Seanie Lee

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Fourier Transform

8 minute read

Definition

We define (if the limit exists)

\[\begin{align*} \int_{-\infty}^\infty f(x)dx = \lim_{N\to\infty} \int_{-N}^N f(x)dx \end{align*}\]

Defintion

Let $f:\mathbb{R}\to\mathbb{C}$ be continuous. If there exists $A>0$ such that

\[\begin{align*} \lvert f(x)\rvert \leq \frac{A}{1+x^2} \end{align*}\]

for all $x\in\mathbb{R}$, then we say $f$ is of moderate decrease and denote by $\mathcal{M}(\mathbb{R})$ the vector space of such functions.

Exercise

Whenever $f\in\mathcal{M}(\mathbb{R})$, its limit

\[\begin{align*} \int_{-\infty}^\infty f(x)dx = \lim_{N\to\infty}\int_{-N}^Nf(x)dx \end{align*}\]

always exists.

<Proof>

For each $N\in\mathbb{N}$, define the integral $I_N=\int_{-N}^Nf(x)dx$. It is well-defined since $f$ is continuous. It suffices to show that $\{I_N\}$ is a Cauchy sequence. Let $M,N\in\mathbb{N}$ with $M>N$ be given. Then

\[\begin{align*} \lvert I_M - I_N\rvert &\leq \left\lvert \int_{N\leq \lvert x \rvert \leq M} f(x) dx\right\rvert \\ &\leq A \int_{N\leq \lvert x \rvert \leq M} \frac{dx}{1+x^2} \\ &\leq A \int_{N\leq \lvert x \rvert \leq M} \frac{dx}{x^2} \\ &=2A\left(\frac{1}{N}-\frac{1}{M}\right) \\ &\leq \frac{2A}{N} \end{align*}\]

which goes to $0$ as $N\to\infty$. \(\tag*{$\square$}\)

Proposition (Properties of the improper integral)

Let $f,g\in\mathcal{M}(\mathbb{R})$, and $\alpha,\beta\in\mathbb{C}$. Then

(1) linearity

\[\begin{align*} \int_{-\infty}^\infty (\alpha f+\beta g)= \alpha \int_{-\infty}^\infty f +\beta\int_{-\infty}^\infty g. \end{align*}\]

(2) translation invariance

\[\begin{align*} \int_{-\infty}^\infty f(x-\alpha)dx = \int_{-\infty}^\infty f(x)dx. \end{align*}\]

(3) scaling under dilations

Given any $\delta>0$,

\[\begin{align*} \delta\int_{-\infty}^\infty f(\delta x) dx = \int_{-\infty}^\infty f(x)dx. \end{align*}\]

(4) continuity

\[\begin{align*} \lim_{h\to 0}\int_{-\infty}^\infty \lvert f(x-h)-f(x)\rvert dx=0. \end{align*}\]

<Proof>

(1)

Since $f,g\in\mathcal{M}(\mathbb{R})$ and \(\begin{align*} \int_{-N}^N \alpha f +\beta g = \alpha \int_{-N}^Nf + \beta \int_{-N}^Ng , \end{align*}\)

\[\begin{align*} \lim_{N\to\infty}\left(\alpha\int_{-N}^Nf + \beta\int_{-N}^Ng \right) &=\lim_{N\to\infty}\alpha \int_{-N}^Nf+\lim_{N\to\infty}\beta\int_{-N}^Ng \\ &=\alpha\lim_{N\to\infty}\int_{-N}^Nf + \beta\lim_{N\to\infty}\int_{-N}^Ng \\ &=\alpha \int_{-\infty}^\infty f+\beta\int_{-\infty}^\infty g. \end{align*}\]

(2) It suffices to show that

\[\lim_{N\to\infty}\left(\int_{-N}^N f(x-\alpha) dx -\int_{-N}^Nf(x) dx \right) =0.\]

With change of variables,

\[\begin{align*} \int_{-N}^N f(x-\alpha)-f(x)dx &= \int_{-N-\alpha}^{N-\alpha} f(x) dx - \int_{-N}^N f(x)dx \\ &= \int_{-N-\alpha}^{-N} f(x) dx + \int_{N-\alpha}^N f(x)dx. \end{align*}\]

Take $N>2\alpha$. Then

\[\begin{align*} \left\lvert\int_{-N-\alpha}^{-N} f(x) dx + \int_{N-\alpha}^N f(x)dx\right\rvert &\leq \int_{-2N}^{-N} \lvert f(x)\rvert dx + \int_{\frac{N}{2}}^N \lvert f(x)\rvert dx \\ &\leq \int_{\frac{N}{2}\leq \lvert x\rvert \leq 2N} \lvert f(x)\rvert dx \\ &\leq \int_{\frac{N}{2}\leq \lvert x\rvert \leq 2N} \frac{A}{x^2}dx \\ &=2A\left(-\frac{1}{2N} + \frac{2}{N}\right) \\ &=\frac{3A}{N} \to 0 \text{ as } N\to\infty. \end{align*}\]

(3) With change of variables,

\[\begin{align*} \delta \int_{-N}^N f(\delta x) dx = \int_{-\delta N}^{\delta N} f(x)dx. \end{align*}\]

Thus,

\[\begin{align*} \lim_{N\to\infty} \delta\int_{-N}^N f(\delta x) dx = \int_{-\infty}^\infty f(x)dx \end{align*}\]

(4) Given $\epsilon>0$, we want to show that there exists $H>0$ such that if $\lvert h \rvert < H$, then

\[\begin{align*} \lim_{N\to\infty} \int_{-N}^N \lvert f(x-h)-f(x)\rvert dx <\epsilon. \end{align*}\]

Without loss of generality, we take $\lvert h \rvert\leq1$.

Note that

\[\begin{align*} \int_{\lvert x \rvert \geq N_0} \lvert f\rvert &\leq 2\int_{x \geq N_0} \frac{A}{1+x^2}dx \\ &\leq2A\int_{x\geq N_0}\frac{dx}{x^2} \\ &=2A \lim_{M\to\infty} \int_{N_0}^M \frac{dx}{x^2} \\ &=2A \lim_{M\to\infty}\left(\frac{1}{N_0}-\frac{1}{M}\right) \\ &=\frac{2A}{N_0}. \end{align*}\]

Similarly,

\[\begin{align*} \int_{N_0}^M \lvert f(x-h)\rvert dx &\leq \int_{ N_0}^M \frac{A}{1+(x-h)^2}dx \\ &=A\int_{N_0-h}^{M-h}\frac{dx}{1+x^2} \\ &\leq A \int_{N_0-h}^{M-h} \frac{dx}{x^2} \\ &=A\left(\frac{1}{N_0-h} -\frac{1}{M-h}\right) \end{align*}\]

We take $N_0$ large enough that

\[\begin{align*} \int_{\lvert x \rvert \geq N_0} \lvert f\rvert \leq \frac{\epsilon}{4} \text{ and } \int_{\lvert x \rvert \geq N_0} \lvert f(x-h)\rvert dx \leq \frac{\epsilon}{4}. \end{align*}\]

Since $f$ is continuous, it is uniformly continuous on $[-N_0-1, N_0+1]$. Thus, we can choose $H$ such that if $h<H$,

\[\begin{align*} \lvert f(x-h) - f(x) \rvert < \frac{\epsilon}{4N_0} \text{ for all } \lvert x\rvert \leq N_0. \end{align*}\]

This implies that

\[\begin{align*} \sup_{\lvert x \rvert \leq N_0}\lvert f(x-h) - f(x) \rvert < \frac{\epsilon}{4N_0}. \end{align*}\]

Then for any $N>N_0$, we get

\[\begin{align*} \int_{-N}^N \lvert f(x-h)-f(x)\rvert dx & = \int_{-N_0}^{N_0} \lvert f(x-h)-f(x)\rvert dx + \int_{N_0 \leq \lvert x \rvert \leq N} \lvert f(x-h)-f(x)\rvert dx \\ &\leq \int_{-N_0}^{N_0} \lvert f(x-h)-f(x)\rvert dx + \int_{\lvert x \rvert \geq N_0} \lvert f \rvert +\int_{\lvert x \rvert \geq N_0} \lvert f(x-h)\rvert dx \\ &\leq \frac{\epsilon}{2} +\frac{\epsilon}{4} + \frac{\epsilon}{4} = \epsilon. \end{align*}\] \[\tag*{$\square$}\]

Definition

Given $f\in\mathcal{M}(\mathbb{R})$, we define the Fourier transform $\hat{f}$ of $f$ by

\[\begin{align*} \hat{f}(\xi) = \int_{-\infty}^\infty f(x) e^{-2\pi ix\xi}dx \end{align*}\]

Definition

We call a function *rapidly decreasing if for every $k\geq0$, we have

\[\begin{align*} \sup_{x\in\mathbb{R}} \lvert x \rvert^k \lvert f(x) \rvert <\infty, \end{align*}\]

the function shrinks faster than the reciprocal of any polynomial function.

Definition

Let $f$ be an infinitely differentiable $(\mathscr{C}^\infty)$ function. If $f$ and all of its derivatives are rapidly decreasing, we call $f$ a Schwartz class function and write $f\in\mathcal{S}(\mathbb{R})$.

Proposition

Let $f\in\mathcal{S}(\mathbb{R}), h\in\mathbb{R}$, and $\delta >0$. Then

(1) $f(x+h)\longrightarrow\hat{f}(\xi)e^{2\pi i h \xi}$ (translation becomes modulation

(2) $f(x)e^{-2\pi i xh}\longrightarrow \hat{f}(\xi+h)$

(3) $f(\delta x) \longrightarrow \delta^{-1} \hat{f}(\delta^{-1}\xi)$ (dilation)

(4) $f^\prime(x) \longrightarrow 2\pi i\xi \hat{f}(\xi)$ (differentiation becomes polynomial multiplication)

(5) $-2\pi i x f(x) \longrightarrow \frac{d}{d\xi} \hat{f}(\xi)$

<Proof>

(1)

\[\begin{align*} \int_{-\infty}^\infty f(x+h)e^{-2\pi i x \xi} dx &= \int_{-\infty}^\infty f(x) e^{2\pi ih\xi} e^{-2\pi i x \xi} dx \text{ (by change of the variable)} \\ &=e^{2\pi i h \xi} \hat{f}(\xi) \end{align*}\]

(2)

\[\begin{align*} \int_{-\infty}^\infty f(x) e^{-2\pi ixh} e^{-2\pi ix \xi}dx &= \int_{-\infty}^\infty f(x) e^{-2\pi ix (h+\xi)}dx = \hat{f}(\xi+h) \end{align*}\]

(3)

\[\begin{align*} \int_{-\infty}^\infty f(\delta x) e^{-2\pi i x \xi}dx = \int_{-\infty}^\infty \delta^{-1}f(x) e^{-2\pi i \delta^{-1} x \xi}dx = \delta^{-1} \hat{f}(\delta^{-1} \xi) \end{align*}\]

(4)

\[\begin{align*} \int_{-N}^N f^\prime(x) e^{-2\pi ix\xi} dx &= [f(x) e^{-2\pi i x\xi}]_{-N}^N + 2\pi i \xi \int_{-N}^N f(x) e^{-2\pi i x \xi}dx \\ &\to 2\pi i \xi \hat{f}(\xi) \text{ as } N\to\infty. \end{align*}\]

(5) We must show that $\hat{f}$ is differentiable and find its derivative. Let $\epsilon$ be given and consider

\[\begin{align*} \left\lvert\frac{\hat{f}(\xi +h) -\hat{f}(\xi)}{h} - \widehat{-2\pi i x f}(\xi)\right\rvert &= \left\lvert \int_\mathbb{R} f(x) \frac{1}{h}\left( e^{-2\pi ix(\xi +h)}-e^{-2\pi i x \xi}\right) dx+ \int_{\mathbb{R}} 2\pi ixf(x)e^{-2\pi i x \xi} dx\right\rvert \\ &= \left\lvert \int_{\mathbb{R}} f(x) e^{-2\pi i x \xi} \left( \frac{2^{-2\pi ixh}-1}{h} + 2\pi ix \right) dx\right\rvert \end{align*}\]

Now, $f(x)$ and $xf(x)$ are both rapidly decreasing, so there exists $N\in\mathbb{N}$ such that

\[\begin{align*} \int_{\lvert x \rvert \geq N} \lvert f \rvert < \epsilon \text{ and } \int_{\lvert x \rvert \geq N} \lvert x \rvert \lvert f(x)\rvert dx < \epsilon. \end{align*}\]

By L’Hôpital’s Rule, for each $x_0 \in [-N, N]$, there exists $H>$ such that for $h <\lvert H \rvert$,

\[\left \lvert \frac{e^{-2\pi i xh -1}}{h} + 2\pi i x \right\rvert < \frac{\epsilon}{N}.\]

For each $x_0\in [-N, N]$, take such small $h_0>0$. Define $g(x) = \frac{e^{-2\pi i xh_0 -1}}{h_0} + 2\pi i x$. Take

\[\begin{align*} \epsilon_0=\min\{\lvert \epsilon /N - g(x_0) \rvert,\lvert g(x_0) + \epsilon / N \rvert \}. \end{align*}\]

Then by continuity of $g$, there exists $\delta_0>0$ such that

\[\begin{align*} x\in N_{\delta_0}(x_0) \Rightarrow \lvert g(x) \rvert < \frac{\epsilon}{N}. \end{align*}\]

Now cover the compact set $[-N, N]$ with the open set $\bigcup_{x_0\in [-N, N]}N_{\delta_0}(x_0)$. By the compactness, we can choose finite open sets to cover $[-N, N]$. Then take the minimum $h_0$. Then for $0< h < h_0$, which is independent of $x$, we get the following inequality

\[\left \lvert \frac{e^{-2\pi i xh -1}}{h} + 2\pi i x \right\rvert < \frac{\epsilon}{N}.\]

for $x\in [-N, N]$.

Note that

\[\begin{align*} \lvert e^{ix} -1 \rvert &= \lvert \cos x -1 + i \sin x \rvert \\ &= \lvert -1 + 1- 2\sin^2(x/2) + i 2\sin(x/2)\cos(x/2) \rvert \\ &=\lvert 2\sin(x/2) (i\cos(x/2) -\sin(x/2)) \rvert \\ &=2\lvert \sin(x/2) \rvert. \end{align*}\]

With the equality, outside of $\lvert x \rvert \leq N$, we have the bound

\[\begin{align*} \left\lvert \frac{e^{-2\pi i x h}-1}{h} + 2\pi i x \right\rvert & \leq \left \lvert \frac{e^{-2\pi i x h}-1}{h} \right \rvert + \lvert 2\pi x \rvert \\ &= \left\lvert \frac{2\sin(-\pi i xh)}{h} \right\rvert + 2\pi \lvert x \rvert \\ &\leq A + 2\pi \lvert x \rvert \end{align*}\]

since $\frac{\sin h}{h}$ is bounded. Thus we have

\[\begin{align*} \left\lvert\frac{\hat{f}(\xi +h) -\hat{f}(\xi)}{h} - \widehat{-2\pi i x f}(\xi)\right\rvert &\leq \int_{-N}^N \left\lvert f(x) e^{-2\pi ix \xi}\left(\frac{e^{-2\pi ix h}-1}{h} + 2\pi i x\right) \right\rvert dx + C\epsilon \\ &\leq C^\prime \epsilon \end{align*}\] \[\tag*{$\square$}\]

Definition

We call $f(x)=e^{-x^2}$ the Gaussian.

Remark

The Gaussian is a Schwartz class function. In fact $e^{-ax^2}$ is in $\mathcal{S}(\mathbb{R})$ for all $a>0$. The choice $a=\pi$ is particular because

\[\begin{align*} \left(\int_{-\infty}^\infty e^{-\pi x^2} dx \right)^2 &= \left(\int_{-\infty}^\infty e^{-\pi y^2}dy \right)\cdot \left(\int_{-\infty}^\infty e^{-\pi x^2}dx \right) \\ &= \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-\pi(x^2 + y^2)} dxdy \\ &=\int_{0}^{2\pi} \int_0^\infty e^{-\pi r^2}r dr d\theta \\ &=\int_0^\infty 2\pi r e^{-\pi r^2} dr \\ &=\int_0^{\infty} e^{-x} dx =1 \end{align*}\]

Theorem

Let $f(x)=e^{-\pi x^2}$. Then $\hat{f}=f$.

<Proof>

Let

\[\begin{align*} F(\xi) :=\hat{f}(\xi)=\int_{-\infty}^\infty e^{-\pi x^2} e^{-2\pi ix\xi}dx. \end{align*}\]

Then by property (5) in Proposition, we have

\[\begin{align*} F^\prime(\xi) &= \int_{-\infty}^\infty (-2\pi i x)e^{-\pi x^2} e^{-2\pi i x \xi} dx \\ &=i \int_{-\infty}^\infty f^\prime(x) e^{-2\pi i x \xi} dx \\ &=i 2\pi i\xi\hat{f}(\xi) =-2\pi\xi F(\xi). \end{align*}\]

If we define $G(\xi)=F(\xi)e^{\pi\xi^2}$, its derivative is

\[\begin{align*} G^\prime(\xi)=F^\prime(\xi)e^{\pi\xi^2} + F(\xi)2\xi\pi e^{\pi\xi^2} =-2\pi\xi F(\xi)e^{\pi\xi^2} + 2\pi\xi F(\xi)e^{\pi\xi^2}=0. \end{align*}\]

That is $G(\xi)$ is constant function. Note that $G(0)=F(0)=1$, which implies that

\[\begin{align*} F(\xi)=e^{-\pi\xi^2}=f(\xi). \end{align*}\] \[\tag*{$\square$}\]

Corollary

For $\delta>0$, let $K_\delta$ denote a dilated (by $\sqrt{\delta}$) Gaussian:

\[\begin{align*} K_\delta(x) &= \delta^{-1/2}f(\delta^{-1/2}x)\\ &=\delta^{-1/2}e^{-\pi x^2/\delta}. \end{align*}\]

Then, by the interaction of dilation and Fourier transform,

\[\begin{align*} \widehat{K_\delta}(\xi) = \hat{f}(\delta^{1/2}\xi)=e^{-\pi\delta\xi^2}. \end{align*}\]

<Proof>

\[\begin{align*} \widehat{K_\delta(\xi)} &= \widehat{\delta^{-1/2}f(\delta^{-1/2}\xi)} \\ &=\delta^{-1/2} \delta^{1/2}\hat{f}(\delta^{1/2}\xi) \\ &=\hat{f}(\delta^{1/2}\xi) \end{align*}\] \[\tag*{$\square$}\]

Theorem

$\{K_\delta\}_{\delta>0}$ is an approximation of the identity as $\delta \to 0$.

\[\begin{align} \int_{-\infty}^\infty K_\delta(x) dx =1 \\ \int_{-\infty}^\infty \lvert K_\delta(x) \rvert dx < M \\ \int_{\lvert x \rvert > \eta} \lvert K_\delta (x)\rvert dx \to 0 \text{ as } \delta \to 0 \end{align}\]

<Proof>

(1) By setting $x/\sqrt{\delta}=t$, we get \(\begin{align*} \int_{-\infty}^\infty \frac{1}{\delta} \exp\left(\frac{-\pi x^2}{\delta}\right) dx &= \int_{-\infty}^\infty \exp(-\pi t^2) dt =1 \end{align*}\)

(2) Since $K_\delta(x) >0$ for all $x\in\mathbb{R}$,

\[\begin{align*} \int_{-\infty}^\infty \left\lvert K_\delta(x)\right\rvert dx=\int_{-\infty}^\infty K_\delta(x) dx=1. \end{align*}\]

(3) Since $e^{-\pi x^2}$ is decreasing rapidly,

\[\begin{align*} \int_{\lvert x \rvert > \eta} \frac{1}{\sqrt{\delta}} \exp\left(-\frac{-\pi x^2}{\delta}\right) = \int_{\lvert y \rvert > \frac{\eta}{\sqrt{\delta}}} e^{-\pi y^2}dy \to 0 \text{ as } \to 0. \end{align*}\]

Definition

Given $f,g\in\mathcal{S}(\mathbb{R})$, we deinfe the convolution $f*g$ by

\[\begin{align*} (f*g)(x) = \int_{-\infty}^\infty f(x-t) g(t) dt \end{align*}\]

Corollary

Given any $f\in\mathcal{S}(\mathbb{R})$,

\[\begin{align*} \lim_{\delta\to0^+} f * K_\delta(x) = f(x) \end{align*}\]

uniformly.

<Proof>

Step 1. we show that $f$ is uniformly continuous on all of $\mathbb{R}$. Given $\epsilon >0$, there exists $N\in\mathbb{N}$ such that

\[\begin{align*} \lvert x \rvert >N \Rightarrow \lvert f(x) \rvert < \frac{\epsilon}{2}. \end{align*}\]

since $f\in\mathcal{S}(\mathbb{R})$. Thus, for any $x,y$ with $\lvert x \rvert, \lvert y \rvert >N$,

\(\begin{align*} \lvert f(x) - f(y) \rvert < \epsilon. \end{align*}\) For $x\in [-N, N]$ (compact set), $f$ is continuous, $f$ is uniformly continuous. Combining these, $f$ is uniformly continuous on all of $\mathbb{R}$.

Step2. Choose $\eta >0$ such that

\[\begin{align*} \lvert t \rvert \leq \eta \Rightarrow \lvert f(x-t) - f(x) \rvert < \epsilon. \end{align*}\]

With the given $\eta>0$,

\[\begin{align*} \left \lvert (f*K_\delta)(x) - f(x)\right\rvert &=\left \lvert \int_{-\infty}^\infty K_\delta(t) \left( f(x-t) - f(x) \right) dt \right\rvert \\ &\leq \int_{\lvert t \rvert < \eta} \underbrace{K_\delta(t)}_\text{bounded} \underbrace{\left\lvert f(x-t) - f(x)\right\rvert}_\text{uniform continuity} dt + \int_{\lvert t \rvert \geq \eta } \underbrace{K_\delta(t)}_\text{good kernel} \underbrace{\left\lvert f(x-t) -f(x)\right\rvert}_\text{bounded} dt \\ &\to 0 \text{ as } \eta \to 0 \text{ and } \delta \to 0^+. \end{align*}\]

Therefore, $f*K_\delta \rightrightarrows f \text{ as } \delta \to0^+$.

\(\tag*{$\square$}\)

Reference

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