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Distributional Convolution Lemma 📂Distribution Theory

Distributional Convolution Lemma

Theorem1

Let $F$ be a distribution, and $\phi,\psi$ be a test function. Then $F \ast \phi$ is a function defined in the real space and is locally integrable. Therefore, there exists a corresponding regular distribution $T$ as follows:

$$ T_{F \ast \phi}(\psi)=F(\tilde{\phi} \ast \psi) $$

Here, $\tilde{\phi}(x)=\phi (-x)$.

Description

The name ‘distribution convolution lemma’ is arbitrarily given as there’s no specific name attached to the content above.

Proof

  • Case 1. If $F$ is a regular distribution

    There exists a corresponding $f \in L_{\mathrm{loc}}^{1}$ for $F$.

    $$ F (\phi) = F_{f} (\phi) = \int f(x)\phi (x) dx $$

    Therefore, the following equation holds.

    $$ \begin{align*} T_{F \ast \phi}(\psi) &= \int (F*\phi)(x)\psi (x)dx \\ &=\int F(\tilde{\phi}_{x})\psi (x) dx \\ &= \int \int f(y)\tilde{\phi}(y-x)dy\psi (x)dx \\ &= \int f(y)\int\tilde{\phi}(y-x)\psi (x)dxdy \\ &= \int f(y)(\tilde{\phi} \ast \psi)(y)dy \\ &= F(\tilde{\phi} \ast \psi) \end{align*} $$

  • Case 2. If $F$ is not a regular distribution

    Since $\tilde{\phi}, \psi \in C^{\infty}$, $\tilde{\phi} \ast \psi$ is Riemann integrable. Then, it is possible to approximate the integral with an infinite series as follows.

    $$ \tilde{\phi} \ast \psi (y)= \int \tilde{\phi}(x-y)\psi (x)dy=\lim \limits_{n\to \infty} \sum \limits _{i=1} ^{n}\phi (x_{i}-y)\psi_ {x_{i}}\Delta x_{i} $$

    Therefore, the following holds:

    $$ \begin{align*} F(\tilde{\phi} \ast \psi) &= F \left( \lim \limits_{n\to \infty} \sum \limits _{i=1} ^{n}\phi (x_{i}-\cdot)\psi_{x_{i}}\Delta x_ {i} \right) \\ &=\lim \limits_{n\to \infty} F \left( \sum \limits _{i=1} ^{n}\phi (x_{i}-\cdot)\psi_{x_{i}}\Delta x_{i} \right) \\ &=\lim \limits_{n\to \infty} \sum \limits _{i=1} ^{n}F \left( \phi (x_{i}-\cdot) \right)\psi_{x_{i}}\Delta x_{i} \\ &=\lim \limits_{n\to \infty} \sum \limits _{i=1} ^{n} (F \ast \phi)(x_{i}) \ast \psi_{x_{i}}\Delta x_{i} \\ &= \int F \ast \phi (x)\psi (x)dx \\ &=T_{F \ast \phi}(\psi) \end{align*} $$

    The second equality is due to the continuity of the distribution, and the third equality is due to its linearity.


  1. Gerald B. Folland, Fourier Analysis and Its Applications (1992), p318 ↩︎