# Applications Stone-Weierstrass Theorem

Problem: Let $$X$$ and $$Y$$ be compact spaces. Prove that for any real-valued function $$f$$ on $$X \times Y$$ and any $$\epsilon > 0$$ we can find continuous real-valued functions $$g_1,g_2,g_3,\dots,g_n$$ on $$X$$ and $$h_1,h_2,h_3,\dots,h_n$$ on $$Y$$ such that

$$\left|f(x,y)-\sum_{i=1}^n g_i(x)h_i(y)\right| < \epsilon \quad (x,y) \in X \times Y$$

From what I understand I need to define $$\mathcal{A} = \left\{\sum_{i=1}^ng_1(x)h_i(y) : g_i \in C(X) \text{ and } h_i \in C(Y) \right\}$$. Then I need to prove that

1. $$\mathcal{A}$$ is an algebra
2. $$1 \in \mathcal{A}$$
3. $$\mathcal{A}$$ separates points.

Work so far:

1. Let $$p(x,y) = \sum_{i=1}^n a_i(x)b_i(y)$$ and $$q(x,y) = \sum_{i=1}^n c_i(x)d_i(y)$$ then we need to show that $$p+q \in \mathcal{A}$$ and $$pq \in \mathcal{A}$$. For the product of $$p$$ and $$q$$ I have that \begin{align*} p(x,y)q(x,y) = \sum_{i=1}^n\sum_{j=1}^n a_i(x)b_i(y)c_j(x)d_j(y) \end{align*} For each $$i$$ define $$m_i(x) = \left(a_i(x)\left[\sum_{j=1}^n c_j(x)\right]\right)$$ and $$n_i(y)= \left(b_i(y)\left[\sum_{j=1}^n d_j(y)\right]\right)$$ then both $$m_i$$ and $$n_i$$ are continuous since they are the sum and products of continuous functions and we get \begin{align*} p(x,y)q(x,y) = \sum_{i=1}^n m_i(x)n_i(y) \in \mathcal{A} \end{align*} As for the sum I cannot find a way to add them both up and get a summation of the form required in the algebra.
2. To show $$1 \in \mathcal{A}$$ we just need to let $$g_i(x) = 1 \; \forall \; i$$ and $$h_i(y) = \frac{1}{n} \; \forall \; i$$ and the sum equals to 1 so $$1 \in \mathcal{A}$$
3. We need to show to there exists functions $$g \in C(X)$$ and $$h \in C(Y)$$ such that if $$(x_1,y_1) \neq (x_2,y_2)$$ then there exist a $$z(x,y) \in \mathcal{A}$$ such that $$z(x_1,y_1) \neq z(x_2,y_2)$$ where $$z(x,y) = \sum_{i=1}^n g_i(x)h_i(y)$$. Would it be allowed to have $$g(x_1) = 1$$ and $$g(x_2) = 0$$ and $$h(y) = 1 \; \forall \; y$$ then $$g(x_1)h(y_1) \neq g(x_2)h(y_2)$$?

Let me know if any additional information that might be needed to complete the problem. Thank you.

• $n$ isn't fixed. Nov 26, 2019 at 19:41
• I am aware, thank you. Nov 26, 2019 at 19:57

In your definition of $$\mathcal A$$ you have a fixed $$n$$, but you have to take all finite sums $$\sum_{i=1}^n g_i h_i$$ with arbitrary $$n$$.
You will see that your proof works smoothly, it even becomes simpler for 2. (take $$n =1$$). Concerning 3. note that not necessarily $$x_1 \ne x_2$$. We only know that $$x_1 \ne x_2$$ or $$y_1 \ne y_2$$. Both cases can be treated as you did.
• What about for the first part of the question, even for $n=1$ I can't get it right for $p(x,y)+q(x,y)$ to be in the desired form Nov 27, 2019 at 0:27
• how do you get $a(x)b(y) + c(x)d(y) = \alpha(x)\beta(y)$ Nov 27, 2019 at 1:21
• I think I got it, if $n=1$ and you take $p(x,y)+q(x,y)$ then using the $n=2$ version of the sum then the problem is trivial correct? Nov 27, 2019 at 1:42