# Equicontinuity and uniform boundedness for “distributions”

Exercise. (Rudin, Functional Analysis, chapter 2, pag. 53). Let us consider the space $$\mathcal D :=\{f \in C^{\infty}(\mathbb R), \, \text{supp}f\subseteq [-1,1] \}$$ with the topology induced by the usual topology of $C^{\infty}(\Omega)$. Consider the linear functionals $$\mathcal D \ni \phi \mapsto \Lambda_n \phi :=\int_{[-1,1]}f_n(t)\phi(t)dt \in \mathbb R$$ where $\{f_n\}$ is a sequence of Lebesgue integrable functions s.t. $\lim_n \Lambda_n\phi$ esixts for every $\phi \in \mathcal D$.

Using the facts that each $\Lambda_n$ is continuous and, moreover, that $\{\Lambda_n\}_{n \in \mathbb N}$ is equicontinuous I would like to prove that :

There exist two numbers $p \in \mathbb N$, $M \in \mathbb R^+$ s.t. $$\left\vert \int_{[-1,1]}f_n(t)\phi(t)dt \right\vert \le M \Vert D^p\phi \Vert_{\infty}$$ for every $n$, for every $\phi \in \mathcal D$.

I think that this is a simple matter of uniform boundedness: we kwow that equicontinuity implies uniform boundedness so we can say $$\forall E \subset \mathcal D \text{ bounded, }\, \exists M > 0 \text{ s.t. } \Lambda_nE \subset [-M,M], \quad \forall n \in \mathbb N.$$ I think that this fact is all we need to solve Rudin'exercise, but I do not kwow how to identify bounded sets in $\mathcal D$...

Denote by $p_i(\phi) = \sup_{x \in [-1,1]} |D^i \phi(x)|$ the usual seminorms on $\mathcal{D}$. Observe that $p_i \leq 2 p_{i+1}$ for all $i\in\mathbb{N}$. By the equicontinuity of the $\Lambda_n$ there is a $\delta>0$ and a neighbourhood $$U = \{ \phi \in \mathcal{D}; p_{i_1}(\phi) < \delta, ..., p_{i_k}(\phi)<\delta\}$$ of $0$ such that $|\Lambda_n(\phi)| \leq 1$ for all $\phi \in U$ and $n\in \mathbb{N}$. Without loss of generality assume $i_1 < ... < i_k$. Now let $\phi\neq 0$ and $M:=\max\{p_{i_1}(\phi),...,p_{i_k}(\phi)\}$. By the observation above $M>0$ and $M\leq 2^{i_k}p_{i_k}(\phi)$. Then $\frac{\delta}{M}\phi \in U$ and hence
$$\vert\Lambda_n(\phi)| = \vert\frac{M}{\delta} \Lambda_n(\frac{\delta}{M}\phi)\vert \leq \frac{M}{\delta} \leq \frac{2^{i_k}}{\delta} p_{i_k}(\phi).$$
• Hi, I know this is a quite old question. But I'm stuck in the same problem, and I still can't manage to solve it yet. I'm taking this question as inspiration for my solution but there're few things I don't understand. First of all the seminorms you're using they don't seem the same as the ones defined in section 1.46 of the book, it's my understanding they're induce the same topology though. Why do you have the bound $p_i \leq 2p_{i+1}$? Can your solution be adapted to the seminorm used in section 1.46 of the book? – user8469759 Dec 18 '19 at 16:08
• The restriction of $\mathcal{D}$ to the subspace $\mathcal{D}_K$ for $K=[-1,1]$ is topologized by the seminorms $p_i$ as above, also in Rudin. And by the mean value theorem you have $p_1(\phi)\geq |f(1)-f(-1)|/2\geq p_0(\phi)$, so you may even forget the $2$. – Vobo Dec 18 '19 at 20:34