for any Banach space $X,$ $C(L \times Q, X) $ is isometric to $C(L \times K, X)$ and so $C(Q,C(L,X))$ is isometric with $C(K,C(L,X))$.

Question

Currently I am reading the book 'Isometries on Banach spaces: Vector valued function spaces and operator spaces, Volume $2$.'

$C(X,Y)$ is the set of continuous functions $f:X \rightarrow Y$ with supremum norm $\| f \|_{\infty} = \sup_{x \in X}\| f(x) \|_Y.$

In chapter $7$, page $1,$ the authors quoted the following sentences:

Suppose that $Q,K$ are non-homeomorphic compact Hausdorff spaces and $L$ s a compact Hausdorff space such that $Q \times L$ is homeomorphic to $K \times L.$ For example, take $$Q = \{ (a,b): 1/2 \leq a^2+b^2\leq1; or \space 0\leq a \leq 2,b=0\},$$ where $(a,b)$ is a point in the Euclidean plane, $$K=\{ (a,b):1/2\leq a^2+b^2 \leq 1;or \space 1\leq a \leq 2,b=0; or \space a=0,1,1\leq b \leq 2 \},$$ and $L$ is the unit interval. Then for any Banach space $X,$ $C(L \times Q, X) $ is isometric to $C(L \times K, X)$ and so $C(Q,C(L,X))$ is isometric with $C(K,C(L,X))$.

Question: I do not understand the last sentence, namely: for any Banach space $X,$ $C(L \times Q, X) $ is isometric to $C(L \times K, X)$ and so $C(Q,C(L,X))$ is isometric with $C(K,C(L,X))$.

How to prove the above statement?

Answer 1

Suppose $$\varphi:C(L\times Q,X)\to C(L\times K,X)$$ is an isometry. Then so is $$\psi:C(Q,C(L,X))\to C(K,C(L,X))$$ given by (for $F\in C(Q,C(L,X))$ and $k\in K$, $\ell\in L$) $$\psi(F)(k)(\ell)=\varphi(G_F)(\ell,k)$$ where $G_F\in C(L\times Q,X)$ is given by $G_F(\ell',q')=F(q')(\ell')$. Because $$\|\psi(F)\|_{\infty}=\|\varphi(G_F)\|_{\infty}=\|G_F\|_{\infty}=\|F\|_{\infty}.$$ At least, this is what will have to be true along the lines of @Hurkyl's comment.

Answer 2

We use two properties ($A,B,U,V$ are assumed to be compact):

1. If $A$ is homeomorphic to $B$, then $C(A,X)$ is isometrically isomorphic to $C(B,X)$,
2. $C(U\times V,X)$ is isometrically isomorphic to $C(U, C(V,X))$

and the fact that a composition of isometric isomorphisms is an isometric isomorphism to obtain

$$C(Q,C(L,X)) \overset{2}{\cong} C(Q\times L, X) \overset{1}{\cong} C(K\times L, X) \overset{2}{\cong} C(K, C(L, X))$$

where "$\cong$" denotes the existence of an isometric isomorphism.

To see 1, let $\varphi \colon A \to B$ a homeomorphism. Then the induced pull-back $\varphi^{\ast} \colon f \mapsto f\circ \varphi$ is an isometric isomorphism $C(B,X) \to C(A,X)$. By continuity of $\varphi$, we have $\varphi^{\ast}(f) \in C(A,X)$ for all $f \in C(B,X)$, and $\varphi^{\ast}$ is easily seen to be linear. By the surjectivity of $\varphi$, we have $f(B) = \varphi^{\ast}(f)(A)$, so $\varphi^{\ast}$ is an isometry. And $(\varphi^{-1})^{\ast}$ is the inverse of $\varphi^{\ast}$, so $\varphi^{\ast}$ is an isometric isomorphism.

To see 2, we use currying. We define $\kappa \colon C(U\times V, X) \to C(U, C(V,X))$ by

$$\kappa(f) \colon u \mapsto f(u,\,\cdot\,).\tag{1}$$

Its inverse $\lambda \colon C(U, C(V,X)) \to C(U\times V,X)$ is given by

$$\lambda(g) \colon (u,v) \mapsto g(u)(v).\tag{2}$$

Once it is shown that $\kappa$ and $\lambda$ are indeed maps between the indicated spaces, the rest is easy to check: We have

$$\bigl(\lambda(\kappa(f))\bigr)(u,v) = \bigl(\kappa(f)(u)\bigr)(v) = \bigl(f(u,\,\cdot\,)\bigr)(v) = f(u,v)$$

and

$$\bigl(\kappa(\lambda(g))\bigr)(u)(v) = \bigl(\lambda(g)(u,\,\cdot\,)\bigr)(v) = \lambda(g)(u,v) = g(u)(v),$$

so $\lambda$ and $\kappa$ are indeed inverses of each other. The linearity is readily verified too. By compactness of the domain, for $f\in C(U\times V,X)$ there is a point $(u_0,v_0)$ with $\lVert f(u_0,v_0)\rVert = \lVert f\rVert_{\infty}$, and so

$$\lVert \kappa(f)(u_0)\rVert_{\infty} = \lVert \kappa(f)(u_0)(v_0)\rVert = \lVert f(u_0,v_0)\rVert = \lVert f\rVert_{\infty},$$

while

$$\lVert \kappa(f)(u)\rVert_{\infty} = \max_{v\in V} \lVert f(u,v)\rVert \leqslant \lVert f\rVert_{\infty},$$

hence $\kappa$ is an isometry. That $\lambda$ is an isometry follows from this, but can also be shown by a similar argument.

It remains to see that $\kappa$ and $\lambda$ are indeed maps between the indicated spaces. Since for every $f\in C(U\times V,X)$ and $u \in U$ the section $\kappa(f)(u) = v \mapsto f(u,v)$ is continuous, clearly $\kappa(f)$ is a map from $U$ to $C(V,X)$. The continuity of $\kappa(f)$ follows from the compactness of $V$. Let $u_0 \in U$ and $\varepsilon > 0$. By continuity of $f$, for every $v \in V$ there are neighbourhoods $M_v$ of $u_0$ and $N_v$ of $v$ such that $\lVert f(u,w) - f(u_0,v)\rVert \leqslant \varepsilon/2$ for $(u,w) \in M_v \times N_v$. By compactness of $V$, there is a finite $F \subset V$ such that

$$\bigcup_{v \in F} N_v = V.$$

Let $M = \bigcap_{v \in F} M_v$. Then $M$ is a neighbourhood of $u_0$, and for $u \in M$ and $w \in V$ we have

$$\lVert f(u,w) - f(u_0,w)\rVert \leqslant \lVert f(u,w) - f(u_0,v)\rVert + \lVert f(u_0,v) - f(u_0,w)\rVert \leqslant \frac{\varepsilon}{2} + \frac{\varepsilon}{2}$$

for a $v \in F$ with $w \in N_v$. Hence $\lVert\kappa(f)(u) - \kappa(f)(u_0)\rVert_{\infty} \leqslant \varepsilon$ for $u \in M$. Thus $\kappa(f)$ is continuous at $u_0$. Since $u_0\in U$ was arbitrary, indeed $\kappa(f) \in C(U, C(V,X))$.

Let finally $g \in C(U, C(V,X))$. To show $\lambda(g) \in C(U\times V,X)$, consider an arbitrary point $(u_0,v_0) \in U \times V$ and let $\varepsilon > 0$ be given. By continuity of $g$, there is a neighbourhood $M$ of $u_0$ such that $\lVert g(u) - g(u_0)\rVert_{\infty} \leqslant \varepsilon/2$ for $u \in M$. By the continuity of $g(u_0)$, there is a neighbourhood $N$ of $v_0$ such that $\lVert g(u_0)(v) - g(u_0)(v_0)\rVert \leqslant \varepsilon/2$ for all $v \in N$. Then

$$\lVert \lambda(g)(u,v) - \lambda(g)(u_0,v_0)\rVert \leqslant \lVert \lambda(g)(u,v) - \lambda(g)(u_0,v)\rVert + \lVert \lambda(g)(u_0,v) - \lambda(g)(u_0,v_0)\rVert \leqslant \frac{\varepsilon}{2} + \frac{\varepsilon}{2}$$

for $(u,v) \in M\times N$. Thus $\lambda(g)$ is continuous at $(u_0,v_0)$. Since $(u_0,v_0)$ was arbitrary, indeed $\lambda(g) \in C(U\times V,X)$.