Convergence of elliptic operators - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T04:06:48Z http://mathoverflow.net/feeds/question/72003 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/72003/convergence-of-elliptic-operators Convergence of elliptic operators Chris Judge 2011-08-03T16:15:29Z 2011-08-03T17:32:24Z <p>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?</p> http://mathoverflow.net/questions/72003/convergence-of-elliptic-operators/72017#72017 Answer by Deane Yang for Convergence of elliptic operators Deane Yang 2011-08-03T17:32:24Z 2011-08-03T17:32:24Z <p>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.</p>