I have asked this question with other problems, but about this part nobody answers. So I want to ask again, and put some details in it.
My question is whether the following equality is correct. If it is wrong, which part of my proof is wrong?
$$ C([0,T];H^{s+1}) \cap C^1([0,T];H^{s})=C^1([0,T];H^{s+1}) ? $$
It is obvious that $$ C^1([0,T];H^{s+1}) \subset C([0,T];H^{s+1}) \cap C^1([0,T];H^{s}) $$
We just need to prove the other direction: $$ C([0,T];H^{s+1}) \cap C^1([0,T];H^{s}) \subset C^1([0,T];H^{s+1}) $$ I think it is easy, too.
For every $f \in C([0,T];H^{s+1}) \cap C^1([0,T];H^{s})$, we have $$ f \in C([0,T];H^{s+1})~ \text{and} ~f \in C^1([0,T];H^{s}) $$ So, $$ f(t) \in H^{s+1}~ \text{and} ~f(t) \in H^{s} $$ Because $H^{s+1} \subset H^{s}$, we get $f(t) \in H^{s+1}$
Then, we have $f \in C([0,T];H^{s+1})$ and $f \in C^1([0,T];H^{s+1})$
Therefore, $$ f \in C^1([0,T];H^{s+1}) $$
I think it is more than clear, but if it is true, why we don't write $C([0,T];H^{s+1}) \cap C^1([0,T];H^{s}) $ as $C^1([0,T];H^{s+1})$ which is more simple (In all books relating to hyperbolic equations they write it as the first form)? But if it is wrong, which part I make a mistake. I think I might make a mistake, but I still don't realise it.
Thank you.
In the 5th displayed equation, it should read $f'(t) \in H^s$. That's what $C^1$ means - one derivative with respect to the variable $t$.
Suppose $\theta:[0,\infty) \to [0,\infty)$ satisfies $\theta(s)/(1+s) \in [\frac12,2]$, but $\theta'(s)$ varies a lot.
Consider $\hat f(t,\xi) = \theta((1+t) |\xi|)^{-s-3/2-\epsilon}$ for some small $\epsilon>0$. Then $f \in C(H^{s+1})$. However $\frac\partial{\partial t} \hat f(t,\xi) = |\xi|\theta'((1+t)|\xi|)\theta((1+t) |\xi|)^{-s-5/2-\epsilon}$. So by suitably thinking of what $\theta$ could be, it should be possible to create an example where $f' \in H^{s}\setminus H^{s+1}$.