Operators as Fourier Transforms 📂Lebesgue Spaces

Operators as Fourier Transforms

Definition¹

The Fourier Transform of the function $f$

$\widehat{f} (\gamma ) := \int_{\mathbb{R}} f(x) e^{-2 \pi i x \gamma} dx, \quad \gamma \in \mathbb{R}$

can also be represented as the following operator $\mathcal{F}$ .

$(\mathcal{F} f) (\gamma ) := \widehat{f} ( \gamma )$

Description

Fourier Transform is widely used throughout analysis and the two expressions $\widehat{f}$ and $\mathcal{F} f$ essentially do not differ, but there is a slight nuance in the use of symbols. When emphasis is on practical calculation, formulas, and quick notation, $\widehat{f}$ is preferred, and when the properties and order of operations as an operator are important, $\mathcal{F}$ is preferred.

Let us define $f,g \in L^{1}$ .

For $a \in \mathbb{R}$

$\mathcal{F} T_{a} = E_{-a} \mathcal{F}$

For $b \in \mathbb{R}$

$\mathcal{F} E_{b} = T_{b} \mathcal{F}$

For $c \ne 0$

$\mathcal{F} D_{c} = D_{1/c} \mathcal{F}$

Convolution:

$\mathcal{F} ( f \ast\ g) = (\mathcal{F} f \cdot \mathcal{F} g)$

1~3: $T_{a}, E_{b}, D_{c}$ refers to Translation, Modulation, Dilation.

4: $\ast$ is Convolution convolution, and $\cdot$ simply means the product of functions. In other words, for $\gamma \in \mathbb{R}$

$\widehat{f \ast\ g} (\gamma) = \widehat{f} (\gamma) \widehat{g} (\gamma)$

Let $f , g \in L^{2}$ .

Norm: $\left\| \mathcal{F} f \right\|_{2} = \left\| f \right\|_{2}$
Inner Product:

$\langle \mathcal{F} f , \mathcal{F} g \rangle = \langle f , g \rangle$

The above properties are well-known ones in Fourier analysis, represented again in the language of operator theory.

Proof

Refer to here for the proofs of 1~4.

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