Approximation of inductive tensor product $C(X) \mathbin{\bar{\otimes}} C(Y)$

Question

The following question is from Banach Algebra Techniques in Operator Theory written by Ronald G. Douglas.

Assume both $X, Y$ are Banach spaces and $X \otimes Y$ is the algebraic tensor product. Let ${X^*}_{\leq1}$ be the closed unit ball in $X^*$. For $w \in X \otimes Y$, define $\|w\|_i = \sup\{\left|\sum_{k = 1}^n \phi(x_k) \psi(y_k)\right| : x_k \in X, y_k \in Y, w = \sum_{k = 1}^n x_k \otimes y_k\}$ (one of $w$'s expression in $X \otimes Y$), $\phi \in {X^*}_{\leq1}, \psi \in {Y^*}_{\leq1}$}. One can check this is a norm on $X \otimes Y$ and we let ($X \mathbin{\bar{\otimes}} Y, \| \cdot \|_i$) be the completion of the set ($X \otimes Y, \| \cdot \|_i$).

Now assume $X, Y$ are both compact Hausdorff topological spaces and hence ($C(X), \| \cdot \|_{\infty}$), ($C(Y), \| \cdot \|_{\infty}$) are Banach spaces. Show that ($C(X) \bar{\otimes} C(Y), \| \cdot \|_i$) is isometrically isomorphic to ($C(X \times Y), \| \cdot \|_{\infty}$). Here $X \times Y$ is equipped with the product topology.

Note that any norm $\|\cdot\|$ in $C(X) \oplus C(Y)$ (the direct sum of two Banach Spaces) is equivalent to $\|\cdot\|_1$ because both $C(X), C(Y)$ are Banach spaces equipped with $\|\cdot\|_{\infty}$ (hence $\|(f_x, f_y)\|_1 = \|f_x\|_{\infty} + \|f_y\|_{\infty}$. Meanwhile, one can find a homeomorphism between $C(X)\oplus C(Y)$ and $C(X\times Y)$ because $\|f\|_{\infty} \leq \|f_x\|_{\infty} + \|f_y\|_{\infty} \leq 2\|f\|_{\infty}$. Hence I directly start finding relation between ($C(X)\oplus C(Y), \|\cdot\|_1$) and ($C(X) \mathbin{\bar{\otimes}} C(Y), \|\cdot\|_i$)

$$\Large Question Part$$

Say $w \in C(X)\bar{\otimes} C(Y)$ and here I have difficulty finding upper bound of $\|w\|_i$ with respect to $\|\cdot\|_1$. Naively I consider partition of unity of $X$, say {$P_i, i \leq n$} and $\sum_{i \leq n}fP_i$ is one to break down $f$. Hence this could be one of the expression of $f$ part in $w$. I do not know if $n$ is the max number of pieces of $f$ I can break down.

According to hints in the book, by Krein-Milman, it suffices to consider extreme points in $X^*$ and $Y^*$. Before using this, I believe I need to collect enough information of $w$.

Answer 1

We will proceed through several steps.

At first consider a linear map $\mu: C(X) \otimes C(Y) \rightarrow C(X \times Y)$ that is induced by bilinear map $(f(x),g(y)) \rightarrow f(x)g(y)$.

$\mathit{Proposition \; 1}.$ $\mu$ is an injective linear map. $\mu(C(X) \otimes C(Y))$ is dense in $C(X \times Y)$.

$\mathit{Proof}$. The injectivity is obvious (it can be checked algebraically). Density follows from Stone-Weierstrass theorem. $\blacksquare$

Now consider a following construction. Let $\alpha \in C(X)'$ be a continuous linear functional and $f \in C(X \times Y)$. Then consider function $(\alpha \bullet f)(y) = \alpha(f(\cdot, y))$. It is a continuous function on $Y$ and $||\alpha \bullet f|| \le ||f|| ||\alpha||$. Therefore we can apply a functional $\beta \in C(Y)'$ to $\alpha \bullet f$. The resulting functional we will denote by $\alpha \otimes_1 \beta$. Also we can at first apply $\beta$ to $f$ (in variable $y$) and then apply $\alpha$. The resulting thing we will denote by $\alpha \otimes_2 \beta$.

$\mathit{Proposition\; 2}.$ $\alpha \otimes_1 \beta = \alpha \otimes_2 \beta$ (some sort of Fubini theorem). Also there holds an equality $||f|| = \sup\limits_{||\alpha|| \le 1, ||\beta|| \le 1} |(\alpha \otimes \beta)(f)|$ for $f \in C(X \times Y)$.

$\mathit{Proof}.$ Consider $f(x,y) = g(x) h(y)$. Then $(\alpha \otimes_1 \beta)(f) = \alpha(g) \beta(h) = (\alpha \otimes_2 \beta)(f)$. From density (proved above) follows equality $\alpha \otimes_1 \beta = \alpha \otimes_2 \beta$.

We have already proved that $|(\alpha \otimes \beta)(f)| \le ||\alpha || \; ||\beta||\; ||f||$ and therefore $||f|| \ge \sup\limits_{||\alpha|| \le 1, ||\beta|| \le 1} |(\alpha \otimes \beta)(f)|$. Now consider functionals $\delta^{X}_{x}(g) = g(x)$ and $\delta^{Y}_{y}(h) = h(y)$. Then $\sup\limits_{||\alpha|| \le 1, ||\beta|| \le 1} |(\alpha \otimes \beta)(f)| \ge \sup\limits_{x \in X, y \in Y} |(\delta^{X}_x \otimes \delta^y_y)(f)| = \sup\limits_{x \in X, y \in Y} |f(x,y)| = ||f||$. $\blacksquare$

Now consider a norm $|| \cdot||_i$ on $C(X) \otimes C(Y)$ that you have defined.

$\mathit{Proposition\; 3}.$ $||w||_i = ||\mu(w)||$ for all $w \in C(X) \otimes C(Y)$.

$\mathit{Proof}.$ This equality easily follows from an expression for norm $||\mu(w)||$ that was obtained in second proposition: $$\mu(w)(x,y) = \sum\limits_{i = 1}^{n}g_i(x) h_i(y) \Rightarrow (\alpha \otimes \beta)(\mu(w)) = \sum\limits_{i = 1}^n \alpha(g_i) \beta(h_i)$$. $\blacksquare$

$\mathit{Theorem}$. $\mu$ induces an isometric isomorphism between the completion of $C(X) \otimes_i C(Y)$ and $C(X \times Y)$.

$\mathit{Proof}$. Indeed, $\mu$ is an isometric map from $C(X) \otimes C(Y)$ onto a dense subspace in $C(X \times Y)$.