Let $\Gamma$ be a simple smooth($C^\infty$) curve in $\mathbb{C}$ parametrized by the injective map $\gamma:[0,1] \to \mathbb{C}$. Assume $f$ is a function defined on $\Gamma$ s.t. $f$ is sufficiently smooth in the sense that $f\circ \gamma (t) \in C^\infty([0,1])$, or $f\in C^\infty(\Gamma)$ where the space $C^\infty(\Gamma)$ is the one given in Chapter 6 of Gilbarg and Trudinger's PDE book.

Aim: I want to approximate $f$ by polynomials $p(z)$ where $z\in\Gamma$. NOTE that I'm not trying to approximate $f\circ\gamma$ by polynomials in $t\in [0,1]$.

1. Is the set of polynomials $p(z), z\in\Gamma$ dense in the space $C(\Gamma)$ of complex-valued continuous functions? What about in the space of $f$ satisfying smoothness conditions mentioned above ? (here we assume the norm is always the 'sup' norm in $C(\Gamma)$ )
2. If the answer to Question 1 is affirmative, can we estimate the error in polynomial interpolation ? More precisely, since the error is controlled by divided difference, we ask: what is the upper bound for $|f[z_0,\dots,z_m]|$ ? Is it bounded independent of choices of $z_i$? Here $z_0,\dots,z_m \in \Gamma$ are distinct points used in the interpolation and $f[\dots]$ denotes divided difference. A nice introduction on interpolation and divided difference can be found here.
3. If the answer to Question 1 is negative, can we succeed in the special case where $f\circ\gamma (t) := |\gamma'(t)|/\gamma'(t) =exp(-i*\text{arg} \gamma'(t))$ ?

1. Since the set of polynomials $p(z),z\in\Gamma$ is not closed under involution, we can not use Stone-Weierstrass to claim that it is dense in $C(\Gamma)$. Some may claim that it is not dense because uniform limit of analytic functions is analytic while $f$ is not. This reasoning fails since $\Gamma$ is not an open set in $\mathbb{C}$.

2. We know that it is almost always the case to simply interpolate $f\circ \gamma$ by polynomials $p(t)$ with $t\in \mathbb{R}$, so the method above is unorthodox and there are few results on estimating divided difference of complex interpolating points. If $f$ is analytic, an integral representation for divided difference was derived by Hermite, referred to as Genocchi-Hermite formula in this survey. However, in my case, $f$ has no analytic continuation. I can not even prove that whether the divided difference is bounded.

Any idea would be appreciated. Thank you.


1 Answer 1


$\Gamma$ is compact, has empty interior, and its complement is connected, so Mergelyan's theorem says that any continuous function on $\Gamma$ can be uniformly approximated by polynomials.

  • $\begingroup$ Ah, thank you for reminding me of this theorem as well as Runge's theorem alike. So the first question is solved. $\endgroup$
    – booksee
    Oct 15, 2015 at 23:34

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