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Properties of Fourier Transform 📂Fourier Analysis

Properties of Fourier Transform

Theorem[^1]

Let’s consider Ff,f^\cal{F}f, \hat{f} as the Fourier transform of ff. Let f∈L1f \in L^{1}. Then, the following properties hold for the Fourier transform:

  • (a) For any real number aa,

F[f(x−a)](ξ)=e−iaξf^(ξ)andF[eiaxf(x)](ξ)=f^(ξ−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 fδ(x):=1δf(xδ)f_\delta (x) := \frac{1}{\delta}f ( \frac{x}{\delta} ) for δ>0\delta >0. Then,

F[fδ](ξ)=(Ff)(δξ)andF[f(δx)](ξ)=(Ff)δ(ξ) \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)

F[f′](ξ)=iξFf(ξ) \mathcal{F} \left[ f^{\prime} \right] (\xi) = i \xi \mathcal{F} f (\xi)

Meanwhile, if xf(x)xf(x) is integrable,

F[xf(x)](ξ)=i(Ff)′(ξ) \mathcal{F} \left[ xf(x) \right] (\xi) = i ( \mathcal{F} f ) ^{\prime} (\xi)

  • (d) If g∈L1g\in L^{1},

F[f∗g](ξ)=f^(ξ)g^(ξ) \mathcal{F} \left[ f \ast g \right] (\xi)= \hat{f} (\xi) \hat{g}(\xi)

Here, f∗gf \ast g is the convolution of ff and gg.

  • (d’) For {fn}⊂L1\left\{ f_{n} \right\} \subset L^{1},

F[f1∗f2∗⋯∗fn]=f1^f2^⋯fn^ \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 δ\delta and taking the operation of δ_\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 iξi\xi.

Proof

(a)

F[f(x−a)](ξ)=∫f(x−a)e−iξxdx=∫f(y)e−iξ(y+a)dy=e−iaξ∫f(y)e−iξydy=e−iaξf^(ξ) \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=yx-a=y.

F[eiaxf(x)](ξ)=∫f(x)e−iξxeiaxdx=∫f(x)e−i(ξ−a)xdx=f^(ξ−a) \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.

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(b)

Similarly to (a), it can be easily proved.

F[fδ](ξ)=∫fδ(x)e−iξxdx=∫1δf(xδ)e−iξxdx=∫f(y)e−i(δξ)ydy=f^(δξ) \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 xδ=y\frac{x}{\delta}=y.

F[f(δx)](ξ)=∫f(δx)e−iξxdx=1δ∫f(y)e−i(ξ/δ)ydy=1δf^(ξ/δ)=f^δ(ξ)=(Ff)δ(ξ) \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 δx=y\delta x=y.

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(c)

First,

∫0∞f′(x)dx=lim⁡t→∞∫0tf′(x)dt=lim⁡t→∞f(t)−f(0) \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′∈L1f^{\prime} \in L^{1}, ∫f′(x)dx\displaystyle \int f^{\prime}(x)dx exists, and therefore lim⁡t→∞f(t)\lim \limits_{t \rightarrow \infty} f(t) exists. By assumption, because f∈L1f \in L^{1}, its value is 00. This is the same even when lim⁡t→−∞f(t)\lim \limits_{t \rightarrow -\infty}f(t), so

lim⁡x→±∞f(x)=0 \begin{equation} \lim \limits_{x \rightarrow \pm \infty} f(x)=0 \label{eq1} \end{equation}

Therefore,

F[f′](ξ)=∫f′(x)e−iξxdx=[e−iξxf(x)]−∞∞+iξ∫f(x)e−iξxdx=iξ∫f(x)e−iξxdx=iξf^(ξ) \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 (eq1)\eqref{eq1}.

F[xf(x)](ξ)=∫xf(x)e−iξxdx=iddξ∫f(x)e−iξxdx=iddξFf(ξ)=i(Ff)′(ξ) \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*}

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(d)

Considering the general definition of convolution, (d) is in fact a definition rather than a property.

F[f∗g](ξ)=∫(f∗g)(x)e−iξxdx=∫[∫f(x−y)g(y)dy]e−iξxdx=∫[∫f(x−y)g(y)dy]e−iξ(x−y)e−iξydx=∫∫f(x−y)g(y)e−iξ(x−y)e−iξydydx=∫∫f(x−y)g(y)e−iξ(x−y)e−iξydxdy=∫[∫f(x−y)e−iξ(x−y)dx]g(y)e−iξydy=∫[∫f(z)e−iξzdz]g(y)e−iξydy=∫f^(ξ)g(y)e−iξydy=f^(ξ)∫g(y)e−iξydy=f^(ξ)g^(ξ) \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=zx-y=z.

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(d')

Since convolution is associative, it immediately follows from (d).

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