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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.

mw19930312
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  • I think that the proof in the posted link does work for single variable complex functions, the issue seems to be multivariate. – N. S. Jan 10 '20 at 23:41

1 Answers1

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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 $|\phi'(\gamma(t))-f'(\gamma(t))| < \epsilon$.

copper.hat
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  • You should make it clear what you assume on $f',K$ or which theorem when saying $g$ exists – reuns Jan 11 '20 at 00:10
  • @reuns: $K$ is the range of the curve $\gamma$, and since Lavrentiev's theorem is quoted, it seemed reasonable to use it without explicit reference. I am not sure what you mean about assumptions on $f'$ other than being continuous on $K$. – copper.hat Jan 11 '20 at 00:12
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    And $K$ has no self-intersection (since $1/z$ cannot be approximated by a polynomial on the unit circle) – reuns Jan 11 '20 at 00:17
  • @reuns: Thanks, missed that. – copper.hat Jan 11 '20 at 00:18
  • @copper.hat Thanks for the answer. What about the case where $\gamma$ is non-injective? Do you make this assumption just to apply the Lavrentiev's theorem? – mw19930312 Jan 13 '20 at 17:17
  • @copper.hat Can we say that if $\mathbb{C}/K$ is not connected, then there exists differentiable function that cannot be approximated by polynomials? (Lavrentiev's theorem only guarantees the existence of such continuous function.) – mw19930312 Jan 13 '20 at 17:22
  • If $\mathbb{C} \setminus K$ is not connected then all bets are off? There could be crossovers as @reuns mentioned. – copper.hat Jan 13 '20 at 17:33
  • @copper.hat Sorry I should have been clearer. What I am asking is if $\mathbb{C} \setminus K$ is not connected, then Lavrentiev's theorem says there exists a continuous function that cannot be approximated. Is there any other theorem saying that there also exists a differentiable function that cannot be approximated? – mw19930312 Jan 13 '20 at 20:27
  • I don't know. ${}{}$ – copper.hat Jan 13 '20 at 20:59