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Let $C(Y,X)$ be the space of continuous functions from $Y$ to $X$ together with the supremums norm. Here $Y$ is a compact space and $X$ a metric space.

Let $a,b,c,d \in \mathbb R$ be finite, with $a<b$ and $c<d$.

Now my question is which of the following 3 spaces are isomorphic:

  1. $C([a,b]\times[c,d],X)$
  2. $C([a,b],C([c,d],X))$
  3. $C([c,d],C([a,b],X))$

I think the first space is not isomorphic to the two others in the same way as discussed here: Continuity and joint continuity

But the other two (2. and 3.) should be isomorphic?

Are these claims true? (is there an easy proof?)

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mimi
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1 Answers1

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If $K$ and $L$ are compact metric spaces and $X$ is some metric space, then $C(K\times L,X)$ is actually isometric to $C(K,C(L,X))$. The natural definition would be the following: $$\phi:C(K\times L,X)\to C(K,C(L,X)),\qquad \phi(F)(k)(l)=F(k,l).$$

In principle, $\phi$ is actually just a function $C(K\times L,X)\to\operatorname{Functions}(K,\operatorname{Functions}(L,X))$. But remember that since we are working in compacts, continuity and uniform continuity are equivalent. Using this, you can show that for every $F\in C(K\times L,X)$ and $k\in K$, $\phi(F)(k)$ is actually a continuous function from $L$ to $X$, and uniform continuity again (of $F$) shows that $\phi(F)$ is continuous, so $\phi$ is well-defined. Also, you can show that $\phi$ is an isometry.

The natural mapping in the other direction is $$\psi:C(K,C(L,X))\to C(K\times L,X),\qquad \psi(f)(k,l)=f(k)(l),$$ and we also have to show that this is well-defined: For $(k,l)\in K\times L$, if $l'$ is sufficiently close to $l$, then $f(k)(l)$ is close to $f(k)(l')$, and if $k'$ is sufficiently close to $k$, then $f(k')$ is (uniformly) close to $f(k)$, hence $f(k')(l')\simeq f(k)(l')\simeq f(k)(l)$. This shows that $\psi$ is well defined, and it is clearly the inverse for $\phi$ as above.

Luiz Cordeiro
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