Context: I am currently working through Chapter $8$ of Anders Vretblad's Fourier Analysis and Its Applications. This particular chapter focuses on distributions, and builds up to the Fourier transform of such. I am currently looking at Chapter $8.7$, focused on periodic distributions.
I am particularly concerned with showing the Dirac comb is a tempered distribution, when interpreted properly.
Relevant Definitions:
-
A sufficiently-nice function $
\newcommand{\R}{\mathbb{R}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\vp}{\varphi}
\newcommand{\SS}{\mathcal{S}}
\newcommand{\CC}{\mathcal{C}}
\newcommand{\FF}{\mathcal{F}}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\f}{\widehat{f}}
\newcommand{\w}{\widehat}
\newcommand{\comb}{\operatorname{comb}}
\newcommand{\d}{\delta}
\newcommand{\abs}[1]{\left| #1 \right|}
f : \R \to \C$ is identified identically with its distribution $T_f : \CC_c^\infty(\R) \to \C$ given by $T_f[\vp] := \int_\R f \vp$, i.e. we may refer to $f$ both as the function $f$ and the distribution $T_f$. -
Translation: As a shorthand, the text likes to – given a function $f$ and $a \in \R$ – define $f_a$ to be the function $f_a(x) := f(x-a)$.
-
Dirac Comb: The Dirac comb (or impulse train or Sha function) here will be defined by
$$
\operatorname{comb}(x) := \sum_{n \in \Z} \delta_{2 \pi n}(x) \equiv \sum_{n \in \Z} \delta(x-2 \pi n)
$$
where $\delta$ denotes the Dirac $\delta$. -
Schwartz Space: The Schwartz space $\mathcal{S} := \mathcal{S}(\R)$ is defined as the set of all $\vp \in \CC^\infty(\R)$ such that, for any $n,k \in \Z_{\ge 0}$, we can find constants $C_{n,k}$ such that
$$
\left| \vp^{(k)}(x) \right| \le \frac{C_{n,k}}{ \big( 1 + |x| \big)^n } \text{ for all } x \in \R
$$
(i.e. $\SS$ consists of rapidly-decaying smooth functions). -
Tempered Distributions: The topological dual of the Schwartz class, $\SS'$, is known as the tempered distributions. Hence $f \in \SS'$ if and only if it is linear as a function $f : \SS \to \C$, and continuous in the sense of sequential convergence. That is, $f : \SS \to \C$ is continuous if and only if $\,\forall\{\vp_n\}_{n \in \N} \subseteq \SS$ and $\vp \in \SS$ such that
$$
\vp_j \xrightarrow[\SS]{j\to\infty} \vp; \;\;\; \text{ i.e., } \;\;\; \forall n,k \in \Z_{\ge 0}, \;
\lim_{j \to \infty} \sup_{x \in \R} \big( 1 + |x| \big)^n \left| \vp_j^{(k)} – \vp^{(k)} \right| = 0
$$
then we have $f[\vp_n] \xrightarrow[\C]{n \to \infty} f[\vp]$. -
Periodic Distributions: A tempered distribution $f$ is said to be $p$-periodic if $f[\vp_p] = f[\vp]$ for every $\vp \in \mathcal{S}$.
Background & Questions:
On the section on periodic functions, the text wishes to show $\comb(x)$ is a $2\pi$-periodic tempered distribution. The calculation for showing periodicity is straightforward from the definition of the Dirac $\delta$. Let $\vp \in \SS$. Then:
$$\begin{align*}
\comb[\vp]
&= \int_\R \comb(x) \vp(x) \, \dd x \\
&= \sum_{n \in \Z} \int_\R \d(x-2\pi n) \vp(x) \, \dd x \\
&= \sum_{n \in \Z}\vp( 2\pi n) \\
\comb[\vp_{2\pi}]
&= \int_\R \comb(x) \vp(x- 2\pi) \, \dd x \\
&= \sum_{n \in \Z} \int_\R \d(x-2 \pi n) \vp(x- 2\pi) \, \dd x \\
&= \sum_{n \in \Z} \vp(2 \pi n – 2\pi) \\
&= \sum_{n \in \Z} \vp (2 \pi (n-1)) \\
&= \comb[\vp]
\end{align*}$$
Question $1$: The textbook claims instead that
$$
\comb[\vp] = \sum_{n \in \Z} \vp(x+2n\pi)
$$
Surely there is no $x$ involved in reality and this is just a typo? To my understanding, being interpreted as a tempered distribution, the result should be some constant in $\C$ dependent on the input Schwartz function $\vp$, and not itself a function of, say, $x$.
This is all well and good, though it remains to show that $\comb[\cdot]$ defines a tempered distribution in the first place. We should show that $\comb[\vp]$'s summations here converge, then.
From the fact that $\vp \in \SS$, then, we can find constants $C_n$ such that
$$
\abs{ \vp(x) } \le \frac{C_n}{(1+|x|)^n}
$$
and so
$$\begin{align*}
\abs{ \comb[\vp] }
&\stackrel{(1)}{\le} \sum_{n \in \Z} \frac{C_n}{\big(1 + 2\pi |n|\big)^n}\\
&\stackrel{(2)}{\le} C_0 + 2\sum_{n=1}^\infty \frac{M_n}{\big(1 + 2\pi n\big)^n}\\
&\stackrel{(3)}{\le} C_0 + 2\sum_{n=1}^\infty \frac{M_n}{1+n^2}\\
&\stackrel{?}{<} \infty
\end{align*}$$
- $(1)$ follows from just the triangle inequality, the fact that the $C_n$ are necessarily nonnegative, and that $|2\pi n| = 2\pi |n|$
- $(2)$ follows by splitting the sum into three pieces: for $n<0,n=0,n>0$. We define $M_n := \max \{C_n,C_{-n}\}$ to allow this representation, which results in an inequality.
- $(3)$ comes at the suggestion of the text to bound by $M/(1+n^2)$. (This $M$ is not specified, and the bound is said to apply "with wide margins" for $\vp \in \SS$, which is very unclear phrasing to me. The prior inequalities were not stated in the text, I'm just trying to justify it all.) This inequality $(3)$ can be said to hold by removing the factor of $2\pi$ from the denominator (hence shrinking it, and making the fraction larger), and then applying Bernoulli's inequality to $(1+n)^n$:
$$
(1+x)^r \ge 1+rx \text{ for } x \ge 1 \text{ and } r \in \R \setminus (0,1)
$$
(Just let $r=x=n \in \N$.)
Question $2$: How is the final summation here justified to be finite?
$$
\sum_{n=1}^\infty \frac{M_n}{1+n^2} \stackrel{?}{<} \infty
$$
One obvious angle would be
$$
\sum_{n=1}^\infty \frac{M_n}{1+n^2} \le \sup_{n \in \N} M_n \cdot \sum_{n \in \N} \frac{1}{1+n^2}
$$
The summation is known to converge (e.g., by comparison and $p$-series tests), and in fact does so to $\frac 1 2(1 + \pi \coth(\pi))$. But how would one show that $\sup M_n < \infty$ (equivalently, $\sup C_n < \infty$)? If this is not the angle to approach this from, how would I show the convergences of the sum? Obviously one needs to appeal to the fact that the $\vp$ lie in $\SS$ (or, less strongly, in $\CC_c^\infty(\R)$) in some way, but the necessary approach is unclear to me.
Best Answer
Answer 1. You are right, there should be no $x.$
Answer 2. You don't need all $C_n$s. Just use $C_2$: $$\sum_{n\in\Bbb Z}|\varphi(2n\pi)|\le C_2\sum_{n\in\Bbb Z} \frac1{\big(1 + 2\pi |n|\big)^2}<\infty.$$