📂Lebesgue Spaces

Translation, Modulation, and Dilation Commutation Relations in L2 Spaces

Theorem¹

For all $a, b \in \mathbb{R}$ and $c > 0$, $T_{a}, E_{b}, D_{c}$ has the following relationship:

$$ \begin{equation} (T_{a} E_{b} f ) (x) = e^{- 2 \pi i b a} (E_{b} T_{a} f ) (x) \end{equation} $$

$$ \begin{equation} (T_{a} D_{c} f ) (x) = (D_{c} T_{a/c} f ) (x) \end{equation} $$

$$ \begin{equation} (D_{c} E_{b} f ) (x) = (E_{b/c} D_{c} f ) (x) \end{equation} $$

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

Proof

(1)

$$ \begin{align*} (T_{a} E_{b} f ) (x) =& T_{a} \left( e^{2 \pi i b x} f(x) \right) \\ =& e^{2 \pi i b (x-a)} f(x-a) \\ =& e^{2 \pi i b (-a)} e^{2 \pi i b x} f(x-a) \\ =& e^{- 2 \pi i b a} (E_{b} T_{a} f ) (x) \end{align*} $$

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(2)

$$ \begin{align*} (T_{a} E_{b} f ) (x) =& T_{a} \left( e^{2 \pi i b x} f(x) \right) \\ =& e^{2 \pi i b (x-a)} f(x-a) \\ =& e^{2 \pi i b (-a)} e^{2 \pi i b x} f(x-a) \\ =& e^{- 2 \pi i b a} (E_{b} T_{a} f ) (x) \end{align*} $$

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(3)

$$ \begin{align*} (T_{a} E_{b} f ) (x) =& T_{a} \left( e^{2 \pi i b x} f(x) \right) \\ =& e^{2 \pi i b (x-a)} f(x-a) \\ =& e^{2 \pi i b (-a)} e^{2 \pi i b x} f(x-a) \\ =& e^{- 2 \pi i b a} (E_{b} T_{a} f ) (x) \end{align*} $$

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