📂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.

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