5.2. Properties of Fourier transform

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Properties of Fourier transform

  1. Basic properties
  2. Convolution
  3. Examples
  4. Poisson summation formula

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)

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 (aka momentum representation) and Inverse Fourier transform is sometimes referred as going to $q$-representation (aka 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. $\hat{g}=\int e^{-i k x}g(x)\,dx = \int e^{-i k x}f(x-a)\,dx= \int e^{-i k (x+a)}f(x)\,dx= e^{-i k a}\hat{f}( k )$. We replaced $x$ by $(x+a)$ in the integral.
  2. $\hat{g}=\int e^{-i k x}g(x)\,dx = \int e^{-i k x}e^{ibx}f(x)\,dx= \int e^{-i ( k -b)x}f(x)\,dx= \hat{f}( k -b)$.
  3. $\hat{g}=\int e^{-i k x}g(x)\,dx = \int e^{-i k x}f'(x)\,dx \overset{\text{by parts}}= -\int \bigl(e^{-i k x}\bigr)'f(x)\,dx= i k \hat{f}( k )$.
  4. $\hat{g}=\int e^{-i k x}g(x)\,dx = \int e^{-i k x}xf(x)\,dx= \int i\partial_ k \bigl(e^{-i k x}\bigr) \, f(x)\,dx= i\hat{f}{}'( k )$.
  5. $\hat{g}=\int e^{-i k x}g(x)\,dx = \int e^{-i k x}f(\lambda x )\,dx = \int e^{-i k |\lambda|^{-1}x}f(x)\,\lambda^{-1}dx= \lambda^{-1}\hat{f}(\lambda^{-1} k )$. Here we replaced $x$ by $\lambda^{-1}x$ in the integral and $|\lambda|^{-1}$ is an absolute value of Jacobian.

Corollary 1.] $f$ is even (odd) iff $\hat{f}$ is even (odd).


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

Theorem 4.

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


  1. \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 $\hat{f} *\hat{g}$ is a Fourier transform of $\frac{\kappa_1}{2\pi}fg$ where $\kappa_1=\frac{2\pi}{\kappa}$.


Example 1. Let $f(x)=e^{-\alpha x}$ as $x >0$ and $f(x)=0$ as $x <0$. Here $\Re \alpha >0$. \begin{equation*} \hat{f}( k )= \int_0^\infty e^{-(\alpha +i k )x}\,dx = -(\alpha +i k )^{-1}e^{-(\alpha +i k )x}\bigr|_{x=0}^{x=\infty}= (\alpha +i k )^{-1} \end{equation*} provided $\kappa=2\pi$.

In the general case $\hat{f}( k )= \frac{\kappa}{2\pi}(\alpha+i k )^{-1}$.

Example 2. 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*} 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}{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 $a>0$ \begin{equation} \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} \end{equation}

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}).

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