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Show that polynomials $P(x,y)=p(x)q(y)$ are dense in $C([0,1]^2,\mathbb{R})$ (i.e. the set of continous functions in two variables).

My attempt:

Based on inspiration from Multivariate Weierstrass theorem?

$M=[0,1]\times[0,1]$, $A=\{p:[0,1]\times[0,1] \rightarrow \mathbb{R}, \text{p are polynomials}\}$

Since $M$ is a compact metric space and $A\subset C([0,1]^2,\mathbb{R})$ is a function algebra that separates points and that vanishes nowehere, then by Stone-Weirestrass, $A$ is dense in $C([0,1]^2,\mathbb{R})$.

That A is closed under addition, multiplication, and scalar multiplication is clear. To show that it separates points, take $(r_1,r_2),(s_1,s_2) \in M $ not equal. Then $P(x,y)=xy$ separates points. The $P(x,y)=1$ vanishes nowhere.

I think that this is a correct proof of the statement.

However, I also think that there is a way of using only the Weirestrass approximation theorem since the special structure of the polynomials are $P(x,y)=p(x)q(y)$. Following Rudin, how would you in that case choose the polynomials that in one variable looks like $Q_n(x)=c_n(1-x^2)^n$,?

user5744148
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  • Well, $xy$ may not separate some points i may pick... – dan_fulea Aug 09 '21 at 13:50
  • @dan_fulea I guess (1,2) and (2,1) would both give P(x,y)=2, right? – user5744148 Aug 09 '21 at 13:52
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    Your $A$ is dense, by S-W, fine. But $A$ is not the same as the set of polynomials of the form $p(x)q(y)$! For example $x+y\ne p(x)q(y)$. And in fact the set of all polynomials of that form is not dense... – David C. Ullrich Aug 09 '21 at 13:53
  • There is not so simple to follow Rudin, since Rudin has written a lot... Which particular construction is meant? Please state a clear question with a clear setting, best without a long introduction, if this introduction has no meaning for the question.... How to choose $Q_n$ so that what happens... ?! – dan_fulea Aug 09 '21 at 13:53
  • @dan_fulea In Baby Rudin, when he proofs Stone-Weierstrass, he chooses polynomials $Q_n$ as above and then convolve them with the continous function f to find a polynomial $P_n(x)$ that is later shown to approximate the original function f – user5744148 Aug 09 '21 at 13:58
  • @DavidC.Ullrich But can't polynomials all polynomials in A be written as \sum_{n,m}cx^{n}y^{m}, so aren't the case when the polynomials can be separated included in A, or maybe I am missing something? – user5744148 Aug 09 '21 at 14:02
  • Say $\prod$ is the set of all polynomials of the form $p(x)q(y)$. Then yes, every polynomial is the sum of finitely many elements of $\prod$. That has no bearing on the fact that $\prod$ is not dense. $\prod$ is a proper subset of $A$. I can't quite parse "so aren't the case when the polynomials can be separated included in A, or maybe I am missing something?", sorry. – David C. Ullrich Aug 09 '21 at 14:06
  • @DavidC.Ullrich Okay, then I am probably completely wrong in my approach then, but it seemed like it should be something in that direction based on other posts I have seen here for example https://math.stackexchange.com/questions/1036439/multivariate-weierstrass-theorem. How would you approach it? – user5744148 Aug 09 '21 at 14:10
  • That other question doesn't even mention $\prod$ as far as I can see, so I don't see the relevance. How would I approach it? How would I approach what??? – David C. Ullrich Aug 09 '21 at 14:12
  • "wrong in your approach": You say you want to show that $\prod$ is dense. Then you give a correct proof that $A$ is dense. Showing $A$ is dense is not a good approach to showing $\prod$ is dense, especially since$\prod$ is actually not dense... – David C. Ullrich Aug 09 '21 at 14:13
  • @DavidC.Ullrich How you would approach to show that polynomials P(x,y)=p(x)q(y) are dense in $C([0,1]^2,R)$ – user5744148 Aug 09 '21 at 14:13
  • What? I've said several times that that is not true. Are you actually meaning toask me how i'd approach proving something false? – David C. Ullrich Aug 09 '21 at 14:14
  • @DavidC.Ullrich Sorry, I thought I had deleted the previous comment, because I wrote it before your later reply appeard. I think I am lost here, so bear with me. I have shown that A is dense, but if I understand you correctly, that doesn't show what I am originally am looking for. So instead what I want to show is that ∏ is dense, but, again if I understand you correctly, it is not dense. So far I am with you. The thing is that this is an exam question from a couple of years back so my assumption was that it is correctly formulated – user5744148 Aug 09 '21 at 14:25
  • Again, it feels like similar questions has appeared here before, another example is, https://math.stackexchange.com/questions/1841253/showing-a-2d-continuous-function-can-be-approximated-by-a-finite-sum-of-simpler, but maybe it is the structure of the polynomials in this specific case that makes it different – user5744148 Aug 09 '21 at 14:28
  • the two most likely possibilitites are (i) the exam question was wrong, (ii) you're not stating the question correctly. For example, the span of $\prod$ is dense; leaving out the one word "span" changes it from true to false... – David C. Ullrich Aug 09 '21 at 14:29

1 Answers1

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For convenience let $\prod$ (for "product") denote the set of all polynomials of the form $p(x)q(y)$. In fact $\prod$ is not dense in $C([0,1]^2)$. Your algebra $A$ is dense, by the S-W theorem, but $\prod$ is a proper subset.

Details: First, note that if $P\in\prod$ then $$P(0,0)P(1,1)=P(0,1)P(1,0).$$So if $P_n\in\prod$ and $P_n\to f$ then $f(0,0)f(1,1)=f(0,1)f(1,0)$; hence $f(x,y)\ne x+y$.

David C. Ullrich
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  • Great. Then this is a proof that the professor must have missed a "not dense" the question. I have one (probably basic) question about this example. Where did $P(0,0)=0$ come from and doesn't that assume that at least one of $p(x)$ and $q(y)$ have no constant, right? So how does the final conclusion that it is not dense follow from this? My knowledge on this area is pretty shaky right now – user5744148 Aug 09 '21 at 15:38
  • and the only way I have seen to show that something is not dense, is to show that an element (in this case a continous founction) is not a limit point of $P(x,y)$. As previous stated, one such example I guess would be $x+y$ which is not in $P(x,y)$ – user5744148 Aug 09 '21 at 15:50
  • @user5744148 I came up with a perhaps simpler argument... – David C. Ullrich Aug 09 '21 at 15:59