The Rellich-Kondrachov Theorem states that, if the support has the cone property, then the embedding of $W^{2,2}(R)$ (the Sobolev space that controls the l2 norm of the derivatives up to the second order) into L^2(R) is a compact embedding. This is to say the embedding operator maps a closed ball in W^{2,2}(R) into a totally bounded set in $L^2(R)$.
But is there a theorem says that the image of the ball is also complete in the $L^2$ space, so that we can conclude the image is actually ‘compact’ in $L^2$?
Thanks in advance for any proofs or counterexamples.
I think the answer is yes, since $W^{2,2}(R)$ is a reflexive Banach space I think you can conclude that the image of the closed ball of $W^{2,2}(R)$ is actually ("norm") compact in $L^2(R)$.
For reference see here.
And also you can see here completely continuous. So the closed unit ball of $W^{2,2}(R)$ is weakly sequence compact, again since $W^{2,2}(R)$ is reflexive and so it follows with the completely continuity (or equivalently compactness) of the embedding - lets call it E - that for any weakly convergent subsequence of $(x_n)$ in the closed unit Ball in $W^{2,2}$, $(Ex_n)$ has a strongly convergent subsequence in $L^2$, hence the image of the closed unit Ball is (norm) compact.