Are these spaces of continuous functions equivalent? I'm wondering if $C([0,T]\times\mathbb{R}^n)$ and $C([0,T];C(\mathbb{R}^n))$ are equivalent. I know I can show $C([0,T];C(\mathbb{R}^n))\hookrightarrow C([0,T]\times\mathbb{R}^n)$ through
$$|f(t,x)|\leq |f(t,x)-f(t',x)|+|f(t',x)|$$ by using uniform continuity in $t$ to $C(\mathbb{R}^n)$. However, when going the other direction, I know for fixed $x\in\mathbb{R}^n$, $f(t,x)$ is uniformly continuous in time, but I don't know if this uniform modulus of continuity is uniform over $x$ as it would be in $C([0,T];C(\mathbb{R}^n))$.
Edit since I can't comment: $C(\mathbb{R}^n)$ is the set of continuous, bounded functions from $\mathbb{R}^n$ to $\mathbb{R}$.
 A: If you consider $C_b([0,T]\times\mathbb{R}^n)$ and $C([0,T],C_b(\mathbb{R}^n))$ then both are Banach algebras with the supremum norm, and they are isometrically isomorphic as Banach algebras.
Namely both are algebras with the pointwise addition and multiplication, and with norms
$$
\|f\|=\sup\{\|f(t)\|:\ t\in[0,T]\},\qquad\qquad f\in C([0,T],C_b(\mathbb R^n)),
$$
$$
\|g\|=\sup\{|g(t,x):\ t\in[0,T],\ x\in\mathbb R^n\},\qquad\qquad g\in C_b([0,T]\times\mathbb R^n).
$$
One defines $\Psi:C([0,T],C_b(\mathbb R^n))\to C_b([0,T]\times\mathbb R^n)$ by
$$
(\Psi f)(t,x)=f(t)(x). 
$$
Because the operations are pointwise, it is trivial to check that this is an algebra homomorphism. We also have
\begin{align}
\|\psi f\|&=\sup\{|f(t)(x)|:\ t\in[0,T],\ x\in\mathbb R^n\}\\[0.2cm]
&=\sup\{\|f(t)\|:\ t\in[0,T]\}\\[0.2cm]
&=\|f\|.
\end{align}
So $\Psi$ is an isometry. In particular, it is injective. Finally, let $g\in C_b([0,T]\times\mathbb R^n)$. We define $f(t)\in C_b(\mathbb R^n)$ by $f(t)(x)=g(t,x)$. Then $f$ is continuous and $\Psi f=g$, showing that $\Psi$ is bijective. That is, we have shown that $\Psi$ is a bijective isometric isomorphism of algebras.
