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Let $A_t$ be family of second order, positive, elliptic differential operator mapping Sobolev $H^2$ of a compact smooth manifold (or bounded domain) to L^2. Suppose that the coefficients of $A_t$ converge uniformly in $C^k$ for every $k$ to the coefficients of a second order, positive, elliptic differential operator $A$. $A$ is invertible (with domain L^2 and range H^2) and so we may consider the sequence $A_t \circ A_0^{-1}$ of operators from $L^2$ to $L^2$. Does this family converge to the identity in the $L^2$ operator norm? Why or why not?

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It suffices to show that the $L_2$ operator norm of $A_t\circ A_0^{-1} - I = (A_t - A_0)\circ A_0^{-1}$ is small if $t$ is sufficiently small. To do this, it suffices to show that the operator norm of $A_t - A_0$, as map from $H^2$ to $L_2$ is small if $t$ is small. But a linear second order operator like this has small operator norm, if the $C^0$ norm of the coefficients are small. So the fact that the coefficients converge in the $C^0$ norm gives what you want.

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Is ellipticity necessary? (Part of the `why' part of question...) –  Chris Judge Aug 3 '11 at 20:36
    
I guess ellipticity has nothing to with it. [ \int |(A-B) u|^2 \leq C \sum_{\alpha} \int |\partial^{\alpha} u|^2] where [ C= \sup |a_{\alpha}- b_{\alpha}|^2] where $\alpha$ is a multi-index and the $a$'s and $b$'s are the coefficients. –  Chris Judge Aug 3 '11 at 20:57
    
Ellipticity is only used in so far as $A_0$ is invertible (and if $A_0$ is elliptic, and $A_t\to A_0$ in coefficients as $t\to 0$, $A_t$ is also elliptic for sufficiently small $t$. –  Willie Wong Aug 3 '11 at 21:27
    
Agreed with comments above by Willie and Chris. The only thing required here is the existence of a right inverse $A_0^{-1}$ that recovers all the regularity lost by the differential operator $A_0$. There is a more general class of differential operators known as hypoelliptic operators for which such inverses exist. –  Deane Yang Aug 3 '11 at 21:44
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