I want to prove the following result

Proposition 1

Let $T \in D'(\mathbb R^n)$ and $S \in D'(\mathbb R^m)$. Then $$ \langle T_x, \langle S_y, \varphi(x,y) \rangle \rangle= \langle S_y \langle T_x, \varphi(x,y) \rangle \rangle \quad \forall \varphi \in C_0^\infty(\mathbb R^n \times \mathbb R^m ),$$

using the fact:

The set $S$ of all linear combinations of tensor products of functions in $C_0^\infty(\mathbb R^n)$ and $C_0^\infty(\mathbb R^m)$ is dense in $C_0^\infty(\mathbb R^n \times \mathbb R^m)$.

My attempt

It's clear that Proposition 1 holds for functions in $S$. Then, given $\varphi \in C_0^\infty(\mathbb R^n \times \mathbb R^m)$, there exists $\{\varphi_k\} \subset S$ such that $\varphi_k \xrightarrow{k\to\infty} \varphi$ in $C_0^\infty(\mathbb R^n \times \mathbb R^m)$. We have

$$ \langle T_x, \langle S_y \varphi_k(x,y) \rangle \rangle= \langle S_y \langle T_x, \varphi_k(x,y) \rangle \rangle \quad \forall k,$$

Since $\varphi_k(x,\cdot) \xrightarrow{k\to\infty} \varphi_k(x,\cdot)$ in $C_0^\infty(\mathbb R^m)$ for every $x \in \mathbb R^n$, we get by the continuity of the distributions, \begin{align}\label{some} \langle S_y, \varphi_k(x,y) \rangle \xrightarrow{k\to\infty}\langle S_y ,\varphi(x,y)\rangle, \quad \quad (1) \end{align} pointwise for every $x \in \mathbb R^n$.

I already know that $x \mapsto \langle S_y, \varphi_k(x,y) \rangle $ and $x \mapsto \langle S_y, \varphi(x,y) \rangle$ belong to $C_0^\infty(\mathbb R^n)$ , and, $\partial^\alpha \langle S_y, \varphi_k(x,y) \rangle= \langle S_y, \partial^\alpha_x \varphi_k(x,y)\rangle $ and $\partial^\alpha \langle S_y, \varphi(x,y) \rangle= \langle S_y, \partial^\alpha_x \varphi(x,y)\rangle$.

I want to show that \begin{align} \langle S_y, \varphi_k(\cdot,y) \rangle \xrightarrow{k\to\infty}\langle S_y ,\varphi(\cdot,y)\rangle, \end{align} in $C_0^\infty(\mathbb R^n)$ to conclude, but I'm not sure how to.

Any help of idea would be appreciated.


$\{\varphi_n\}\subset C_0^\infty(\mathbb R^p)$ converges to $\varphi \in C_0^\infty(\mathbb R^p)$ if $\partial^\alpha \varphi_k$ converges to $\partial^\alpha \varphi$ uniformly for every multiindex $\alpha$.

  • $\begingroup$ You basically need to prove the continuity of $\varphi\mapsto T_x(S_y(\varphi(x,y))$. For this assume $\varphi_k \to 0$ and so the support of all $\varphi_k$ are contained in one compact set $K\times L$. How can you conclude from pointwise convergence in $x$ to uniform convergence? $\endgroup$ – Vobo Aug 2 '18 at 18:24

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