Properties of Fourier Transform
Theorem[^1]
Let’s consider $\cal{F}f, \hat{f}$ as the Fourier transform of $f$. Let $f \in L^{1}$. Then, the following properties hold for the Fourier transform:
- (a) For any real number $a$,
$$ \mathcal{F} \left[ f(x-a) \right] ( \xi ) = e^{-ia\xi}\hat{f}(\xi) \quad \mathrm{and} \quad \mathcal{F} \left[ e^{iax}f(x)\right] (\xi) = \hat{f}(\xi-a) $$
- (b) Define $f_\delta (x) := \frac{1}{\delta}f ( \frac{x}{\delta} )$ for $\delta >0$. Then,
$$ \mathcal{F}\left[ f_\delta \right] (\xi ) = (\mathcal{F}f)(\delta \xi) \quad \mathrm{and} \quad \mathcal{F} \left[ f(\delta x) \right] (\xi) = ( \mathcal{F} f ) _{\delta} (\xi) $$
- (c) Assume $f$ is continuous and piecewise smooth. Then,
$$ \mathcal{F} \left[ f^{\prime} \right] (\xi) = i \xi \mathcal{F} f (\xi) $$
Meanwhile, if $xf(x)$ is integrable,
$$ \mathcal{F} \left[ xf(x) \right] (\xi) = i ( \mathcal{F} f ) ^{\prime} (\xi) $$
- (d) If $g\in L^{1}$,
$$ \mathcal{F} \left[ f \ast g \right] (\xi)= \hat{f} (\xi) \hat{g}(\xi) $$
Here, $f \ast g$ is the convolution of $f$ and $g$.
- (d’) For $\left\{ f_{n} \right\} \subset L^{1}$,
$$ \mathcal{F}\left[ f_{1} \ast f_{2} \ast \cdots \ast f_{n} \right]=\hat{f_{1}} \hat{f_{2}} \cdots \hat{f_{n}} $$
Explanation
(a) This means the operations of translation and multiplication by an exponential function switch under the transformation. Translating and then transforming multiplies by an exponential function, and multiplying by an exponential function and then transforming results in translation. (b) Similarly, multiplying the variable by $\delta$ and taking the operation of $_\delta$ on the function switch under the transformation. (c) The derivative’s Fourier transform is the same as multiplying the Fourier transform by a constant $i\xi$.
Proof
(a)
$$ \begin{align*} \mathcal{F} \left[ f(x-a) \right] (\xi) &= \int f(x-a)e^{-i\xi x} dx \\ &= \int f(y)e^{-i\xi (y+a)} dy \\ &= e^{-ia\xi} \int f(y)e^{-i\xi y}dy \\ &= e^{-ia\xi} \hat{f}(\xi) \end{align*} $$
The second equality holds by substitution with $x-a=y$.
$$ \begin{align*} \mathcal{F}\left[ e^{iax}f(x) \right] (\xi) &= \int f(x)e^{-i\xi x}e^{iax} dx \\ &= \int f(x) e^{-i(\xi-a)x}dx \\ &= \hat{ f }(\xi-a) \end{align*} $$
The third equality follows from the definition of the Fourier transform.
■
(b)
Similarly to (a), it can be easily proved.
$$ \begin{align*} \mathcal{F} \left[ f_\delta \right] (\xi) &= \displaystyle {\int} f_\delta (x) e^{-i\xi x} dx \\ &= {\displaystyle \int} \dfrac{1}{\delta}f \left( \frac{x}{\delta} \right)e^{-i\xi x} dx \\ &= \displaystyle{ \int} f(y) e^{-i(\delta \xi )y} dy \\ &= \hat{f}(\delta\xi) \end{align*} $$
The second equality holds by substitution with $\frac{x}{\delta}=y$.
$$ \begin{align*} \mathcal{F} \left[ f(\delta x) \right] (\xi) &= \displaystyle{ \int} f(\delta x)e^{-i\xi x}dx \\ &= \dfrac{1}{\delta} \displaystyle{ \int} f(y)e^{-i(\xi / \delta)y} dy \\ &= \dfrac{1}{\delta} \hat{f} ( \xi / \delta) \\ &= \hat{f}_{\delta}(\xi) = ( \mathcal{F }f )_{\delta} (\xi) \end{align*} $$
The third equality holds by substitution with $\delta x=y$.
■
(c)
First,
$$ \int_{0}^\infty f^{\prime}(x)dx=\lim \limits_{t \rightarrow \infty} \int_{0}^tf^{\prime}(x)dt=\lim \limits_{t \rightarrow \infty} f(t)-f(0) $$
and since $f^{\prime} \in L^{1}$, $\displaystyle \int f^{\prime}(x)dx$ exists, and therefore $\lim \limits_{t \rightarrow \infty} f(t)$ exists. By assumption, because $f \in L^{1}$, its value is $0$. This is the same even when $\lim \limits_{t \rightarrow -\infty}f(t)$, so
$$ \begin{equation} \lim \limits_{x \rightarrow \pm \infty} f(x)=0 \label{eq1} \end{equation} $$
Therefore,
$$ \begin{align*} \mathcal{F} \left[ f^{\prime} \right] (\xi) &= \int f^{\prime}(x)e^{-i\xi x} dx \\ &= \left[ e^{-i\xi x} f(x)\right]_{-\infty}^\infty + i\xi \int f(x) e^{-i \xi x} dx \\ &= i \xi \int f(x) e^{-i\xi x}dx \\ &= i \xi \hat{f}(\xi) \end{align*} $$
The second equality follows from partial integration. The third equality holds by $\eqref{eq1}$.
$$ \begin{align*} \mathcal{F} \left[ xf(x) \right] (\xi) &= \int x f(x)e^{-i \xi x}dx \\ &= i\dfrac{d}{d\xi} \int f(x) e^{-i \xi x}dx \\ &= i\dfrac{d}{d \xi} \mathcal{F} f (\xi) \\ &= i (\mathcal{F} f )^{\prime}(\xi) \end{align*} $$
■
(d)
Considering the general definition of convolution, (d) is in fact a definition rather than a property.
$$ \begin{align*} \mathcal{F} \left[ f \ast g \right] (\xi) &= \int (f \ast g)(x)e^{-i \xi x}dx \\ &= \int \left[ \int f(x-y)g(y)dy\right]e^{-i \xi x}dx \\ &= \int \left[ \int f(x-y)g(y)dy\right]e^{-i \xi (x-y)}e^{-i\xi y}dx \\ &= \int \int f(x-y)g(y)e^{-i \xi (x-y)}e^{-i\xi y}dydx \\ &= \int \int f(x-y)g(y)e^{-i \xi (x-y)}e^{-i\xi y}dxdy \\ &= \int \left[ \int f(x-y)e^{-i \xi (x-y)}dx \right] g(y)e^{-i \xi y} dy \\ &= \int \left[ \int f(z)e^{-i \xi z}dz \right] g(y)e^{-i \xi y} dy \\ &= \int \hat{f}(\xi) g(y)e^{-i \xi y} dy \\ &= \hat{f}(\xi)\int g(y)e^{-i \xi y} dy \\ &= \hat{f}(\xi) \hat{g}(\xi) \end{align*} $$
The seventh equality holds by substitution with $x-y=z$.
■
(d')
Since convolution is associative, it immediately follows from (d).
■