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Let $K$ be a compact subset of $\mathbb{R}^2$. Let $P$ be a polynomial in the variables $x$ and $y$. Given $\epsilon>0$, can we find two polynomials $P_1=P_1(x)$ and $P_2=P_2(y)$ such that $ \sup_{x,y\in K}|P(x,y) - (P_1(x)+P_2(y))| \leq \epsilon ?$

My feeling is No, but I don't see how to prove it...

EDIT : Concerning my feeling, I had in mind the case when $K$ is the unit disc. As mentioned by Nate Eldredge in the comment below, the question depends on $K$. I consequently ask the question assuming that there exists no relation like $y=f(x)$ parametrizing $K$, where $f$ is continuous where $x$ runs.

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    You're right, that's also true if $K$ is the unit circle... Let me edit my post to specify the question. Thx !2012-01-03

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If $K$ contains the corners of the unit square, try taking $P$ with $P(0,0)=0, P(0,1)=0, P(1,0)=1, P(1,1)=-1.$ An appropriate transformation should cover any $K$ with nonempty interior.

Edit: Davide Giraudo asks about the (uniform) closure of the polynomials of the form $P(x) + Q(y)$ (denote the set of all such polynomials by $\mathcal{P}$). It is indeed given by the functions of the form $F(x,y) = f(x) + g(y)$ where $f,g$ are continuous. Let $\mathcal{A}$ denote the set of such functions. The Stone-Weierstrass theorem gives $\mathcal{A} \subset \overline{\mathcal{P}}$, as Davide mentions. For the other direction, note that $F \in \mathcal{A}$ if and only if it satisfies the equation $F(x,y) = F(x,y_0) + F(x_0,y) - F(x_0,y_0)$ for some arbitrary fixed $(x_0, y_0) \in K$. Since the right side is continuous in $F$ with respect to the uniform norm, it is easy to see that $\mathcal{A}$ is uniformly closed. Thus $\overline{\mathcal{P}} = \mathcal{A}$. This argument works for any compact $K$.

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    @DavideGiraudo: See my edit.2012-01-04