polynomial approximation on compacts Let's say $f:\mathbb{R}^d\rightarrow \mathbb{R}$ is of class $C^k$ with $k \geq 0$.  How do I know that I can find a sequence of polynomials such that all its derivatives up to order $k$ converge uniformly on compacts to the corresponding derivatives of $f$?  I can do this if d=1 by first solving the easier problem of approximating on a fixed compact set using the FTC, then using a diagonal argument to turn this into convergence on compacts for all of $\mathbb{R}^1$  The trouble is the FTC step doesn't work in multiple dimensions because you match the polynomial to f for one of the derivatives of order k, and then that says nothing about the other ones because taking derivatives of sup norm-close functions does not leave them close in sup norm.
If this is false, the motivation can be found here on p. 46 proof of theorem 3.23 http://www.statslab.cam.ac.uk/~beresty/teach/StoCal/sc3.pdf.  How else can the proof continue in the case of general d?
 A: This is a general theorem for locally compact spaces called the Stone-Weierstrass theorem. Polynomials are in fact dense in $C_0(X)$, the space of continuous functions on a locally compact space $X$. In particular, this theorem guarantees uniform convergence in all derivatives up to order $k$ (depending on the approximation).
There is a rather nice probabilistic proof of this statement. Since your tag includes probability, I'll sketch the argument for the continuous case on the compact interval $[0,1]$.  
Consider $f$ continuous on $[0,1]$ and real-valued. We claim that $f$ can be approximated by Bernstein polynomials of the form
$$(B_n f)(x) = \sum_{j=0}^n \binom{n}{j} x^j(1-x)^{n-j} f(j/n).$$
In particular, $B_nf$ is a polynomial of degree at most $n$ for all $n \geq 1$. 
To prove this, pick any $p \in [0,1]$ and define a countable family of Bernoulli random variables $X_i$. Let $S_n = \sum_{i=1}^n X_i$ so that $S_n$ is distributed according to $Bin(n,p)$. Note that $B_nf(p)$ is expectation of $f(S_n/n)$ (awesome!). 
One can bound $|(B_n f)(p) - f(p)|$ by the conditional expectation of $|f(S_n/n) - f(p)|$ given that $|S_n/n - p| \leq \delta$ and $|S_n/n - p| > \delta$, then use the weak law of large numbers to conclude that
$$ \sup_{p \in [0,1]} |B_nf(p) - f(p)| \rightarrow 0$$
as $n \rightarrow \infty$ and $\delta \rightarrow 0$. 
