Refers to Fourier Analysis by Stein, analysis by Zorich

# Fourier transform

# The Fourier transform on the space of functions of moderate decrease

Before extend the integral of a function on a closed and bounded interval, we should check whether the value exists or not. A useful conditino is as follows.

Definition 5.1.1 A function ff is said to be of moderate decrease if fC(R)f\in C(\mathbb R) and A>0,ϵ>0\exists A>0,\epsilon>0 so that

f(x)A1+x1+ϵ,xR|f(x)|\leq \dfrac {A}{1+x^{1+\epsilon}},\ \forall x\in \mathbb R

We shall denote M(R)\mathcal M(\mathbb R) the set of functions of moderate decrease on R\mathbb R.

Example

There are two examples of functions of moderate decrease

f(x)=11+xn,n2f(x)=\dfrac 1{1+|x|^n},\ n\geq 2

and

f(x)=eax,a>0f(x)=e^{-a|x|},\ a>0

Remark

Under the usual addition of functions and multiplication by scalars, M(R)\mathcal M(\mathbb R) forms a vector space over C\mathbb C.

Definition 5.1.2 For all fM(R)f\in\mathcal M(\mathbb R), we may define its integral

f(x)dx:=limNNNf(x)dx\int^\infty_{-\infty}f(x)\mathrm dx:=\lim_{N\to\infty}\int^N_{-N}f(x)\mathrm dx

We summarize some elementary properties of integration over R\mathbb R in a property

Property 5.1.2 The integral of a function of moderate decrease defined above satisfies the following properties, if f,gM(R)f,g\in \mathcal M(\mathbb R) and a,bCa,b\in\mathbb C, then

  1. Linearity:

(af(x)+bg(x))dx=af(x)dx+bg(x)dx\int^\infty_{-\infty}(af(x)+bg(x))\mathrm dx=a\int^\infty_{-\infty}f(x)\mathrm dx+b\int^\infty_{-\infty}g(x)\mathrm dx

  1. Translation invariance: hR\forall h\in\mathbb R we have

f(xh)dx=f(x)dx\int^\infty_{-\infty}f(x-h)\mathrm dx=\int^\infty_{-\infty}f(x)\mathrm dx

  1. Scaling under dilations: if δ0\delta \neq 0, then

δf(δx)dx=sgn(δ)f(x)dx\delta\int^\infty_{-\infty}f(\delta x)\mathrm dx=\mathrm{sgn}(\delta )\cdot \int^\infty_{-\infty}f(x)\mathrm dx

  1. Continuity:

f(xh)f(x)dx0,ash0\int ^\infty_{-\infty}|f(x-h)-f(x)|\mathrm dx\to 0,\ \text{ as }h\to 0

Proof

Prove the properties by using integral on the bounded interval instead of using indefinite integral directly.

Definition 5.1.3 If fM(R)f\in\mathcal M(\mathbb R), we define its Fourier transform for ξR\xi\in\mathbb R by

f^(ξ)=f(x)e2πixξdx\hat f(\xi)=\int^\infty_{-\infty}f(x)e^{-2\pi i x\xi}\mathrm dx

Obviouly the integral makes sense. However, nothing in the definition above guarantees that f^\hat f is of moderate decrease.

Roughly speaking, the Fourier transform is a continuous version of the Fourier coefficients.

# The Fourier transform on the Schwartz space

Definition 5.1.4 The Schwartz space on R\mathbb R consists of the se t of all fC(R)f\in C^\infty(\mathbb R) so that ff and its derivatives f,...,f(l),...f',...,f^{(l)},... are rapidly decreasing, in the sense that

supxRxkf(l)(x)<,k,l0\sup_{x\in\mathbb R}|x|^k|f^{(l)}(x)|<\infty,\ \forall k,l\geq 0

We denote this space by S=S(R)\mathcal S=\mathcal S(\mathbb R)

Remark

Under the usual addition of functions and multiplication by scalars, S(R)\mathcal S(\mathbb R) forms a vector space over C\mathbb C.

Moreover, if fS(R)f\in \mathcal S(\mathbb R), we have

f(x)=dfdxS(R)f'(x)=\dfrac{\mathrm df}{\mathrm dx}\in \mathcal S(\mathbb R)

xf(x)S(R)xf(x)\in\mathcal S(\mathbb R)

This expresses the important fact that the Schwartz space is closed under differentiation and multiplication by polynomials.

Example

The Gaussian eax2S(R)e^{-ax^2}\in\mathcal S(\mathbb R) whenever a>0a>0, while exe^{-|x|} does not for not differentiable at 00.

Definition 5.1.5 The Fourier transform of a function fS(R)f\in \mathcal S(\mathbb R) is defined by

f^(ξ)=f(x)e2πixξdx\hat f(\xi)=\int^\infty_{-\infty}f(x)e^{-2\pi i x\xi}\mathrm dx

We use the notation

f(x)f^(ξ)f(x)\to \hat f(\xi)

to mean that f^(ξ)\hat f(\xi) denotes the Fourier transform of ff.

Property 5.1.6 If fS(R)f\in\mathcal S(\mathbb R), then

  1. Whenever hRh\in\mathbb R

f(x+h)f^(ξ)e2πihξf(x+h)\to \hat f(\xi)e^{2\pi i h\xi}

  1. Whenever hRh\in\mathbb R

f(x)e2πixhf^(ξ+h)f(x)e^{-2\pi ix h}\to \hat f(\xi +h)

  1. Whenever δ>0\delta >0

f(δx)δ1f^(δ1ξ)f(\delta x)\to \delta^{-1}\hat f(\delta^{-1}\xi)

  1. Derivatives:

f(x)2πiξf^(ξ)f'(x)\to 2\pi i\xi\hat f(\xi)

2πixf(x)ddξf^(ξ)-2\pi ixf(x)\to \dfrac {\mathrm d}{\mathrm d\xi}\hat f(\xi)

Proof

Tips: Uniformly convergence

Theorem 5.1.7 If fS(R)f\in\mathcal S(\mathbb R), then f^S(R)\hat f\in\mathcal S(\mathbb R).

Proof

Note that

1(2πi)k(ddx)k[(2πix)lf(x)]ξk(ddξ)lf^(ξ)\dfrac 1{(2\pi i)^k}\left(\dfrac{\mathrm d}{\mathrm dx}\right)^k[(-2\pi ix)^lf(x)]\to \xi^k\left(\dfrac {\mathrm d}{\mathrm d\xi}\right)^l \hat f(\xi)

Theorem 5.1.8 If fR(R)|f|\in\mathcal R(\mathbb R), then f^\hat f is uniformly continuous in R\mathbb R.

Proof

Tips: Interpolation

It is the same to Property 5.1.2

# Fourier inversion

# The Gaussians as good kernels

Theorem 5.2.1 If f(x)=eπx2f(x)=e^{-\pi x^2}, then f^(ξ)=f(ξ)\hat f(\xi)=f(\xi)

Proof

Consider

ddξf^(ξ)=if(x)e2πixξdx\dfrac {\mathrm d}{\mathrm d\xi}\hat f(\xi)=i\int^\infty_{-\infty}f'(x)e^{-2\pi ix\xi}\mathrm dx

Remark

Note that f(x)=eπx2f(x)=e^{-\pi x^2} is of normalization

eπx2dx=1\int^\infty_{-\infty}e^{-\pi x^2}\mathrm dx=1

because of the Gaussian

ex2dx=π\int^\infty_{-\infty}e^{-x^2}\mathrm dx=\sqrt \pi

Corollary 5.2.2 If δ>0\delta>0 and Kδ(x)=δ1/2eπx2/δK_\delta(x)=\delta^{1/2}e^{-\pi x^2/\delta}, then K^δ(ξ)=eπδξ2\hat K_\delta(\xi)=e^{-\pi \delta \xi^2}.

We have now constructed a family of good kernels on the real line.

# Good kernel and convolution

Theorem 5.2.3 The collection {Kδ}δ>0\{K_\delta\}_{\delta>0} is a family of good kernels as δ0\delta \to 0. With

Kδ=δ1/2eπx2/δK_\delta=\delta^{-1/2}e^{-\pi x^2/\delta}

we have:

  1. For every δ>0\delta>0, we have

Kδ(x)dx=1\displaystyle \int^\infty_{-\infty}K_\delta(x)\mathrm dx=1

  1. δ>0,M>0\forall \delta>0, \exists M>0, we have

KδdxM\displaystyle \int ^\infty_{-\infty}|K_\delta|\mathrm dx\leq M

  1. For every η>0\eta>0, we have

x>ηKδ(x)dx0,asδ0\displaystyle \int_{|x|>\eta}|K_\delta(x)|\mathrm dx\to 0, \text{ as } \delta \to 0

Definition 5.2.4 If f,gS(R)f,g\in \mathcal S(\mathbb R), their convolution is defined by

(fg)(x)=f(xt)g(t)dt(f*g)(x)=\int^\infty_{-\infty}f(x-t)g(t)\mathrm dt

Corollary 5.2.5 If fS(R)f\in \mathcal S(\mathbb R), then

(fKδ)(x)f(x),asδ0(f*K_\delta)(x)\rightrightarrows f(x), \text{ as } \delta\to 0

# Fourier inversion

Property 5.2.6 If f,gS(R)f,g \in\mathcal S(\mathbb R), then we have the multiplication formula

f(x)g^(x)dx=f^(y)g(y)dy\int^\infty_{-\infty}f(x)\hat g(x)\mathrm dx=\int^\infty_{-\infty}\hat f(y)g(y)\mathrm dy

Proof

Tips: Uniformly convergence, Fubini

Uniformly convergence guarantees the interchange of the order of integration for double integrals.

Property 5.2.7 If f,gS(R)f,g\in\mathcal S(\mathbb R), then

  1. fgS(R)f*g\in \mathcal S(\mathbb R)
  2. fg=gff*g=g*f
  3. fg^(ξ)=f^(ξ)g^(ξ)\widehat{f*g}(\xi)=\hat f(\xi)\hat g(\xi)
Proof

Note that

supxxlg(xt)Al(1+tl)\sup_x|x|^l|g(x-t)|\leq A_l(1+|t|^l)

Theorem 5.2.7 If fS(R)f\in \mathcal S(\mathbb R), then

f(x)=f^(ξ)e2πixξdξf(x)=\int^\infty_{-\infty}\hat f(\xi)e^{2\pi ix\xi}\mathrm d\xi