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First of all I must apologize for the vague title and am open to suggestions.

This is not a Homework Assignment but something I once again encountered while reading a very compactly written paper.

$\tilde{\mathcal{C}}([0,1]^2,\mathbb{R}))$ is defined as follows for $f\in \tilde{\mathcal{C}}([0,1]^2,\mathbb{R}))$ the following holds:

  1. $f$ is continuous
  2. $f(x)=0 \quad \forall x\in [0,1]^2\setminus [0,1)^2$

I am looking for a countable family of continuous functions $\mathcal{F}:=(f_m)_{m\in\mathbb{N} }$ on $[0,1]^2$ so that the following requirements are satisfied

  1. $f\in \mathcal{F}\implies f(x)=0 \quad \forall x\in [0,1]^2\setminus [0,1)^2$
  2. the closure of the linear span of $(f_m)_{m\in\mathbb{N}}$ consists of all continuous Functions on $[0,1]^2$

According to the authors such a sequence is supposed to exist. But I do not know what the $f_m$ would like or why the family with the aforementioned properties exists. So I am looking for an explicit construction of the $f_m$ or a theorem which proofs the existence of such a sequence.

My first guess was $f_{n.m}(x_1,x_2):=x_1^nx_2^m$ for with Stone-Weierstrass it fullfills the second requirement. Unfortunately it does not meet the first one and I am somewhat at a loss.

Has anybody encountered a similar construction or knows of a theorem that might help?

As always thanks in advance :)

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I don't understand: the functions in $\mathcal F$ vanish on the set $\{1\}\times [0,1]$. So will do a function in the linear span of these functions, and a uniform limit on $[0,1]^2$ of such functions. – Davide Giraudo Jul 30 '12 at 11:22
If you take the closure wrt $||.||_{\infty}$ you will only get functions which satisfy the conclusion of condition 1. – user20266 Jul 30 '12 at 11:22
Functions which satisfy the first condition can have the form $(1-x)(1-y)f(x,y)$, where $f$ is continuous on $[0,1]^2$. As a wrote in my first comment, it seems that such a family cannot exist. Do you have a like for the paper you are reading? – Davide Giraudo Jul 30 '12 at 11:32
Thank you Davide - I edited the question. In it's initial form it was unlcear. I actually meant the space $\tilde{\mathcal{C}}([0,1]^2,\mathbb{R}))$ (as described above) and not $\mathcal{C}([0,1]^2,\mathbb{R}))$ like initially written. As I see it $(1-x)(1-y)f(x,y)$ seems already to answer my question. Just write you comment as an answer so that I can accept it :) – Probabilitnator Jul 30 '12 at 11:39
Another solution is $(1-x_1)^n(1-x_2)^m$. – Yuval Filmus Jul 30 '12 at 11:40
up vote 0 down vote accepted

We have that $\widetilde{\mathcal C}$ is a (closed) subspace of $\mathcal C[0,1]^2$. The latter is separable, and so will be $\widetilde{\mathcal C}([0,1]^2,\Bbb R)$. So a countable dense family $\mathcal F$ will do the job.

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