If $u(.,t) \in H^2(\mathbb{R})$, then $u (.,t) \in C^1(\mathbb{R})$?

Question

The statement is as such:

$u \in L^2([0,T]; H^2(\mathbb{R}))$ for any $T>0$, thus $u (.,t) \in > C^1(\mathbb{R})$ a.e. in $t>0$.

• My guess was that this must be a result of Sobolev embedding. We know $W^{2,2} (\mathbb{R})$ can be continuously embedded in the space $C^{1,1/2}(\mathbb{R})$ since $W^{1,2} (\mathbb{R})$ continuously embeds into $C^{0,1/2}(\mathbb{R})$. However, I do not know how to show from here that this would mean that for any $t>0$, $u (.,t) \in C^1(\mathbb{R})$ a.e.

• Also, why "almost everywhere?"

In this, they are working on a torus, and I have not yet encountered an embedding theorem like this in my course. Additionally, in my setting, I am not on a compact or bounded domain.

Answer 1

You have $$\int_0^T\|u(\cdot, t)\|_{H^2(\mathbb{R})}^2dt<\infty$$ and so for a.e. $t$ it follows that $$\|u(\cdot, t)\|_{H^2(\mathbb{R})}^2<\infty.$$ This just follows from the fact that if you have a non-negative function $f$ whose integral is finite, then the function $f$ can assume the value $\infty$ at most on a set of Lebesgue measure zero.

Now for every good $t$, you use the embedding theorem.