📂Lebesgue Spaces

Translation, Modulation, and Dilation Operators in L2 Space

Theorem¹

$T_{a}, E_{b}, D_{c}$ is unitary, and its inverse operator is as follows.

$$ T_{a}^{-1} = T_{-a} = \left( T_{a} \right)^{ \ast } $$

$$ E_{b}^{-1} = E_{-b} = \left( E_{b} \right)^{ \ast } $$

$$ D_{c}^{-1} = D_{1/c} = \left( D_{c} \right)^{ \ast } $$

Here, $T_{a}, E_{b}, D_{c}$ is defined respectively in $L^{2}$ as translation, modulation, dilation.

Proof

Translation

By substituting with $t := x - a$,

$$ \begin{align*} \langle T_{a} f , g \rangle =& \int_{-\infty}^{\infty} f \left( x - a \right) \overline{g \left( x \right)} dx \\ =& \int_{-\infty}^{\infty} f \left( t \right) \overline{g \left( t + a \right)} dt \\ =& \langle f , T_{-a} g \rangle \\ =& \langle f , T_{a}^{ \ast } g \rangle \end{align*} $$

since $T_{-a} = T_{a}^{ \ast }$,

$$ T_{a} T_{a}^{ \ast } = T_{a} T_{-a} = I = T_{-a} T_{a} = T_{a}^{ \ast } T_{a} $$

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Modulation

Since $\displaystyle e^{2 \pi i b x} = \overline{e^{-2 \pi i b x}}$,

$$ \begin{align*} \langle E_{b} f , g \rangle =& \int_{-\infty}^{\infty} e^{2 \pi i b x} f \left( x \right) \overline{g \left( x \right)} dx \\ =& \int_{-\infty}^{\infty} f \left( x \right) \overline{e^{-2 \pi i b x}} \overline{ g \left( x \right)} \\ =& \int_{-\infty}^{\infty} f \left( x \right) \overline{ e^{2 \pi i (-b) x} g \left( x \right)} \\ =& \langle f , E_{-b} g \rangle \\ =& \langle f , E_{b}^{ \ast } g \rangle \end{align*} $$

since $E_{-b} = E_{b}^{ \ast }$,

$$ E_{b} E_{b}^{ \ast } = E_{b} E_{-b} = I = E_{-b} E_{b} = E_{b}^{ \ast } E_{b} $$

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Dilation

By substituting with $\displaystyle t := {{ x } \over { c }}$,

$$ \begin{align*} \langle D_{c} f , g \rangle =& \int_{-\infty}^{\infty} {{ 1 } \over { \sqrt{c} }} f \left( {{ x } \over { c }} \right) \overline{g \left( x \right)} dx \\ =& \int_{-\infty}^{\infty} f \left( t \right) {{ 1 } \over { \sqrt{c} }} \overline{g \left( ct \right)} c dt \\ =& \int_{-\infty}^{\infty} f \left( t \right) \overline{ \sqrt{c} g \left( ct \right)} c dt \\ =& \langle f , D_{1/c} g \rangle \\ =& \langle f , D_{c}^{ \ast } g \rangle \end{align*} $$

since $D_{1/c} = D_{c}^{ \ast }$,

$$ D_{c} D_{c}^{ \ast } = D_{c} D_{1/c} = I = D_{1/c} D_{c} = D_{c}^{ \ast } D_{c} $$

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