$f\in C([0,T];C^1(\Omega))$ implies $f \in C([0,T];W^{1,p}(\Omega))$?

Question

Let $f \in C([0,T];C^1(\Omega))$ where $\Omega$ is compact.

Am I correct that this implies

$f \in C([0,T];W^{1,p}(\Omega))$ for all $p > 1$?? Because we have $L^\infty$ estimates on $f(t)$.

Answer 1

As @Siminore pointed out, $f\in W^{1,p}(\Omega)$ is immediate. Let's prove continuity. i.e. if $t_n\to t$ then $f(t_n)\to f(t)$, or equivalently $$\tag{1}\|f(t_n)-f(t)\|_{1,p}\to 0$$

Note that $$\tag{2}\|f(t_n)-f(t)\|_{1,p}=\|f(t_n)-f(t)\|_p+\|\nabla f(t_n)-\nabla f(t)\|_p$$

On the other hand, by hypothesis we have that $$\tag{3}\|f(t_n)-f(t)\|_\infty+\|\nabla f(t_n)+\nabla f(t)\|_\infty\to 0$$

To conclude, combine $(2)$, $(3)$ and the fact that (for bounded $\Omega$) $L^\infty(\Omega)$ is continuously embedded in $L^p(\Omega)$.