Let be $\{ \varphi_m \}$ a sequence of mollifiers in $\mathbb{R}^n$ and $u \in D'(\mathbb{R}^n)$. Show that $\varphi_m \ast u \rightarrow u$ in $D'(\mathbb{R}^n)$.

I would like to know if what I did is right, because I know that if $u \in L_{loc}^1(\mathbb{R}^n)$, then I can see $u$ as a distribution in the following way:

$$\langle u, \psi \rangle := \int_{\mathbb{R}^n} u(x)\psi(x) dx,$$

but I only know that $u \in D'(\mathbb{R}^n)$, then I don't sure if I can see the distribution $u$ as a function which is associated to a distribution as defined above because not every distribution is in $L_{loc}^1(\mathbb{R}^n)$. Thus, I would like to know if makes sense $u(x)$ in the what I did. I think this exercise is not difficult, but I'm confuse if makes sense $u(x)$.

This is what I did:

Let be $\psi \in D(\mathbb{R}^n)$.

$\begin{align*} |\left\langle \varphi_m \ast u - u, \psi \right\rangle | &= \left|\int_{\mathbb{R}^n} \left( \int_{\mathbb{R}^n} \varphi_m(x-y) u(y) dy - u(x) \right) \psi(x) dx \right|\\ &= \left| \int_{\mathbb{R}^n} \left( \int_{\mathbb{R}^n} \varphi_m(x-y) (u(y) - u(x)) dy \right) \psi(x) dx \right|\\ &= \left| \int_{\text{supp} \ \varphi_m \ \cap \ \text{supp} \ u \ \cap \ \text{supp} \ \psi } \left( \int_{\text{supp} \ \varphi_m \ \cap \ \text{supp} \ u } \varphi_m(x-y) (u(y) - u(x)) dy \right) \psi(x) dx \right|\\ &\leq \int_{\text{supp} \ \varphi_m \ \cap \ \text{supp} \ u \ \cap \ \text{supp} \ \psi} \left( \int_{\text{supp} \ \varphi_m \ \cap \ \text{supp} \ u } C(\varphi_m,\psi) |u(y) - u(x)| dy \right) dx, \end{align*}$ where $C(\varphi_m,\psi) > 0$ is a constant depending on $\varphi_m$ and $\psi$.

As integration occurs in the compact sets

$$(\text{supp} \ \varphi_m \ \cap \ \text{supp} \ u \ \cap \ \text{supp} \ \psi) \subset \text{supp} \ \varphi_m \ \text{and} \ (\text{supp} \ \varphi_m \ \cap \ \text{supp} \ u) \subset \text{supp} \ \varphi_m \ (*)$$

and $u \in D'(\mathbb{R}^n)$, $u$ is uniformly continuous on $\text{supp} \ \varphi_m$, then

$$\forall \varepsilon >0, \exists\delta_m > 0; \forall x,y \in \text{supp} \ \varphi_m, |y - x| < \delta_m \Longrightarrow |u(y) - u(x)| < \varepsilon \ (**)$$


$\int_{\text{supp} \ \varphi_m \ \cap \ \text{supp} \ u \ \cap \ \text{supp} \ \psi} \left( \int_{\text{supp} \ \varphi_m \ \cap \ \text{supp} \ u } C(\varphi_m,\psi) |u(y) - u(x)| dy \right) dx \leq \int_{\text{supp} \ \varphi_m} \left( \int_{\text{supp} \ \varphi_m} C(\varphi_m,\psi) |u(y) - u(x)| dy \right) dx \ (***)$

by (*).

Remembering that $\text{supp} \ \varphi_m = \overline{B}_{\frac{1}{m}} (0)$ and combining $(**)$ with $(***)$, exists $n_0 \in \mathbb{N}$ such that

\begin{align*} \int_{\text{supp} \ \varphi_m \ \cap \ \text{supp} \ u \ \cap \ \text{supp} \ \psi} \left( \int_{\text{supp} \ \varphi_m \ \cap \ \text{supp} \ u } C(\varphi_m,\psi) |u(y) - u(x)| dy \right) dx &\leq \int_{\text{supp} \ \varphi_m} \left( \int_{\text{supp} \ \varphi_m} C(\varphi_m,\psi) |u(y) - u(x)| dy \right) dx\\ &< \int_{\text{supp} \ \varphi_m} \left( \int_{\text{supp} \ \varphi_m} C(\varphi_m,\psi) \varepsilon dy \right) dx\\ &= C(\varphi_m,\psi) \varepsilon \ (\text{vol} \ (\text{supp} \ \varphi_m))^2\\ &< C(\varphi_m,\psi) \varepsilon^3, \end{align*} $\forall m \geq n_0$, therefore $\varphi_m \ast u \rightarrow u$ in $D'(\mathbb{R}^n)$.

  • $\begingroup$ Did you mean $\varphi_m = m^n \varphi(m.)$ or something more general (if so what do you assume about $\varphi_m$) ? $\endgroup$ – reuns Sep 19 at 16:43
  • $\begingroup$ $\varphi_m$ is the a sequence of mollifiers defined in a similar way as defined in Evans' book (see here) $\endgroup$ – Math enthusiast Sep 20 at 15:59
  • $\begingroup$ It seems you don't understand why it is important. Do you follow my answer ? $\endgroup$ – reuns Sep 20 at 16:43

Let $(\varphi_m)$ be a sequence of continuous functions all supported on some common compact such that for some $e_1,\ldots,e_n$, $\varphi_m = \partial_{x_1}^{e_1+1} \ldots \partial_{x_n}^{e_n+1} f_m$ and $f_m \to \prod_{l=1}^n 1_{x_l > 0} \frac{x_l^{e_l}}{e_l!}$ uniformly.

Then for all $\psi \in C^\infty_c(\Bbb{R}^n)$, $\varphi_m \ast \psi \to \psi$ in the $ C^\infty_c(\Bbb{R}^n)$ topology

proof: let $\partial^b = \partial_{x_1}^{e_1+1} \ldots \partial_{x_n}^{e_n+1}$ then $$\partial^a (\varphi_m \ast \psi )= f_m \ast \partial^{a+b}\psi \to f_\infty\ast \partial^{a+b}\psi=\partial^b f_\infty \ast \partial^a \psi= \delta \ast \partial^a \psi= \partial^a \psi$$ uniformly and all staying supported on some common compact

And hence for $u \in D'(\Bbb{R}^n)$, by definition of distributions as continuous linear functionals,

we obtain $$\forall \psi \in C^\infty_c(\Bbb{R}^n),\qquad \langle\varphi_m\ast u,\psi\rangle\overset{def}=\langle u,\varphi_m\ast\psi\rangle \to \langle u,\psi\rangle\qquad \implies\qquad \varphi_m\ast u\to u$$

The convergence in strong semi-norms topologies follows similarly.

  • $\begingroup$ I didn't understand some steps: 1. Why you assume that "$\varphi_m = \partial_{x_1}^{e_1+1} \ldots \partial_{x_n}^{e_n+1} f_m$ and $f_m \to \prod_{l=1}^n 1_{x_l > 0} \frac{x_l^{e_l}}{e_l!}$ uniformly."? 2. Why $\partial^b f_\infty = \delta$? Is $\delta$ the Dirac delta seen as a distribution? I'm asking this because I didn't understand why $\delta \ast \partial^a \psi= \partial^a \psi$. $\endgroup$ – Math enthusiast Sep 21 at 23:35
  • $\begingroup$ If you don't know how to define the distribution $\varphi_m \ast u$ you'll have some problems. If you take the simplest mollifier $\varphi_m = m^n \varphi(m.), \varphi \in C^\infty_c,\int_{R^n} \varphi(x)dx=1$ then you only need to show $\partial^a(\varphi_m \ast \psi)=\varphi_m \ast \partial^a \psi\to \partial^a \psi$ uniformly which is more or less obvious. The Dirac $\delta$ is the identity element of the convolution, try with $n=1$ and $n=2$ to see why $\partial_{x_1}^{e_1+1} \ldots \partial_{x_n}^{e_n+1} \prod_{l=1}^n 1_{x_l > 0} \frac{x_l^{e_l}}{e_l!} = \delta$. $\endgroup$ – reuns Sep 21 at 23:41

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