📂Lebesgue Spaces

Translation in L2 Spaces: Translations, Modulations, Dilations

Definition¹

• Translation is defined as $T_{a} : L^{2} \to L^{2}$ for $a \in \mathbb{R}$ as follows.

$$ \left( T_{a} f \right) (x) := f(x-a) $$

• Modulation is defined as $E_{b} : L^{2} \to L^{2}$ for $b \in \mathbb{R}$ as follows.

$$ \left( E_{b} f \right) (x) := e^{2 \pi i b x} f(x) $$

• Dilation is defined as $D_{c} : L^{2} \to L^{2}$ for $c > 0$ as follows.

$$ \left( D_{c} f \right) (x) := {{ 1 } \over { \sqrt{c} }} f \left( {{ x } \over { c }} \right) $$

Explanation

The linear operators mentioned above are commonly used in the space of $L^{2}$. Although in Korean they are translated to translation, modulation, and dilation respectively, it might be easier to understand them directly in English due to the mathematical context.

The $e^{2 \pi i b x}$ multiplied in modulation is literally an abstracted rotation.

The $\displaystyle {{ 1 } \over { \sqrt{c} }}$ multiplied in dilation can be seen as being squared to match the norm $\left\| \cdot \right\|_{2}$. Especially for $c = 1/2$, $D$ defined as follows plays a special role.

$$ ( D f ) (x) := \sqrt{2} f (2x) $$

$D$ is written as follows for convenience with respect to $j \in \mathbb{Z}$.

$$ ( D^{j} f ) (x) := \sqrt{2}^{j} f \left( 2^{j} x \right) $$

Properties

For all $a, b \in \mathbb{R}$, $c > 0$, and $f,g \in L^{1}$,

1. $T_{a} , E_{b}, D_{c}$ is a bounded linear operator.

2. Inverse operator: $T_{a} , E_{b}, D_{c}$ is unitary.

3. Commutation relations:

$$ (T_{a} E_{b} f ) (x) = e^{- 2 \pi i b a} (E_{b} T_{a} f ) (x) \\ (T_{a} D_{c} f ) (x) = (D_{c} T_{a/c} f ) (x) \\ (D_{c} E_{b} f ) (x) = (E_{b/c} D_{c} f ) (x) $$

• Relation with Fourier transform:

  $$ \mathcal{F} T_{a} = E_{-a} \mathcal{F} \\ \mathcal{F} E_{b} = T_{b} \mathcal{F} \\ \mathcal{F} D_{c} = D_{1/c} \mathcal{F} $$

  Regarding $D$, the following can be obtained about $j, k \in \mathbb{Z}$ as corollaries of the above theorems.

  $$ T_{k} D^{j} = D^{j} T_{2^{j} k } \\ D^{j} T_{k} = T_{2^{-j}k} D^{j} \\ \left( D^{j} \right)^{ \ast } = D^{-j} $$

Proof

1.

• Part 1. Linear

  For all $f,g \in L^{2}$ and $\alpha , \beta \in \mathbb{C}$,

  $$ \begin{align*} T_{a} \left( \alpha f + \beta g \right)(x) =& \left( \alpha f + \beta g \right)(x-a) \\ =& \alpha f (x-a) + \beta g (x-a) \\ =& \alpha T_{a} f (x) + \beta T_{a} g (x) \end{align*} $$

  Therefore, $T_{a}$ is linear.

  $$ \begin{align*} E_{b} \left( \alpha f + \beta g \right)(x) =& e^{ 2 \pi i b x } \left( \alpha f + \beta g \right)(x) \\ =& \alpha e^{ 2 \pi i b x } f (x) + \beta e^{ 2 \pi i b x } g (x) \\ =& \alpha E_{b} f (x) + \beta E_{b} g (x) \end{align*} $$

  Therefore, $E_{b}$ is linear.

  $$ \begin{align*} D_{c} \left( \alpha f + \beta g \right)(x) =& {{ 1 } \over { \sqrt{c} }} \left( \alpha f + \beta g \right) \left( {{ x } \over { c }} \right) \\ =& \alpha {{ 1 } \over { \sqrt{c} }} f (x) + \beta {{ 1 } \over { \sqrt{c} }} g (x) \\ =& \alpha D_{c} f (x) + \beta D_{c} g (x) \end{align*} $$

  Therefore, $D_{c}$ is linear.

• Part 2. Bounded

  Substitute as $t := x - a$,

  $$ \begin{align*} \left\| T_{a} f \right\|_{2} =& \int_{-\infty}^{\infty} \left| T_{a} f \left( x \right) \right|^{2} dx \\ =& \int_{-\infty}^{\infty} \left| f \left( x - a \right) \right|^{2} dx \\ =& \int_{-\infty}^{\infty} \left| f \left( t \right) \right|^{2} dt \\ =& \left\| f \right\|_{2} \end{align*} $$

  Therefore, $T_{a}$ is bounded. Since $\left| e^{2 \pi i b x } \right| =1$,

  $$ \begin{align*} \left\| E_{b} f \right\|_{2} =& \int_{-\infty}^{\infty} \left| E_{b} f \left( x \right) \right|^{2} dx \\ =& \int_{-\infty}^{\infty} \left| e^{2 \pi i b x } f \left( x \right) \right|^{2} dx \\ =& \int_{-\infty}^{\infty} 1 \cdot \left| f \left( t \right) \right|^{2} dt \\ =& \left\| f \right\|_{2} \end{align*} $$

  Therefore, $E_{b}$ is bounded. Substitute as $t := x/c$,

  $$ \begin{align*} \left\| D_{c} f \right\|_{2} =& \int_{-\infty}^{\infty} \left| D_{c} f \left( x \right) \right|^{2} dx \\ =& \int_{-\infty}^{\infty} \left| {{ 1 } \over { \sqrt{c} }} f \left( {{ x } \over { c }} \right) \right|^{2} dx \\ =& \int_{-\infty}^{\infty} {{ 1 } \over { c }} \left| f \left( t \right) \right|^{2} c dt \\ =& \int_{-\infty}^{\infty} \left| f \left( t \right) \right|^{2} dt \\ =& \left\| f \right\|_{2} \end{align*} $$

  Therefore, $D_{c}$ is bounded.

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