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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? – t.b. May 10 '11 at 23:04
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. – Giuseppe Negro May 10 '11 at 23:47
The line up there got cut off it should be |f - p|<e – user10742 May 10 '11 at 23:55
Okay, very good then. This proof should thus be accessible for you. – t.b. May 11 '11 at 0:22
@dissonance, the proof of the Weierstrass Approximation Theorem by Bernstein is quite explicit. See… – lhf May 11 '11 at 2:09

Here is a way to exhibit the polynomials.

First step: on the unit cube As already said by others, if $f$ was defined and continuous on the cube $K=[0,1]^3$, a straightforward modification of Bernstein construction would do. Namely, one could approximate $f$ at $u=(x,y,z)$ in $K$ by $$ E\left(f\left(\frac{X^u_1+\cdots+X^u_n}n\right)\right), $$ for an i.i.d. sequence $(X^u_n)$ of random variables with mean $E(X_n^u)=u$ and support in $K$. For example, using Bernoulli random variables, the sequence of polynomials $(B_n(f))$ would converge uniformly to $f$ on $K$, with $$ B_n(f)(x,y,z)=\sum_{1\le i,j,k\le n}b^n_{i,j,k}(x,y,z)f\left(\frac{i}n,\frac{j}n,\frac{k}n\right), $$ where $B_n(f)$ is based on the elementary polynomials $b^n_{i,j,k}$, defined as $$ b^n_{i,j,k}(x,y,z)={n\choose i}x^i(1-x)^{n-i}{n\choose j}y^j(1-y)^{n-j}{n\choose k}z^k(1-z)^{n-k}. $$ Second step: on another cube Likewise, to approximate a continuous function $f$ defined on $K_3=[-3,3]^3$, at every point $(x,y,z)$ in $K_3$ one could use $$ B^{(3)}_n(f)(x,y,z)=\sum_{1\le i,j,k\le n}b^n_{i,j,k}\left(\frac{x+3}6,\frac{y+3}6,\frac{z+3}6\right)f\left(\frac{6i-3n}n,\frac{6j-3n}n,\frac{6k-3n}n\right), $$ Third step: on subsets of a cube For any $Y\subset K_3$ and any continuous function $f$ on $Y$, an obvious idea is to use the polynomial $B^{(3)}_n(g)$, where $g:K_3\to\mathbb R$ is defined by $g(x,y,z)=f(x,y,z)$ on $Y$ and $g(x,y,z)=0$ on $K_3\setminus Y$. The function $g$ is not continuous everywhere but $g$ is bounded and this is enough to ensure that $B^{(3)}_n(g)(x,y,z)$ converges to $g(x,y,z)$ at every point $(x,y,z)$ where $g$ is continuous. In particular $B^{(3)}_n(g)(x,y,z)$ converges to $f(x,y,z)$ at every point $(x,y,z)$ in the interior of $Y$. One can furthermore show that the convergence is uniform on every compact set included in this interior.

Last step: on $X$ All this yields the existence of polynomials approaching any function defined and continuous on $X$, uniformly on every $(1-t)X$. Now, if $f$ can be extended to a continuous function $\tilde f$ defined on $(1+t)X$ for a given positive $t$, since $X$ is a compact subset of the interior of $(1+t)X$, applying the preceding step to $\tilde f$ would yield polynomials $B^{(3)}_n(\tilde g)$ approximating $f$ uniformly on $X$.

To complete the proof, we now explain how to extend any continuous function $f:X\to\mathbb R$ to a continuous function $\tilde f:K_3\to\mathbb R$ in such a way that $\tilde f$ and $f$ coincide on $X$. Define $\tilde f$ at $u$ in $K_3\setminus X$ as follows. Start a 3D Brownian motion $(W^u_t)$ at $W^u_0=u$ and consider the hitting times $\tau_X$ and $\tau_K$ of the boundaries of $X$ and $K_3$ respectively by $(W^u_t)$. Then the random time $\min\{\tau_X,\tau_K\}$ is almost surely finite and one considers $$ \tilde f(u)=E(f(W^u_{\tau_X});\tau_X<\tau_K). $$ The continuity of $\tilde f$ on $K_3\setminus X$ and on the interior of $X$ are obvious and its continuity at the boundary of $X$ is standard. This completes the proof.

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