Proof that $C(X)C(Y)$ is dense in $C(X \times Y)$

Question

I know from Stone-Weierstrass that the span of $C(X)C(Y)$ is in fact dense in $C(X \times Y)$. Nevertheless, I saw from an old guide from my university that they ask to prove $C(X)C(Y)$ (without taking the span) is dense in $C(X \times Y)$. The statement is the following:

Let $\phi \in C(X \times Y)$. Then prove that there exists $\{f_k\}_k \in C(X)$ and $\{g_k\}_k \in C(Y)$ sequences such that $f_k g_k \rightarrow \phi$ under the norm $\lVert \cdot \rVert_{\infty}$. $X$ and $Y$ are compact subsets of $\mathbf{R}^n$ and $\mathbf{R}^m$, respectively.

Is this statement true? Because as far as I saw from the proof of the Stone-Weierstrass theorem, it seems important for $\min\{h_1,h_2\}$ or $\max\{h_1,h_2\}$ to stay in the set for $h_1, h_2$ in the set, or that the set is a vector subspace closed under taking products in $C(X \times Y)$. The second condition implies the first up to taking limits because of the fact that there is $\{p_k(x)\}_k \in \mathbf{R}[x]$ such that $p_k\rightarrow |x|$ uniformly on any compact set.

Answer 1

This is not true. Indeed, take $X = Y = [0, 1]$ and $\phi(x, y) = \max\{x, y\}$. Choose $\epsilon > 0$ small enough s.t. $\epsilon < \frac{1}{4}$ and $1 - \epsilon > \frac{4\epsilon^2}{1 - 4\epsilon}$. Suppose $f \in C(X), g \in C(Y)$ are s.t. $\|fg - \phi\|_\infty < \epsilon$. Then $|f(0)g(1) - 1| < \epsilon$ and $|f(0)g(0)| < \epsilon$. We also have $|f(1)g(1) - 1| < \epsilon$, so $f(1), g(1) \neq 0$ and $|f(0) - f(1)| < \frac{2\epsilon}{|g(1)|}$. On the other hand, we also have $|f(1)g(0) - 1| < \epsilon$, so $g(0) \neq 0$, $|g(0) - g(1)| < \frac{2\epsilon}{|f(1)|}$, and,

$$\frac{2\epsilon}{|g(1)|} > |f(0) - f(1)| > \frac{1 - 2\epsilon}{|g(0)|} > \frac{1 - 2\epsilon}{|g(1)| + \frac{2\epsilon}{|f(1)|}}$$

So,

$$\begin{split} &\frac{2\epsilon}{|f(1)g(1)|} > \frac{1 - 2\epsilon}{|f(1)g(1)| + 2\epsilon}\\ \Longrightarrow\; &2\epsilon|f(1)g(1)| + 4\epsilon^2 > (1 - 2\epsilon)|f(1)g(1)|\\ \Longrightarrow\; &|f(1)g(1)| < \frac{4\epsilon^2}{1 - 4\epsilon} \end{split}$$

At the same time, $|f(1)g(1)| > 1 - \epsilon$, so,

$$1 - \epsilon < \frac{4\epsilon^2}{1 - 4\epsilon}$$

which is a contradiction with our choice of $\epsilon$.

Answer 2

The statement is not true for any non single point compact spaces $X$ and $Y$. To this end we will show that if a function $h(x,y),$ nonvanishing identically, is a pointwise limit of functions of the form $f_n(x)g_n(y),$ then $$ h(x_0,y_0)h(x,y)=h(x,y_0)h(x_0,y)\quad (*)$$ for any $(x,y)$ and any $(x_0,y_0)$ such that $h(x_0,y_0)\neq 0.$ Indeed we have $$ f_n(x)g_n(y_0)\to h(x,y_0),\ \ f_n(x_0)g_n(y_0)\to h(x_0,y_0)$$ Therefore $$ {f_n(x)\over f_n(x_0)}\to {h(x,y_0) \over h(x_0,y_0)}$$ Similarly $$ {g_n(y)\over g_n(y_0)}\to {h(x_0,y)\over h(x_0,y_0)}$$ We obtain $${h(x,y)\over h(x_0,y_0)}=\lim_n {f_n(x)\over f_n(x_0)}{g_n(y)\over g_n(y_0)}={h(x,y_0)h(x_0,y)\over h(x_0,y_0)^2}$$

Example $X=Y=[0,1]$ and $h(x,y)=x+y.$ Then $h$ does not satisfy $(*).$ Indeed for $x>0$ we have $$h(1,0)h(x,y)=x+y\neq x(y+1)=h(x,0)h(1,y)$$