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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 $\lim_{y \rightarrow z} ||S(y)u - S(z)u||_\infty = 0$?

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up vote 4 down vote accepted

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.

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Here is a little more: It suffices to prove this for an operator of order $0$. So to get control of the $L_\infty$ norm, you need at least $n$ derivatives in $L_2$, where $n = \dim M$. So you need to control the operator norm of $A(z): W^{k,2} \rightarrow W^{k,2}$, where $k \ge n/2$ and show it is a continuous function of $z$. To do this, it suffices to assume that $\sup_{i,j < M} \|\partial_x^i\partial_\xi^ja(z,x,\xi)\|_\infty$ is a continuous function of $z$ for some appropriately chosen $M$. –  Deane Yang Oct 20 '11 at 14:40
    
The "$n$ derivatives in $L_2$" should be "$n/2$ derivatives in $L_2$". –  Deane Yang Oct 20 '11 at 14:41
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