Let $(M,g)$ be a closed, Riemannian manifold. Let $S(z)$ be a holomorphic family of pseudo-differential operators, with $z \in \Bbb{C}$. Let $u$ be a smooth function. Does it follow that $\lim\_{y \rightarrow z} ||S(y)u - S(z)u||\_\infty = 0$?



meromorphic family of pseudo-differential operators
If you set $A(y) = S(y) - S(z)$, your question is equivalent to asking if $A(z)$ is a holomorphic family of pseudo-differential operators such that $A(0) = 0$, then does $\|A(z)u\|\_\infty \rightarrow 0$ as $z \rightarrow 0$.


Here's what I think is true: Let $a(z, x, \xi)$ be the symbol of $A(z)$. Assume $a$ is a smooth function of $(z,x,\xi) \in \mathbb{C}\times T^\*M$ such that $a(0,x,\xi)$ is identically zero. Then $\|A(z)u\|\_\infty \rightarrow 0$ as $z \rightarrow 0$.


The proof should be relatively straightforward, because you just go through the usual proof that a pseudodifferential operator is bounded with respect to the appropriately chosen Sobolev norms but keep track of how the bound depends on the symbol and its derivatives with respect to $x$ and $\xi$. These estimates should then imply that if the symbol and its derivatives converge to zero uniformly, then the operator norm does, too.


