12.3 Fourier transforms

中文

For any function $\phi (x)$ its Fourier transform is the function $\mathcal{F}\phi$ defined by

$$\mathcal{F}\phi (y)=\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }{\mathrm{e}}^{-ixy}\phi (x)\mathrm{d}x.$$

The inverse Fourier transform is defined by

$${\mathcal{F}}^{-1}\phi (y)=\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }{\mathrm{e}}^{ixy}\phi (x)\mathrm{d}x.$$

Fourier’s integral theorem, applicable to all functions $\phi$ such that $|\phi |$ is integrable over $[-\infty ,\infty ]$ and is of bounded variation, says that ${\mathcal{F}}^{-1}\mathcal{F}\phi =\phi$ , expressed in integral form as

$$\begin{array}{l}\phi (a)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }\mathrm{d}y{\mathrm{e}}^{iay}{\int }_{-\infty }^{\infty }{\mathrm{e}}^{-iyx}\phi (x)\mathrm{d}x\\ =\frac{1}{2\pi }{\int }_{-\infty }^{\infty }\mathrm{d}x\phi (x){\int }_{-\infty }^{\infty }{\mathrm{e}}^{iy(a-x)}\mathrm{d}y.\end{array}$$

The proofofthis theorem can be found in many books on real analysis. The reader is referred to [6, chap. 7] or [2, $p$. 88].

Applying standard rules of integrals applied to delta functions, we expect

$${\delta }_a(x)=\delta (x-a)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }{\mathrm{e}}^{iy(a-x)}\mathrm{d}y,\text{\hspace{1em}(12.10)}$$

or, on setting $a=0$ and using $\delta (x)=\delta (-x)$ ,

$$\delta (x)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }{\mathrm{e}}^{-iyx}\mathrm{d}y=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }{\mathrm{e}}^{iyx}\mathrm{d}y.\text{\hspace{1em}(12.11)}$$

Similarly, the Fourier transform of the delta function should be

$$\mathcal{F}\delta (y)=\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }{\mathrm{e}}^{-ixy}\delta (x)\mathrm{d}x=\frac{1}{\sqrt{2\pi }}\text{\hspace{1em}(12.12)}$$

and Eq. (12.11) agrees with

$$\delta (x)={\mathcal{F}}^{-1}\frac{1}{\sqrt{2\pi }}=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }{\mathrm{e}}^{ixy}\mathrm{d}y.\text{\hspace{1em}(12.13)}$$

Mathematical consistency can be achieved by defining the Fourier transform of a distribution $T$ to be the distribution $\mathcal{F}T$ given by

$$\mathcal{F}T(\phi )=T(\mathcal{F}\phi ),\text{\hspace{1em}(12.14)}$$

for all test functions $\phi .$ . For regular distributions we then have the desired result,

$$T_{\mathcal{F}f}(\phi )=\mathcal{F}{T}_f(\phi ),$$

since

$$\begin{array}{l}\mathcal{F}{T}_f(\phi )=T_f(\mathcal{F}\phi )\\ =\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }({\int }_{-\infty }^{\infty }{\mathrm{e}}^{-iyx}\phi (x)\mathrm{d}x)f(y)\mathrm{d}y\\ =\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }\phi (x)({\int }_{-\infty }^{\infty }{\mathrm{e}}^{-iyx}f(y)\mathrm{d}y)\mathrm{d}x\\ ={\int }_{-\infty }^{\infty }\phi (x)\mathcal{F}f(x)\mathrm{d}x\\ =T_{\mathcal{F}f}(\phi ).\end{array}$$

If the inverse Fourier transform is defined on distributions by ${\mathcal{F}}^{-1}T(\phi )=T({\mathcal{F}}^{-1}\phi )$ ), then

$${\mathcal{F}}^{-1}\mathcal{F}T=T,$$

for

$${\mathcal{F}}^{-1}\mathcal{F}T(\phi )=\mathcal{F}T({\mathcal{F}}^{-1}\phi )=T(\mathcal{F}{\mathcal{F}}^{-1}\phi )=T(\phi ).$$

There is, however, a serious problem with these definitions. If $\phi$ is a function of bounded support then $\mathcal{F}\phi$ is generally an entire analytic function and cannot be of bounded support, since an entire function that vanishes on any open set must vanish everywhere. Hence the right-hand side ofEq. (12.14) is not in general well-defined. A way around this is to define a more general space of test functions $S(\mathbb{R})$ called the space of rapidly decreasing functions – functions that approach 0 as $|x|\to \infty$ faster than any inverse power $|x{|}^{-n}$ ,

$$\mathcal{S}(\mathbb{R})=\{\phi |\underset{x\in \mathbb{R}}{\sup }|x^m{\phi }^{(p)}(x)|<\infty \text{ for all integers }m,p>0\}.$$

Convergence in $\mathcal{S}(\mathbb{R})$ is defined by ${\phi }_n\to \phi$ if and only if

$$\underset{n\to \infty }{\lim }\underset{x\in \mathbb{R}}{\sup }|x^m({\phi }^{(p)}(x)-{\phi }^{(p)})|=0\text{ for all integers }m,p>0.$$

The space of continuous linear functions on $S(\mathbb{R})$ is denoted $S^{\prime }(\mathbb{R})$ , and they are called tempered distributions. Since every test function is obviously a rapidly decreasing function, $\mathcal{D}(\mathbb{R})\subset S(\mathbb{R})$ . If $T$ is a tempered distribution in Eq. (12.14), the Fourier transform $\mathcal{F}T$ is well-defined, since the Fourier transform of any rapidly decreasing function may be shown to be a function of rapid decrease.

Example 12.8

The Fourier transform of the delta distribution is defined by

$$\begin{array}{l}\mathcal{F}{\delta }_a(\phi )={\delta }_a(\mathcal{F}\phi )\\ =\mathcal{F}\phi (a)\\ =\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }{\mathrm{e}}^{-iax}\phi (x)\mathrm{d}x\\ =T_{(2\pi {)}^{-1/2}{\mathrm{e}}^{iax}}(\phi ).\end{array}$$

Similarly

$${\mathcal{F}}^{-1}{T}_{{\mathrm{e}}^{-iax}}=\sqrt{2\pi }{\delta }_a.$$

The delta function versions of these distributional equations are

$$\mathcal{F}{\delta }_a(y)=\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }{\mathrm{e}}^{-iyx}\delta (x-a)\mathrm{d}x=\frac{{\mathrm{e}}^{-iay}}{\sqrt{2\pi }}$$

and

$${\mathcal{F}}^{-1}{\mathrm{e}}^{-iax}=\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }{\mathrm{e}}^{ixy}{\mathrm{e}}^{-iax}\mathrm{d}x=\sqrt{2\pi }\delta (x-a),$$

in agreement with Eqs. Eq. (12.10)–Eq. (12.13) above.

Problems

Problem 12.10

Find the Fourier transforms of the functions

$$f(x)=\{\begin{array}{ll}1& \text{ if }-a\le x\le a\\ 0& \text{ otherwise }\end{array}$$

and

$$g(x)=\{\begin{array}{ll}1-\frac{|x|}{2}& \text{ if }-a\le x\le a\\ 0& \text{ otherwise. }\end{array}$$

Problem 12.11

Show that

$$\mathcal{F}\left({\mathrm{e}}^{-a^2{x}^2/2}\right)=\frac{1}{|a|}{\mathrm{e}}^{-k^2/2a^2}.$$

Problem 12.12

Evaluate Fourier transforms of the following distributional functions:

(a) $\delta (x-a).$

(b) ${\delta }^{\prime }(x-a).$

(c) ${\delta }^{(n)}(x-a)$

(d) $\delta (x^2-a^2)$

(e) ${\delta }^{\prime }(x^2-a^2)$

Problem 12.13

Prove that

$$x^m{\delta }^{(n)}(x)=(-1{)}^m\frac{n!}{(n-m)!}{\delta }^{(n-m)}(x)\text{ for }n\ge m.$$

Hence show that the Fourier transform of the distribution

$$\sqrt{2\pi }\frac{k!}{(m+k)!}{x}^m{\delta }^{(m+k)}(-x)(m,k\ge 0)$$

is $(-iy{)}^k$

Problem 12.14

Show that the Fourier transform of the distribution

$${\delta }_0+{\delta }_a+{\delta }_{2a}+\cdots +{\delta }_{(2n-1)a}$$

is a distribution with density

$$\frac{1}{\sqrt{2\pi }}\frac{\sin (nay)}{\sin (\frac{1}{2}ay)}{\mathrm{e}}^{-(n-\frac{1}{2})iay}.$$

Show that

$${\mathcal{F}}^{-1}(f(y){\mathrm{e}}^{iby})=({\mathcal{F}}^{-1}f)(x+b).$$

Hence find the inverse Fourier transform of

$$g(y)=\frac{\sin nay}{\sin (\frac{1}{2}ay)}.$$