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.
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.
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 $(0,0)$.