transform so long as the roots of the characteristic polynomial are purely imaginary. In all other cases, solutions can grow exponentially as
or
and so are not tempered. There are other theories of generalised functions that can handle these objects (e.g. hyperfunctions) but we will not discuss them here.
Now we turn to PDE. To illustrate the method, let us focus on solving Poisson’s equation
in , where
is a Schwartz function and
is a distribution, where
is the Laplacian. (In some texts, particularly those using spectral analysis, the Laplacian is occasionally defined instead as
, to make it positive semi-definite, but we will eschew that sign convention here, though of course the theory is only changed in a trivial fashion if one adopts it.)
We first settle the question of uniqueness:
Exercise 34 Let
. Using the Fourier transform, show that the only tempered distributions
which are harmonic (by which we mean that
in the sense of distributions) are the harmonic polynomials. (Hint: use Exercise 22.) Note that this generalises Liouville’s theorem. There are of course many other harmonic functions than the harmonic polynomials, e.g.
, but such functions are not tempered distributions.
From the above exercise, we know that the solution to (6), if tempered, is defined up to harmonic polynomials. To find a solution, we observe that it is enough to find a fundamental solution, i.e. a tempered distribution
solving the equation
Indeed, if one then convolves this equation with the Schwartz function , and uses the identity
(which can either be seen directly, or by using Exercise 31), we see that
will be a tempered distribution solution to (6) (and all the other solutions will equal this solution plus a harmonic polynomial). So, it is enough to locate a fundamental solution
. We can take Fourier transforms and rewrite this equation as
(here we are treating the tempered distribution as a function to emphasise that the dependent variable is now
). It is then natural to propose to solve this equation as
though this may not be the unique solution (for instance, one is free to modify by a multiple of the Dirac delta function, cf. Exercise 16).
A short computation in polar coordinates shows that is locally integrable in dimensions
, so the right-hand side of (7) makes sense. To then compute
explicitly, we have from the distributional inversion formula that
so we now need to figure out what the Fourier transform of a negative power of (or the adjoint Fourier transform of a negative power of
) is.
Let us work formally at first, and consider the problem of computing the Fourier transform of the function in
for some exponent
. A direct attack, based on evaluating the (formal) Fourier integral
does not seem to make much sense (the integral is not absolutely integrable), although a change of variables (or dimensional analysis) heuristic can at least lead to the prediction that the integral (8) should be some multiple of . But which multiple should it be? To continue the formal calculation, we can write the non-integrable function
as an average of integrable functions whose Fourier transforms are already known. There are many such functions that one could use here, but it is natural to use Gaussians, as they have a particularly pleasant Fourier transform, namely
for (see Exercise 42 of Notes 2). To get from Gaussians to
, one can observe that
is invariant under the scaling
for
. Thus, it is natural to average the standard Gaussian
with respect to this scaling, thus producing the function
, then integrate with respect to the multiplicative Haar measure
. A straightforward change of variables then gives the identity
where
is the Gamma function. If we formally take Fourier transforms of this identity, we obtain
Another change of variables shows that
and so we conclude (formally) that
thus solving the problem of what the constant multiple of should be.
Exercise 35 Give a rigorous proof of (9) for
(when both sides are locally integrable) in the sense of distributions. (Hint: basically, one needs to test the entire formal argument against an arbitrary Schwartz function.) The identity (9) can in fact be continued meromorphically in
, but the interpretation of distributions such as
when
is not locally integrable is somewhat complicated (cf. Exercise 12) and will not be discussed here.
Specialising back to the current situation with , and using the standard identities
we see that
and similarly
and so from (7) we see that one choice of the fundamental solution is the Newton potential
leading to an explicit (and rigorously derived) solution
to the Poisson equation (6) in for Schwartz functions
. (This is not quite the only fundamental solution
available; one can add a harmonic polynomial to
, which will end up adding a harmonic polynomial to
, since the convolution of a harmonic polynomial with a Schwartz function is easily seen to still be harmonic.)
Exercise 36 Without using the theory of distributions, give an alternate (and still rigorous) proof that the function
defined in (10) solves (6) in
.
Exercise 37
- Show that for any
, a fundamental solution
to the Poisson equation is given by the locally integrable function
where
is the volume of the unit ball in
dimensions.
- Show that for
, a fundamental solution is given by the locally integrable function
.
- Show that for
, a fundamental solution is given by the locally integrable function
.
This we see that for the Poisson equation,
is a “critical” dimension, requiring a logarithmic correction to the usual formula.
Similar methods can solve other constant coefficient linear PDE. We give some standard examples in the exercises below.
Exercise 38 Let
. Show that a smooth solution
to the heat equation
with initial data
for some Schwartz function
is given by
for
, where
is the heat kernel
(This solution is unique assuming certain smoothness and decay conditions at infinity, but we will not pursue this issue here.)
Exercise 39 Let
. Show that a smooth solution
to the Schrödinger equation
with initial data
for some Schwartz function
is given by
for
, where
is the Schrödinger kernel
and we use the standard branch of the complex logarithm (with cut on the negative real axis) to define
. (Hint: You may wish to investigate the Fourier transform of
, where
is a complex number with positive real part, and then let
approach the imaginary axis.) (The close similarity with the heat kernel is a manifestation of Wick rotation in action. However, from an analytical viewpoint, the two kernels are very different. For instance, the convergence of
to
as
follows in the heat kernel case by the theory of approximations to the identity, whereas the convergence in the Schrödinger case is much more subtle, and is best seen via Fourier analysis.)
Exercise 40 Let
. Show that a smooth solution
to the wave equation
with initial data
for some Schwartz functions
is given by the formula
for
, where
is the distribution
where
is Lebesgue measure on the sphere
, and the derivative
is defined in the Newtonian sense
, with the limit taken in the sense of distributions.
Remark 5 The theory of (tempered) distributions is also highly effective for studying variable coefficient linear PDE, especially if the coefficients are fairly smooth, and particularly if one is primarily interested in the singularities of solutions to such PDE and how they propagate; here the Fourier transform must be augmented with more general transforms of this type, such as Fourier integral operators. A classic reference for this topic is the four volumes of Hörmander’s “The analysis of linear partial differential operators”. For nonlinear PDE, subspaces of the space of distributions, such as Sobolev spaces, tend to be more useful.
151 comments
Comments feed for this article
19 April, 2009 at 6:14 pm
Anonymous
WONDERFUL POST.
THANK YOU PROF. TAO
20 April, 2009 at 11:22 am
Successful Researcher
Superb! Thank you!
20 April, 2009 at 4:38 pm
bird
Dear Prof. Tao,
in some places, you write ” absolutely integrable and compactly supported”…
but if a function is compactly supported, then isn’t it absolutely integrable? is there a specific reason that you are writing in that way?
thanks
20 April, 2009 at 4:46 pm
Anonymous
When we say ” a function f is supported in a set V”, do we mean that the set where f is nonzero is a subset of V or the closure of the set where f is nonzero is a subset of V?
thanks
20 April, 2009 at 5:06 pm
bird
in exercise 16, on the right hand side is it
?
20 April, 2009 at 5:25 pm
Anonymous
Dear Prof. Tao,
in execise 17,
is that right
integral value of f?
if it is right, why the answer is 1, I did not understand?
thanks
20 April, 2009 at 9:03 pm
Dale Roberts
Typo: The display eq. between (7) & (8), shouldn’t RHS be K (without hat)?
Like all your (analysis course) posts, your exposition really clarifies some subtle points that were glossed over in a course that I took as an undergrad and provide a lot of great intuition.
IMHO, your blog posts really help fill the lack of good graduate-level analysis courses in Australia. It is true that we can sit down and plough through the books but your style really brings it to life. Thanks, I personally really appreciate it.
21 April, 2009 at 11:19 am
Terence Tao
Not all compactly supported functions are absolutely integrable, e.g. the restriction of 1/x to the interval [0,1].
Either definition of support would work here, but one can for sake of concreteness take the closure of the set where the function is non-zero.
For exercise 16, I forgot to mention that to make the various notions of multiplication consistent, distributions should be given the anti-complex structure rather than the complex structure; thus if
is a distribution and
is a complex number, then
is the distribution given by
rather than
. This conjugation is always something of an irritant (the same technicality shows up when considering duals of complex Hilbert spaces), though if one wishes one can avoid this issue by restricting attention to real-valued functions (but then it becomes harder to implement Fourier analysis).
For exercise 17, 1 is indeed the distribution that maps any test function f to its integral value:
.
Dale: thanks for the correction!
21 April, 2009 at 3:23 pm
lutfu
Dear Prof. Tao,
for Exc 11,12,13 showing that they define a distribution is not too hard but the second part, i.e showing that they does not arise from either a locally integrable function or a Radon measure is not easy.
Could you please give some hint for anyone of them?
thanks
23 April, 2009 at 12:11 pm
Terence Tao
Dear Lutfu,
In all of these exercises you should try arguing by contradiction, assuming that, say,
arises from a Radon measure
and then trying to deduce as much about
as possible. For instance, in that specific example one can look at lower bounds for the total variation of
on the interval
for small
.
30 April, 2009 at 9:34 pm
245C, Notes 4: Sobolev spaces « What’s new
[…] dual space of , the distributional limit of any sequence bounded in remains in , by Exercise 28 of Notes 3.) To prove (1), observe from the fundamental theorem of calculus […]
2 May, 2009 at 5:55 pm
Student
I’m sorry if it’s a dumb question, but I don’t understand how you can take a Fourier Transform of the right hand side of the equation that is 2lines after (8). It’s not in L^1 as it blows up at the origin.
If you could specify the change of variables you are using, that would be appreciated too.
2 May, 2009 at 9:01 pm
Terence Tao
Ah, several of the powers of
in the text should have been
instead. This should be fixed now…
6 May, 2009 at 6:41 pm
At the Fefferman conference « What’s new
[…] some explicit constant (this can be seen by computations similar to those in my recent lecture notes on distributions, or by analytically continuing such computations; see also Stein’s “Singular integrals […]
11 May, 2009 at 4:09 pm
maxbaroi
There is a disparity in Exercise 19’s hint. Your hint’s text references Exercise 3, but links to Lemma 3. [Fixed, thanks – T.]
13 May, 2009 at 7:28 pm
Yasser Taima
Dear Professor Tao,
In lemma 2, isn’t it that the Riemann sum can be interchanged with the inner product because of the linearity of the inner product, rather than the compactness of the support of f?
It was mentioned in lecture that
converges to
as
pointwise. We then said that the convergence is uniform. Why is that?
This appears to be useful to obtain the Lipschitz condition needed for the convergence of the Riemann sum. Is that right?
Thanks and apologies as I’m new to latex.
13 May, 2009 at 7:40 pm
Terence Tao
Dear Yasser,
The compactness of f is needed to make the Riemann sum finite, at which point the sum and inner product can be interchanged by linearity.
The uniform convergence of the Newton quotients to the derivative (i.e. the uniform differentiability of
in x) is a consequence of the smoothness of h. For instance, one can use Taylor’s theorem with remainder to get a uniform bound on the error between
and
in terms of the
norm of h. (Actually, this is overkill; having h be
would already be sufficient thanks to the fundamental theorem of calculus or mean value theorem and the uniform continuity of h’, although one wouldn’t get as quantitative a decay rate as
with that approach.)
19 May, 2009 at 9:24 pm
maxbaroi
I believe there’s a typo in Exercise 16’s hint.
. [Corrected, thanks, -T.]
I believe it should read
19 May, 2009 at 11:04 pm
Solutions to Selected Exercises from 245C, Notes 3: Distributions « Less Incompetence
[…] Of course, the actual exercises can be found at Professor Tao’s blog. […]
21 May, 2009 at 8:36 am
Cloclo
Thank you very much for this very very interesting post, as usual !!
I would like to ask you the following question since I couldn’t find an answer in the references I looked at:
Given two functions, f in
, g in
(
), let h be their convolution in the "classical" sense (without the help of the theory of distribution)
. Thus h is a function in
and its Fourier transform is a tempered distribution.
Can we relate it to the Fourier transform of f and g ?
Or in other words is it possible under maybe given assumptions to have the formula
?
Thank you by advance !
21 May, 2009 at 8:44 am
Terence Tao
Dear Cloclo,
Yes, this is true for
. The easiest way to prove this is to express f and g as the limit (in
and
respectively) of test functions, which by Young’s inequality gives h as the limit (in
) of the convolution of those test functions. In particular, all the convergences here are also in the sense of tempered distributions, and so their Fourier transforms will also converge in this sense. Meanwhile, one gets a similar convergence to
by their test function counterparts using Hausdorff-Young and Holder’s inequality. Since the Fourier transform is already known to intertwine convolution and multiplication for test functions, one can now take distributional limits to conclude the claim (note that the space of distributions is Hausdorff and so limits are unique).
For
the situation is more subtle, because the expression
does not necessarily make sense (
need not be a locally integrable function, instead being merely a distribution in general; meanwhile,
is continuous but need not be smooth). One could define this product by fiat to be
, but this is a tautological way to solve the problem and does not have any actual content.
This general strategy (establishing identities for rough functions by taking limits of the corresponding identity for classical functions) is generally quite a good way to establish identities for rough functions, especially if they are somehow “linear” or otherwise likely to obey good continuity properties.
21 May, 2009 at 10:07 am
Cloclo
Thank you so much for your answer.
and
), then
, indeed
may be just a continuous but not a smooth function.
And I am really sorry for bothering you once again. But I am not sure I understand how you prove that (if
Thank you so much once again and sorry !
21 May, 2009 at 10:14 am
Terence Tao
If
converges to f in
, then
will converge to
in
by Hausdorff-Young; similarly,
will converge to g in
(here we need p to be at most 2). From Holder’s inequality we conclude that
converges to
in
and thus in the sense of tempered distributions.
21 May, 2009 at 10:14 am
Cloclo
Thank you very much. I have now to study carefully your answer (I didn’t get it in time). So don’t take into account my last questions and sorry for that.
with best regards.
10 June, 2009 at 4:07 pm
student
Dear Prof Tao,
in dimension 1, for the wave equation, if the given initial data are distributions, how do we define the solution for the equation? how do we interpret the equation?
thanks
11 June, 2009 at 6:36 pm
Terence Tao
Dear student,
There are a number of different ways to define “weak” solutions to these sorts of equations, which are mostly equivalent for linear equations such as the free wave equation, but which are subtly different for nonlinear equations. For instance, one can integrate the equation against a test function in spacetime in the half-space
, and interpret the resulting integral equation in a distributional sense. Or, one can take Fourier transforms in space, solve the equation in time using the fundamental solution, and then interpret everything distributionally. A closely related approach is to use the Duhamel formula to define the notion of a solution. Yet another approach is to interpret a rough solution as a limit of smooth solutions to the same equation (or a regularised or “viscosity” version of that equation) in suitable topologies (e.g. the distributional topology). Then there are other approaches based on comparison with subsolutions or supersolutions, or based on variational characterisations of the equation, or on entropy or kinetic formulations of the equation; as one can imagine, the topic of how to properly define a rough solution is quite a subtle one in general, particularly for nonlinear equations.
For wave equations, there is some discussion of this in Sogge’s book “nonlinear wave equations”; I also discuss this a little in my own PDE book.
11 June, 2009 at 6:50 pm
student
Thank you very much Prof Tao, you are really very kind person. you are always trying to help everyone as much as you can….
thanks again….
2 August, 2009 at 12:56 pm
George
Why isn’t the following a counterexample to exercise 2 section 1?
Assume that d=1 and the limit function f is identically 0.
Take a positive bump function b(x) with support [-1,1] . And let f_n(x) be b(x/n)/n.
The support of f_n is [-n,n] (so that the union of supports is all R).
And on each compact K, (f_n)^(j) tends uniformly to 0 for each j.
3 August, 2009 at 5:41 am
Terence Tao
Dear George,
This sequence does not converge to zero in the
topology. Indeed, consider the set of all f in
such that
(say) for all x. It is not hard to see that the restriction of this set to
is open in
for every compact K, so this set is an open neighbourhood of the origin in the final topology of
; however, it does not absorb the sequence
.
One can modify this argument to show that in order to get convergence in this topology, the supports of the
must be uniformly bounded (indeed, this is the bulk of the work in establishing Exercise 2).
3 August, 2009 at 8:57 am
George
Dear Prof. Tao,
Thank you for your explanation.
7 January, 2010 at 8:42 pm
Mean field games « What’s new
[…] will evolve as time goes forward. There are several ways to find the answer, but we will take a distributional viewpoint and test the density against various test functions – smooth, compactly supported […]
18 January, 2010 at 6:29 pm
254A, Notes 3b: Brownian motion and Dyson Brownian motion « What’s new
[…] the sense of (tempered) distributions (see e.g. my earlier notes on this topic). In other words, is a (tempered distributional) solution to the heat equation (3). […]
2 February, 2010 at 1:35 pm
254A, Notes 4: The semi-circular law « What’s new
[…] of approximations to the identity, and thus converges in the vague topology to (see e.g. my notes on distributions). Thus we see […]
5 March, 2010 at 1:46 pm
2010 Mar 消磨时间 « 逝去日子
[…] then use the fundamental theorem of calculus to strengthen the topology. Another is to use Exercise 5 and then discretise the convolution. Another is to embed into a torus and use Fourier series, […]
19 September, 2010 at 7:22 pm
245A, Notes 2: The Lebesgue integral « What’s new
[…] functions are all dense subsets of with respect to the (semi-)metric. Much later in the course (in 245C), we will see that a similar statement holds if one replaces continuous, compactly supported […]
16 October, 2010 at 8:29 pm
245A, Notes 5: Differentiation theorems « What’s new
[…] and its derivative is continuous, then we say that is continuously differentiable. Remark 1 Much later in this sequence, when we cover the theory of distributions, we will see the notion of a weak derivative or […]
8 November, 2010 at 7:46 pm
245A, Notes 4: Modes of convergence « What’s new
[…] (not to be confused with convergence in the sense of distributions, which we will study later in this sequence) is commonly used in probability; but, as the above exercise demonstrates, it is quite a weak […]
16 December, 2010 at 4:30 am
245A, Notes 2: The Lebesgue integral « mathTHÍCHinTOÁNmyHỌCbrain
[…] functions are all dense subsets of with respect to the (semi-)metric. Much later in the course (in 245C), we will see that a similar statement holds if one replaces continuous, compactly supported […]
30 April, 2011 at 7:29 pm
Anonymous
In Exercise 23, I used integral by part and finally get
How can I get rid of the first term on the right hand side? Since
, one may get
. But how to deal with the
part?
1 May, 2011 at 12:02 am
Terence Tao
Try the mean value theorem.
23 September, 2011 at 12:11 am
fedfue
A technical question: Don’t you need that the topology on $C_c^\infty(\R^d)$ be defined as a locally convex topology (and not simply as the final topology)? My question is because Rudin and other books have this requirement, but I’m not really sure. If not, why do these books make this more complex construction, when a much simpler and natural one can be defined (as done here)?
16 December, 2011 at 10:14 am
254B, Notes 3: Quasirandom groups, expansion, and Selberg’s 3/16 theorem « What’s new
[…] See for instance this blog post for a very brief introduction to Riemannian geometry, and these two previous posts for an introduction to distributions and Sobolev […]
29 February, 2012 at 4:49 pm
Rex
For Exercise 5, it seems to me that, because of the fact that
is not compact, we need something a little stronger in the definition of “approximations to the identity” than just uniformly converging to zero away from the origin for the convolution
o converge pointwise to
. Perhaps we need to require that the support of
is also shrinking to the origin as well.
Here is the situation I am worried about: suppose that
is a sequence of kernels which actually has increasing support as
goes to infinity. We can arrange
to have a big “spike” of width
centered at the origin, but also a long “tail” of height
along the interval of length
centered at the origin. Thus
is supported on an interval of length
. For most of this interval,
(which is close to zero), with the exception of the spike in the center. Set up correctly, we should be able to make
into a sequence of approximations of the identity satisfying the definition you’ve given.
However, suppose we now choose a continuous
such that
and
increases very rapidly as
moves away from the origin. So rapidly, in fact, that if we convolve
with
, then the “tail” portion of the function
collects a very large contribution to the integral
because
is very large when
is around
. Then it seems that
will not even converge to
.
This can’t happen if
has compact support, but in Exercise 5,
is only assumed to be continuous.
[Thanks, I’ve altered the definition of approximation to the identity suitably. -T]
17 July, 2012 at 6:19 pm
zuchongzhi
Professor, what is a good way to define the convolution of multiple distributions? Do we need at least one of them having compact support? I have been thinking about this for a while but feel uncertain what is the best way to define it. The textbook (Friedlander) used the notion of proper maps, but unfortunately his ‘proof’ has a technical flow that made the support of convolutions of distributions closed but could be non-compact. So I am looking an alternative.
17 July, 2012 at 7:06 pm
zuchongzhi
Sorry I made a mistake, Friedlander is right.
8 March, 2013 at 6:39 am
Sriyan Wickramasuriya
Dear Professor Tao,
Your notes are just brilliant! When we restrict the metric defined on Schwartz Space(pg 204 of Mike Taylor’s PDE book) to C_c^{infinity}(R^n) we do not get a complete metric topology and therefore we put the inductive limit topology on C_c^{infinity}(R^n). But this inductive limit topology is not metrizable and the proof of this fact involves Baire’s Category theorem.
Is there any way we prove that the inductive limit topology on C_c^{infinity}(R^n) is not metrizable without using the Baire’s Category Theorem?
8 March, 2013 at 8:31 am
Terence Tao
See exercise 3 (metrizable topologies are necessarily first countable).
8 March, 2013 at 6:38 pm
Sriyan Wickramasuriya
Dear Professor Tao,
Thank you very very much!
Best Wishes,
Sriyan
20 May, 2013 at 6:34 pm
Sara
Respected Terence Tao,
How will prove that sin(nx)—>0 as n goes to infinity in D'(R),
but sin^2(nx) –/->0 as n goes to infinity.
That is, multiplication of distn. is not a continuous operation even it is defined.
4 July, 2013 at 6:56 pm
Debdeep
Dear Prof. Tao
I want to construct a function f which is in Holder \alpha space for some \alpha > 0 and compactly supported such that the tail of the L_2 integral of the Fourier transform is lower bounded by |x|^{-\beta} for some \beta i.e.
int_{|t| > x} |\hat{f}(t)|^2 dt > = |x|^{-\beta}
What is the smallest \beta (related to \alpha) (tightest lower bound) we can achieve?
Can \beta = \alpha achievable? Thanks for your reply
11 March, 2014 at 8:23 pm
Federico
Dear Professor Tao,
These notes are great! However, I’m having a hard time to believe (and prove) the last part of exercise 3. You seem to suggest that the space of test functions is sequential with that topology (despite not being first countable). However I do not think this is true. A question I asked here and an article by Dudley seem to suggest otherwise. Indeed, sequential continuity is enough for proving continuity only in the case of linear maps, but to me this does not imply the space itself is sequential. Naturally, I could be wrong, but could you please clarify this subtle point?
11 March, 2014 at 10:48 pm
Terence Tao
Yes, you’re right; this issue had been pointed out to me before but I had neglected to update the exercise. But now I’ve altered it to just assert that continuity is equivalent to sequential continuity (which is actually relatively easy to prove, and is what is actually needed in applications).
12 March, 2014 at 12:26 pm
Federico
Thanks for the update Professor Tao! It is a relief. Just to finish clearing things up in my head: Isn’t sequential continuity (for the space of test functions) equivalent to continuity only in the case of linear maps to another locally convex topological vector space? Or is it true for arbitrary maps as well?
12 March, 2014 at 1:26 pm
Terence Tao
Linearity is not required (in fact the range can be an arbitrary topological space). If
is given the final topology from topological spaces
, then a map
to any topological space
is continuous if and only if its restrictions to each
are continuous. From this it is not difficult to see the equivalence of continuity and sequential continuity for maps from
to any topological space.
12 March, 2014 at 1:49 pm
Federico
Thank you for your reply! Yes, I agree with you regarding the final topology. Indeed this was my first approach to the exercise. But isn’t the test function space an inductive limit in the category of locally convex topological vector spaces (which is different from the typical inductive limit, which in turn is equivalent to the final topology induced by these identity maps)? I mean, isn’t the test function topology the finest (largest) locally convex topology on
which makes the identity maps
continuous as opposed to simply the final topology induced by the identity maps
? I think this subtle difference in the inductive limit topology (of LF spaces in general, I believe) might make a difference in regard to the sequential properties of the test function space. This is the true source of my doubt.
12 March, 2014 at 4:31 pm
Terence Tao
You’re right, I had not realised this subtlety between the two different notions of a final topology. I’ve modified the text (using the locally convex final topology rather than just the plain final topology), and hopefully all these issues are fixed now…
12 March, 2014 at 5:04 pm
Federico
Professor Tao, I really appreciate your answers very much! Thank you for your attention to this issue and for making this clearer for everyone.
30 March, 2014 at 9:48 pm
Anonymous
Dear Prof. Tao,
I am struggling to solve all the exercises from section 1. I am pretty sure I solved from 1 through 5 but I am lost in Exercise 6. There I am trying to use Exercise 5 but I don’t understand what you mean by discretise the convolution and how we can discretise the convolution?
Thanks!
31 March, 2014 at 8:53 am
Terence Tao
If one approximates an integral by a Riemann sum, one can approximate
by a finite linear combination of translates of
, which is one way to approach separability. (There are many others.)
31 March, 2014 at 10:05 pm
Anonymous
Thanks Prof. Tao!
Yes, i know f*\phi_n is infinitely differential but we we discretise with Riemann Sum then the Riemann Sum not gonna be in C_c^\infty. How do we make this precise?
Thanks in advance!
31 March, 2014 at 10:30 pm
Terence Tao
If one replaces the convolution
with a Riemann sum approximant
, one will obtain a function in
.
2 April, 2014 at 11:28 pm
Anonymous
Thank you very much, Professor Tao for your guidance.
22 April, 2014 at 10:38 pm
Anonymous
Dear Dr. Tao, I am not sure what you mean by ‘dualise’ the Fundamental Theorem of Calculus in Exercise 25 (iii). So, would you mind giving me more details?
Thanks in advance!
23 April, 2014 at 7:37 am
Terence Tao
As a model case, try k=0 and assume the “cheat” that the antiderivative of a test function supported on an interval is again a test function supported on that interval. (In this case, the additional distribution
would not be required.) The fundamental theorem of calculus, which guarantees the existence of an antiderivative for any test function, is what will be needed to construct
in this case.
24 April, 2014 at 1:09 am
Anonymous
So, do you mean to define
where
?
Then order of
not gonna be
, right?
1 May, 2014 at 8:13 pm
Anonymous
Professor T. Tao,
I am teaching your notes for the first time and I am also getting better understanding with my students. But I have almost don’t have knowledge of PDE and we are struggling to solve Exercise 40. So , it would be great help if you direct us in this Exercise.
Thanks in advance!
3 May, 2014 at 9:36 pm
Anonymous
Dear Dr. Tao,
In Exercise 40, is it $t/4\pi$ or 1/4\pi t$ ?
3 May, 2014 at 10:43 pm
Terence Tao
It is
, due to the rescaling of the integral to the unit sphere, rather than the sphere of radius t.
4 May, 2014 at 8:12 pm
Anonymous
Thanks Dr. Tao for your response.
I have again couple of confusions and it would be great if you clarify these too.
For Exercise 38 an 39, in most of the text book of PDE I found just {e^{-|x|^2/4t}} and {e^{i|x|^2/4t}} respectively instead of {e^{-|x-y|^2/4t}} and {e^{i|x-|^2/4t}}. Is there any special reason subtracting {y} from {x}?
In Exercise 40, you said just wave equation {-\partial_{tt} u + \Delta u}. Do you mean {-\partial_{tt} u + \Delta u =0}?
Thanks in advance!
[Corrected, thanks – T.]
8 May, 2014 at 10:22 pm
Anonymous
Dear Dr. Tao,
In the process of solving Exercise 40, how do we define the in Fourier transform of a tempered distribution? For, example, let \lambda be a tempered distribution then (F\lambda) (f)=\lambda(Ff) or (F\lambda) (f)=\lambda(F^{-1}f), where F is denoting the Fourier transform?
9 May, 2014 at 7:25 am
Terence Tao
See equation (5).
9 May, 2014 at 8:09 pm
Anonymous
Dear Dr. Tao,
Are we supposed to use Kirchhoff’s formula to get the distribution K_{t} in Exercise 40?
Thanks
.
14 May, 2014 at 8:13 pm
Anonymous
Dear Dr. Tao,
In Exercise 40, what is the meaning of “Newtonian Sense” and how to use it in solving this Exercise?
Thanks!
14 May, 2014 at 8:28 pm
Terence Tao
http://en.wikipedia.org/wiki/Newton_quotient
14 May, 2014 at 10:47 pm
Anonymous
Just a minor comment for correction, in Exercise 40 isn’t it “… for some Schwartz functions {f} and {g} …” instead of “… for some Schwartz functions {f} is given by the formula …?”
[Corrected, thanks – T.]
20 August, 2015 at 5:04 am
KE operator and eigenfunctions
[…] I could explain what's going on but it will involve a long sojourn into distribution theory: https://terrytao.wordpress.com/2009/04/19/245c-notes-3-distributions/#more-2072 Really your teacher needs to explain it – post here with what he/she says – it should prove […]
26 August, 2015 at 6:52 am
Anonymous
In some books, a distribution on
is defined as a linear functional
with the following property
Suppose
is a sequence in
such that
There exists a compact subset
of
such that
for all
and
as
Then
as
.
Is it exactly the same as the distributions defined in this note?
[Yes; see Exercise 4(iii). -T.]
27 August, 2015 at 10:52 am
Anonymous
What is the weak derivative of the distribution defined in Exercise 12? It seems that p.v.(1/x^2) is not well defined…
27 August, 2015 at 11:17 am
Terence Tao
If one chases the definitions and integrates by parts, one has
So the derivative of p.v. 1/x is a renormalised principal value of -1/x^2.
27 August, 2015 at 12:10 pm
Anonymous
Hmm, how can one get the last line?
[Taylor expansion -T.]
28 August, 2015 at 7:21 am
Anonymous
It seems that this relates to https://en.wikipedia.org/wiki/Hadamard_regularization
But I don’t see why the limit exists…
28 August, 2015 at 7:35 am
Anonymous
Can one say that the limit in the last line exists since it is “equal” to
, the existence of which has been proven in Ex. 12? It looks like a circular argument to me since in the very beginning one assumes the existence of
.
[Yes, the existence of the earlier limits (which indeed follows from the easily verified fact that the distributional derivative of a distribution is again a distribution, together with Exercise 12) can be used to establish the existence of the later limits. It is also instructive to establish (again, using Taylor expansion) directly that the later limit is of a Cauchy sequence and thus convergent. -T.]
29 August, 2015 at 1:25 pm
Anonymous
I read somewhere before but I don’t remember exactly in what book:
Let $(a_n)$ be a strictly decreasing (I don’t remember if it is increasing or decreasing) positive sequence such that
. If we defines
where
, then
converges (uniformly) to a function in
.
Has anybody here seen a reference for such constructions of a nontrivial test function before? It seems that Exercise 1(i) is more standard.
18 September, 2015 at 1:25 pm
Anonymous
Stop cheating on math M541 homework.
2 November, 2015 at 7:06 pm
275A, Notes 4: The central limit theorem | What's new
[…] in the sense of distributions that arises in distribution theory (discussed for instance in this previous blog post), however strictly speaking the two notions of convergence are distinct and should not be confused […]
9 November, 2015 at 6:52 am
Anonymous
When construct an approximation to the identity, is there a particular reason that one chooses sequence
instead of a continuous version
?
9 November, 2015 at 7:00 am
Anonymous
In Exercise 5(ii), does one need to assume that
first in order to argue that “
converges in
to
“?
9 November, 2015 at 10:25 am
Terence Tao
No, this is implicitly part of the conclusion (but this follows easily from Young’s inequality or Minkowski’s inequality).
If one only assumes local integrability or
on
, then one needs compact support of the approximating sequence
, as otherwise the tail of
could interact with an arbitrarily large growth of
and it is not even immediate that
is anywhere finite, let alone convergent to anything interesting.
In most applications there is not much difference between using a discrete sequence of approximations rather than a continuous sequence, but using a countable discrete sequence makes it essentially trivial to establish measurability of any limit objects obtained, and also there are some sequential compactness results one can exploit when working with a discrete sequence that are not easily available in the continuous setting. (Of course in many situations one can use topological compactness as a substitute for sequential compactness, e.g. use topological Banach-Alaoglu in place of sequential Banach-Alaoglu.)
9 November, 2015 at 7:12 am
Anonymous
If
is locally integrable or more generally
, can one have the similar fact in Exercise 5?
9 November, 2015 at 11:55 am
Anonymous
In Exercise 1(iv), what is
? I’ve looked it up in your textbook and I didn’t find it there.
[See Exercise 1.10.7 (or Exercise 7 of 245B Notes 12), or also Exercise 1.10.17, Remark 1.10.16, etc..]
9 November, 2015 at 12:03 pm
Anonymous
In Exercise 9
if a sequence of locally integrable functions converge in
to a limit…
What is the topology defined for
?
[The topology of
is the topology generated by the seminorms
for compact
, as in Example 1.9.5. -T]
30 November, 2015 at 5:49 pm
Anonymous
Can one safely replace
with any open set
in this note to get the theory of distribution on
?
30 November, 2015 at 6:53 pm
Terence Tao
The theory of distributions localises fairly easily (basically just replace
with
). But the theory of tempered distributions is significantly more difficult to work with on domains, because it is not obvious how to define the Schwartz class on a domain (and because the main tool that makes the tempered distribution concept useful, namely the distributional Fourier transform, is not obviously available).
9 March, 2016 at 12:26 pm
Anonymous
I’m confused with the distributional derivatives in the setting of linear PDE. When one uses the formula
to define
, is
just the “formal adjoint” of
? For general
, when one works in a boundary value problem in PDE (say Dirichlet problem or the Neumann problem for the Laplacian equations), should the adjoint
depend also on the boundary conditions so that one has different definitions of the distributional derivative of the same differential operator ?
9 March, 2016 at 1:58 pm
Terence Tao
For distributions on a domain
, the test functions
involved are in
, so they vanish to infinite order on the boundary. In particular, the precise notion of adjoint used is not relevant, as they will all agree on these test functions; in particular, the formal adjoint suffices. But one certainly has to take boundary issues into account when attempting to extend a distribution on
to all of
; the situation here is more subtle than the corresponding situation with measurable functions, in which one can simply extend the function by zero outside of the domain. In general, the extension operation on distributions is not unique, which is related to the non-uniqueness of the adjoint operator when applied to more general functions than
functions.
31 December, 2015 at 2:39 pm
Anonymous
Given a distribution
, can we in general find a distribution
with
as its distributional derivative?
[Yes; this is a good exercise for you to establish. The key point is that the test functions of mean zero form a hyperplane in the space of all test functions, in that any test function can be made mean zero by subtraction of a scalar multiple of a fixed reference test function. -T.]
11 May, 2016 at 4:48 pm
Anonymous2
A result due to de Rham says the following:
Let
,
, be distributions, where
is a domain in
. Then
for some
iff
for all
with
.
Does this contradicts the positive answer to the question above?
11 May, 2016 at 5:57 pm
Terence Tao
No; the previous question concerned only the one-dimensional case
(or of a single partial derivative, rather than the full gradient), in which case the condition
forces
to vanish identically.
20 January, 2016 at 11:50 am
Anonymous
How do we see that ”
embeds ‘continuously’ into
”
[Easiest way is via nets – show that every net that converges in
also converges in the Schwartz topology. Or one can explicitly show that pullbacks of basic open sets are open. -T.]
17 February, 2016 at 5:01 pm
Debanjana Kundu
I was wondering if when we’re in
and we know that
, also
; for a test function
if
is not in the support of
can we say
for all
?
10 March, 2016 at 3:09 am
Anonymous
i am in grade 7 and i knew about your iq
30 March, 2016 at 10:16 am
Anonymous
In Exercise 5(i), I get
and
But I don’t see how to go on. Do you have a hint?
30 March, 2016 at 10:37 am
Terence Tao
30 March, 2016 at 1:56 pm
jack
Instead of
, one needs a bounded for
, I think?
Consider the standard symmetric mollifier
and
. Fix
. For large enough
,
uniformly in 
Why do we need "chop up the integral into two pieces"? Am I doing something wrong here?
30 March, 2016 at 2:10 pm
Terence Tao
Thanks for the correction. I had in mind a more general notion of approximation to the identity than the one that is in this post (in which some leakage of mass outside of a small ball is permitted), but you are right that with the notion of approximation to the identity used here, one does not need to decompose the integral.
4 April, 2016 at 3:56 pm
Anonymous
When one tries to upgrade the convergence to uniform convergence on
, the proof above certainly would not work. But do you have a counterexample that one cannot upgrade the compact convergence? Why is compactness crucial here?
30 March, 2016 at 2:14 pm
Anonymous
What can one say in general for Exercise 5? Let
for some function space
such that
makes sense. Can one always expect that
in some convergence mode $latex $Y$ (or there are cases that not convergent in any mode at all)?
31 March, 2016 at 12:28 pm
Anonymous
In Exercise 5(ii), do you have a quick counterexample for
?
31 March, 2016 at 1:26 pm
Terence Tao
Uniform limits of continuous functions are necessarily continuous.
4 April, 2016 at 10:59 am
Anonymous
In Exercise 5(iii), what is the definition of
when
is a vector value function? Perhaps you mean
?
4 April, 2016 at 1:15 pm
Terence Tao
Yes. By default, operations on scalar-valued functions are understood to act componentwise on vector-valued functions unless otherwise stated.
5 July, 2016 at 3:21 pm
Anonymous
Since
embeds continuously into
(with a dense image), we see that the space of tempered distributions can be embedded into the space of distributions.
How “good” is the embedding? In other words, is every tempered distribution locally integrable? Or is it in the “basic two examples of distributions”?
[No. For instance, the Dirac delta distribution is tempered, but not locally integrable. -T.]
30 August, 2016 at 6:33 am
Anonymous
In your definition of the “approximation to the identity”, is the third condition “whose supports shrink to the origin” equivalent to the following condition?
where the limit is understood in the space of Schwartz distributions.
22 September, 2016 at 9:53 pm
246A, Notes 1: Complex differentiation | What's new
[…] by the way, can be largely explained using the theory of distributions, as covered for instance in this previous post, but this is beyond the scope of the current […]
15 October, 2016 at 6:31 am
Anonymous
I don’t see in the notes if the following is true:
the weak derivative, if exists, must be unique.
I found that I end up with examining the following
if
is locally integrable and
for all test functions
, then
a.e.
I don’t find a proof here. Would you give me a hint that how this can be proved?
15 October, 2016 at 10:48 pm
Terence Tao
Use a limiting argument to show that if
is orthogonal to all test functions, then it is also orthogonal to indicator functions of bounded measurable sets. Then argue by contradiction.
1 February, 2018 at 8:27 am
Anonymous
Would anyone elaborate the hint? (1)Where does this set of notes show that the set of test functions is dense in the set of indicator functions of bounded measurable sets? (2) What does the “contradiction” mentioned above refer to?
17 February, 2017 at 9:57 pm
254A, Notes 2: The central limit theorem | What's new
[…] in the sense of distributions that arises in distribution theory (discussed for instance in this previous blog post), however strictly speaking the two notions of convergence are distinct and should not be confused […]
2 August, 2017 at 5:42 am
LR
Dear Prof. Tao,
I suggest you to take a look at “Multiplication of the Distributions”, by Colombeau, which have some thoughts about applying it to nonlinear problems and the “Impossibility of multiplication” cited by Schwartz.
11 February, 2018 at 4:45 am
Sébastien
In your excellent PCM note on distributions, you mention that distributions can be composed both ways by suitably smooth functions. Was this a typo or not? Indeed, here and on the internet, I can only find composition in the “apply the true function first” order, and whenever people speak of the exponential of the Gaussian Free Field, they insist on the “one cannot a priori take the exponential of a distribution” issue, so I am confused now.
11 February, 2018 at 11:00 am
Terence Tao
Oops, that is indeed a typo. Distributions can be composed on the right with smooth functions but not on the left in general.
11 February, 2018 at 12:00 pm
Sébastien
Thank you very much!
19 March, 2018 at 8:49 am
Anonymous
Probably too late but I am curious. Would you define |delta(x)|=delta(x)?
I think there might be reasons for doing it but also reasons for not doing it.
19 March, 2018 at 6:26 pm
Terence Tao
Sure; one can for instance view this as a (somewhat degenerate) special case of the concept of a total variation of a measure. In general, though, it is only the measures of finite total variation for which one has a meaningful absolute value; more general distributions, such as the (distributional) derivative
of
, do not have any particularly useful notion of an absolute value.
20 March, 2018 at 2:41 am
Anonymous
[sorry I am messing up with latex]
Thank you very much for your answer. What bothers me is that the reasoning used to explain why we do not define \delta^2 also might be invoked for |\delta|. If one takes the functions f_n(x)= n*c_n*\mbox{sinc}(n*x)/\mbox{rect}(x)
where c_n is an appropriate constant, then it seems to me that f_n\to \delta, but |f_n| does not converge. Am I missing something here?
20 March, 2018 at 7:42 am
Terence Tao
I do not know what the rect(x) function is, but it is certainly the case that the operation
(as defined on “nice” functions) is not continuously extendible to the space of all distributions, much as is the case with
. So there is no meaningful notion of the absolute value of an arbitrary distribution. However, if one restricts to the subclass of distributions that are measures of finite total variation, then
becomes a continuous operation (using now the total variation topology), and so one can define
for
in this class. (The sequence you provide will likely not converge in the total variation topology, but only in the distributional topology.)
20 March, 2018 at 3:12 pm
Anonymous
Thank you very much, I think I understand although I am a bit surprised by this idea of considering a subclass of measures of finite total variation. I wrote the functions in a wrong way but I meant to multiply n*sinc(n*x) by the indicator function of an interval say like [-1,1] and rescale appropriately to have integral =1. But actually even n*sinc(n*x) would have worked for my concern.
; if I consider again n*sinc(n*x) then this converges to
in distribution and its square does converge to \delta too… so we might consider L_2 and define \delta^2?… mm…. ok I don’t want to bother further, you helped already enough, thank you very much again! :)
I thought (I am not a professional mathematician) one should usually require nice behaviour over any converging sequence in the distributional topology. I will think more about it.
– Actually, now I am puzzled by
28 April, 2018 at 5:55 pm
Anonymous
Since Wolff’s notes are used in 271A, I think my following question is related to this post.
In an estimate, the author gets (the appendix of Chapter 2 The Schwartz Space) a pointwise bound:


instead of
. It is said in Wolff’s notes that one can simply do it using rescaling. Can any one elaborate how the rescaling is done?
where
But what one really needs is the bound with
28 April, 2018 at 5:59 pm
Anonymous
The original goal is to prove by induction that

where
.
with
The course should be MATH 247A: http://www.math.ucla.edu/~tao/247a.1.06f/
29 April, 2018 at 1:32 pm
Terence Tao
Apply the previous bound to the function
.
29 April, 2018 at 1:50 pm
Anonymous
In the first comment, it should be that “But what one really needs is the bound with
instead of
.
I find that I messed up with the change of variables calculations. In the simplest case, one has
and hence
Thanks!
12 May, 2018 at 3:22 pm
Anonymous
Considering both Exercise 1 and 5, I still can’t see how to answer the question in https://terrytao.wordpress.com/2009/04/06/the-fourier-transform/#comment-498013
If one uses an approximation to the identity,
, then

and
for
. This seems to be quite close to what I want. But there is no guarantee that

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

.
where
Since
as well, similarly one has

But then one has two different sequences, $\phi_n*h and \phi_n*g$.
where
12 May, 2018 at 3:32 pm
Anonymous
In the first case, I have already had
in both the
where
But it is known that
. If one can control the decay (and retain the differentiability) of
to get
, then the proof would be complete…
13 May, 2018 at 7:45 am
Terence Tao
In addition to convolving with a mollifier, one should also multiply by a smooth cutoff such as
to obtain compact support (which, together with the smoothness provided by the mollifier, place one in the Schwartz class without difficulty).
5 July, 2018 at 8:39 pm
Josh Chen
Is there any reference or literature you might point to for generalized functions/distributions/Schwartz functions in infinite dimensions? It’s interesting to consider tempered distributions rather than for example Gaussian measures for PDEs with random coefficient fields or time stochastic PDE processes..
13 August, 2018 at 3:24 am
Daniel Pires
Hi professor Tao! I was trying to solve this one problem and I thought it was related to this notes
than
if
(here
denotes the inverse Fourier Transform).
Do you have any hint for how can I solve this?
13 August, 2018 at 7:33 am
Terence Tao
Basically one needs some decay estimates on the Fourier transform (it turns out in this case that it decays like
). One can see this for instance by dyadic decomposition (splitting
into a bunch of rescaled bump functions). In this particular case there may also be some exact formulae that could be helpful (there are some computations of related kernels for instance in Stein’s “singular integrals”, possibly also in Stein-Weiss’s “Fourier analysis on Euclidean spaces”).
3 September, 2018 at 3:13 pm
254A, Notes 0: Physical derivation of the incompressible Euler and Navier-Stokes equations | What's new
[…] but we will adopt the viewpoint of the theory of distributions (as reviewed for instance in these old lecture notes of mine) and consider approximation against test functions in spacetime, thus we assume […]
2 October, 2018 at 3:47 pm
254A, Notes 2: Weak solutions of the Navier-Stokes equations | What's new
[…] some key aspects of the theory. A more comprehensive discussion of distributions may be found in this previous blog post. To avoid some minor subtleties involving complex conjugation that are not relevant for this post, […]
26 October, 2018 at 12:21 am
Rajnikant Snha
Dear Professor Tao,
On p 172, line number 5, Rudin’s Functional Analysis, there is a limit in the space of test functions. It seems to me wrong. Can you supply its intelligible proof.
With thanks in advance.
28 November, 2018 at 1:58 pm
254A, Notes 3: Local well-posedness for the Euler equations | What's new
[…] some key aspects of the theory. A more comprehensive discussion of distributions may be found in this previous blog post. To avoid some minor subtleties involving complex conjugation that are not relevant for this post, […]
7 February, 2019 at 7:14 am
Laszlo
Does the convolution algebra of compactly supported distributions on $R^n$ have any delicate algebraic properties (like Noether ring, etc.)?
7 February, 2019 at 8:34 am
Terence Tao
I did a bit of poking around and found this recent paper by Vogt on the subject: https://mathscinet.ams.org/mathscinet-getitem?mr=3858282 . Presumably the references cited in the introduction will describe the current state of knowledge in this direction.
7 February, 2019 at 11:11 am
Laszlo
Thx!
28 June, 2019 at 4:57 pm
acx01bc
You should mention
which works everywhere even for things such as
, the only point is that the limit doesn’t need to converge in the sense of distributions.
29 August, 2019 at 9:34 am
Victor
Dear Professor Tao, I am trying to prove the following statement:
Let
be the space of the distributions with compact support. Suppose that
in
(In the sense
for all
.Then, there exist a compact
such that
for all
.
Is there an easy way to prove it?
30 August, 2019 at 7:56 am
Terence Tao
Perhaps I am misreading your question, but it appears that “moving bump” type examples such as
or “spreading bump” examples such as
would give counterexamples to your claim.
6 November, 2019 at 4:35 pm
254A, Supplement 2: A little bit of complex and Fourier analysis | What's new
[…] exercise is intended for readers familiar with distribution theory, as discussed for instance in this previous blog post.) If is holomorphic on an open set , establish the […]
8 April, 2020 at 11:13 am
Tingzhou
Hi Prof. Tao
Can you give more hints about Exercise 37: “This we see that for the Poisson equation, {d=2} is a “critical” dimension, requiring a logarithmic correction to the usual formula?” How can we get the “logarithmic correction” to the Fourier transform of
8 April, 2020 at 12:40 pm
Terence Tao
The logarithmic correction basically arises from the need to add a correction term (in the spirit of Exercise 13) to interpret
as a distribution as it is no longer absolutely integrable in dimensions
.
29 October, 2020 at 12:15 pm
Anonymous
(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 function.)
There is no Exercise 46 in Notes 2. Typo?
[Corrected, thanks – T.]