A $C^1$ function and uniform continuity (Bochner type spaces)

Question

Let $\Omega \subset \mathbb{R}^n$ be a compact domain and let $f \in C^1([0,T]\times\Omega).$

Is it obvious that this implies that $$f \in C([0,T];C^1(\Omega)) \cap C^1([0,T];C(\Omega))$$ ?

Answer 1

For each $t\in [0,T]$ define $(\overline{f}(t))(x)=f(t,x)$. We want to show that $$\overline{f} \in C([0,T];C^1(\Omega)) \cap C^1([0,T];C(\Omega))$$

This is equivalently to show that if $t_n\to t$ in $[0,T]$, then $$\|\overline{f}(t_n)-\overline{f}(t)\|_{C^1(\Omega)}\to 0\tag{1}$$

and $$\|\overline{f}(t_n)-\overline{f}(t)\|_{C(\Omega)}+\|\frac{d}{dt}\overline{f}(t_n)-\frac{d}{dt}\overline{f}(t)\|_{C(\Omega)}\to 0\tag{2}$$

Note that $$\|\overline{f}(t_n)-\overline{f}(t)\|_{C^1(\Omega)}=\sup_{x\in\Omega}\|f(t_n,x)-f(t,x)\|_{C(\Omega)}+\sup_{x\in\Omega}\|\nabla_xf(t_n,x)-\nabla_xf(t,x)\|_{C(\Omega)}$$

By using the compactness of $\Omega$ ad the fact that $f\in C^1([0,T]\times\Omega)$, we can conclude from the last equality that $(1)$ holds. A similar argument works for $(2)$ as well.