Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

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.

share|improve this answer
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
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.