Is C^{k+1}(X) compactly contained in C^{k}(X) for a closed manifold X? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T19:57:29Z http://mathoverflow.net/feeds/question/106081 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/106081/is-ck1x-compactly-contained-in-ckx-for-a-closed-manifold-x Is C^{k+1}(X) compactly contained in C^{k}(X) for a closed manifold X? trex 2012-09-01T00:14:08Z 2012-09-03T12:42:17Z <p>Hi all,</p> <p>I apologize if this question is too low level for mathoverflow. I'm happy to move it to math.stackexchange if so.</p> <p>Let \$X\$ be a closed manifold, let \$k\$ be a nonnegative integer and let \$C^k(X)\$ denote the space of \$k\$-times continuously differentiable functions equipped with the \$C^k\$ norm.</p> <p>Is \$C^{k+1}(X)\$ compactly contained in \$C^k(X)\$? Does this follow from Arzela-Ascoli?</p> <p>Thanks.</p> http://mathoverflow.net/questions/106081/is-ck1x-compactly-contained-in-ckx-for-a-closed-manifold-x/106098#106098 Answer by jbc for Is C^{k+1}(X) compactly contained in C^{k}(X) for a closed manifold X? jbc 2012-09-01T06:48:56Z 2012-09-03T12:42:17Z <p>The answer depends on what you mean by a closed manifold and a compact operator. If the manifold is compact without boundary then the spaces involved are Banach spaces and the answer is yes and this is indeed proved using the Arzela-Ascoli theorem, after a standard localisation argument.</p> <p>If the manifold is without boundary but not compact (think real line), then the spaces are Frechet spaces and the answer depends on what you mean by a compact operator on a locally convex space. If you define these as taking bounded sets into relatively compact ones, then the answer is yes, but if you mean those which take a neighbourhood of zero to a relativeley compact set (the usual definition), then the answer is no.</p>