5.2. Properties of Fourier transform

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## Properties of Fourier transform

### Basic properties

In the previous Section 5.1 we introduced Fourier transform and Inverse Fourier transform \begin{align} & \hat{f}( k )= \frac{\kappa}{2\pi}\int_{-\infty}^\infty f(x)e^{-i k x}\,dx \tag{FT}\\ & \check{F}(x)= \frac{1}{\kappa} \int_{-\infty}^\infty F( k ) e^{i k x}\,d k \tag{IFT} \end{align} with $\kappa=1$ (but here we will be a bit more flexible):

Theorem 1. $F= \hat{f} \iff f=\check{F}$. (Already "proved"--formally)

Theorem 2.

1. Fourier transform: $f\mapsto \hat{f}$ is a linear operator $L^2(\mathbb{R},\mathbb{C})\to L^2(\mathbb{R},\mathbb{C})$;
2. Inverse Fourier transform: $F\mapsto \check{F}$ is an inverse operator (and also a linear operator) $L^2(\mathbb{R},\mathbb{C})\to L^2(\mathbb{R},\mathbb{C})$;
3. If $\kappa=\sqrt{2\pi}$ these operators are unitary i.e. preserve norm and an inner product: \begin{gather} \|f\|=\Bigl(\int_{\mathbb{R}} |f(x)|^2\,dx\Bigr)^{\frac{1}{2}}, \label{eq-5.2.1}\\ (f,g)= \int_{\mathbb{R}} f(x)\bar{g}(x)\,dx. \label{eq-5.2.2} \end{gather}

Proof. Easy. Preservation of inner product follows from preservation of norm.

Remark 1.

1. Here $L^2(\mathbb{R},\mathbb{C})$ is a space of square integrable complexample-valued functions. Accurate definition requires a measure theory (studied in the course of Real Analysis). Alternatively one can introduce this space as a closure of the set of square integrable continuous functions but it also require a certain knowledge of Real Analysis.
2. Properties [1] and [2] are obvious and [3] is due to Plancherel's theorem.
3. In Quantum Mechanics Fourier transform is sometimes referred as "going to $p$-representation" (a.k.a. momentum representation) and Inverse Fourier transform is sometimes referred as "going to $q$-representation" (a.k.a. coordinate representation). In this case $\pm i k x$ is replaced by $\pm i\hbar^{-1} k x$ and $2\pi$ by $2\pi\hbar$.

Theorem 3.

1. $g(x)=f(x-a)\implies \hat{g}( k )=e^{-i k a}\hat{f}( k )$;
2. $g(x)=f(x)e^{ibx}\implies \hat{g}( k )=\hat{f}( k -b)$;
3. $g(x)=f'(x) \implies \hat{g}( k )=i k \hat{f}( k )$;
4. $g(x)=xf (x) \implies \hat{g}( k )=i \hat{f}{}'( k )$;
5. $g(x)=f(\lambda x)\implies \hat{g}( k )=|\lambda|^{-1} \hat{f}(\lambda^{-1} k )$;

Proof. Here for brevity we do not write that all integrals are over $\mathbb{R}$ and set $\kappa=2\pi$.

1. $\displaystyle{\hat{g}=\int e^{-i k x}g(x)\,dx = \int e^{-i k x}f(x-a)\,dx= \int e^{-i k (x+a)}f(x)\,dx= e^{-i k a}\hat{f}( k )}$.

We replaced $x$ by $(x+a)$ in the integral.

2. $\displaystyle{\hat{g}=\int e^{-i k x}g(x)\,dx = \int e^{-i k x}e^{ibx}f(x)\,dx= \int e^{-i ( k -b)x}f(x)\,dx= \hat{f}( k -b)}$.

3. $\displaystyle{\hat{g}=\int e^{-i k x}g(x)\,dx = \int e^{-i k x}f'(x)\,dx \overset{\text{by parts}}= -\int \bigl(e^{-i k x}\bigr)'f(x)\,dx= i k \hat{f}( k )}$.

4. $\displaystyle{\hat{g}=\int e^{-i k x}g(x)\,dx = \int e^{-i k x}xf(x)\,dx= \int i\partial_ k \bigl(e^{-i k x}\bigr) \, f(x)\,dx= i\hat{f}{}'( k )}$.

5. $\displaystyle{\hat{g}=\int e^{-i k x}g(x)\,dx = \int e^{-i k x}f(\lambda x )\,dx = \int e^{-i k \lambda^{-1}x}f(x)\,|\lambda|^{-1}dx= |\lambda|^{-1}\hat{f}(\lambda^{-1} k )}$.

Here we replaced $x$ by $|\lambda|^{-1}x$ in the integral and $|\lambda|^{-1}$ is an absolute value of Jacobian.

Remark 2.

1. Differentiating [1] by $a$ and taking $a=0$ we get $-\widehat{(f')}(k)=-ik\hat{f}(k)$ (another proof of [3].
2. Differentiating [2] by $b$ and taking $b=0$ we get $i\widehat{(xf)}(k)=-\hat{f}'(k)$ (another proof of [4]).

Corollary 4. $f$ is even (odd) iff $\hat{f}$ is even (odd).

### Convolution

Definition 1. Convolution of functions $f$ and $g$ is a function $f *g$: $$(f*g)(x):=\int f(x-y)g(y)\,dy. \label{eq-5.2.3}$$

Theorem 4.

1. $h=f*g\implies \hat{h}( k )=\frac{2\pi}{\kappa}\hat{f}( k )\hat{g}( k )$;
2. $h(x)=f(x)g(x)$ $\implies \hat{h}=\kappa \hat{f} *\hat{g}$;

Proof. Again for brevity we do not write that all integrals are over $\mathbb{R}$.

1. Obviously \begin{equation*} \hat{h}(k)=\frac{\kappa}{2\pi} \int e^{-ix k }h(x)\,dx = \frac{\kappa}{2\pi} \iint e^{-ix k }f(x-y)g(y)\,dxdy; \end{equation*} replacing in the integral $x:=y+z$ we arrive to \begin{equation*} \frac{\kappa}{2\pi} \iint e^{-i(y+z) k }f(z)g(y)\,dzdy= \frac{\kappa}{2\pi} \int e^{-iz k }f(z)\,dz \times \int e^{-iy k }g(y)\,dz \end{equation*} which is equal to $\frac{2\pi}{\kappa}\hat{f}( k )\hat{g}( k )$.
2. Similarly, using inverse Fourier transform we prove that $\hat{f} *\hat{g}$ is a Fourier transform of $\frac{\kappa_1}{2\pi}fg$ where $\kappa_1=\frac{2\pi}{\kappa}$.

### Examples

Example 1.

1. Let f(x)=\left\{\begin{aligned} &e^{-\alpha x} &&x >0,\\ &0 &&x <0. \end{aligned}\right. Here $\Re( \alpha) >0$. \begin{equation*} \hat{f}( k )= \int_0^\infty e^{-(\alpha +i k )x}dx = -\frac{\kappa}{2\pi (\alpha +i k )}e^{-(\alpha +i k )x}\Bigr|_{x=0}^{x=\infty}= \frac{\kappa}{2\pi (\alpha +i k )}. \end{equation*}
2. Correspondingly, f(x)=\left\{\begin{aligned} &e^{\alpha x} &&x <0,\\ &0 &&x >0 \end{aligned}\right.\implies \hat{f}(k)= \dfrac{\kappa}{2\pi (\alpha +i k )}.

3. Then $f(x)=e^{-\alpha |x|}\implies \hat{f}(k)= \dfrac{\kappa\alpha}{\pi (\alpha^2 + k ^2 )}$.

Example 2. Let $f(x)= \frac{1}{\alpha ^2+x^2}$ with $\Re (\alpha)>0$. Then
$\hat{f}(k)=\kappa e^{-\alpha |k|}$. Indeed, using Example~1[3] we conclude that $f(x)$ is an inverse Fourier transform of $\kappa e^{-\alpha |k|}$ since we need only to take into account different factors and replace $i$ by $-i$ (for even/odd functions the latter step could be replaced by multiplication by $\pm 1$).

Using Complex Variables one can calculate it directly (residue should be calculated at $-\alpha i\operatorname{sign}(k)$).

Example 3. Let $f(x)=e^{-\frac{\alpha}{2}x^2}$ with $\Re(\alpha)\ge 0$. Here even for $\Re (\alpha)=0$ Fourier transform exists since integrals are converging albeit not absolutely.

Note that $f'=-\alpha x f$. Applying Fourier transform and Theorem 3 [3],[4] to the left and right expressions, we get $i k \hat{f}= -i\alpha \hat{f}'$; solving it we arrive to $\hat{f}=Ce^{-\frac{1}{2\alpha} k ^2}$.

To find $C$ note that $C=\hat{f}(0)= \frac{\kappa}{2\pi}\int e^{-\frac{\alpha}{2}x^2}\,dx$ and for real $\alpha >0$ we make a change of variables $x=\alpha^{-\frac{1}{2}}z$ and arrive to $C=\frac{\kappa}{\sqrt{2\pi \alpha}}$ because $\int e^{-z^2/2}\,dz=\sqrt{2\pi}$. Therefore \begin{equation*} \hat{f}( k )= \frac{\kappa}{\sqrt{2\pi\alpha}}e^{-\frac{1}{2\alpha} k ^2}. \end{equation*}

Remark 3. Knowing Complex Variables one can justify it for complex $\alpha$ with $\Re(\alpha)\ge 0$; we take a correct branch of $\sqrt{\alpha}$ (condition $\Re(\alpha)\ge 0$ prevents going around origin).

In particular, $(\pm i )^{\frac{1}{2}}=e^{\pm \frac{i\pi}{4}}=\frac{1}{\sqrt{2}}(1\pm i)$ and therefore for $\alpha=\pm i\beta$ with for $\beta >0$ we get $f=e^{\mp\frac{i\beta}{2 }x^2}$ and \begin{equation*} \hat{f}( k )=\frac{\kappa}{\sqrt{2\pi\beta}}e^{\pm \frac{\pi i}{4}} e^{\pm\frac{i}{2\beta} k ^2} =\frac{\kappa}{2\sqrt{\pi\beta}} (1\mp i)e^{\pm\frac{i}{2\beta} k ^2}. \end{equation*}

### Poisson summation formula

Theorem 5. Let $f(x)$ be a continuous function on the line $(-\infty,\infty)$ which vanishes for large $|x|$. Then for any $a>0$ $$\sum_{n=-\infty}^\infty f(an) = \sum_{n=-\infty}^\infty \frac{2\pi }{a}\hat{f}(\frac{2\pi }{a}n) . \label{eq-5.2.4}$$

Proof. Observe that function \begin{equation*} g(x)= \sum_{n=-\infty}^\infty f(x+an) \end{equation*} is periodic with period $a$. Note that the Fourier coefficients of $g(x)$ on the interval $(-\frac{a}{2}, \frac{a}{2})$ are $b_m=\frac{2\pi}{a}\hat{f}(\frac{2\pi}{a})$, where $\hat{f}(k)$ is the Fourier transform of $f(x)$.

Finally, in the Fourier series of $g(x)$ on $(-\frac{a}{2}, \frac{a}{2})$ plug $x = 0$ to obtain $g(0)=\sum_m b_m$ which coincides with (\ref{eq-5.2.4}).

### Final remarks

Remark 4. 1. Properties of Multidimensional Fourier transform and Fourier integral are discussed in Subsection 5.2.A. 2. It is very important to do all problems from Subsection 5.2.P: instead of calculating Fourier transforms directly you use Theorem 3 to expand the "library'' of Fourier transforms obtained in Examples 1--3.