Extension of pseudodifferential operators

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 D(M) as the dual space to the smooth densities with compact support, and E(M) as the dual space to the smooth densities. Let A be a pseudodifferential operator A:C0(M)C(M).

We start by noting that we may imbed C in D(M) by Muv, where uC and v is a density with compact support, and thereby also C0(M) in E(M). We also note that C0(M) is dense in E(M).
We wish to extend A by continuity to an operator A:E(M)D(M).

What I’ve been thinking is that if (un)nN is a sequence converging to some uE(M), then we wish to show that Aun converges to some vD(M), i.e. Aun,ϕ converges to v,phi for every ϕ a smooth density with compact support, since 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.

EDIT:

I just realized that I (think) just need to show continuity on C0(M) since it is dense in E(M). So if we let (un)nNuC0 we have Aun,ϕAu,ϕ for all ϕ smooth densities with compact support on M, and therefore A:E(M)D(M) is continuous on C0, and thereby also on E since C0 is dense therein as mentioned.

Answer

Your operator A is continuous from C0(M) into C(M), so the adjoint A
is continuous from E(M) into 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 Rn, 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).

Attribution
Source : Link , Question Author : Ukhrir , Answer Author : Bazin

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