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.


$\Gamma$ 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 '15 at 23:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.