Let $u_n\in\mathcal{D}'(\mathbb{R}^n)$ have $u_n\to0$ where $u_n\in C_c^\infty$ have uniformly compact support. Does $u_n\to0$ in $C_c^\infty$? [closed]

Question

Suppose we have functions $u_n\in C_c^\infty(\mathbb{R}^d)$ with support all lying in $B(0,R)$, and suppose $u_n\to 0 $ in $\mathcal{D}'(\mathbb{R}^n)$, i.e. for all $\eta\in C_c^\infty(\mathbb{R}^d)$, $$\int_{\mathbb{R}^n} u_n \eta\,\mathrm{d}x\xrightarrow{n\to\infty}0$$ then is it true that $\partial^\alpha u_n\to 0$ uniformly for all $\alpha$?

The reason I'm asking this is because in Hörmander's The Analysis of Linear Partial Differential Operators I, in the paragraphs immediately above Definition 4.2.1, we have the following statement:

Let $u\in\mathcal{D}'(\mathbb{R}^n)$ have compact support, and then recall that $u*\phi\in C_c^\infty(\mathbb{R}^n)$ for all $\phi\in C_c^\infty(\mathbb{R}^n)$. Then in fact, the map $$\phi\mapsto u*\phi,\quad C_c^\infty(\mathbb{R}^n)\to C_c^\infty(\mathbb{R}^n)$$ is continuous.

However, I'm struggling to prove this fact.

Answer 1

A comment but I am not entitled. As regards the question at the end, the comment of Nate Eldredge points in the right direction---the result follows from the closed graph theorem since the convolution is continuous on the space of distributions. Note however that we do not have a Fréchet space here. However it is a strict $LF$ space (Dieudonné and Schwartz) and Grothendieck showed in his thesis that the CGT holds in this context. As regards the first question, let me add that a necessary and sufficient condition for this to hold is that the sequence be bounded in the sense that it is uniformly bounded as well as all sequences obtained by successive differentiation.

Answer 2

For a relatively simple proof of the continuity of $\phi \mapsto u * \phi$, use the fact that any distribution is locally a derivative of a bounded function.

To be specific: Consider $u$ supported in a compact set $K_1$ and a sequence $\phi_n \in \mathcal{D}$ convergent to $0$ in $\mathcal{D}$, all supported in some compact set $K_2$. Clearly, $u * \phi_n$ is supported in $K_1 + K_2$. Write $u = D^\alpha f$ for some multi-index $\alpha$ and some bounded function $f$ on a large ball (a neighbourhood $K_1 + K_2 - K_2$). Then $$ u * \phi_n(x) = f * D^\alpha \phi_n(x) = \int_{K_2} f(x - y) D^\alpha \phi_n(y) dy $$ for $x \in K_1 + K_2$. Since $D^\alpha \phi_n$ converges uniformly to zero, and $f$ is integrable, $u * \phi_n$ converges to zero uniformly on $K_1 + K_2$ as $n \to \infty$. The same argument works if $\phi_n$ are replaced by their derivatives, so $u * \phi_n$ converges to zero in $\mathcal{D}$.

Answer 3

Here are two alternative methods:

• You have an explicit formula for $u * \phi$ : it is given by $u* \phi(x) = \langle u,\phi(x-y)\rangle_y$, and you can "differentiate under the bracket". Now if $(\phi_n)_n$ goes to $0$ in $C^\infty_c(\mathbb{R}^d)$, you have for any $\alpha\in\mathbb{N}^d$ that $\|u* \partial^\alpha \phi_n\|_\infty = \|\langle u,\partial^\alpha \phi_n(\cdot-y)\rangle_y\|_\infty$ which converges to $0$ (you can work on a fixed compact $K$ because the sequence converges in $C^\infty_c(\mathbb{R}^d)$ and $u$ is assumed compactly supported).
• You can first try to prove the (weaker) fact : $(u* \phi_n)_n$ is bounded in $C^\infty_c(\mathbb{R}^d)$ whenever $(\phi_n)_n$ is. Once you know that the conclusion follows thanks to the Montel property satisfied in $C^\infty_c(\mathbb{R}^d)$ : bounded sets are relatively compact (this is simply an iterate use of Ascoli's Theorem). Since $(u * \phi_n)_n$ obviously converges to $u*\phi$ for weaker topologies (for instance the one of $\mathcal{D}'(\mathbb{R}^d)$), the latter is the only possible limit point: the whole sequence converges to it.