Application of the Stone Weierstrass Theorem

[CCMarie]

How do I prove that:
If X and Y are two compact Hausdorff spaces and f : X × Y → R is a continuous function, then f is approximable by ∑ fi gi , wheret  f1, ...,  fn  in X and g1, ..., gn in Y are continuous functions.

As far as I read I need to use the Stone-Weierstarss Theorem to prove this.

I know that if X and Y are compact, X × Y is compact.

I should prove that {∑ fi gi } is a Banach algebra that separates points... and I don't know what I need to prove?

 

[fresh_42]

You first have to construct families of $\{\,f_i\,\}$ and $\{\,g_i\,\}$ from your given function.

 

[WWGD]

Just show that the family {figi} satisfy the conditions of Stone-Weirstrass. IIRC, they are an algebra, contain the constants and one I can't remember.

 

[Infrared]

@fresh_42 I think $f_i$ (resp. $g_i$) are arbitrary continuous real-valued functions defined on $X$ (resp. $Y$). You are given the algebra of functions on $X\times Y$ generated by products of functions on $X$ and $Y$. You don't need to construct anything.

@WWGD The last condition is that the algebra separates points.

Since compact Hausdorff spaces are normal, Urysohn's lemma implies that the algebra of continuous real-valued functions on such a space separates points. You can use this to check that the hypotheses of Stone-Weierstrass are met.

 

[fresh_42]

@fresh_42 I think $f_i$ (resp. $g_i$) are arbitrary continuous real-valued functions defined on $X$ (resp. $Y$). You are given the algebra of functions on $X\times Y$ generated by products of functions on $X$ and $Y$. You don't need to construct anything.

Yes, but we have to find them, i.e. their existence. Given is only $f$. Stone-Weierstraß has to provide the $f_i,g_i$. That's what it is used for. So first we have to split $f$ and then use the theorem for the algebra of the component functions.

The other way round is to find $\sum f_i = f$ and then the problem is to split them afterwards. I think before is easier, but maybe not.

 

[Infrared]

Why do you have to find them? The problem is only to prove that $f$ can be approximated by functions of the given form, not actually to construct an approximation. The hypotheses of SW are satisfied, so I don't see the issue.

 

[fresh_42]

Why do you have to find them? The problem is only to prove that $f$ can be approximated by functions of the given form, not actually to construct an approximation. The hypotheses of SW are satisfied, so I don't see the issue.

Yes, existence. But direct application gives $\sum f_i =f$ with $f_i\, : \,X\times Y \longrightarrow \mathbb{R}$ and we want $f_i\, : \,X\longrightarrow \mathbb{R}\, , \,g_i\, : \,Y\longrightarrow \mathbb{R}$ with $\sum f_ig_i =f$. So some split has to be done, before or after.

 

[Infrared]

Sorry, I don't understand your point. Let $\mathcal{A}$ be the algebra of functions $X\times Y\to\mathbb{R}$ of the form $(x,y)\mapsto \sum_{i} f_i(x)g_i(y)$, where $f_i,g_i$ are continuous on $X,Y$ and the sum is finite (but not of any given fixed length). This algebra satisfies the hypothesis of SW so for any $\varepsilon>0$ there exists a function $g\in\mathcal{A}$ such that $|g(x,y)-f(x,y)|<\varepsilon$ for all $(x,y)\in X\times Y$. This is what we wanted to show.

Also, Stone-Weierstrass does not say that we can write $f$ as being exactly equal to an infinite sum of the form $\sum f_ig_i$, just that we can (uniformly) approximate by finite sums. These are not equivalent.

 

[fresh_42]

Sorry, I don't understand your point. Let $\mathcal{A}$ be the algebra of functions $X\times Y\to\mathbb{R}$ of the form $(x,y)\mapsto \sum_{i=1}^N f_i(x)g_i(y)$, where $f_i,g_i$ are continuous on $X,Y$. This algebra satisfies the hypothesis of SW so for any $\varepsilon>0$ there exists a function $g\in\mathcal{A}$ such that $|g(x,y)-f(x,y)|<\varepsilon$ for all $(x,y)\in X\times Y$. This is what we wanted to show.

I thought that we cannot start with the special form, but that we had to use SW on the components. If you define the algebra already by products of functions "without mixed terms", then yes, the split is hidden in the construction of the algebra.

 

[CCMarie]

Thank you very much!