The Lavrentiev's theorem is stated as follows
Let $K \subset \mathbb{C}$ be a compact set. Then every continuous function $f: K\to \mathbb{C}$ can be approximated uniformly by polynomials if and only if $K$ has no interior and $\mathbb{C}/ K$ is connected.
My question is, suppose $K$ is a one-dimensional curve on complex plane, under what condition every analytic function $f:K\to \mathbb{C}$ and its derivative $f'$ can be uniformly approximated by polynomials? That is to say, under what condition, we have $\forall f\in C^1(K, \mathbb{C}), \epsilon>0, \exists c_1, \ldots, c_N\in \mathbb{C}$ such that $\|f(z)-\sum_{i=1}^N c_k z^k\|_{\infty} < \epsilon$ and $\|f'(z)-\sum_{i=1}^N c_k k z^{k-1}\|_{\infty} < \epsilon$.
I saw there is a similar question on the extension of Stone-Weierstrass theorem. I'm wondering how this could be generalized to complex case. Any reference would also be good for me.
Best Answer
Let the curve be $\gamma:[0,1] \to K$. I am assuming that $\gamma$ is injective and Lipschitz so $K$ satisfies the conditions of the stated theorem (thanks to @ruens for pointing this out).
Note that $f(\gamma(t)) = f(\gamma(0)) + \int_0^t f'(\gamma(x)) \gamma'(x) dx$.
Choose $\epsilon>0$ and let $g$ be the $\epsilon$-uniform polynomial approximation to $f'$ on $K$ and define $\eta(t) = f(\gamma(0)) + \int_0^t g(\gamma(x)) \gamma'(x) dx $.
Note that $|f(\gamma(t))-\eta(t)| \le \int_0^t | f'(\gamma(x)) -g(\gamma(x))|\gamma'(x) dx \le \epsilon l(\gamma)$.
Let $g(z) = \sum_{k=0}^n g_k z^k$ and let $\phi(z) = f(\gamma(0))+\sum_{k=0}^n {g_k \over k+1} z^{k+1}$. Observe that $\eta(t) = \phi(\gamma(t))$.
Note that $|\phi(\gamma(t))-f(\gamma(t))| < \epsilon l(\gamma)$ and $|\phi'(\gamma(t))-f'(\gamma(t))| < \epsilon$.