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Why this statement true?

If $f \in C^0([0,1], W^{2,2}(K)) $ then $ f \in C^0 ([0,1] \times K)$.

$ K \subset R^n $ , W : Sobolev Space.

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Is $K$ an arbitrary subset of $\mathbb R^n$? Probably not, otherwise how would you define $W^{2,2}(K)$... – user31373 Jun 5 '12 at 18:30
up vote 1 down vote accepted

I don't think this is true. The Sobolev embedding theorem suggests this should only work for small $n$; in this case I think it should be $n < 4$ if I have the parameters right.

To try for a counterexample, let's take, say, $K$ to be the unit ball of $\mathbb{R}^6$, and let $f(t,x) = |x|^t$. One should be able to show that $$f(t, \cdot), \partial_i f(t, \cdot), \partial_i \partial_j f(t,\cdot)$$ are continuous as functions $[0,1] \to L^2(K)$, which would make $f$ continuous as a map from $[0,1]$ to $W^{2,2}(K)$. However, as a function on $[0,1] \times K$, $f$ is clearly not continuous at $(0,0)$.

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Oh then how can I apply Sobolev embedding thm to conclude that n<4 ? – Misaj Jun 5 '12 at 19:19
The Sobolev embedding theorem says in this case that for $n < kp$, $W^{k,p}$ is continuously embedded in the Holder space $C^{k - [n/p]-1,\gamma}(K)$ for some $\gamma > 0$, which in turn is continuously embedded in $C(K)$. (Here $k=p=2$ so we need $n < 4$.) So in this case if $f \in C([0,1], W^{2,2})$ then $f \in C([0,1], C(K)) = C([0,1] \times K)$. This isn't a proof that it fails when $n \ge 4$, but I think an example like the one I gave should show it. – Nate Eldredge Jun 5 '12 at 19:30

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