Post on 22-Dec-2021
Haim Grebnev Last saved: October 18, 2021
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G. Friedlander and M. Joshi Introduction to the Theory of
Distributions Notes
Table of Contents
Notations and Conventions .......................................................................................................... 1
Chapter 1 ....................................................................................................................................... 3
Principal Value Distribution ๐โ๐ (Problem 1.3) (9/25/2020) ................................................. 3
A Distribution ๐ โ ๐โฒ(โ) such that ๐ = ๐/๐ on (๐,โ) and ๐ = ๐ on (โโ, ๐) (Problem
1.4) (9/25/2020) ........................................................................................................................... 4
An Interesting Example of a Non-Extendible Distribution (Problem 1.5) [2/26/2021]. ...... 5
Chapter 5 ....................................................................................................................................... 7
Convolution Equations on Forward Cones (Problem 5.5) (1/2/2021) .................................. 7
Chapter 6 ....................................................................................................................................... 9
Schwartz Kernel Theorem [Maintenance planned] (11/4/2020) ........................................... 9
Chapter 8 ..................................................................................................................................... 15
Structure Theorem for Tempered Distributions [Theorem 8.3.1] (5/11/2021) .................. 15
Poissonโs Summation Formula [Theorem 8.5.1] (5/18/2021) .............................................. 21
Notations and Conventions
Notation: For any point ๐ฅ โ โ๐, we will denote its Euclidean length by |๐ฅ|:
|๐ฅ| = โ๐ฅ12 + โฏ+ ๐ฅ๐
2.
Notation: For any ๐ฅ โ โ๐ and any ๐ > 0, we let ๐ต๐(๐ฅ) denote the open ball of radius ๐ centered
at ๐ฅ with respect to the Euclidean distance:
๐ต๐(๐ฅ) = {๐ฆ โ โ๐ โถ |๐ฆ โ ๐ฅ| < ๐}.
Notation: Suppose that ๐ โ ๐ โ โ๐ are open sets. For any function ๐ โ ๐ถ๐โ(๐), we let ๐๐ โ
๐ถ๐โ(๐) denote the smooth extension of ๐ to ๐ obtained by setting ๐ โก 0 on ๐ โ ๐. Itโs trivial to
see then that supp๐ = supp๐๐ and hence ๐๐ is indeed compactly supported as well.
Notation: We let โค+ stand for the positive integers: โค+ = {1,2,3, โฆ }.
Notation: For any ๐ โ โค+, let โ(๐) denote the set of multi-indices of length ๐:
โ(๐) = {(๐ผ1, โฆ , ๐ผ๐) โ โค๐ โถ each ๐ผ๐ โฅ 0}.
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Notation: Let ๐ผ, ๐ฝ โ ๐ผ(๐). Then
1.) ๐ผ โค ๐ฝ (resp. ๐ผ < ๐ฝ) means that each ๐ผ๐ โค ๐ฝ๐ (resp. ๐ผ๐ < ๐ฝ๐).
2.) ๐ผ! denotes ๐ผ1! โ โฆ โ ๐ผ๐!.
3.) |๐ผ| denotes ๐ผ1 + โฏ+ ๐ผ๐.
4.) For any ๐ฅ โ โ๐, ๐ฅ๐ผ denotes ๐ฅ1๐ผ1 โ โฆ โ ๐ฅ๐
๐ผ๐ .
5.) For any sufficiently differentiable function or distribution ๐, ๐๐ผ๐ denotes ๐๐ผ1 โฆ๐๐ผ๐๐.
Notation: Let ฮฉ โ โ๐ be an open subset. We let the following denote the following spaces of
complex-valued functions:
1.) ๐ถ๐(ฮฉ) denotes the space of ๐-times continuously differentiable functions over ฮฉ. In
particular, ๐ถโ(ฮฉ) denotes the space of smooth functions.
2.) ๐ถ๐๐(ฮฉ) denotes the space of ๐-times continuously differentiable functions over ฮฉ with
compact support. Sometimes ๐ถ๐โ(ฮฉ) is also denoted by ๐(ฮฉ).
3.) We let ๐ฎ(โ๐) denotes the space of rapidly decreasing functions:
๐ฎ(โ๐) = {๐ โ ๐ถโ(โ๐) โถ |๐ฅ๐ผ๐๐ฝ๐(๐ฅ)| < โ โ๐ผ, ๐ฝ โ โ(๐)}.
This space is called the Schwartz space.
Notation: Let ฮฉ โ โ๐ be an open subset. We let the following denote the following space of
distributions:
1.) ๐โฒ(ฮฉ) denotes the space of distributions over ฮฉ.
2.) โฐโฒ(ฮฉ) denotes the space of distributions over ฮฉ with compact support.
3.) ๐ฎโฒ(โ๐) denotes the space of tempered distributions over โ๐.
Definition: A function ๐ โ โ๐ โ โ is said to be of polynomial growth if there exist ๐ถ,๐ โฅ 0
such that
|๐(๐ฅ)| โค ๐ถ(1 + |๐ฅ|)๐ โ๐ฅ โ โ๐.
Notation: If ๐ is a function or distribution over โ๐ and โ โ โ๐, then ๐โ๐ stands for ๐ shifted in
the direction โ:
๐โ๐ = ๐(๐ฅ โ โ)
(in the case ๐ is a distribution, this means that ๐โ๐ is the distribution ๐ โฆ โจ๐, ๐(๐ฅ + โ)โฉ).
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Chapter 1
Principal Value Distribution ๐โ๐ (Problem 1.3) (9/25/2020)
The principle value distribution ๐ฅโ1 in ๐โฒ(โ) is defined by:
โจ๐ฅโ1, ๐โฉ = p.v. โซ๐(๐ฅ)
๐ฅ๐๐ฅ = lim
๐โ0+( โซ
๐(๐ฅ)
๐ฅ๐๐ฅ
โ๐
โโ
+ โซ๐(๐ฅ)
๐ฅ๐๐ฅ
โ
๐
).
To see that this is indeed a distribution, take any compact subset ๐พ โ โ. Let ๐ > 0 be such that
๐พ โ [โ๐, ๐]. For any ๐ โ ๐ถ๐โ(๐พ), we have that:
lim๐โ0+
( โซ๐(๐ฅ)
๐ฅ๐๐ฅ
โ๐
โโ
+ โซ๐(๐ฅ)
๐ฅ๐๐ฅ
โ
๐
) = lim๐โ0+
( โซ๐(๐ฅ)
๐ฅ๐๐ฅ
โ๐
โ๐
+ โซ๐(๐ฅ)
๐ฅ๐๐ฅ
๐
๐
)
= lim๐โ0+
(โซ๐(๐ฅ) โ ๐(โ๐ฅ)
๐ฅ๐๐ฅ
๐
๐
).
Using the fact that:
๐(๐ฅ) = ๐(0) + โซ๐โฒ(๐ )๐๐
๐ฅ
0
= ๐(0) + ๐ฅ โซ๐โฒ(๐ฅ๐ก)๐๐ก
1
0
(the rightmost expression is in fact just the 1st order Taylor expression for ๐ based at 0), we can
rewrite the previous expression as:
= lim๐โ0+
(โซ(๐(0) + ๐ฅ โซ ๐โฒ(๐ฅ๐ก)๐๐ก
1
0) โ (๐(0) โ ๐ฅ โซ ๐โฒ(โ๐ฅ๐ก)๐๐ก
1
0)
๐ฅ๐๐ฅ
๐
๐
)
lim๐โ0+
(โซ๐ฅ โซ (๐โฒ(๐ฅ๐ก) + ๐โฒ(โ๐ฅ๐ก))๐๐ก
1
0
๐ฅ๐๐ฅ
๐
๐
) = lim๐โ0+
(โซโซ(๐โฒ(๐ฅ๐ก) + ๐โฒ(โ๐ฅ๐ก))๐๐ก
1
0
๐๐ฅ
๐
๐
)
= โซโซ(๐โฒ(๐ฅ๐ก) + ๐โฒ(โ๐ฅ๐ก))๐๐ก
1
0
๐๐ฅ
๐
0
.
Where in the last equality Iโve used the Dominated Convergence Theorem with the observation
that the integrand of the outside integral is bounded by 2 sup๐โฒ < โ and that its integration
domain is bounded (itโs overkill to cite DCT here though). Thus, for any ๐ โ ๐ถ๐โ(โ) we have
the estimate:
|โจ๐ฅโ1, ๐โฉ| โค 2๐ sup๐โฒ
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and so ๐ฅโ1 is indeed a distribution.
Now, what is this distributionโs order? It turns out to be one. To prove this, first letโs observe that
because of the above inequality we know that its order is less than or equal to one. So if we
prove that itโs not equal to zero, then weโll be done. To do this, for any positive numbers ๐, ๐, ๐ โ
โ โถ ๐, ๐, ๐ > 0 let ๐๐,๐,๐ โถ โ โ โ be a compactly supported smooth bump function that satisfies:
1. supp๐ โ [0,โ),
2. ๐ โก ๐ on the interval [๐, ๐],
3. 0 โค ๐ โค ๐ everywhere.
Jack Leeโs Smooth Manifolds book shows how to construct such a smooth bump function. A
typical example looks like:
Notice that with these functions:
โจ๐ฅโ1, ๐๐,๐,๐โฉ = lim๐โ0+
โซ๐๐,๐,๐(๐ฅ)
๐ฅ๐๐ฅ
โ
๐
โฅ โซ๐
๐ฅ๐๐ฅ
๐
๐
= ๐ ln (๐
๐).
Now, consider the sequence of functions {๐๐๐,๐๐,๐๐}๐=1
โ with ๐๐ = 1 ๐โ , ๐๐ = 1, and ๐๐ = ๐โ๐2
.
Then:
โจ๐ฅโ1, ๐๐๐,๐๐,๐๐โฉ =
1
๐ln (
1
๐โ๐2) = ๐ โ โ as ๐ โ โ.
But if ๐ฅโ1 was an order 0 distribution, then the distribution โseminorm estimateโ would tell us
that โจ๐ฅโ1, ๐๐๐,๐๐,๐๐โฉ โ 0 as ๐ โ โ since:
sup|๐๐๐,๐๐,๐๐| =
1
๐โ 0 as ๐ โ โ.
So ๐ฅโ1 must indeed be an order 1 distribution.
A Distribution ๐ โ ๐โฒ(โ) such that ๐ = ๐/๐ on (๐,โ) and ๐ = ๐ on (โโ, ๐) (Problem 1.4)
(9/25/2020)
An example of a distribution ๐ข โ ๐โฒ(โ) such that ๐ข = 1 ๐ฅโ on (0,โ) and ๐ข = 0 on (โโ, 0) is:
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โจ๐ข, ๐โฉ = lim๐โ0+
โซ๐(๐ฅ) โ ๐(0)
๐ฅ๐๐ฅ
โ
๐
.
It obviously has the desired properties and itโs obvious that the limit here exists (since the
integrand tends to ๐โฒ(0) as ๐ฅ โ 0+). To see that itโs a distribution, take any compact ๐พ โ โ and
any ๐ > 0 such that ๐พ โ [โ๐, ๐]. Then taking any ๐ โ ๐ถ๐โ(๐พ) and using the equation ๐(๐ฅ) =
๐(0) + ๐ฅ โซ ๐โฒ(๐ก๐ฅ)๐๐ก1
0, we can rewrite the above quantity as:
lim๐โ0+
โซ๐(๐ฅ) โ ๐(0)
๐ฅ๐๐ฅ
๐
๐
= lim๐โ0+
โซโซ๐โฒ(๐ก๐ฅ)
1
0
๐๐ฅ
๐
๐
= โซโซ๐โฒ(๐ก๐ฅ)
1
0
๐๐ฅ
๐
0
,
and so we get the distribution โseminorm estimateโ |โจ๐ข, ๐โฉ| โค ๐ sup|๐โฒ|. Itโs interesting to note
that by the mathematics in the section discussing the principle value distribution ๐ฅโ1, itโs easy to
see that this distribution has order 1. Gunther Uhlmann also observed that if we want to solve the
same problem but instead require that ๐ข = 1 ๐ฅ๐โ on (0,โ), then we can use:
โจ๐ข, ๐โฉ = lim๐โ0+
โซ๐(๐ฅ) โ โ (๐(๐)(0) ๐!โ )๐ฅ๐๐โ1
๐=0
๐ฅ๐๐ฅ
โ
๐
.
The inner sum is of course just the (๐ โ 1)th Taylor polynomial of ๐ based at 0. I wonder what
the order of this distribution is?
An Interesting Example of a Non-Extendible Distribution (Problem 1.5) [2/26/2021].
Consider the linear form ๐ข โถ ๐ถ๐โ(0,โ) โ โ given by
โจ๐ข, ๐โฉ = โ ๐๐๐(1 ๐โ )
โ
๐=1
.
The claim is that this is a distribution and that it cannot be extended to all of โ (i.e. there does
not exist a ๐ฃ โ ๐โฒ(โ) such that the restriction of ๐ฃ to (0,โ) is ๐ข). First letโs show that itโs a
distribution. Take any compact subset ๐พ โ (0,โ) of (0,โ). Since ๐พ is compact we have that
the sequence 1 ๐โ for ๐ โ โค+ escapes ๐พ eventually. More precisely this means that there exists
an ๐ โ โค+ such that for any integer ๐ > ๐, 1 ๐โ โ ๐พ. Thus over ๐พ we have that ๐ข is given by
the finite sum
โจ๐ข, ๐โฉ = โ ๐๐๐(1 ๐โ )
๐
๐=1
โ๐ โ ๐ถ๐โ(๐พ).
Thus ๐ข satisfies the following distribution โsemi-normโ estimate over ๐พ:
|โจ๐ข, ๐โฉ| โค โ sup|๐๐๐|
๐
๐=1
โ๐ โ ๐ถ๐โ(๐พ).
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So ๐ข is indeed a distribution. Next letโs prove that ๐ข cannot be extended to all of โ. Letโs prove
this by contradiction. Suppose not. Then there exists a distribution ๐ฃ โ ๐โฒ(โ) such that
๐ฃ|(0,โ) = ๐ข. Take the compact subset ๐พ = [โ1, 1] of โ and let ๐ถ,๐ > 0 be such that
|โจ๐ฃ, ๐โฉ| โค ๐ถ โ sup|๐๐๐|
๐
๐=0
โ๐ โ ๐ถ๐โ(๐พ).
Let ๐ โ ๐ถ๐โ(โ) be a smooth bump function thatโs identically one on a neighborhood of zero and
whose support is contained in ๐พ = [โ1, 1]. For any ๐ โ โค+ โถ ๐ โฅ 2, let ๐๐ โ ๐ถ๐โ(โ) be the
test function
๐๐(๐ฅ) = ๐(2๐(๐ + 1)๐ฅ) โ ๐ฅ๐ .
(the condition ๐ โฅ 2 is for convenience so that as explained below supp๐๐ โ ๐พ). Graphically
speaking, this test function is equal to ๐ฅ๐ in a neighborhood of 1 ๐โ and whose compact support
is contained in an interval thatโs situated between 1 (๐ + 1)โ and 1 (๐ โ 1)โ (not including
these two points):
For those interested for the more precise statement, the support of ๐๐ is contained in the closed
interval centered at 1 ๐โ with radius half the distance from 1 (๐ + 1)โ and 1 ๐โ . On one hand
we have that (in the last inequality here I bound ๐ฅ by 1 (๐ โ 1)โ since the ๐๐ โก 0 past ๐ฅ >1 (๐ โ 1)โ )
|โจ๐ฃ, ๐๐โฉ| โค ๐ถ โ sup|๐๐๐๐|
๐
๐=0
โค ๐ถ โ โ(๐๐) sup|๐๐[๐(2๐(๐ + 1)๐ฅ)]| โ sup|๐๐โ๐(๐ฅ๐)|
๐
๐=0
๐
๐=0
โค ๐ถ โ โ(๐๐) sup|๐๐๐| (2๐(๐ + 1))
๐๐ โ โฆ โ (๐ โ ๐ + 1) (
1
๐ โ 1)๐โ๐๐
๐=0
๐
๐=0
.
By choosing a big enough constant ๐ท > 0, we can estimate this bound further by
|โจ๐ฃ, ๐๐โฉ| โค ๐ท๐๐(๐ + 1)๐๐๐โ1
(๐ โ 1)๐โ๐.
On the other hand, since supp๐๐ โ (0,โ) we have that:
โจ๐ฃ, ๐๐โฉ = โจ๐ข, ๐๐|(0,โ)โฉ = ๐๐๐๐(1 ๐โ ) = ๐!
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But we then have a contradiction since the previous inequality implies that โจ๐ฃ, ๐๐โฉ โ 0 as ๐ โ
โ while the above equation implies that โจ๐ฃ, ๐๐โฉ โ โ as ๐ โ โ. So indeed no such extension of
๐ข can exist.
Chapter 5
Convolution Equations on Forward Cones (Problem 5.5) (1/2/2021)
Note: Here I will write the components of my points/vectors as superscripts. For example, a
point ๐ฅ โ โ๐ will be explicitly written out as ๐ฅ = (๐ฅ1, โฆ , ๐ฅ๐).
Here I discuss the following result which appears as a problem in the book.
Theorem: Let ๐ โฅ 2 be an integer and for any point ๐ฅ โ โ๐ let ๏ฟฝฬ๏ฟฝ denote the point obtained by
projecting it down to its first ๐ โ 1 components: ๏ฟฝฬ๏ฟฝ = (๐ฅ1, โฆ , ๐ฅ๐โ1). In addition, let ๐ค be the
forward cone:
๐ค = {๐ฅ โ โ๐ โถ ๐ฅ๐ โฅ ๐|๏ฟฝฬ๏ฟฝ|}
where ๐ > 0 is some fixed positive constant. Let ๐๐คโฒ (โ๐) denote the set of all distributions over
โ๐ whose support is contained in ๐ค:
๐๐คโฒ (โ๐) = {๐ข โ ๐โฒ(โ๐) โถ ๐ ๐ข๐๐ ๐ข โ ๐ค}.
Then the following are true:
a) If ๐ข1, โฆ , ๐ข๐ โ ๐๐คโฒ (โ๐), then the convolution ๐ข1 โ โฆ โ ๐ข๐ is well defined.
b) If ๐ข1, โฆ , ๐ข๐ โ ๐๐คโฒ (โ๐) and ๐ฃ โ ๐โฒ(โ๐) is such that ๐ ๐ข๐๐ ๐ฃ โ {๐ฅ๐ โฅ ๐} for some fixed
๐ โ โ, then the convolution ๐ข1 โ โฆ โ ๐ข๐ โ ๐ฃ is also well defined.
c) Suppose that ๐ โ ๐๐คโฒ (โ๐) is a convolution operator that has a fundamental solution ๐ธ โ
๐๐คโฒ (โ๐). Then for any ๐ฃ โ ๐โฒ(โ๐) such that ๐ ๐ข๐๐ ๐ฃ โ {๐ฅ๐ โฅ ๐}, there exists a unique
solution ๐ข to the equation ๐ โ ๐ข = ๐ฃ such that ๐ ๐ข๐๐ ๐ข โ {๐ฅ๐ โฅ ๐} as well. Furthermore,
for any such solution ๐ข, ๐ ๐ข๐๐(๐ข) โ ๐ ๐ข๐๐ ๐ธ + ๐ ๐ข๐๐ ๐ฃ.
Proof: Letโs start with (a). Take any ๐ข1, โฆ , ๐ข๐ โ ๐ฮโฒ (โ๐). We need to show that the addition
function is proper over supp๐ข1 ร โฆร supp๐ข๐. We will do this by using the criterion for such
properness described in Definition 5.3.1 in the book. Take any ๐ฟ > 0 and suppose that
๐ฅ1, โฆ , ๐ฅ๐ โ โ๐ are such that each ๐ฅ๐ โ supp๐ข๐ and |๐ฅ1 + โฏ+ ๐ฅ๐| โค ๐ฟ. Since each supp๐ข๐ โ
ฮ, we have that each ๐ฅ๐๐ โฅ 0 and so the previous inequality implies that each ๐ฅ๐
๐ โค ๐ฟ. By
definition of ฮ we then get that each |๏ฟฝฬ๏ฟฝ๐| โค ๐ฟ ๐โ and so each:
|๐ฅ๐| โค โ(๐ฟ ๐โ )2 + ๐ฟ2 = ๐ฟโ(1 ๐โ )2 + 1.
Setting ๐ฟโฒ = ๐ฟโ(1 ๐โ )2 + 1 in Definition 5.3.1 then proves the desired properness. This proves
(a).
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Onwards to (b)! This is proved similarly: take any ๐ข1, โฆ , ๐ข๐ โ ๐ฮโฒ (โ๐) and any ๐ฃ โ ๐โฒ(โ๐)
such that supp ๐ฃ โ {๐ฅ๐ โฅ ๐} for some fixed ๐ โ โ. Take any ๐ฟ > 0 and suppose that ๐ฅ๐ โ
supp๐ข๐ and ๐ฆ โ supp ๐ฃ are such that |๐ฅ1 + โฏ+ ๐ฅ๐ + ๐ฆ| โค ๐ฟ. The last inequality implies that
both |๏ฟฝฬ๏ฟฝ1 + โฏ+ ๏ฟฝฬ๏ฟฝ๐ + ๏ฟฝฬ๏ฟฝ| โค ๐ฟ and |๐ฅ1๐ + โฏ+ ๐ฅ๐
๐ + ๐ฆ๐| โค ๐ฟ. The latter coupled with the facts
that each ๐ฅ๐๐ โฅ 0 and ๐ฆ๐ โฅ ๐ gives:
|๐ฅ๐๐| โค ๐ฅ1
๐ + โฏ+ ๐ฅ๐๐ + ๐ โ ๐ โค ๐ฅ1
๐ + โฏ+ ๐ฅ๐๐ + ๐ฆ๐ โ ๐ โค ๐ฟ โ ๐,
|๐ฆ๐| โค |๐ฅ1๐ + โฏ+ ๐ฅ๐
๐ + ๐ฆ๐| + |๐ฅ1๐ + โฏ+ ๐ฅ๐
๐ | โค ๐ฟ + ๐(๐ฟ โ ๐).
By the definition of ฮ we have that each |๏ฟฝฬ๏ฟฝ๐| โค (๐ฟ โ ๐) ๐โ . The inequality |๏ฟฝฬ๏ฟฝ1 + โฏ+ ๏ฟฝฬ๏ฟฝ๐ + ๏ฟฝฬ๏ฟฝ| โค
๐ฟ then gives us that:
|๏ฟฝฬ๏ฟฝ| โค |๏ฟฝฬ๏ฟฝ1 + โฏ+ ๏ฟฝฬ๏ฟฝ๐ + ๏ฟฝฬ๏ฟฝ| + |๏ฟฝฬ๏ฟฝ1 + โฏ+ ๏ฟฝฬ๏ฟฝ๐| โค ๐ฟ + ๐(๐ฟ โ ๐) ๐โ .
In total we get the results:
|๐ฅ๐| โค โ[(๐ฟ โ ๐) ๐โ ]2 + (๐ฟ โ ๐)2,
|๐ฆ| โค โ[๐ฟ + ๐(๐ฟ โ ๐) ๐โ ]2 + [๐ฟ + ๐(๐ฟ โ ๐)]2.
Setting ๐ฟโฒ > 0 in Definition 5.3.1 to be any constant bigger than the two constants on the right-
hand sides above then proves (b).
Finally we come to (c). Take any ๐ฃ โ ๐โฒ(โ๐) such that supp ๐ฃ โ {๐ฅ๐ โฅ ๐} for some fixed ๐ โ
โ. Obviously ๐ข = ๐ธ โ ๐ฃ is a desired solution since by Theorem 5.3.2 (iii) in the book,
supp๐ข โ supp๐ธ + supp ๐ฃ โ {๐ฅ๐ โฅ ๐}
and
๐ โ ๐ข = ๐ โ ๐ธ โ ๐ฃ = ๐ฟ โ ๐ฃ = ๐ฃ
(all of these convolutions make sense by part (b)). To prove that this is the unique desired
solution, suppose that ๏ฟฝฬ๏ฟฝ is another such solution. Convoluting both sides of the equation ๐ โ ๏ฟฝฬ๏ฟฝ =
๐ฃ by ๐ธ from the left gives:
๐ธ โ ๐ โ ๏ฟฝฬ๏ฟฝ = ๐ธ โ ๐ฃ.
The left-hand side here is equal to:
๐ธ โ ๐ โ ๏ฟฝฬ๏ฟฝ = ๐ โ ๐ธ โ ๏ฟฝฬ๏ฟฝ = ๐ฟ โ ๏ฟฝฬ๏ฟฝ = ๏ฟฝฬ๏ฟฝ.
Plugging this into the previous equation recovers our previous solution: ๏ฟฝฬ๏ฟฝ = ๐ธ โ ๐ฃ. So indeed the
desired solution to our equation is unique.
โ
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9
Chapter 6
Schwartz Kernel Theorem [Maintenance planned] (11/4/2020)
Note: I plan to do much needed maintenance on this entry. In particular, I want to make it a lot
more concise.
Note: This entry is hard.
Here I give my version of a proof to the Schwartz Kernel Theorem. Itโs exactly the same as the
proof in this book except that I fill in all of the details. Since this proof involves a lot of steps, I
think the best approach for any reader would be to first read Friedlanderโs shorter proof in order
to get the main idea, and then read this proof if they want to see the details filled in.
Before we prove the theorem, letโs first observe a lemma:
Lemma: Suppose that ๐ โ ๐ถโ(โ๐) is a smooth function such that it and all of its partials are ๐-
periodic where ๐ > 0. Then its Fourier series converges uniformly to ๐:
๐(๐ฅ) = โ ๐๐๐โโค
๐2๐๐(๐ฅโ ๐) ๐โ
where each
๐๐ =1
๐๐โซ๐(๐ )๐โ2๐๐(๐ โ ๐) ๐โ ๐๐
๐ ๐
,
where ๐ ๐ is any ๐-periodic box (for example: ๐ ๐ = [0, ๐]๐). Furthermore, for any ๐ผ โ โ(๐) (see
โnotations and conventionsโ), the ๐๐ผ partial of the Fourier seriesโ partial sums converge
uniformly to ๐๐ผ๐.
Proof: Iโd say that this is a rather standard fact from the theory of Fourier series.
โ
Now for the Schwartz Kernel Theorem:
Schwartz Kernel Theorem: Suppose that ๐ โ โ๐ and ๐ โ โ๐ are open subsets. A linear map
๐ โถ ๐ถ๐โ(๐) โ ๐โฒ(๐) is sequentially continuous if and only if itโs generated by a Schwartz Kernel
๐ โ ๐โฒ(๐ ร ๐):
โจ๐๐, ๐โฉ = โจ๐, ๐ โ ๐โฉ.
Furthermore, ๐ here is uniquely determined by ๐.
Proof: We already proved in the book that if such a map ๐ is generated by a Schwartz kernel ๐ โ
๐โฒ(๐ ร ๐), then itโs sequentially continuous. So letโs prove the forward implication. The fact
that ๐ is uniquely determined by ๐ is obvious since the above equation determines what ๐ is
equal to on a dense subset of ๐ถ๐โ(๐ ร ๐) (dense with respect to convergence in ๐ถ๐
โ(๐ ร ๐) of
course). So letโs prove the existence of such a ๐ โ ๐โฒ(๐ ร ๐).
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Let {๐พ๐}๐=1
โ and {๐๐}๐=1
โ be compact exhaustions of ๐ and ๐ respectively (we will need that fact
that each ๐พ๐ โ ๐พ๐+1int and ๐๐ โ ๐๐+1
int later). Consider the bilinear form ๐ต โถ ๐ถ๐โ(๐) ร ๐ถ๐
โ(๐) โ โ
given by:
๐ต(๐, ๐) = โจ๐๐, ๐โฉ.
Now, fix any ๐0 โ โค+. Take any ๐ โ ๐ถ๐โ(๐๐0). Since ๐๐ โ ๐โฒ(๐), we have that there exist
๐ถ๐, ๐๐ > 0 such that
|๐ต(๐, ๐)| = |โจ๐๐, ๐โฉ| โค ๐ถ๐ โ sup|๐๐ผ๐|
|๐ผ|โค๐๐
โ๐ โ ๐ถ๐โ(๐พ๐0).
Now take any ๐ โ ๐ถ๐โ(๐พ๐0). The map ๐ โฆ ๐๐ being sequentially continuous implies that the
map ๐ โฆ โจ๐๐, ๐โฉ is also sequentially continuous. Thus the latter map is a distribution in ๐ทโฒ(๐)
and so there exist ๐ถ๐, ๐๐ > 0 such that:
|๐ต(๐, ๐)| = |โจ๐๐, ๐โฉ| โค ๐ถ๐ โ sup|๐๐ฝ๐||๐ฝ|โค๐๐
โ๐ โ ๐ถ๐โ(๐๐0).
These two inequalities show that the restriction ๐ต๐0 โถ ๐ถ๐โ(๐พ๐0) ร ๐ถ๐
โ(๐๐0) โ โ of ๐ต is a
separately continuous bilinear form over the Frรฉchet spaces ๐ถ๐โ(๐พ๐0) and ๐ถ๐
โ(๐๐0). Now, noting
that Frรฉchet spaces are also Banach spaces we have by a quick corollary of the uniform
boundedness principle that ๐ต๐0 is continuous with respect to the product topology. I prove this
corollary in my electronic diary about Follandโs โReal Analysisโ book. In the that same
electronic diary, I also prove an equivalent condition for a bilinear map between Frรฉchet spaces
being continuous which in our case gives that there exist ๐ถ๐0 , ๐๐0 > 0 such that:
|๐ต(๐, ๐)| = |๐ต๐0(๐, ๐)| โค ๐ถ๐0 ( โ sup|๐๐ผ๐|
|๐ผ|โค๐๐0
)( โ sup|๐๐ฝ๐||๐ฝ|โค๐๐0
)
โ๐ โ ๐ถ๐โ(๐พ๐0) and โ๐ โ ๐ถ๐
โ(๐๐0).
Ok, with this in hand we are now ready to begin constructing the ๐ โ ๐โฒ(๐ ร ๐ ) that we want.
We will do this by defining continuous linear forms ๐๐ โถ ๐ถ๐โ(๐พ๐ ร ๐๐) โ โ for ๐ โ โค+ and then
use their values to define ๐. Fix any ๐0 โ โค+. Let ๐ โ ๐ถ๐โ(๐พ๐0+1) and ๐ โ ๐ถ๐
โ(๐๐0+1) be such
that ๐ โก 1 and ๐ โก 1 on neighborhoods of ๐พ๐0 and ๐๐0 respectively. Let ๐ > 0 be such that
๐พ๐0+1 ร ๐๐0+1 โ [โ๐, ๐]๐+๐. Now, take any ๐ โ ๐ถ๐โ(๐พ๐0 ร ๐๐0). Define the value of ๐๐0 at ๐ to
be:
โจ๐๐0 , ๐โฉ = โ ๐ต(๏ฟฝฬ๏ฟฝ(๐,โ)๐(๐ฅ)๐2๐๐(๐ฅโ ๐) (2๐)โ , ๐(๐ฆ)๐2๐๐(๐ฆโ โ) (2๐)โ )(๐,โ)โโค๐รโค๐
where ๏ฟฝฬ๏ฟฝ(๐,โ) are the Fourier coefficients of ๐:
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๏ฟฝฬ๏ฟฝ(๐,โ) =1
(2๐)๐+๐โซ ๐(๐ฅ, ๐ฆ)๐โ2๐๐(๐ฅโ ๐+๐ฆโ โ) (2๐)โ ๐๐ฅ๐๐ฆ
[โ๐,๐]๐+๐
.
Technically I should be writing ๐โ๐รโ๐ (see notations and conventions) rather than ๐ inside the
above integral since ๐ is not necessarily defined over [โ๐, ๐]. Ok, in order for the previous
expression to make sense letโs prove that the sum on the right-hand side converges uniformly.
Before we do that though, Iโd like to point out that once we prove that the above sum makes
sense and is finite, then it will be clear that ๐๐0 is a linear form since the equation for the Fourier
coefficient is linear in ๐ and ๐ต is linear in its first argument.
Ok, letโs prove that that sum is absolutely convergent. First letโs create a bound to estimate how
large the above Fourier coefficients of ๐ can get. Fix any index (๐, โ) โ โค๐ ร โค๐. First a piece
of notation: for an any multi-index ๐ผ โ โ(๐), let โ(๐ผ) be the multi-index whose ๐ th component
is equal to 1 if ๐ผ๐ > 0 and is equal to 0 if ๐ผ๐ = 0. Let ๐ = ๐๐0+1. Notice that since
supp๐ โ ๐พ๐0 ร ๐๐0 โ ๐พ๐0+1int ร ๐๐0+1
int โ [โ๐, ๐]๐+๐,
we have that ๐ and all of its partials are zero on the boundary of the box [โ๐, ๐]๐+๐. So for any
(๐, โ) โ โค๐ ร โค๐ we have by many integrations by parts that
๏ฟฝฬ๏ฟฝ(๐,โ) =1
(2๐)๐+๐(
2๐
โ2๐๐)
(๐+2)|โ(๐,โ)|1
โ ๐๐๐+2๐
๐=1,๐๐โ 0 โ โ๐ ๐+2๐
๐ =1,โ๐ โ 0
โ โซ ๐(๐+2)โ(๐,โ)[๐(๐ฅ, ๐ฆ)]๐โ2๐๐(๐ฅโ ๐+๐ฆโ โ) (2๐)โ ๐๐ฅ๐๐ฆ
[โ๐,๐]๐+๐
.
Let ๐ธ1 > 0 be a constant such that:
|1
(2๐)๐+๐(
2๐
โ2๐๐)
(๐+2)|โ(๐,โ)|
| โค ๐ธ1.
Furthermore, notice that we can remove the unpleasant ๐๐ โ 0 and โ๐ โ 0 indexing rules above
by writing the estimate (here I use the fact that 1 ๐โ โค 2 (1 + ๐)โ if ๐ > 0 is an integer):
|1
โ ๐๐๐+2๐
๐=1,๐๐โ 0 โ โ๐ ๐+2๐
๐ =1,โ๐ โ 0
| โค2๐+๐
โ (1 + |๐๐|๐+2)๐๐=1 โ (1 + |โ๐ |๐+2)๐
๐ =1
Setting ๐ธ2 = 2๐+๐๐ธ1, we get the following estimate on our Fourier coefficient:
|๏ฟฝฬ๏ฟฝ(๐,โ)| โค ๐ธ2
1
โ (1 + |๐๐|๐+2)๐๐=1 โ (1 + |โ๐ |๐+2)๐
๐ =1
Vol([โ๐, ๐]๐+๐) sup|๐(๐+2)โ(๐,โ)๐|
Setting ๐ธ3 = ๐ธ2 Vol([โ๐, ๐]๐+๐), notice that we can furthermore estimate our Fourier
coefficient as:
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|๏ฟฝฬ๏ฟฝ(๐,โ)| โค ๐ธ3 ( โ sup|๐๐พ๐|
|๐พ|โค๐+2
)1
โ (1 + |๐๐|๐+2)๐๐=1 โ (1 + |โ๐ |๐+2)๐
๐ =1
,
which is written in a bit more familiar terms. Great! Weโre going to use this estimate below.
Meanwhile, going back to our definition of โจ๐๐0 , ๐โฉ, notice that we can bound each summand
term on the right by:
|๐ต(๏ฟฝฬ๏ฟฝ(๐,โ)๐(๐ฅ)๐2๐๐(๐ฅโ ๐) (2๐)โ , ๐(๐ฆ)๐2๐๐(๐ฅโ โ) (2๐)โ )|
โค ๐ถ๐0 ( โ sup|๐๐ผ[๏ฟฝฬ๏ฟฝ(๐,โ)๐(๐ฅ)๐2๐๐(๐ฅโ ๐) (2๐)โ ]||๐ผ|โค๐
)( โ sup|๐๐ฝ[๐(๐ฆ)๐2๐๐(๐ฆโ โ) (2๐)โ ]||๐ฝ|โค๐
)
Now apply the product rule on each ๐๐ผ and ๐๐ฝpartials (recall that by convention 00 = 1 when
raising something to a multi-index):
|๐ต(๏ฟฝฬ๏ฟฝ(๐,โ)๐(๐ฅ)๐2๐๐(๐ฅโ ๐) (2๐)โ , ๐(๐ฆ)๐2๐๐(๐ฅโ โ) (2๐)โ )|
โค ๐ถ๐0|๏ฟฝฬ๏ฟฝ(๐,โ)| ( โ โ sup |๐ผ!
๐! (๐ผ โ ๐)!(2๐๐
2๐)|๐|
๐๐๐2๐๐(๐ฅโ ๐) (2๐)โ ๐๐ผโ๐๐|
๐โค๐ผ|๐ผ|โค๐
)
โ ( โ โ sup |๐ฝ!
๐! (๐ฝ โ ๐)!(2๐๐
2๐)|๐|
โ๐๐2๐๐(๐ฆโ โ) (2๐)โ ๐๐ฝโ๐๐|
๐โค๐ฝ|๐ฝ|โค๐
)
If we distribute above sum, we merely get a big linear combination of terms of the form ๐๐โ๐.
So for some collection of coefficients ๐ด(๐,๐),
|๐ต(๏ฟฝฬ๏ฟฝ(๐,โ)๐(๐ฅ)๐2๐๐(๐ฅโ ๐) (2๐)โ , ๐(๐ฆ)๐2๐๐(๐ฅโ โ) (2๐)โ )| โค ๐ถ๐0|๏ฟฝฬ๏ฟฝ(๐,โ)| โ ๐ด(๐,๐)๐๐โ๐
|๐|โค๐|๐|โค๐
.
Now, if we plug in the above estimate for |๏ฟฝฬ๏ฟฝ(๐,โ)| into this inequality we get that for some
constant ๐ธ4 > 0,
|๐ต(๏ฟฝฬ๏ฟฝ(๐,โ)๐(๐ฅ)๐2๐๐(๐ฅโ ๐) (2๐)โ , ๐(๐ฆ)๐2๐๐(๐ฅโ โ) (2๐)โ )|
โค ๐ธ4 โ๐๐โ๐
โ (1 + |๐๐|๐+2)๐๐=1 โ (1 + |โ๐ |๐+2)๐
๐ =1|๐|โค๐|๐|โค๐
โ sup|๐๐พ๐|
|๐พ|โค๐+2
Ok, since each |๐| โค ๐ in the first sum, we have that
๐๐
โ (1 + |๐๐|๐+2)๐๐=1
โค๐๐
โ ๐๐๐+2๐
๐=1,๐๐โ 0
โค1
โ ๐๐2๐
๐=1,๐๐โ 0
โค2๐
โ (1 + |๐๐|2)๐๐=1
.
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By a similar calculation we have that
โ๐
โ (1 + |โ๐ |๐+2)๐๐ =1
โค2๐
โ (1 + |โ๐ |2)๐๐ =1
.
So we get that for some constant ๐ธ5 > 0,
|๐ต(๏ฟฝฬ๏ฟฝ(๐,โ)๐(๐ฅ)๐2๐๐(๐ฅโ ๐) (2๐)โ , ๐(๐ฆ)๐2๐๐(๐ฅโ โ) (2๐)โ )|
โค ๐ธ5
1
โ (1 + |๐๐|2)๐
๐=1 โ (1 + |โ๐ |2)๐
๐ =1
โ sup|๐๐พ๐|
|๐พ|โค๐+2
.
Tracing the above calculation again, notice that the value of the constant ๐ธ5 is independent of
(๐, โ) (and even ๐). So the above estimate holds for all (๐, โ) โ โค๐ ร โค๐ (and all ๐). Thus the
above inequality shows that that the sum in the definition of โจ๐๐0 , ๐โฉ decays like ๐๐2 and โ๐
2 in all
direction. Hence that sum is indeed absolutely convergent and thus makes sense. The above
inequality shows more: it gives us the estimate:
|โจ๐๐0 , ๐โฉ| โค ๐ธ5 ( โ1
โ (1 + |๐๐|2)๐๐=1 โ (1 + |โ๐ |2)
๐๐ =1(๐,โ)โโค๐รโค๐
) โ sup|๐๐พ๐|
|๐พ|โค๐+2
.
Again, since the ฮฃ(๐,โ)โโค๐รโค๐ sum here is finite and the value of the constant ๐ธ5 is independent
of ๐, this shows that ๐๐0 is in fact a continuous linear form over ๐ถ๐โ(๐พ๐0 ร ๐๐0).
Great, letโs show one last property of ๐๐0 . Letโs show that for any ๐ โ ๐ถ๐โ(๐พ๐0) and any ๐ โ
๐ถ๐โ(๐๐0),
โจ๐๐0 , ๐ โ ๐โฉ = ๐ต(๐,๐) = โจ๐๐, ๐โฉ.
The Fourier coefficients of (๐ โ ๐)โ๐รโ๐ are given by:
(๐ โ ๐)ฬ(๐,โ) =
1
(2๐)๐+๐โซ ๐(๐ฅ)๐(๐ฆ)๐โ2๐๐(๐ฅโ ๐+๐ฆโ โ) (2๐)โ ๐๐ฅ๐๐ฆ
[โ๐,๐]๐+๐
=1
(2๐)๐โซ ๐(๐ฅ)๐โ2๐๐(๐ฅโ ๐) (2๐)โ ๐๐ฅ
[โ๐,๐]๐
โ 1
(2๐)๐โซ ๐(๐ฆ)๐โ2๐๐(๐ฆโ โ) (2๐)โ ๐๐ฆ
[โ๐,๐]๐
.
Setting ๏ฟฝฬ๏ฟฝ๐ and ๏ฟฝฬ๏ฟฝโ to be the Fourier coefficients of ๐โ๐ and ๐โ๐
respectively:
๏ฟฝฬ๏ฟฝ๐ =1
(2๐)๐โซ ๐(๐ฅ)๐โ2๐๐(๐ฅโ ๐) (2๐)โ ๐๐ฅ
[โ๐,๐]๐
,
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๏ฟฝฬ๏ฟฝโ =1
(2๐)๐โซ ๐(๐ฅ)๐โ2๐๐(๐ฆโ โ) (2๐)โ ๐๐ฆ
[โ๐,๐]๐
,
we then get the trivial tensor Fourier coefficient relation:
(๐ โ ๐)ฬ(๐,โ) = ๏ฟฝฬ๏ฟฝ๐ โ ๏ฟฝฬ๏ฟฝโ.
Using this in the definition of โจ๐๐0 , ๐ โ ๐โฉ then gives us that
โจ๐๐0 , ๐ โ ๐โฉ = โ ๐ต(๏ฟฝฬ๏ฟฝ๐๐(๐ฅ)๐2๐๐(๐ฅโ ๐) (2๐)โ , ๏ฟฝฬ๏ฟฝโ๐(๐ฆ)๐2๐๐(๐ฆโ โ) (2๐)โ )(๐,โ)โโค๐รโค๐
.
Now, itโs easy to see by the lemma stated before this theorem that the series
โ ๏ฟฝฬ๏ฟฝ๐๐(๐ฅ)๐2๐๐(๐ฅโ ๐) (2๐)โ
๐โโค๐
= ๐(๐ฅ) โ ๏ฟฝฬ๏ฟฝ๐๐2๐๐(๐ฅโ ๐) (2๐)โ
๐โโค๐
.
converges to ๐ โ ๐ in ๐ถ๐โ(โ๐). So this series converges to ๐ in ๐ถ๐
โ(๐พ๐0+1) since ๐ โ ๐ = ๐ on
๐พ๐0+1 (recall that supp๐ โ ๐พ๐0 and ๐ โก 1 on ๐พ๐0). For the same reason, we have that the series
โ ๏ฟฝฬ๏ฟฝโ๐(๐ฆ)๐2๐๐(๐ฆโ โ) (2๐)โ
โโโค๐
converges to ๐ in ๐ถ๐โ(๐๐0+1) as well. Using the fact that ๐ต๐0+1 โถ ๐ถ๐
โ(๐พ๐0+1) ร ๐ถ๐โ(๐๐0+1) โ โ
is separately continuous we then have that
โจ๐๐0 , ๐ โ ๐โฉ = โ โ ๐ต๐0+1(๏ฟฝฬ๏ฟฝ๐๐(๐ฅ)๐2๐๐(๐ฅโ ๐) (2๐)โ , ๏ฟฝฬ๏ฟฝโ๐(๐ฆ)๐2๐๐(๐ฆโ โ) (2๐)โ )
โโโค๐๐โโค๐
= โ ๐ต๐0+1(๏ฟฝฬ๏ฟฝ๐๐(๐ฅ)๐2๐๐(๐ฅโ ๐) (2๐)โ , ๐)
๐โโค๐
= ๐ต๐0(๐, ๐) = ๐ต(๐,๐) = โจ๐๐, ๐โฉ.
We are now finally ready for the last step in the proof of this theorem. Define the function ๐ โถ
๐ถ๐โ(๐ ร ๐) โ โ as follows. Take any test function ๐ โ ๐ถ๐
โ(๐ ร ๐) and let ๐0 โ โค+ be such that
supp๐ โ ๐พ๐0 ร ๐๐0. Then set:
โจ๐, ๐โฉ = โจ๐๐0 , ๐โฉ.
Letโs prove that this is well defined by showing that this is independent of the ๐0 that we chose
that satisfies the above property. Let ๐0 โ โค+ be any other such index and letโs suppose without
loss of generality that ๐0 โค ๐0. As before, let ๐ โ ๐ถ๐โ(๐พ๐0+1) and ๐ โ ๐ถ๐
โ(๐๐0+1) be such that
๐ โก 1 and ๐ โก 1 on neighborhoods of ๐พ๐0 and ๐๐0 respectively. Letting ๐ > 0 be such that
๐พ๐0+1 ร ๐๐0+1 โ [โ๐, ๐]๐+๐ we have by a similar discussion as above that the series
โ ๏ฟฝฬ๏ฟฝ(๐,โ)๐(๐ฅ)๐2๐๐(๐ฅโ ๐) (2๐)โ ๐(๐ฆ)๐2๐๐(๐ฆโ โ) (2๐)โ
(๐,โ)โโค๐รโค๐
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converges to ๐ in ๐ถ๐โ(๐พ๐0+1 ร ๐๐0+1) and thus in ๐ถ๐
โ(๐พ๐0+1 ร ๐๐0+1) (since ๐พ๐0+1 ร ๐๐0+1 โ
๐พ๐0+1 ร ๐๐0+1). The continuity and linearity of both ๐๐0 and ๐๐0 implies that their values at ๐ are
given by:
โจ๐๐0 , ๐โฉ = โ ๐๐0(๏ฟฝฬ๏ฟฝ(๐,โ)๐(๐ฅ)๐2๐๐(๐ฅโ ๐) (2๐)โ โ ๐(๐ฆ)๐2๐๐(๐ฆโ โ) (2๐)โ )(๐,โ)โโค๐รโค๐
,
โจ๐๐0 , ๐โฉ = โ ๐๐0(๏ฟฝฬ๏ฟฝ(๐,โ)๐(๐ฅ)๐2๐๐(๐ฅโ ๐) (2๐)โ โ ๐(๐ฆ)๐2๐๐(๐ฆโ โ) (2๐)โ )(๐,โ)โโค๐รโค๐
,
both of which are equal to
โ ๐ต(๏ฟฝฬ๏ฟฝ(๐,โ)๐(๐ฅ)๐2๐๐(๐ฅโ ๐) (2๐)โ , ๐(๐ฆ)๐2๐๐(๐ฆโ โ) (2๐)โ )(๐,โ)โโค๐รโค๐
.
So โจ๐๐0 , ๐โฉ = โจ๐๐0 , ๐โฉ, and thus ๐ is indeed well defined. Itโs easy to see from the definition of ๐
that itโs linear. To see that itโs a distribution, we also have to show that it satisfies the distribution
โseminormโ like inequalities. Take any compact set ๐ โ ๐ ร ๐. Let ๐0 โ โค+ be such that ๐ โ๐พ๐0 ร ๐๐0. Then since ๐๐0 is continuous we have that there exist ๐ถ, ๐ > 0 such that for any ๐ โ
๐ถ๐โ(๐ ) โ ๐ถ๐
โ(๐พ๐0 ร ๐๐0),
|โจ๐, ๐โฉ| = |โจ๐๐0 , ๐โฉ| โค ๐ถ โ sup|๐๐พ๐|
|๐พ|โค๐
.
So ๐ is indeed a distribution over ๐ ร ๐ (i.e. ๐ โ ๐โฒ(๐ ร ๐)). The last thing to show is that it
satisfies the property that we desire. Take any ๐ โ ๐ถ๐โ(๐) and any ๐ โ ๐ถ๐
โ(๐). Let ๐0 โ โค+ be
such that supp๐ ร supp๐ โ ๐พ๐0 ร ๐๐0. Then we have that:
โจ๐, ๐ โ ๐โฉ = โจ๐๐0 , ๐ โ ๐โฉ = โจ๐๐, ๐โฉ.
So ๐ is the Schwartz kernel of our map ๐. With this weโve proven the theorem.
โ
Chapter 8
Structure Theorem for Tempered Distributions [Theorem 8.3.1] (5/11/2021)
In this entry I work through the proof of the structure theorem for tempered distributions
(Theorem 8.3.1 in the book) by putting it into my own words and filling in the details.
Theorem: A distribution is a tempered distribution if and only if it is the derivative of a
continuous function of polynomial growth.
Proof: First suppose that a distribution ๐ข โ ๐โฒ(โ๐) is the derivative of a continuous function
๐ โ ๐ถ0(โ๐) of polynomial growth:
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๐ข = ๐๐ผ๐,
where ๐ผ โ โ(๐) of course. We want to show that ๐ข is a tempered distribution. Since ๐ is of
polynomial growth, there exist ๐ถ > 0 and ๐ โ โค+ such that |๐(๐ฅ)| โค ๐ถ(1 + |๐ฅ|)๐. Now, I
claim that the linear functional ๐ฎ(โ๐) โ โ given by
(Eq 1) ๐ โผ (โ1)|๐ผ| โซ๐๐๐ผ๐
is a well-defined continuous extension of the distribution ๐๐ผ๐ โถ ๐ถ๐โ(โ๐) โ โ to ๐ฎ(โ๐). If we
prove this, then this will show that ๐ข is indeed a tempered distribution. Take any ๐ โ ๐ฎ(โ๐).
First letโs show that the above integral even exists. We estimate (here ๐ is the Lebesgue
measure)
โซ|๐๐๐ผ๐| โค ๐ถ โซ(1 + |๐ฅ|)๐|๐๐ผ๐| = ๐ถ [ โซ (1 + |๐ฅ|)๐|๐๐ผ๐|
๐ต1(0)ฬ ฬ ฬ ฬ ฬ ฬ ฬ ฬ
+ โซ(1 + |๐ฅ|)๐|๐ฅ|2๐|๐๐ผ๐|
|๐ฅ|2๐
[๐ต1(0)]๐
]
โค ๐ถ [๐(๐ต1(0)) sup{(1 + |๐ฅ|)๐|๐๐ผ๐|} + โซ1
|๐ฅ|2๐๐๐ฅ
[๐ต1(0)]๐
โ sup{(1 + |๐ฅ|)๐|๐ฅ|2๐|๐๐ผ๐|}].
The last quantity is finite since ๐ โ ๐ฎ(โ๐) and thus the integral in (Eq 1) is indeed well defined.
Itโs clear that (Eq 1) extends ๐๐ผ๐ โถ ๐ถ๐โ(โ๐) โ โ to ๐ฎ(โ๐). So all thatโs left to prove is that the
linear functional defined by (Eq 1) is also continuous. By exactly the same computation as
above, for some constant ๐ถ2 > 0 we obtain the estimate:
|(โ1)|๐ผ| โซ๐๐๐ผ๐| โค ๐ถ2[sup{(1 + |๐ฅ|)๐|๐๐ผ๐|} + sup{(1 + |๐ฅ|)๐|๐ฅ|2๐|๐๐ผ๐|}].
By expanding the (1 + |๐ฅ|)๐ via the binomial theorem and then doing some further estimates,
we see that the above inequality implies that there exist ๐ถ3 > 0 and ๐ โ โค+ such that
|(โ1)|๐ผ| โซ๐๐๐ผ๐| โค ๐ถ3 โ sup|๐ฅ๐ฝ๐๐ผ๐|
|๐ฝ|โค๐
.
Thus the linear functional defined by (Eq 1) is indeed continuous. As discussed above, this
proves that ๐ข is a tempered distribution.
Now letโs prove the other direction. Suppose that ๐ข โ ๐โฒ(โ๐) is a tempered distribution. We will
assume that the support of ๐ข is contained in โ+๐ = {๐ฅ โ โ๐ โถ each ๐ฅ๐ > 0}. The general case will
then follow from the easy-to-see fact that any tempered distribution can be represented as a finite
sum of translations and reflections of such tempered distributions and that reflections and
translations of functions of polynomial growth are still of polynomial growth.
Alright, since ๐ข is a tempered distribution we have that there exist ๐ถ > 0 and ๐ โ โค+ such that
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|โจ๐ข, ๐โฉ| โค ๐ถ โ sup|๐ฅ๐ผ๐๐ฝ๐||๐ผ|,|๐ฝ|โค๐
โ๐ โ ๐ฎ(โ๐).
Let ๐ธ๐+2 โถ โ๐ โ โ denote the function
๐ธ๐+2(๐ฅ) = {(๐ฅ1)
๐+1 โ โฆ โ (๐ฅ๐)๐+1
[(๐ + 1)!]๐ if each ๐ฅ๐ โฅ 0
0 otherwise
Observe that ๐ธ๐+2 is in ๐ถ๐(โ๐) and that
๐(๐+2)1โโ ๐ธ๐+2 = ๐ฟ
in the sense of distributions (here 1โ โ โ(๐) denotes the multi-index with all ones). Let ๐ โ
๐ถ๐โ(โ๐) be a smooth function such that
1.) ๐ โฅ 0
2.) โซ๐ = 1
3.) supp ๐ โ ๐ต1(0).
For each ๐ โ โค+, let ๐๐ โ ๐ถ๐โ(โ๐) denote the functions ๐๐ = ๐๐๐(๐๐ฅ). Observe that ๐๐ โ ๐ฟ in
๐โฒ(โ๐). Now, we will prove that ๐ธ๐+2 โ ๐ข is a well-defined continuous function of polynomial
growth such that
๐(๐+2)1โโ (๐ธ๐+2 โ ๐ข) = ๐ข.
The fact that ๐ธ๐+2 โ ๐ข is well defined comes from the fact that the addition function โ๐ ร โ๐ โ
โ๐ given by (๐ฅ, ๐ฆ) โฆ ๐ฅ + ๐ฆ is clearly proper on โ+๐ฬ ฬ ฬ ฬ ร โ+
๐ฬ ฬ ฬ ฬ and hence proper on
supp๐ธ๐+2 ร supp๐ข (recall that supp๐ธ๐+2 and supp๐ข are both contained in โ+๐ฬ ฬ ฬ ฬ ). The fact that
the above equation holds follows immediately from
๐(๐+2)1โโ (๐ธ๐+2 โ ๐ข) = ๐(๐+2)1โโ (๐ธ๐+2) โ ๐ข = ๐ฟ โ ๐ข = ๐ข.
Next letโs show that ๐ธ๐+2 โ ๐ข is a continuous function. We will do this by showing that itโs the
uniform limit over compact sets of the continuous functions described in the following claim.
Claim: For any ๐ โ โค+, the following is a well-defined smooth function over โ๐:
๐ธ๐+2 โ ๐๐ โ ๐ข.
Furthermore, if ๐ โ ๐ถโ(โ๐) is such that ๐ โก 1 on a neighborhood of ๐ ๐ข๐๐(๐ข) and ๐ ๐ข๐๐ ๐ โ
โ+๐ , then this function is explicitly given by
(Eq 2) ๐ธ๐+2 โ ๐๐ โ ๐ข(๐ฅ) = โจ๐ข(๐ฆ), ๐(๐ฆ)(๐ธ๐+2 โ ๐๐)(๐ฅ โ ๐ฆ)โฉ.
Proof: Fix any ๐ โ โค+. The fact that the convolution ๐ธ๐+2 โ ๐๐ โ ๐ข is well defined follows from
the not-hard-to-prove fact that the restriction of the addition function โ๐ ร โ๐ ร โ๐ โ โ๐ given
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by (๐ฅ, ๐ฆ, ๐ง) โฆ ๐ฅ + ๐ฆ + ๐ง to โ+๐ฬ ฬ ฬ ฬ ร ๐ต1 ๐โ (0)ฬ ฬ ฬ ฬ ฬ ฬ ฬ ฬ ฬ ฬ ร โ+
๐ฬ ฬ ฬ ฬ is proper (recall that supp ๐๐ โ ๐ต1 ๐โ (0)ฬ ฬ ฬ ฬ ฬ ฬ ฬ ฬ ฬ ฬ ). Now,
letโs see why the right-hand side of the above equation even makes sense. Specifically we have
to show that argument of ๐ข on the right-hand side (i.e. ๐(๐ฆ)(๐ธ๐+2 โ ๐๐)(๐ฅ โ ๐ฆ)) is smooth and
of compact support. The compactness of its support follows from:
๐(๐ฆ)(๐ธ๐+2 โ ๐๐)(๐ฅ โ ๐ฆ) โ 0 โน ๐ฆ โ supp๐ and (๐ฅ โ ๐ฆ) โ supp๐ธ๐+2 + supp๐๐
โน ๐ฆ โ โ+๐ and (๐ฅ โ ๐ฆ) โ โ+
๐ + ๐ต1 ๐โ (0) โน ๐ฆ โ โ+๐ and ๐ฆ โ ๐ฅ โ โ+
๐ โ ๐ต1 ๐โ (0)
โน |๐ฆ| โค |๐ฅ| + 1 ๐โ .
The smoothness of the right-hand side of (Eq 2) also follows from this and Theorem 4.1.1 in the
book.
To finish proving the claim, all we have to do now is show that equality holds in (Eq 2). To start,
let ๐ = ๐ธ๐+2 โ ๐๐. Let ๐๐ , ๐๐ข โ ๐ถโ(โ๐) be such that
1.) ๐๐ โก 1 on a neighborhood of supp(๐) and supp๐๐ โ supp(๐) + ๐ต1(0),
2.) ๐๐ข โก 1 on a neighborhood of supp๐ข and supp๐๐ข โ supp๐ข + ๐ต1(0).
Take any test function ๐ โ ๐ถ๐โ(โ๐). Analogously, let ๐๐ โ ๐ถ๐
โ(โ๐) be such that ๐๐ โก 1 on a
neighborhood of supp๐ (note the requirement for supp๐๐ to be compact). By definition, we
then have that
โจ๐ โ ๐ข(๐ง), ๐(๐ง)โฉ = โจ๐(๐ฅ) โ ๐ข(๐ฆ), ๐๐(๐ฅ)๐๐ข(๐ฆ)๐(๐ฅ + ๐ฆ)โฉ
= โจ๐ข(๐ฆ),โซ ๐(๐ฅ)๐๐(๐ฅ)๐๐ข(๐ฆ)๐(๐ฅ + ๐ฆ)๐๐ฅโฉ = โจ๐ข(๐ฆ),โซ๐(๐ง โ ๐ฆ)๐๐ข(๐ฆ)๐๐(๐ง)๐(๐ง)๐๐ฅโฉ
= โจ๐ข(๐ฆ), โจ๐(๐ง), ๐๐ข(๐ฆ)๐๐(๐ง)๐(๐ง โ ๐ฆ)โฉโฉ = โจ๐ข(๐ฆ) โ ๐(๐ง), ๐๐ข(๐ฆ)๐๐(๐ง)๐(๐ง โ ๐ฆ)โฉ
= โซ๐(๐ง)โจ๐ข(๐ฆ), ๐๐ข(๐ฆ)๐๐(๐ง)๐(๐ง โ ๐ฆ)โฉ๐๐ง = โจโจ๐ข(๐ฆ), ๐๐ข(๐ฆ)๐(๐ง โ ๐ฆ)โฉ, ๐(๐ง)โฉ.
Since ๐ โ ๐ถ๐โ(โ๐) was chosen arbitrarily, this shows that
๐ โ ๐ข(๐ง) = โจ๐ข(๐ฆ), ๐๐ข(๐ฆ)๐(๐ง โ ๐ฆ)โฉ.
From here (Eq 2) follows immediately.
End of Proof of Claim.
Let ๐ โ ๐ถโ(โ๐) be as in the above claim. First letโs show that the sequence of functions ๐ธ๐+2 โ
๐๐ โ ๐ข is uniformly Cauchy over compact sets and thus converges to some continuous function.
Fix any compact subset ๐พ โ โ๐ and let ๐ > 0 be such that ๐พ โ ๐ต๐ (0). Then, by the above
claim we have that
(Eq 3) sup๐ฅโ๐พ
|๐ธ๐+2 โ ๐๐ โ ๐ข(๐ฅ) โ ๐ธ๐+2 โ ๐๐ โ ๐ข(๐ฅ)|
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= sup๐ฅโ๐พ
|โจ๐ข(๐ฆ), ๐(๐ฆ)[๐ธ๐+2 โ ๐๐(๐ฅ โ ๐ฆ) โ ๐ธ๐+2 โ ๐๐(๐ฅ โ ๐ฆ)]โฉ|
โค ๐ถ โ sup๐ฅโ๐พ
sup๐ฆโโ๐
|๐ฆ๐ผ๐๐ฆ๐ฝ(๐(๐ฆ)[๐ธ๐+2 โ ๐๐(๐ฅ โ ๐ฆ) โ ๐ธ๐+2 โ ๐๐(๐ฅ โ ๐ฆ)])|
|๐ผ|,|๐ฝ|โค๐
.
Itโs not hard to see that there exists a compact subset ๐ โ โ๐ such that for any fixed ๐ฅ โ ๐พ, the
support of the function
๐(๐ฆ)[๐ธ๐+2 โ ๐๐(๐ฅ โ ๐ฆ) โ ๐ธ๐+2 โ ๐๐(๐ฅ โ ๐ฆ)]
is contained in ๐ (for instance, ๐ = ๐ต๐ +1(0)ฬ ฬ ฬ ฬ ฬ ฬ ฬ ฬ ฬ ฬ ฬ will work). Because of this, we can change the
โ๐ฆ โ โ๐โ to โ๐ฆ โ ๐โ in the last supremum above without changing the value of the supremum.
Itโs not hard to see then by the product rule that for some ๏ฟฝฬ๏ฟฝ > 0 the last quantity in (Eq 3) is
further bounded by
๏ฟฝฬ๏ฟฝ โ sup๐ฅโ๐พ
sup๐ฆโ๐
|๐๐ฆ๐พ[๐ธ๐+2 โ ๐๐(๐ฅ โ ๐ฆ) โ ๐ธ๐+2 โ ๐๐(๐ฅ โ ๐ฆ)]|
|๐พ|โค๐
โค ๏ฟฝฬ๏ฟฝ โ sup๐งโ๐พโ๐
|๐๐ฆ๐พ[๐ธ๐+2 โ ๐๐(๐ง) โ ๐ธ๐+2 โ ๐๐(๐ง)]|
|๐พ|โค๐
.
Since ๐ธ๐+2 โ ๐ถ๐(โ๐), itโs well known that for ๐พ โ โ(๐) such that |๐พ| โค ๐, ๐๐พ(๐ธ๐+2 โ ๐๐)
converges uniformly to ๐๐พ๐ธ๐+2 over compact sets (this is a slight variant of Theorem 1.2.1 in the
book). Thus we see that the above quantity goes to zero as ๐, ๐ โ โ. So the first quantity in
(Eq 3) also goes to zero as ๐, ๐ โ โ:
sup๐ฅโ๐พ
|๐ธ๐+2 โ ๐๐ โ ๐ข(๐ฅ) โ ๐ธ๐+2 โ ๐๐ โ ๐ข(๐ฅ)| โ 0 as ๐, ๐ โ โ.
So indeed ๐ธ๐+2 โ ๐๐ โ ๐ข are uniformly Cauchy over compact sets and thus converge pointwise to
some continuous function. I claim that that continuous function that they converge to is ๐ธ๐+2 โ
๐ข. To see this, take any test function ๐ โ ๐ถ๐โ(โ๐) and do (here I interchange โlimโ and โโซ ,โ
which I can do because of uniform convergence over compact sets)
โจ lim๐โโ
(๐ธ๐+2 โ ๐๐ โ ๐ข) , ๐โฉ = โซ lim๐โโ
(๐ธ๐+2 โ ๐๐ โ ๐ข(๐ฅ))๐(๐ฅ)๐๐ฅ
= lim๐โโ
โซ๐ธ๐+2 โ ๐๐ โ ๐ข(๐ฅ)๐(๐ฅ)๐๐ฅ = lim๐โโ
โจ๐ธ๐+2 โ ๐๐ โ ๐ข, ๐โฉ = โจ๐ธ๐+2 โ ๐ฟ โ ๐ข, ๐โฉ
= โจ๐ธ๐+2 โ ๐ข, ๐โฉ.
Hence ๐ธ๐+2 โ ๐ข is indeed equal to the function that is the pointwise limit of ๐ธ๐+2 โ ๐๐ โ ๐ข and
thus is a continuous function.
The last thing left to do is to show that ๐ธ๐+2 โ ๐ข is of polynomial growth. This is again just a
game of bounding things. For any ๐ฅ โ โ๐ we have that
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|๐ธ๐+2 โ ๐ข(๐ฅ)| = lim๐โโ
|๐ธ๐+2 โ ๐๐ โ ๐ข(๐ฅ)| = lim๐โโ
|โจ๐ข(๐ฆ), ๐(๐ฆ) โ ๐ธ๐+2 โ ๐๐(๐ฅ โ ๐ฆ)โฉ|
โค ๐ถ โ lim๐โโ
sup๐ฆโโ๐
|๐ฆ๐ผ๐๐ฆ๐ฝ(๐(๐ฆ) โ ๐ธ๐+2 โ ๐๐(๐ฅ โ ๐ฆ))|
|๐ผ|,|๐ฝ|โค๐
As before, itโs not hard to see that the support of the function ๐(๐ฆ) โ ๐ธ๐+2 โ ๐๐(๐ฅ โ ๐ฆ) as a
function of ๐ฆ is contained in โ+๐ฬ ฬ ฬ ฬ โฉ ๐ต|๐ฅ|+1(0)ฬ ฬ ฬ ฬ ฬ ฬ ฬ ฬ ฬ ฬ ฬ ฬ . Thus we can change the โ๐ฆ โ โ๐โ to โ๐ฆ โ โ+
๐ฬ ฬ ฬ ฬ โฉ
๐ต|๐ฅ|+1(0)ฬ ฬ ฬ ฬ ฬ ฬ ฬ ฬ ฬ ฬ ฬ ฬ โ in the above supremum without changing its value. By the product rule, itโs then not
hard to see that for some ๐ถ2 > 0 the above quantity can further be bounded by
๐ถ2 โ lim๐โโ
sup|๐ฆ|โค|๐ฅ|+1each ๐ฆ๐โฅ0
|๐ฆ๐ผ๐๐ฆ๐ฝ(๐ธ๐+2 โ ๐๐(๐ฅ โ ๐ฆ))|
|๐ผ|,|๐ฝ|โค๐
.
Now, we have that |๐ฆ๐ผ| โค (|๐ฅ| + 1)|๐ผ| for any ๐ฆ in the domain of the above supremum. In
addition, as before, for any ๐ฝ โ โ(๐) such that |๐ฝ| โค ๐ we have that ๐๐ฝ(๐ธ๐+2 โ ๐๐) converges
to ๐๐ฝ๐ธ๐+2 uniformly over compact sets and so we can interchange the โlimitโ and โsupremumโ
in the above quantity without changing its value. Thus we see that the above quantity is further
bounded by
๐ถ2(|๐ฅ| + 1)|๐ผ| โ sup|๐ฆ|โค|๐ฅ|+1each ๐ฆ๐โฅ0
| lim๐โโ
๐๐ฆ๐ฝ(๐ธ๐+2 โ ๐๐(๐ฅ โ ๐ฆ))|
|๐ผ|,|๐ฝ|โค๐
= ๐ถ2(|๐ฅ| + 1)|๐ผ| โ sup|๐ฆ|โค|๐ฅ|+1each ๐ฆ๐โฅ0
|๐๐ฆ๐ฝ๐ธ๐+2(๐ฅ โ ๐ฆ)|
|๐ผ|,|๐ฝ|โค๐
= ๐ถ2(|๐ฅ| + 1)|๐ผ| โ sup|๐ฆ|โค|๐ฅ|+1each ๐ฆ๐โฅ0
|๐๐ฆ๐ฝ((๐ฅ โ ๐ฆ)(๐+1)1โโ
[(๐ + 1)!]๐)|
|๐ผ|,|๐ฝ|โค๐
=๐ถ2
[(๐ + 1)!]๐(|๐ฅ| + 1)|๐ผ| โ sup
|๐ฆ|โค|๐ฅ|+1each ๐ฆ๐โฅ0
|(๐ฅ โ ๐ฆ)(๐+1)1โโ โ๐ฝ||๐ผ|,|๐ฝ|โค๐
๐ฝโค(๐+1)1โโ
(note the change under the last ฮฃ symbol). Since each
|(๐ฅ โ ๐ฆ)|(๐+1)1โโ โ๐ฝ โค (2|๐ฅ| + 1)|(๐+1)1โโ โ๐ฝ|
for all ๐ฆ in the domain of the above supremum, the last quantity in the previous equation is
further bounded by
๐ถ2
[(๐ + 1)!]๐(|๐ฅ| + 1)|๐ผ| โ (2|๐ฅ| + 1)|(๐+1)1โโ โ๐ฝ|
|๐ผ|,|๐ฝ|โค๐
๐ฝโค(๐+1)1โโ
.
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Since this is a bound on |๐ธ๐+2 โ ๐ข(๐ฅ)|, it is clear from here that ๐ธ๐+2 โ ๐ข is indeed of
polynomial growth. This finally proves the theorem.
โ
Poissonโs Summation Formula [Theorem 8.5.1] (5/18/2021)
In this entry I work through the proof of the special case of the Poisson summation formula when
the distribution in question is the Dirac delta function (Theorem 8.5.1 in the book) by putting it
into my own words and filling in the details.
Theorem: The following equality holds in ๐ฎโฒ(โ๐):
(Eq 4) โ ๐๐๐ฟ
๐โโค๐
= โ ๐2๐๐๐โ ๐ฅ
๐โโค๐
,
where both sides are interpreted as limits of partial sums in ๐โฒ(โ๐) that exhaust โค๐ in any way.
Remark: The above equation requires a bit of interpretation. The left-hand side of (Eq 4) denotes
the functional ๐ฎ(โ๐) โ โ given by
(Eq 5) ๐ โผ โ ๐(๐)
๐โโค๐
while the right-hand side of (Eq 4) denotes the functional ๐ฎ(โ๐) โ โ given by
(Eq 6) ๐ โผ โ ๏ฟฝฬ๏ฟฝ(2๐๐)
๐โโค๐
.
The claim of the theorem then is that these are well defined tempered distributions, the partial
sums of โ ๐๐๐ฟ๐โโค๐ and โ ๐2๐๐๐โ ๐ฅ๐โโค๐ converge to these two distributions in ๐ฎโฒ(โ๐) respectively,
and that these two distributions are equal.
Proof: For brevity, let โ๐ขโ denote the left-hand side of (Eq 4) and let โ๐ฃโ denote the right-hand
side of (Eq 4). First letโs show that that both ๐ข and ๐ฃ are well defined tempered distributions and
that the partials sums on both sides of (Eq 4) converge to ๐ข and ๐ฃ respectively in ๐ฎโฒ(โ๐). One
way to do this of course would be to say that this follows from the uniform boundedness
principle and the fact that the limit of their respective partial sumsโ actions on members of
๐ฎ(โ๐) exist. But I will take a more elementary approach here by directly showing that both ๐ข
and ๐ฃ are tempered distributions and that their respective partial sums converge to them in
๐ฎโฒ(โ๐).
Letโs start with ๐ข. Observe that since any ๐ โ ๐ฎ(โ๐) decays faster than any polynomial, we
have the sum in (Eq 5) is absolutely convergent and thus the action of ๐ข on any ๐ โ ๐ฎโฒ(โ๐) is
well defined (and obviously linear). Next letโs show that itโs a tempered distribution. For any
๐ โ ๐ฎ(โ๐) we have that (here |๐| denotes the usual Euclidian length of ๐)
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|โจ๐ข, ๐ฃโฉ| = |๐(0) + โ ๐(๐)
๐โโค๐โ{0}
| โค sup|๐| + ( โ1
|๐|2๐
๐โโค๐โ{0}
)sup||๐ฅ|2๐๐|.
Since โ 1 |๐|2๐โ๐โโค๐โ{0} is finite, this shows that ๐ข โถ ๐ฎ(โ๐) โ โ is indeed continuous and hence
a tempered distribution. Finally, letโs show that the partial sums on the left-hand side of (Eq 4)
converge to ๐ข in ๐ฎโฒ(โ๐). Let {๐๐}๐=1โ โ โค๐ be any sequence that exhausts โค๐ (i.e. the map ๐ โฆ
๐๐ is a bijective โค+ โ โค๐ map). Then for any ๐ โ ๐ฎ(โ๐) we have that
lim๐โโ
โจโ ๐ข๐๐
๐
๐=1
, ๐โฉ = lim๐โโ
โ ๐(๐๐)
๐
๐=1
= โ ๐(๐)
๐โโค๐
= โจ๐ข, ๐โฉ.
Thus indeed โ ๐ข๐๐
๐๐=1 โ ๐ข in ๐ฎโฒ(โ๐).
Now letโs show the same thing for ๐ฃ. The proof for showing that itโs well defined and that the
partial sums on the right-hand side of (Eq 4) converge to ๐ฃ in ๐ฎโฒ(โ๐) is done exactly the same
way as for ๐ข by remembering that the Fourier transform maps ๐ฎ(โ๐) to itself. So Iโll omit
repeating those details and instead focus on showing that ๐ฃ โถ ๐ฎ(โ๐) โ โ is continuous. Since the
Fourier transform โฑ โถ ๐ฎ(โ๐) โ ๐ฎ(โ๐) is continuous, we know that there exist ๐ถ1, ๐ถ2 > 0 and
๐1, ๐2 โ โค+ such that
sup|๏ฟฝฬ๏ฟฝ| โค ๐ถ1 โ sup|๐ฅ๐ผ๐๐ฝ๐||๐ผ|,|๐ฝ|โค๐1
,
sup||๐|2๐๏ฟฝฬ๏ฟฝ| โค ๐ถ2 โ sup|๐ฅ๐ผ๐๐ฝ๐||๐ผ|,|๐ฝ|โค๐2
,
for all ๐ โ ๐ฎ(โ๐). Thus, proceeding as with ๐ข, we have that for any ๐ โ ๐ฎ(โ๐),
|โจ๐ฃ, ๐โฉ| โค sup|๏ฟฝฬ๏ฟฝ| + ( โ1
|๐|2๐
๐โโค๐โ{0}
)sup||๐|2๐๏ฟฝฬ๏ฟฝ|
โค ๐ถ1 โ sup|๐ฅ๐ผ๐๐ฝ๐||๐ผ|,|๐ฝ|โค๐1
+ ( โ1
|๐|2๐
๐โโค๐โ{0}
)๐ถ2 โ sup|๐ฅ๐ผ๐๐ฝ๐||๐ผ|,|๐ฝ|โค๐2
.
Hence indeed ๐ฃ โถ ๐ฎ(โ๐) โ โ is continuous and thus is a tempered distribution.
Having dealt with these technical setup details, letโs start proving the theorem. I claim that ๐ฃ is
periodic over โค๐:
๐ฃ = ๐โ๐ฃ โโ โ โค๐.
To see this, take any โ โ โค๐ and observe that
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๐โ๐ฃ = โ ๐โ๐2๐๐๐โ ๐ฅ
๐โโค๐
= โ ๐2๐๐๐โ โ๐2๐๐๐โ ๐ฅ
๐โโค๐
= โ ๐2๐๐๐โ ๐ฅ
๐โโค๐
= ๐ฃ.
Now, let ๐ โ ๐ถ๐โ(โ๐) be such that ๐ โฅ 0, supp๐ โ (โ1, 1)๐ and โ ๐๐๐๐โโค๐ = 1 (the
existence of such a ๐ is proven in Lemma 8.5.1 in the book). I claim that
(Eq 7) ๐ฃ = โ (๐๐๐)๐ฃ
๐โโค๐
= โ ๐๐(๐๐ฃ)
๐โโค๐
in the sense that the partial sums here converge to ๐ฃ in ๐โฒ(โ๐) (with respect to any exhaustion
of โค๐). These partial sums turn out to converge to ๐ฃ in ๐ฎโฒ(โ๐) as well, but we donโt need that at
the moment. Observe that the second equality here simply holds because of the fact that ๐ฃ is
periodic over โค๐. So let me prove the first equality. Take any ๐ โ ๐ถ๐โ(โ๐). As before, let
{๐๐}๐=1โ โ โค๐ be any sequence that exhausts โค๐. Then we have that
lim๐โโ
โโจ(๐๐๐)๐ฃ, ๐โฉ
๐
๐=1
= lim๐โโ
โจ๐ฃ, โ(๐๐๐)๐
๐
๐=1
โฉ.
Since ๐ is of compact support, each supp ๐๐๐ โ (โ1, 1)๐ + ๐, and โ ๐๐๐๐โโค๐ = 1, we have
that โ (๐๐๐)๐๐๐=1 is eventually equal to ๐ as ๐ โ โ. So the above limit is equal to โจ๐ฃ, ๐โฉ and
thus indeed we have that the partial sums of โ (๐๐๐)๐ฃ๐โโค๐ converge to ๐ฃ in ๐โฒ(โ๐).
Now, letโs investigate what ๐๐ฃ is equal to. Focusing on ๐ฃ first, I claim that for any ๐ โ {1,โฆ , ๐},
(๐2๐๐๐ฅ๐ โ 1)๐ฃ = 0.
To see this, observe that for any ๐ โ {1,โฆ , ๐} we have that (here ๐๐ is the standard unit vector
with all zeros except a one in the ๐th position)
๐2๐๐๐ฅ๐๐ฃ = โ ๐2๐๐๐ฅ๐๐2๐๐๐โ ๐ฅ
๐โโค๐
= โ ๐2๐๐(๐+๐๐)โ ๐ฅ
๐โโค๐
= โ ๐2๐๐โโ ๐ฅ
โโโค๐
= ๐ฃ.
We obviously then have that for any ๐ โ {1,โฆ , ๐},
(Eq 8) (๐2๐๐๐ฅ๐ โ 1)๐๐ฃ = 0.
We can in fact write a better version of this equation by linearizing the leading coefficient:
Claim: Our ๐๐ฃ above satisfies the following: for any ๐ โ {1,โฆ , ๐},
(Eq 9) ๐ฅ๐๐๐ฃ = 0.
Proof: Our approach will be to prove the above equality locally at any point. Fix any ๐ โ
{1,โฆ , ๐}. Take any point ๐ฆ โ โ๐. First suppose that the function (๐2๐๐๐ฅ๐ โ 1) does not vanish at
the point ๐ฆ (i.e. ๐ฆ๐ โ โค). Choose a small enough neighborhood ๐ of ๐ฆ such that (๐2๐๐๐ฅ๐ โ 1)
does not vanish on ๐. Then equality in (Eq 9) over ๐ follows from (Eq 8) by writing (Eq 8) as
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๐2๐๐๐ฅ๐ โ 1
๐ฅ๐๐ฅ๐๐๐ฃ = 0
and then dividing through by the nonzero (๐2๐๐๐ฅ๐ โ 1) ๐ฅ๐โ . Now suppose that itโs the case that
(๐2๐๐๐ฅ๐ โ 1) does vanish at ๐ฆ. If ๐ฆ โ 0, then we can choose a neighborhood ๐ of ๐ฆ such that
0 โ ๐ and ๐ is disjoint from ๐. Then (Eq 9) clearly holds on ๐ because of the mere fact that
๐ โก 0 on ๐. Lastly, suppose that ๐ฆ = 0. Let ๐ be any neighborhood of ๐ฆ such that ๐ โ(โ1, 1)๐. Then as before we can rewrite (Eq 8) over ๐ as
๐2๐๐๐ฅ๐ โ 1
๐ฅ๐๐ฅ๐๐๐ฃ = 0
where (๐2๐๐๐ฅ๐ โ 1) ๐ฅ๐โ is interpreted to be equal to its limit value at ๐ฅ = 0, which observe makes
it smooth. Since (๐2๐๐๐ฅ๐ โ 1) ๐ฅ๐โ is nonzero over ๐, we again get that equality in (Eq 9) over ๐
follows by dividing trough by (๐2๐๐๐ฅ๐ โ 1) ๐ฅ๐โ in the above equation. Having proved the equality
in (Eq 9) in a neighborhood of any point ๐ฆ in โ๐, the claim follows.
End of Proof of Claim
Now, take any ๐ โ ๐ถโ(โ๐). By Taylorโs theorem we know that there exist ๐๐ โ ๐ถโ(โ๐) for ๐ โ
{1,โฆ , ๐} such that
๐(๐ฅ) = ๐(0) + โ๐ฅ๐๐๐(๐ฅ)
๐
๐=1
.
By (Eq 9) we then get that
โจ๐๐ฃ, ๐โฉ = โจ๐๐ฃ, ๐(0)โฉ = โจโจ๐๐ฃ, 1โฉ๐ฟ, ๐โฉ.
Thus weโve obtained that
๐๐ฃ = ๐ถ๐ฟ
for some constant ๐ถ โ โ. By (Eq 7) we then get that
(Eq 10) ๐ฃ = ๐ถ โ ๐๐๐ฟ
๐โโค๐
in ๐โฒ(โ๐). Since we already proved that both ๐ฃ and โ ๐๐๐ฟ๐โโค๐ are tempered distributions, this
equality in fact holds in ๐ฎโฒ(โ๐). Thus, the only task left to do in order to prove the theorem is to
show that ๐ถ = 1. To do this, let ๐ โถ โ๐ โ โ denote the indicator function over (0, 1)๐:
๐(๐ฅ) = {1 if ๐ฅ โ (0, 1)๐
0 if ๐ฅ โ (0, 1)๐.
Now letโs take the convolution of both sides of (Eq 10) with ๐. The left-hand side becomes
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๐ฃ โ ๐ = โ ๐2๐๐๐โ ๐ฅ โ ๐
๐โโค๐
= โ {1 if ๐ = 00 if ๐ โ 0
๐โโค๐
= 1.
Taking the convolution of the right-hand side of (Eq 10) with ๐ gives
(๐ถ โ ๐๐๐ฟ
๐โโค๐
) โ ๐ = ๐ถ โ (๐๐(๐ฟ) โ ๐)
๐โโค๐
= ๐ถ โ (๐ฟ โ ๐๐(๐))
๐โโค๐
= ๐ถ โ ๐๐๐
๐โโค๐
,
which is equal to the constant function ๐ถ almost everywhere (with respect to the Lebesgue
measure). Hence we must have that ๐ถ = 1. As discussed above, this proves the theorem.
โ