TASK: given an interval $[a,b]\subset \mathbb R ^n$ and $f \in C^0[a,b]$, show that $f$ can be uniformly approximated by polynomials.
HINT: Exercise is given in the context of PDE (so I guess one isn't asked to copycat Weierstrass polynomial approximation theorem) and the task is given the following hint: use continuation of $f$ and corresponding solution of the heat equation.
IDEA: Let $\Phi$ be heat kernel, i.e.: $$\Phi(t,x) = \frac{1}{\sqrt{4\pi t}} \exp (\frac{-|x|^2}{4t}).$$ Let's extend $f$ by
$$\tilde f(x)=\begin{cases} f(x) &x\in[a,b]\\0 &\text{otherwise}\end{cases}$$
Then we can define $u$ as the solution of the heat equation
$$\begin{cases}\partial_t u - \partial_{xx}u = 0 & t>0\\u=\tilde f &t=0\end{cases}$$
so $u$ is given as the convolution $u(t,x)=\Phi(t)\ast \tilde f$.
Then a natural approach would be to try to approximate $f$ by going backwards, for example:
$$g_n(x) = u\left(\frac 1n, x\right).$$
Since $\tilde f$ is continuous and has compact support, we know that $ g_n \rightarrow \tilde f$ in $C^0[a,b]$.
So now the question is, how do I relate this problem with the polynomials, considering that $g_n$ is not a polynomial? So if I understand correctly, at this point I have to find some mapping $P:g_n\mapsto h_n$ s.t. $g_n$ is like the one above and $h_n$ is polynomial of $n$-th degree. Naturally Taylor series come to mind (notice that it wasn't possible to approximate via Taylor series from the beginning since $f$ is only in $C^0$), so we could take the Taylor series at $0$ up to $n$-th degree. But for each function $h_n$ the polynomials will have variable constants if proceed in a straightforward way, so I don't seem to find the exact way to define the polynomial using the solution of the heat equation. Hence asking you for hints and clues.
Thank you in advance!
Try approximating the heat kernel $\Phi$ by a truncated Taylor series in $x$. Then you have an approximation of $u$ that is a convolution of a polynomial with $f$ (which turns out to be again a polynomial).