Fourier Transform on R

We developed the theory of Fourier series on periodic functions, but is it possible to consider an analogous theory on functions of the real line?

For this to work, we must consider functions which vanish towards infinity in some sense, and instead of a series we will try transforming $f$ into another function $\hat{f}$

In our Fourier series definitions, we give
$a_n={\int }_0^{1}f(x)e^{-2\pi inx}dx$
$f(x)={\sum }_{n=-\infty }^{\infty }{a}_n{e}^{2\pi inx}$

The continuous counterpart is then
$\hat{f}(\xi )={\int }_{-\infty }^{\infty }f(x)e^{-2\pi ix\xi }dx$
$f(x)={\int }_{-\infty }^{\infty }\hat{f}(\xi )e^{2\pi ix\xi }d\xi$

This is an important tool in the study of partial differential equations. How do we prove it’s valid and under what assumptions? In this course, we limit ourselves to the Schwartz space of functions

To begin, we clarify ${\int }_{-\infty }^{\infty }f(x)dx={\lim }_{N\to \infty }{\int }_{-N}^{N}f(x)dx$, which is only defined if $f$ decays as $x\to \pm \infty$

Definition: A function $f$ defined on $\mathbb{R}$ is said to be of moderate decrease if $f$ is continuous and $|f(x)|\le \frac{A}{1+x^2}$ for some $A>0$, in which case we write $f\in \mathcal{M}(\mathbb{R})$

• This implies $f$ is bounded and decays at least as fast as $\frac{1}{x^2}$
• We could just as well use a more lenient definition of $\frac{A}{1+x^{1+ϵ}}$

For $f\in \mathcal{M}(\mathbb{R})$, ${\int }_{-\infty }^{\infty }f(x)dx$ exists and satisfies

1. Linearity: ${\int }_{-\infty }^{\infty }(af(x)+bg(x))dx=a{\int }_{-\infty }^{\infty }f(x)dx+b{\int }_{-\infty }^{\infty }g(x)dx$
2. Translation invariance: ${\int }_{-\infty }^{\infty }f(x-h)dx={\int }_{-\infty }^{\infty }f(x)dx$
3. Scaling under dilations: $\delta {\int }_{-\infty }^{\infty }f(\delta x)dx={\int }_{-\infty }^{\infty }f(x)dx$
4. Continuity: ${\int }_{-\infty }^{\infty }|f(x-h)-f(x)|dx\to 0$ as $h\to 0$

These are mostly trivial, although continuity relies on the uniform continuity of $f$ in a compact interval

Definition: The Fourier transform of $f\in \mathcal{M}(\mathbb{R})$ is $\hat{f}(\xi )={\int }_{-\infty }^{\infty }f(x)e^{-2\pi ix\xi }dx$

Because of the conditions on $f$, the integral is of moderate decrease and we can prove that $\hat{f}$ is well-defined, bounded, continuous, and tends to $0$ as $|\xi |\to \infty$

This last point is known as the Reimann-Lebesgue Lemma

But this does not tell us how $\hat{f}$ decays, and whether the resulting integral ${\int }_{-\infty }^{\infty }\hat{f}(\xi )e^{2\pi ix\xi }d\xi$ makes sense

One way we can learn more about $\hat{f}$ is by requiring some more properties from $f$

Definition: The Schwartz space on $\mathbb{R}$, denoted $\mathcal{S}=\mathcal{S}(\mathbb{R})$, consists of the set of all indefinitely differentiable functions such that $f$ and all its derivatives are rapidly decreasing, in the sense that ${\sup }_{x\in \mathbb{R}}|x{|}^k|f^{(l)}(x)|<\infty$, i.e. faster than polynomial decay

If $f\in \mathcal{S}(\mathbb{R})$ then $f^{\prime }(x)=\frac{df}{dx}\in \mathcal{S}(\mathbb{R})$ and $xf(x)\in \mathcal{S}(\mathbb{R})$
Also $\mathcal{S}(\mathbb{R})$ is a vector space over $\mathbb{C}$

An important example is $f(x)=e^{-x^2}$, or more generally $e^{-a{x}^2}$ for $a>0$

The bump functions also belong to $\mathcal{S}(\mathbb{R})$

Although $e^{-|x|}$ decreases rapidly at infinity, it is not differentiable at $0$ so it does not belong to $\mathcal{S}(\mathbb{R})$, which is a subset of $C^{\infty }$

We denote the Fourier transform of a function $f\in \mathcal{S}(\mathbb{R})\subset \mathcal{M}(\mathbb{R})$ as before, $\hat{f}(\xi )={\int }_{-\infty }^{\infty }f(x)e^{-2\pi ix\xi }dx$, and we use the notation $f(x)\to \hat{f}(\xi )$

1. $f(x+h)\to \hat{f}(\xi )e^{2\pi ih\xi }$
2. $f(x)e^{-2\pi ixh}\to \hat{f}(\xi +h)$
3. $f(\delta x)\to {\delta }^{-1}\hat{f}({\delta }^{-1}\xi )$
4. $f^{\prime }(x)\to 2\pi i\xi \hat{f}(\xi )$
5. $-2\pi ixf(x)\to \frac{d}{d\xi }\hat{f}(\xi )$

Only the fifth property is non-trivial to prove

The fact that the Fourier transform interchanges differentiation with multiplication by $x$ makes it a key tool in the theory of differential equations

Theorem: If $f\in \mathcal{S}(\mathbb{R})$, then $\hat{f}\in \mathcal{S}(\mathbb{R})$

For any pair $l$ and $k$ we have ${\xi }^k{\left(\frac{d}{d\xi }\right)}^l\hat{f}(\xi )$ is bounded, because by properties 4 and 5 this is the Fourier transform of $\frac{1}{(2\pi i{)}^k}(\frac{d}{dx}{)}^k[(-2\pi ix{)}^{l}f(x)]$

So under $\mathcal{S}(\mathbb{R})$, our inversion theorem is well-defined, although still not necessarily true

To actually prove this theorem, we will use the Gaussian, which has some special properties

First, ${\int }_{-\infty }^{\infty }{e}^{-\pi x^2}dx=1$

${\left({\int }_{-\infty }^{\infty }{e}^{-\pi x^2}dx\right)}^2={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }{e}^{-\pi (x^2+y^2)}dxdy={\int }_0^{2\pi }{\int }_0^{\infty }{e}^{-\pi r^2}rdrd\theta$
$={\int }_0^{\infty }2\pi {re}^{-\pi r^2}dr$
$=[-e^{-\pi r^2}{]}_0^{\infty }$
$=1$

Theorem: If $f(x)=e^{-\pi x^2}$, then $\hat{f}(\xi )=f(\xi )$

$F(\xi )=\hat{f}(\xi )={\int }_{-\infty }^{\infty }{e}^{-\pi x^2}{e}^{-2\pi ix\xi }dx$
$f^{\prime }(x)=-2\pi xf(x)$

$F^{\prime }(\xi )={\int }_{-\infty }^{\infty }(-2\pi ix)f(x)e^{-2\pi ix\xi }dx$
$=i{\int }_{-\infty }^{\infty }{f}^{\prime }(x)e^{-2\pi ix\xi }dx$
$=i(2\pi i\xi )\hat{f}(\xi )=-2\pi \xi F(\xi )$

We define $G(\xi )=F(\xi )e^{\pi {\xi }^2}$
$G^{\prime }(\xi )=F^{\prime }(\xi )e^{\pi {\xi }^2}+F(\xi )\cdot -2\pi \xi e^{\pi {\xi }^2}=0$
This means $G(\xi )$ is a constant, so $G(\xi )=G(0)=1$, so $F(\xi )=e^{-\pi {\xi }^2}$

As a natural corollary, if $\delta >0$ and $K_{\delta }(x)={\delta }^{-1/2}{e}^{-\pi x^2/\delta }$ then $\widehat{K_{\delta }}(\xi )=e^{-\pi \delta {\xi }^2}$

The book notes that as $\delta \to 0$, $K_{\delta }$ peaks at the origin while $\widehat{K_{\delta }}$ gets flatter, which is related to the Heisenberg uncertainty principle

Actually, this function $K_{\delta }(x)$ satisfies our properties of good kernels established in the context of the discrete Fourier series,

1. ${\int }_{-\infty }^{\infty }{K}_{\delta }(x)dx=1$
2. ${\int }_{-\infty }^{\infty }|K_{\delta }(x)|dx\le M$
3. For ever $\eta >0$, we have ${\int }_{|x|>\eta }^{}|K_{\delta }(x)|dx\to 0$ as $\delta \to 0$

Theorem: The collection $\{K_{\delta }{\}}_{\delta >0}$ is a family of good kernels as $\delta \to 0$

Similar to in the previous material, we apply these good kernels via convolution

Definition: If $f,g\in \mathcal{S}(\mathbb{R})$, their convolution is defined as $(f\ast g)(x)={\int }_{-\infty }^{\infty }f(x-t)g(t)dt$

Since $f(x-t)g(t)$ is of rapid decrease for any fixed $x$, this integral converges

Corollary: If $f\in \mathcal{S}(\mathbb{R})$, then $(f\ast K_{\delta })(x)\to f(x)$ uniformly in $x$ as $\delta \to 0$

Since $f$ is rapidly decreasing, we can establish that it is uniformly continuous

$(f\ast K_{\delta })(x)-f(x)={\int }_{-\infty }^{\infty }{K}_{\delta }(t)[f(x-t)-f(x)]dt$
And since $K_{\delta }\ge 0$,
$|(f\ast K_{\delta })(x)-f(x)|\le {\int }_{|t|>\eta }^{}+{\int }_{|t|\le \eta }^{}{K}_{\delta }(t)|f(x-t)-f(x)|dx$

We can choose $\eta$ to lower this bound as much as we please; the first integral shrinks because of the third property of good kernels, and the second integral shrinks because of the uniform continuity of $f$

Now if $f,g\in \mathcal{S}(\mathbb{R})$, then ${\int }_{-\infty }^{\infty }f(x)\hat{g}(x)dx={\int }_{-\infty }^{\infty }\hat{f}(y)g(y)dy$

Proving this relies on some intuitive properties of double integrals, which is if we have a continuous function in two variables $F(x,y)$ with $F_1(x)={\int }_{-\infty }^{\infty }F(x,y)dy$ and $F_2(y)={\int }_{-\infty }^{\infty }F(x,y)dx$ both of moderate decrease then ${\int }_{-\infty }^{\infty }{F}_1(x)dx={\int }_{-\infty }^{\infty }{F}_2(y)dy$ (essentially calculating the same double integral in two ways)

So we can apply this to $F(x,y)=f(x)g(y)e^{-2\pi ixy}$, where $F_1(x)=f(x)\hat{g}(x)$ and $F_2(y)=\hat{f}(y)g(y)$

Theorem: If $f\in \mathcal{S}(\mathbb{R})$, then $f(x)={\int }_{-\infty }^{\infty }\hat{f}(\xi )e^{2\pi ix\xi }d\xi$

This is a key result! How do we prove it?

Let $G_{\delta }(x)=e^{-\pi \delta x^2}$, such that $\widehat{G_{\delta }}(\xi )=K_{\delta }(\xi )$
${\int }_{-\infty }^{\infty }f(x)K_{\delta }(x)dx={\int }_{-\infty }^{\infty }\hat{f}(\xi )G_{\delta }(\xi )d\xi$

As $\delta \to 0$, the left hand side goes to $f(0)$ and $G_{\delta }(\xi )$ goes to $1$, so we have $f(0)={\int }_{-\infty }^{\infty }\hat{f}(\xi )d\xi$

Now let $F(y)=f(y+x)$
$f(x)=F(0)={\int }_{-\infty }^{\infty }\hat{F}(\xi )d\xi ={\int }_{-\infty }^{\infty }\hat{f}(\xi )e^{2\pi ix\xi }d\xi$, which proves our result

We can define two mappings $\mathcal{F}:\mathcal{S}(\mathbb{R})\to \mathcal{S}(\mathbb{R})$ and ${\mathcal{F}}^{\ast }:\mathcal{S}(\mathbb{R})\to \mathcal{S}(\mathbb{R})$ by $\mathcal{F}(f)(\xi )={\int }_{-\infty }^{\infty }f(x)e^{-2\pi ix\xi }dx$ and ${\mathcal{F}}^{\ast }(g)(x)={\int }_{-\infty }^{\infty }g(\xi )e^{2\pi ix\xi }d\xi$, such that $\mathcal{F}$ is the Fourier transform, and we’ve proven ${\mathcal{F}}^{\ast }\circ \mathcal{F}=I$ on $\mathcal{S}(\mathbb{R})$

Since these transforms differ only by a sign in the exponential, we see $\mathcal{F}(f)(y)={\mathcal{F}}^{\ast }(f)(-y)$, so $\mathcal{F}\circ {\mathcal{F}}^{\ast }=I$

Therefore, the Fourier transform is a bijective mapping on the Schwartz space

To get to our final identity, we also need a few more properties,

1. $f\ast g\in \mathcal{S}(\mathbb{R})$
2. $f\ast g=g\ast f$
3. $\widehat{(f\ast g)}(\xi )=\hat{f}(\xi )\hat{g}(\xi )$

To show $f\ast g$ is rapidly decreasing, note that any $l\ge 0$ we have ${\sup }_x|x{|}^l|g(x-y)|\le A_l(1+|y|{)}^l$, from which we see that ${\sup }_x|x^l(f\ast g)(x)|\le A_l{\int }_{-\infty }^{\infty }|f(y)|(1+|y|{)}^{l}dx$, meaning $x^l(f\ast g)(x)$ is bounded

We can show this bound also carries over to the derivatives of $f\ast g$ by proving the identity ${\left(\frac{d}{dx}\right)}^k(f\ast g)(x)=\left(f\ast {\left(\frac{d}{dx}\right)}^{k}g\right)(x)$ for $k=1,2,\dots$ (not trivial, but going to omit here cause I’m tired)

To prove our second property, we look at a change of variables $x-y=u$,
$(f\ast g)(x)={\int }_{-\infty }^{\infty }f(x-u)g(u)du=(g\ast f)(x)$

For the third property, we consider $F(x,y)=f(y)g(x-y)e^{-2\pi ix\xi }$ which yields $F_1(x)=(f\ast g)(x)e^{-2\pi ix\xi }$ and $F_2(y)=f(y)e^{-2\pi iy\xi }\hat{g}(\xi )$, and ${\int }_{-\infty }^{\infty }{F}_1(x)dx={\int }_{-\infty }^{\infty }{F}_2(y)dy$ is our desired statement

Finally, we equip the Schwartz space with a Hermitian inner product,
$(f,g)={\int }_{-\infty }^{\infty }f(x)\overline{g(x)}dx$
$\Vert f\Vert ={\left({\int }_{-\infty }^{\infty }|f(x){|}^{2}dx\right)}^{1/2}$

We now have the tools to simply prove the Plancherel formula (the continuous analogue to Parseval’s identity)

Theorem: If $f\in \mathcal{S}(\mathbb{R})$ then $\Vert \hat{f}\Vert =\Vert f\Vert$

To prove this, we define $f^{♭}(x)=\overline{f(-x)}$, so $\widehat{f^{♭}}(\xi )=\overline{\hat{f}(\xi )}$
$h=f\ast f^{♭}$

We clearly have $\hat{h}(\xi )=|\hat{f}(\xi ){|}^2$ and $h(0)={\int }_{-\infty }^{\infty }|f(x){|}^{2}dx$, so our formula follows directly from the inversion formula with $x=0$, ${\int }_{-\infty }^{\infty }\hat{h}(\xi )d\xi =h(0)$

We ultimately are able to achieve the results so far on general $L^2(\mathbb{R})$, whose proof relates to the fact that the Schwartz functions are a dense subset of the $L^2$ functions

We can also use good kernels to prove the Weierstrass approximation theorem

Theorem: Let $f$ be a continuous function on the closed and bounded interval $[a,b]\subset \mathbb{R}$, then for any $ϵ>0$, there exists a polynomial $P$ such that ${\sup }_{x\in [a,b]}|f(x)-P(x)|<ϵ$

Let $[-M,M]$ denote an interval containing $[a,b]$ and let $g$ be a continuous function on $\mathbb{R}$ that equals $0$ outside $[-M,M]$ and equals $f$ in $[a,b]$, bounded by $B$

Since $\{K_{\delta }\}$ is a family of good kernels and $g$ is continuous with compact support, $g\ast K_{\delta }$ converges uniformly to $g$ as $\delta$ tends to $0$

We choose ${\delta }_0$ such that $|g(x)-(g\ast K_{{\delta }_0})(x)|<ϵ/2$

Now, since $e^x={\sum }_{n=0}^{\infty }\frac{x^n}{n!}$ converges uniformly, we can bound $|K_{{\delta }_0}(x)-R(x)|\le \frac{ϵ}{4MB}$ for all $x\in [-2M,2M]$, where $R(x)={\delta }_0^{-1/2}{\sum }_{n=0}^N\frac{(-\pi x^2/{\delta }_0{)}^n}{n!}$

This is enough to construct our bound, since
$|(g\ast K_{{\delta }_0})(x)-(g\ast R)(x)|=\left|{\int }_{-M}^{M}g(t)[K_{{\delta }_0}(x-t)-R(x-t)]dt\right|$
$\le {\int }_{-M}^M|g(t)||K_{{\delta }_0}(x-t)-R(x-t)|dt$
$\le 2MB\underset{z\in [-2M,2M]}{\sup }|K_{{\delta }_0}(z)-R(z)|$
$<ϵ/2$

Combined with our earlier bound, we get $|g(x)-(g\ast R)(x)|<ϵ$ for $x\in [-M,M]$ and therefore $|f(x)-(g\ast R)(x)|<ϵ$ for $x\in [a,b]$

Now does this bound prove our theorem? Because $g\ast R$ is a polynomial in the $x$ variable by definition, since $(g\ast R)(x)={\int }_{-M}^{M}g(t)R(x-t)dt$ and $R(x-t)$ is ultimately a polynomial in $x$