# tempered distribution convergence

I have a question that looks somehow very easy, but I cannot find a proof. We say that a sequence $(\psi_k)$ in the space of tempered distributions $\mathcal{S}'(\mathbb{R}^d)$ converges to $\psi$, if $\psi_k(\phi) \to \psi(\phi)$ for every $\phi \in \mathcal{S}(\mathbb{R}^d)$.

Assume now $\psi_k \to \psi$ in $\mathcal{S}'(\mathbb{R}^d)$ and $\varphi_k \to \varphi$ in the Schwartz-space $\mathcal{S}(\mathbb{R}^d)$. Then the following holds: $\psi_k(\phi_k) \to \psi(\phi)$.

There is a hint that one should consider the Banach-Steinhaus theorem. I can prove this result if one replaces $\mathcal{S}(\mathbb{R}^d)$ by some Banachspace. But in this case, I am stuck.

I tried it this way: $|\psi_k(\phi_k)-\psi(\phi))|\leq |\psi_k(\phi_k-\phi)|+|\psi_k(\phi)-\psi(\phi)|$. Now the last term vanishes, but what can I do with the first term?

• What formulation(s) of the Banach-Steinhaus theorem do you know? Commented Jan 11, 2017 at 20:27

Theorem. Let $X$ be a Frechet space and $Y$ a normed space and let $\{\psi_\alpha \}_{\alpha \in A}$ be a collection of continuous linear maps $\psi_{\alpha}: X \to Y$. If $\sup_{\alpha \in A} \| \psi_\alpha(x) \|_{Y} < \infty$ for each $x \in X$, then $\{ \psi_\alpha\}_{\alpha \in A}$ is equicontinuous.
Here we see $\sup_{k \in \mathbb N}\lvert\psi_k(\phi) \rvert$ is bounded for each $\phi \in S(\mathbb R^d)$ since $\{ \psi_k(\phi)\}$ is a convergent sequence. Thus $\{\psi_k \}$ is an equicontinuous family.
Let $\epsilon > 0$. By equicontinuity, there is $N \in \mathbb N$ so that $n \ge N$ gives $$\lvert \psi_k(\phi_n - \phi)\rvert < \epsilon$$ for all $k \in \mathbb N$. By convergence $\psi_k(\phi) \to \psi(\phi)$, there is $K \in \mathbb N$ so that $k \ge K$ gives $$\lvert \psi_k(\phi) - \psi(\phi) \rvert < \epsilon.$$ Hence for $k \ge \max\{K,N\}$, we have $$\lvert \psi_k(\phi_k) - \psi(\phi)\rvert \le \lvert \psi_k(\phi_k - \phi)\rvert + \lvert \psi_k(\phi) - \psi(\phi) \rvert < 2\epsilon.$$