Prove sets of continuous mappings are the same

Question

Let $C([0,T];C(\overline{U}))$ denote the set of all continuous functions $u:[0,T]\rightarrow C(\overline{U})$ with $$\|u\|_{C([0,T];C(\overline{U}))}:=\max_{0\leqslant t \leqslant T} \|u(t)\|<\infty$$

Prove that $C([0,T];C(\overline{U}))=C([0,T]\times \overline{U})$

I am skeptical this is even true. I feel like we could apply a theorem from topology regarding the product space, but am having little success. Not really sure how to approach such a problem. Are there any counter examples that disprove the above? Any help would be much appreciated.

Answer 1

I presume the norm $\ \|u(t)\| \ $ is intended to be $\ \sup_{v\in\overline{U}}\left|u(t)(v)\right|\ $. Unless the set $\ \overline{U}\ $ were suitably restricted this would not necessarily be finite. In what follows, I assume for concreteness that $\ \overline{U}\ $ is a compact subset of some Euclidean space, $\ \mathbb{R}^n\ $.

Strictly speaking, your suspicion that $\ C([0,T];C(\overline{U}))\ne$$C([0,T]\times \overline{U})\ $ is quite correct. The members of the first space are functions from $\ [0,T]\ $ to $\ C(\overline{U})\ $, while the members of the second are functions from $\ [0,T]\times \overline{U}\ $ to $\ \mathbb{R}\ $, which are completely different objects. However, there's a fairly obvious candidate, $\ \varphi:C([0,T];C(\overline{U}))\rightarrow C([0,T]\times \overline{U})\ $, for an isometric isomorphism between the two spaces, given by $$ \varphi(v)\left(t,u\right)=v(t)(u) $$ for $\ v\in C([0,T];C(\overline{U}))\ $, $\ t\in[0,T]\ $, and $\ u\in \overline{U}\ $. Therefore, since differently defined isomorphic mathematical objects are often thought of as being merely different representations of the same underlying structure, I expect the request to "[p]rove that $\ C([0,T];C(\overline{U}))=C([0,T]\times \overline{U})$" is using what is commonly called an "abuse of notation" to ask you to prove that the two spaces are isometrically isomorphic.

While I'm fairly sure that the mapping $\ \varphi\ $ I define above is indeed an isometric isomorphism, I must stress that I have not confirmed this belief by checking carefully that it satisfies all the conditions necessary for that to be the case. You will therefore need to do that for yourself

Answer 2

For the norm part, just note that \begin{align*} \|u\|_{C([0,T];C(\overline{U}))}&=\sup_{t\in[0,T]}\|u(t)\|\\ &=\sup_{t\in[0,T]}\sup_{w\in\overline{U}}|u(t)(w)|\\ &=\sup_{(t,w)\in[0,T]\times\overline{U}}|\varphi_{u}(t,w)|\\ &=\|\varphi_{u}\|_{C([0,T]\times\overline{U})}, \end{align*} where $u\rightarrow\varphi_{u}$ is defined by $\varphi_{u}(t,w)=u(t)(w)$ for $u\in C([0,T];C(\overline{U}))$.