Can you recover a distribution from mollification?

Question

Let $f\in \mathcal S'(\mathbb R)$ be a Schwartz distribution. Given $\rho \in C^\infty_c(\mathbb R)$ define the convolution as the function $$x\mapsto (f\ast\rho)(x):=\langle f, \rho (\cdot -x)\rangle.$$

I believe this function should be smooth. Anyway my question is: suppose further that $\rho\geq 0$ and $\int \rho = 1$ and for $\varepsilon>0$, define $$\rho_\varepsilon(x):=\frac{1}{\varepsilon}\rho(x\varepsilon^{-1}).$$ Then is it the case that $f\ast\rho_\varepsilon\to f$ in $\mathcal S'(\mathbb R)$ as $\varepsilon\to 0$?

Answer 1

To answer your question, yes. For a proof, first require a few prerequisites. From F. G. Friedlander and M. Joshi's Introduction to the Theory of Distributions, we have

Theorem 1.2.1. Let $f \in C_c^k(\mathbb R)$, let $\rho \in C_c^\infty(\mathbb R)$ be such that $\rho \geq 0$ and $\int \rho = 1$, let $\rho_\varepsilon = \varepsilon^{-1} \rho( x / \varepsilon)$, and let $f_\varepsilon = f * \rho_\varepsilon$. Then $\partial^\alpha f_\varepsilon \to \partial^\alpha f$ uniformly as $\varepsilon \to 0$ for all $|\alpha| \leq k$.

We borrow the notation $$ f_\varepsilon = f * \rho_\varepsilon $$ for your tempered distribution $f$. We also define $\mathcal D(\mathbb R) = C_c^\infty(\mathbb R)$ to be a set of "test functions" with dual space "the distributions" $\mathcal D'(\mathbb R)$. The relation to the Schwartz functions $\mathcal S(\mathbb R)$ and the tempered distributions $\mathcal S'(\mathbb R)$ is $$ \mathcal D \subset \mathcal S \subset \mathcal S' \subset \mathcal D'; $$ it's true that

Theorem. $\mathcal D$ is dense in $\mathcal D'$.

from which it follows that all inclusions are dense.

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We make two key observations from the above two theorems before finally attacking the proof.

• $f \in \mathcal S'(\mathbb R) \subset \mathcal D'(\mathbb R)$, and since $\mathcal D(\mathbb R) = C_c^\infty(\mathbb R)$ is dense in $\mathcal D'(\mathbb R)$, there exists a sequence of $\psi_n \in \mathcal D(\mathbb R)$ such that $$ \text{$\psi_n \to f \qquad$ in $\mathcal S'(\mathbb R)$ as $n\to\infty$.} \tag{1} $$ It follows that $$ \text{for each $\varepsilon > 0, \qquad \psi_n * \rho_\varepsilon \to f_\varepsilon\qquad$ in $\mathcal S'(\mathbb R)$ as $n\to\infty$.} \tag{2} $$
• By Theorem 1.2.1, for each fixed $n$, we have $\partial^\beta (\psi_n * \rho_\varepsilon) \to \partial^\beta \psi_n$ uniformly as $\varepsilon \to 0$ for all multi-indices $\beta$. Since $\psi_n$ and $\rho_\varepsilon$ are compactly supported, it follows that for each $n$, $\| \psi_n * \rho_\varepsilon - \psi_n \|_{\alpha, \beta} \to 0$ for all multi-indices $\alpha, \beta$ as $\varepsilon \to 0$. The same conclusion follows when considered as tempered distributions: $$ \text{for each $n$, $\qquad \psi_n * \rho_\varepsilon \to \psi_n\qquad$ in $\mathcal S'(\mathbb R)$ as $\varepsilon \to 0$.} \tag{3} $$

Finally, $f_\varepsilon \to f$ in $\mathcal{S}'(\mathbb R)$ as $\varepsilon \to 0$ iff for all $\varphi \in \mathcal S(\mathbb R)$, we have $\langle f_\varepsilon, \varphi \rangle \to \langle f, \varphi \rangle$ as $\varepsilon \to 0$. To prove this is the case, let $\varepsilon' > 0$. Then

\begin{align*} f - f_\varepsilon &= f - \psi_n + \psi_n - f_\varepsilon \\ &= (f - \psi_n) + (\psi_n - \psi_n * \rho_\varepsilon) + (\psi_n * \rho_\varepsilon - f_\varepsilon) \\ \implies \left| \left\langle f - f_\varepsilon, \varphi \right\rangle \right| & \leq \left| \left\langle f - \psi_n, \varphi \right\rangle \right| + \left| \left\langle \psi_n - \psi_n * \rho_\varepsilon, \varphi \right\rangle \right| + \left| \left\langle \psi_n * \rho_\varepsilon - f_\varepsilon, \varphi \right\rangle \right| \end{align*} Regardless of $\varepsilon$, we may bound the first and third terms of the latter expression by $\varepsilon'$ if we choose $n$ sufficiently large according to $(1)$ and $(2)$. This gives us $$ \left| \left\langle f - f_\varepsilon, \varphi \right\rangle \right| \leq 2 \varepsilon' + \left| \left\langle \psi_n - \psi_n * \rho_\varepsilon, \varphi \right\rangle \right|, $$ and then applying $(3)$, we have $$ \left| \left\langle f - f_\varepsilon, \varphi \right\rangle \right| \leq 3 \varepsilon' $$ for sufficiently small $\varepsilon$, which proves the claim.