Fourier Transform and Convolution

Some nice reading: https://people.math.gatech.edu/~heil/7338/fall09/approxid.pdf

The Fourier Transform

proposition (?):

If $\widehat{f} = \widehat{g}$ then $f=g$ almost everywhere.

proposition (Riemann-Lebesgue: Fourier transforms have small tails.):

$\begin{align*} f\in L^1 \implies \widehat{f}(\xi) \rightarrow 0 \text { as }|\xi| \rightarrow \infty ,\end{align*}$

if $f \in L^1$ , then $\widehat{f}$ is continuous and bounded.

remark:

Note that this implies there can be no identity for convolution: if there existed a function $\delta$ with $\delta(x) = 0$ for $x\neq 0$ and $\int\delta = 1$ , then $\begin{align*} \widehat{\delta}(\xi)=\int \delta(x) e^{-2 \pi i \xi x} d x=\int \delta(x) d x=1 ,\end{align*}$ contradicting Riemann-Lebesgue.

proof (?):

Boundedness: $\begin{align*} {\left\lvert {\widehat{f}(\xi)} \right\rvert} \leq \int {\left\lvert {f} \right\rvert}\cdot {\left\lvert {e^{2\pi i x\cdot \xi }} \right\rvert} = {\left\lVert {f} \right\rVert}_{1} .\end{align*}$
Continuity:
${\left\lvert {\widehat{f}(\xi_{n}) - \widehat{f} (\xi) } \right\rvert}$
Apply DCT to show $a\overset{n\to\infty}\to 0$ .

theorem (Fourier Inversion):

$\begin{align*} f(x)=\int_{\mathbb{R}^{n}} \widehat{f}(x) e^{2 \pi i x \cdot \xi} d \xi .\end{align*}$

warnings:

Fubini-Tonelli does not work here!

proof (?):

Idea: Fubini-Tonelli doesn’t work directly, so introduce a convergence factor, take limits, and use uniqueness of limits.

Take the modified integral:

$\begin{align*} I_{t}(x) &= \int \widehat{f}(\xi) ~e^{2\pi i x \cdot \xi} ~e^{-\pi t^2 {\left\lvert {\xi} \right\rvert}^2} \\ &= \int \widehat{f}(\xi) \phi(\xi) \\ &= \int f(\xi) \widehat{\phi}(\xi) \\ &= \int f(\xi) \widehat{\widehat{g}}(\xi - x) \\ &= \int f(\xi) g_{t}(x - \xi) ~d\xi \\ &= \int f(y-x) g_{t}(y) ~dy \quad (\xi = y-x)\\ &= (f \ast g_{t}) \\ &\to f \text{ in $L^1$ as }t \to 0 .\end{align*}$

We also have $\begin{align*} \lim_{t\to 0} I_{t}(x) &= \lim_{t\to 0} \int \widehat{f}(\xi) ~e^{2\pi i x \cdot \xi} ~e^{-\pi t^2 {\left\lvert {\xi} \right\rvert}^2} \\ &= \lim_{t\to 0} \int \widehat{f}(\xi) \phi(\xi) \\ &=_{DCT} \int \widehat{f}(\xi) \lim_{t\to 0} \phi(\xi) \\ &= \int \widehat{f}(\xi) ~e^{2\pi i x \cdot \xi} \\ .\end{align*}$
So $\begin{align*} I_{t}(x) \to \int \widehat{f}(\xi) ~e^{2\pi i x \cdot \xi} ~\text{ pointwise and }~{\left\lVert {I_{t}(x) - f(x)} \right\rVert}_{1} \to 0 .\end{align*}$
So there is a subsequence $I_{t_{n}}$ such that $I_{t_{n}}(x) \to f(x)$ almost everywhere
Thus $f(x) = \int \widehat{f}(\xi) ~e^{2\pi i x \cdot \xi}$ almost everywhere by uniqueness of limits.

proposition (Eigenfunction of the Fourier transform):

$\begin{align*} g(x) \coloneqq e^{-\pi {\left\lvert {t} \right\rvert}^2} \implies \widehat{g}(\xi) = g(\xi) {\quad \operatorname{and} \quad} \widehat{g}_{t}(x) = g(tx) = e^{-\pi t^2 {\left\lvert {x} \right\rvert}^2} .\end{align*}$

proposition (Useful properties of the Fourier transform):

$\begin{align*} \widehat{f\ast g}(\xi) &= \widehat{f}(\xi) \cdot \widehat{g} (\xi) \\ \widehat{\tau_h f}(\xi) &= e^{2\pi i \xi \cdot h}\widehat{f}(\xi) \\ \widehat{e^{2\pi i \xi \cdot h}f(\xi)} &= \tau_{-h}\widehat f(\xi) \\ \widehat{f \circ T}(\xi) &= {\left\lvert {\operatorname{det}T} \right\rvert}^{-1}(\widehat{f} \circ T^{-t})(\xi) \\ {\frac{\partial }{\partial \xi}\,} \widehat{f}(\xi) &= -2\pi i \cdot \widehat {\xi f} (\xi) \\ \widehat{{\frac{\partial }{\partial \xi}\,} f}(\xi) &= 2\pi i \xi \cdot \widehat{f}(\xi) .\end{align*}$

example (Some transform pairs):

$\begin{align*} \text{Dirichlet:} && \chi_{\left\{{-\frac{1}{2} \leq x \leq \frac{1}{2}}\right\}} &\iff \mathrm{sinc}(\xi) \\ \text{Fejer:} && \chi_{\left\{{-1 \leq x \leq 1}\right\}} (1 - {\left\lvert {x} \right\rvert}) &\iff \mathrm{sinc}^2(\xi) \\ \text{Poisson:} && \frac{1}{\pi} \frac{1}{1+x^2} &\iff e^{2\pi {\left\lvert {\xi} \right\rvert}} \\ \text{Gauss-Weierstrass:} && e^{-\pi x^2} &\iff e^{-\pi \xi^2} .\end{align*}$

Approximate Identities

example (of an approximation to the identity.):

$\begin{align*} \phi(x) \coloneqq e^{-\pi x^2} .\end{align*}$

theorem (Convolving against an approximate identity converges in

$L^1$ .):

$\begin{align*} {\left\lVert {f \ast \phi_{t} - f} \right\rVert}_{1} \overset{t\to 0}\to 0 .\end{align*}$

proof (?):

$\begin{align*} {\left\lVert {f - f\ast \phi_{t}} \right\rVert}_1 &= \int f(x) - \int f(x-y)\phi_{t}(y) ~dy dx \\ &= \int f(x)\int \phi_{t}(y) ~dy - \int f(x-y)\phi_{t}(y) ~dy dx \\ &= \int \int \phi_{t}(y)[f(x) - f(x-y)] ~dy dx \\ &=_{FT} \int \int \phi_{t}(y)[f(x) - f(x-y)] ~dx dy \\ &= \int \phi_{t}(y) \int f(x) - f(x-y) ~dx dy \\ &= \int \phi_{t}(y) {\left\lVert {f - \tau_{y} f} \right\rVert}_1 dy \\ &= \int_{y < \delta} \phi_{t}(y) {\left\lVert {f - \tau_{y} f} \right\rVert}_1 dy + \int_{y \geq \delta} \phi_{t}(y) {\left\lVert {f - \tau_{y} f} \right\rVert}_1 dy \\ &\leq \int_{y < \delta} \phi_{t}(y) \varepsilon + \int_{y \geq \delta} \phi_{t}(y) \left( {\left\lVert {f} \right\rVert}_1 + {\left\lVert {\tau_{y} f} \right\rVert}_1 \right) dy \quad\text{by continuity in } L^1 \\ &\leq \varepsilon + 2{\left\lVert {f} \right\rVert}_1 \int_{y \geq \delta} \phi_{t}(y) dy \\ &\leq \varepsilon + 2{\left\lVert {f} \right\rVert}_1 \cdot \varepsilon \quad\text{since $\phi_{t}$ has small tails} \\ &\overset{{\varepsilon}\to 0}\to 0 .\end{align*}$

theorem (Convolutions vanish at infinity):

$\begin{align*} f,g \in L^1 \text{ and bounded} \implies \lim_{|x| \rightarrow \infty} (f * g)(x) = 0 .\end{align*}$

proof (?):

Choose $M \geq f,g$ .
By small tails, choose $N$ such that $\int_{B_{N}^c} {\left\lvert {f} \right\rvert}, \int_{B_{n}^c} {\left\lvert {g} \right\rvert} < \varepsilon$
Note $\begin{align*} {\left\lvert {f \ast g} \right\rvert} \leq \displaystyle\int {\left\lvert {f(x-y)} \right\rvert} ~{\left\lvert {g(y)} \right\rvert} ~dy \coloneqq I .\end{align*}$
Use ${\left\lvert {x} \right\rvert} \leq {\left\lvert {x-y} \right\rvert} + {\left\lvert {y} \right\rvert}$ , take ${\left\lvert {x} \right\rvert}\geq 2N$ so either $\begin{align*} {\left\lvert {x-y} \right\rvert} \geq N \implies I \leq \int_{\left\{{x-y \geq N}\right\}} {\left\lvert {f(x-y)} \right\rvert}M ~dy\leq \varepsilon M \to 0 \end{align*}$ then $\begin{align*} {\left\lvert {y} \right\rvert} \geq N \implies I \leq \int_{\left\{{y \geq N}\right\}} M{\left\lvert {g(y)} \right\rvert} ~dy\leq M \varepsilon \to 0 .\end{align*}$

proposition (Corollary of Young's inequality):

Take $q = 1$ in Young’s inequality to obtain $\begin{align*} {\left\lVert {f \ast g} \right\rVert}_{p} \leq {\left\lVert {f} \right\rVert}p {\left\lVert {g} \right\rVert}1 .\end{align*}$

proposition (

$L^1$ is closed under convolution.):

If $f, g \in L^1$ then $f\ast g\in L^1$ .