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$\displaystyle X= \left \{\frac{x^2}{2} + \frac{y^2}{3} + \frac{z^2}{6} \leq 1 \right \}$ is a compact set

If $f(x,y,z)$ is continuous on $X$, then for any $\epsilon \gt 0$, there exists a polynomial $p(x,y,z)$ such that $|f - p|\lt \epsilon$ on $X$.

I need to prove this and I have no idea how.

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    Do you know the Weierstrass approximation theorem?2011-05-10
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    Do you need to exhibit such $p$ *explicitly*? This is a difficult problem in approximation theory (as far as I know). On the contrary, proving the lone existance of $p$ is easy if you have some tools - the Stone-Weierstrass theorem, for example.2011-05-10
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    The line up there got cut off it should be |f - p|2011-05-10
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    but what we went over in class was how to prove this with only the x variable,, and now I have to do it in three dimensions. I do Know the Weierstrass theorem but im not sure how to adapt it into three dimensions.2011-05-10
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    Can you expand a little bit on how exactly you did it for one variable? There are many proofs of that theorem.2011-05-11
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    it was a lot like the general proof at the bottom of this page http://planetmath.org/encyclopedia/ProofOfWeierstrassApproximationTheorem.html2011-05-11
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    Okay, very good then. [This proof](http://planetmath.org/encyclopedia/ProofOfWeierstrassApproximationTheoremInRn.html) should thus be accessible for you.2011-05-11
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    @dissonance, the proof of the Weierstrass Approximation Theorem by Bernstein is quite explicit. See http://en.wikipedia.org/wiki/Bernstein_polynomial#Approximating_continuous_functions2011-05-11
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    @lhf: Oh yes, I had heard of that proof during a course in probability theory. However, I do not see an immediate generalization of that construction from the interval $[0, 1]$ to the set $X$ the OP refers to. Should we use vector-valued random variables, maybe...? This is the difficulty I referred to.2011-05-11
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    @dissonance, I haven't checked any details, but doesn't tensor products work? I mean, $B_n(x)B_n(y)B_n(z)$?2011-05-11
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    @lhf: I think that there is some more work to do. For a univariate $f\colon [0, 1] \to \mathbb{R}$, its $n$th Bernstein polynomial is $$B_n[f](x)=\sum_{k=0}^n {n \choose k} x^k (1-x)^{n-k}f\left(\frac{x}{k}\right).$$ How to extend this to a multivariate $F$? If it is defined on $[0, 1] \times [0,1] \times [0, 1]$, we can build a tensor product of Bernstein polynomials, maybe like this: $$B_{n+m+l}[F](x, y, z)=B_n[F(\cdot, y, z)](x)B_m[F(x, \cdot, z)](y)B_l[F(x, y, \cdot)](z).$$ Will this thing converge to $F$? Also, what to do if $F$ is not defined on a cube, but on something different?2011-05-11
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    A reference on multivariate Bernstein polynomials: [Korovkin-type theorems and approximation by positive linear operators](http://www.math.technion.ac.il/sat/papers/13/13.pdf) pag.114 (23). The author talks about Bernstein polynomials on the hypercube and on the $n$-simplex. From this last construction, one could build an explicit approximation by means of Bernstein polynomials over any compact convex subset of $\mathbb{R}^n$.2011-05-11

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