# Is a ball in Sobolev space $W^{2,2}(\Bbb R)$ closed in $L^2(\Bbb R)$?

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 $$L^2(R)$$.
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.