I'm very sorry if this is the wrong place to ask this question, but I've asked it on StackExchange and received no answers. ( https://math.stackexchange.com/questions/813063/convergence-to-a-schwartz-distribution )

Let $M$ be a smooth manifold with countable atlas, and define the distributions $\mathscr{D}'(M)$ as the dual space to the smooth densities with compact support, and $\mathcal{E}'(M)$ as the dual space to the smooth densities. Let $A$ be a pseudodifferential operator $A:C_0^{\infty}(M)\to C^{\infty}(M)$.

We start by noting that we may imbed $C^{\infty}$ in $\mathscr{D}'(M)$ by $\int_M uv$, where $u\in C^{\infty}$ and $v$ is a density with compact support, and thereby also $C_0^{\infty}(M)$ in $\mathcal{E}'(M)$. We also note that $C_0^{\infty}(M)$ is dense in $\mathcal{E}'(M)$. We wish to extend $A$ by continuity to an operator $A: \mathcal{E}'(M)\to \mathscr{D}'(M)$.

What I've been thinking is that if $(u_n)_{n\in \mathbb{N}}$ is a sequence converging to some $u\in \mathcal{E}'(M)$, then we wish to show that $Au_n$ converges to some $v\in \mathscr{D}'(M)$, i.e. $\langle Au_n,\phi \rangle$ converges to $\langle v,phi \rangle$ for every $\phi$ a smooth density with compact support, since $\mathscr{D}'(M)$ is equipped with the weak$^*$ topology. However, I seem to be stuck at this point, so any hint or idea at all would be nice.


I just realized that I (think) just need to show continuity on $C_0^{\infty}(M)$ since it is dense in $\mathcal{E}'(M)$. So if we let $(u_n)_{n\in \mathbb{N}} \to u \in C_0^{\infty}$ we have $\langle Au_n, \phi \rangle \to \langle Au ,\phi \rangle$ for all $\phi$ smooth densities with compact support on $M$, and therefore $A:\mathcal{E}'(M) \to \mathscr{D}'(M)$ is continuous on $C_0^{\infty}$, and thereby also on $\mathcal{E}'$ since $C_0^{\infty}$ is dense therein as mentioned.


Your operator $A$ is continuous from $C_0^\infty(M)$ into $C^\infty(M)$, so the adjoint $A^*$ is continuous from $\mathcal E'(M)$ into $\mathscr D'(M)$. Now, the operator $A^*$ is also a pseudodifferential operator of the same order than $A$, whose principal symbol is the complex conjugate of the symbol of $A$: this fact can be established by looking at the chart expression of the pseudodifferential operator.

More precisely in $\mathbb R^n$, if $a$ is the symbol of $A$, the symbol of $A^*$ is $$ a^*=\exp(2iπ D_\xi\cdot D_x)\bar a, $$ so that starting with the continuity property of $op(a^*)$ (the operator with symbol $a^*$), you obtain the sought property of $op(a^{**})$ which is $op(a)$.

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