# Elliptic regularity of second order pseudos

Let $M$ be a compact manifold. Choose a point $q \in M$. Let $P$ be a second order positive self-adjoint pseudodifferential operator such that $\text{Spec}(P) \subset (0, \infty)$. Also, we know that $P$ is elliptic on $M \setminus \{q\}$. Can we say that $(Pu, u) \cong \Vert u\Vert^2_{H^1}$, or at least $(P(\varphi u), \varphi u) \cong \Vert \varphi u\Vert_{H^1}^2$ for all $\varphi \in C_c^\infty (M \setminus \{q\})$? If $P$ were elliptic on $M$, this is standard material. But here the lack of ellipticity at a point is throwing me off.

Let $\varphi\in C^\infty_c(M\backslash\{q\})$ be given. Choose a function $\tau\in C^\infty_c(M,[0,1])$ which is equal to $1$ in a neighbourhood of $q$ and which has a support disjoint from the support of $\varphi$. Put a Riemannian metric on $M$, so that we have the negative Laplacian $-\Delta$, a positive elliptic second order differential operator. Let $\widetilde{P}:=(1-\tau)P+\tau (-\Delta)$. Then $\widetilde P$ is a positive second order pseudodifferential operator that is elliptic everywhere on $M$ and one has$$\widetilde P\varphi u=(1-\tau)P\varphi u+\tau (-\Delta)\varphi u=(1-\tau)P\varphi u=P\varphi u -\tau P\varphi u\qquad \forall\;u\in H^2(M),$$ because $\Delta$ is a local operator. Now, as the supports of $\varphi$ and $\tau$ are disjoint, one gets \begin{align}(P(\varphi u),\varphi u)_{L^2(M)}&= (\widetilde{P}(\varphi u),\varphi u)_{L^2(M)}+(\tau P\varphi u,\varphi u)_{L^2(M)}\\&= (\widetilde{P}(\varphi u),\varphi u)_{L^2(M)}+(P\varphi u,\underbrace{\tau\varphi}_{\equiv0} u)_{L^2(M)}\\&=(\widetilde{P}(\varphi u),\varphi u)_{L^2(M)}.\end{align} Thus, the problem for functions supported away from $q$ reduces to the situation of an elliptic operator, for which the result is standard, as you say.
• Nice answer. Since the two norms are equivalent with equivalence constants independent of $\varphi$, is it possible to make a limiting argument to answer the first question affirmatively? – user74880 Jun 12 '15 at 14:12
• One last detail: could you please elaborate a little on why $C^\infty_c(M\setminus \{q\})$ is dense in $H^2(M)$? It seems to me (intuitively) that any sequence of approximating functions would have to go to $0$ sharply at $q$ resulting in a high gradient norm. What am I missing here? – user74880 Jun 12 '15 at 21:03
• On second thoughts, in dimensions $n \geq 3$, I am convinced. But the case $n = 2$ seems a bit odd to me. – user74880 Jun 12 '15 at 22:18
• Also, it is not clear to me that $\varphi_n \to 1_M$ in $H^2(M)$-norm implies $\varphi_n u \to u$ in $H^1(M)$-norm. Is this obvious? – user74880 Jun 12 '15 at 22:47