# Does compact imbedding of $X$ in $Y$ imply there is an element $u\in X$ with $\|u\|_X=1$ and $\|u\|_Y < \varepsilon$?

In section 6 chapter II of Finite Elements: Theory, Fast Solvers, and Applications in Solid Mechanics by Dietrich Braess (Cambridge University Press, 2007), there is an appendix on the optimality of the estimates on page 85-86. And a conclusion is given by the author like this:

Suppose the complete normed linear space $X$ is compactly imbedded in $Y$. Then there exists an element $u\in X$ with $\|u\|_X=1$ and $\|u\|_Y < \varepsilon$.

Is this true? And how to prove it?

Suppose this is not true, then we have $\|u\|_Y \ge \varepsilon \|u\|_X, \forall u\in X$. Now we can say $\|\cdot\|_X$ and $\|\cdot\|_Y$ is equivalent because of the imbedding. Is this a contradiction with compact imbedding?

• I believe you need your spaces to be infinite dimensional here, otherwise the identity map in a finite dimensional normed space provides a counterexample.
– Neal
Commented Nov 30, 2016 at 2:21
• @Neal, yes, $X$ and $Y$ is infinite dimensional space, usually Sobolev spaces like $H^1(\Omega)$, $H^2(\Omega)$
– xyz
Commented Nov 30, 2016 at 2:56

Suppose that it holds $\|u\|_Y\ge \epsilon \|u\|_X$. Then the identity as mapping from $Y$ to $X$ is bounded. By assumption, the identity as mapping from $X$ to $Y$ is compact. Then the product of these two mappings is compact. The product is the identity on $X$ (or $Y$). Hence $X$ and $Y$ are finite-dimensional, a contradiction.