15 Fourier Transform

Definition

The Fourier transform $\hat{f}(x)$ of a function $f$ on ${\mathbb{R}}^n,n\ge 1$, defined formally as

$$\hat{f}(x)=\frac{1}{(2\pi {)}^n}{\int }_{{\mathbb{R}}^n}f(y)e^{-ix\cdot y}dy.$$

We first list some properties of $\hat{f}(x)$ in $L^1(\mathbb{R})$.

• Note that $|\hat{f}(x)|=\left|\frac{1}{(2\pi {)}^n}{\int }_{{\mathbb{R}}^n}f(y)e^{-ix\cdot y}dy\right|⩽\frac{1}{(2\pi {)}^n}{\int }_{{\mathbb{R}}^n}|f(y)|dy,x\in {\mathbb{R}}^n$, then we have ${\sup }_{x\in {\mathbb{R}}^n}|\hat{f}(x)|⩽(2\pi {)}^{-n}\Vert f{\Vert }_1$.
• $f\to \hat{f}$ is linear on $L^1({\mathbb{R}}^n)$.
• $\hat{f}(x)$ is a uniformly continuous function of $x$ on ${\mathbb{R}}^n$ if $f\in L^1\left({\mathbb{R}}^n\right)$. (By considering $(2\pi {)}^n|\hat{f}(x+h)-\hat{f}(x)|=\left|{\int }_{{\mathbb{R}}^n}f(y)e^{-ix\cdot y}\left\{e^{-iy\cdot h}-1\right\}dy\right|$.)
• Riemann Lebesgue theorem. See here.
• (Translation) $\hat{f}(x+h)=e^{ixh}\hat{f}(x)$.
• (Dilation) Define ${\delta }_{\lambda }f(x)=f(\lambda x)$, then $\widehat{{\delta }_{\lambda }f}(x)=1/|\lambda {|}^n\hat{f}(x/\lambda )$.
• (Rotation) Let $\mathcal{O}$ be an orthogonal linear transformation of ${\mathbb{R}}^n$, then $\widehat{\mathcal{O}f}(x)=\hat{f}(\mathcal{O}(x))$.
• The Fourier transform of an integrable radial function is a radial function.
• Let $f(x)=e^{-|x{|}^2},x\in {\mathbb{R}}^n$, then $\hat{f}(x)=\frac{1}{(2\sqrt{\pi }{)}^n}{e}^{-|x{|}^2/4}$.
• Gauss-Weierstrass kernel
• (Shifting hats) ${\int }_{{\mathbb{R}}^n}\hat{f}(x)g(x)dx={\int }_{{\mathbb{R}}^n}\hat{g}(x)f(x)dx$.

Theorem

Suppose $1\le p\le \infty$ and $f\in L^p\left({\mathbb{R}}^n\right)$.

• (1) For each $ϵ>0,f\ast {\varphi }_{ϵ}$ is infinitely differentiable. For each ${\alpha }_1,{\alpha }_2,\dots ,{\alpha }_n$ non-negative integers

$$\frac{{\partial }^{{\alpha }_1+\cdots {\alpha }_n}\left(f\ast {\varphi }_{ϵ}\right)}{{\partial }_{x_1}^{{\alpha }_1}\cdots {\partial }_{x_n}^{{\alpha }_n}}=f\ast \frac{{\partial }^{{\alpha }_1+\cdots {\alpha }_n}\left({\varphi }_{ϵ}\right)}{{\partial }_{x_1}^{{\alpha }_1}\cdots {\partial }_{x_n}^{{\alpha }_n}}$$

We use the convention that the $0^{\text{th }}$ order derivative of a function is just the function itself.

• (2) $f\ast {\varphi }_{ϵ}\to f$ a.e. as $ϵ\to 0$.
• (3) If $f$ is continuous, then $f\ast {\varphi }_{ϵ}\to f$ uniformly on compact sets as $ϵ\to 0$.
• (4) If $1\le p<\infty$ and $f\in L^p\left({\mathbb{R}}^n\right)$, then $f\ast {\varphi }_{ϵ}\to f$ in $L^p$.

\begin{proof} See here. \end{proof}

Proposition

Suppose $\varphi \in L^1\left({\mathbb{R}}^n\right)$ and ${\int }_{{\mathbb{R}}^n}\varphi (x)dx=1$. Let ${\varphi }_{\delta }(x)=$ ${\delta }^{-n}\varphi (x/\delta )$.

• If $g$ is continuous with compact support, then $g\ast {\varphi }_{\delta }$ converges to $g$ pointwise as $\delta \to 0$.
• If $g$ is continuous with compact support, then $g\ast {\varphi }_{\delta }$ converges to $g$ in $L^1$ as $\delta \to 0$.
• If $f\in L^1$, then ${\Vert f\ast {\varphi }_{\delta }-f\Vert }_1\to 0$ as $\delta \to 0$.

\begin{proof} See here. （我并没有看。 \end{proof}

Inversion of the Fourier transform on $L^1$

If $f$ and $\hat{f}$ are in $L^1\left({\mathbb{R}}^n\right)$, then

$$f(x)={\int }_{{\mathbb{R}}^n}\hat{f}(y)e^{ix\cdot y}dy\text{, a.e. }$$

\begin{proof} See here.

Uniqueness

If $f,g\in L^1\left({\mathbb{R}}^n\right)$ and $\hat{f}(x)=\hat{g}(x)$ for all $x\in {\mathbb{R}}^n$, then $f=g$ a.e. in ${\mathbb{R}}^n$.

The Convolution Property

The convolution of any two integrable functions is also integrable and therefore has a well-defined Fourier transform.

Theorem

If $f,g\in L^1\left({\mathbb{R}}^n\right)$, then

$$\widehat{f\ast g}(x)=(2\pi {)}^n\hat{f}(x)\hat{g}(x)\text{ for all }x\in {\mathbb{R}}^n.$$

This follows easily from Fubini’s theorem and is left as an exercise.

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Theorem

• Let $x_k,k=1,\dots ,n$, denote the $k$ th coordinate of $x\in {\mathbb{R}}^n$. If $f$ and $x_{k}f(x)$ belong to $L^1\left({\mathbb{R}}^n\right)$, then $\hat{f}$ has a partial derivative with respect to $x_k$ everywhere in ${\mathbb{R}}^n$, and

$$\frac{\partial \hat{f}}{\partial x_k}=\left(\widehat{-i{x}_{k}f}\right)=\frac{1}{(2\pi {)}^n}{\int }_{{\mathbb{R}}^n}\left(-i{y}_{k}f(y)\right)e^{-ix\cdot y}dy.$$

• Fix $k=1,\dots ,n$ and let $h=\left(0,\dots ,0,h_k,0,\dots ,0\right)\ne 0$ lie on the $k$ th coordinate axis. Suppose that $f\in L^1\left({\mathbb{R}}^n\right)$ and that there is a function $g$ such that

$$\underset{h_k\to 0}{\lim }{\int }_{{\mathbb{R}}^{\mathrm{n}}}\left|\frac{f(x+h)-f(x)}{h_k}-g(x)\right|dx=0.$$

Then $g\in L^1\left({\mathbb{R}}^n\right)$ and

$$\hat{g}(x)=i{x}_k\hat{f}(x)\text{ for all }x\in {\mathbb{R}}^n.$$

\begin{proof} See here. \end{proof}

Fourier Transform on $L^2$

Lemma

Let $f,g\in L^2\left({\mathbb{R}}^n\right)$. Then $f\ast g$ is uniformly continuous and bounded on ${\mathbb{R}}^n$, and $\Vert f\ast g{\Vert }_{\infty }\le \Vert f{\Vert }_2\Vert g{\Vert }_2$.

\begin{proof} By Cauchy inequality. See here. \end{proof}

Plancherel

If $f$ is continuous with compact support, then $\hat{f}\in L^2\left({\mathbb{R}}^n\right)$ and

$$\Vert \hat{f}{\Vert }_2=(2\pi {)}^{-n/2}\Vert f{\Vert }_2.$$

\begin{proof} By Gauss-Weierstrass kernel, we have

$${\int }_{{\mathbb{R}}^n}{e}^{-iuy}\hat{f}(u)H_a(u)du=(f\ast \widehat{H_a})(y).$$

Take $f=f\ast \overline{f(-x)}$ and take $y=0$, then

$$(2\pi {)}^n\int |\hat{f}(u){|}^2{H}_a(u)du=(f\ast g)\ast \widehat{H_a}(0).$$

Take $a\to \infty$ and we finish the proof.

Also see here. \end{proof}

Remark. If $\{f_k\}$ satisfying $||f_k-f|{|}_2\to 0$, then $\{\widehat{f_k}\}$ is Cauchy in $L^2$ and converges to a function $\hat{f}$ in $L^2$. We can check that $\hat{f}$ is not depending on the choice of $f_k$, and so it is well-defined. (That is the way how we define $\hat{f}$?)