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:C∞0(M)→C∞(M).

We start by noting that we may imbed C∞ in D′(M) by ∫Muv, where u∈C∞ and v is a density with compact support, and thereby also C∞0(M) in E′(M). We also note that C∞0(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)n∈N is a sequence converging to some u∈E′(M), then we wish to show that Aun converges to some v∈D′(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 C∞0(M) since it is dense in E′(M). So if we let (un)n∈N→u∈C∞0 we have ⟨Aun,ϕ⟩→⟨Au,ϕ⟩ for all ϕ smooth densities with compact support on M, and therefore A:E′(M)→D′(M) is continuous on C∞0, and thereby also on E′ since C∞0 is dense therein as mentioned.

**Answer**

Your operator A is continuous from C∞0(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*