# Chebyshev series on the complex plane

Denote $$T_n(x) := \cos(n \arccos(x)),\,\, n\in \mathbb{N}$$ the Chebyshev polynomials. Let $$f$$ be a continuous function on $$[-1, 1].$$ It is well known that $$f$$ can be written in its Chebyshev series defined by $$f(x) = \sum_{n=0}^\infty a_n T_n(x),$$ where the coefficients $$\{a_n\}$$ are given by the formulas $$a_0 := \frac{1}{\pi}\int_{-1}^1\frac{f(x)}{\sqrt{1-x^2}}dx, \,\, a_n := \frac{2}{\pi}\int_{-1}^1\frac{f(x) T_n(x)}{\sqrt{1-x^2}}dx,\,\, n>1.$$

For $$r\geq 1,$$ Denote $$\Gamma_r := \left\{z \in \mathbb{C},\,\, z = \frac 12(r e^{i\theta}+ r^{-1} e^{- i\theta}), \theta \in [0, 2\pi]\right \}.$$ $$\Gamma_r$$ is called the Bernstein ellipse which has foci $$±1$$ and its major and minor semiaxis lengths summing to $$r.$$

I'm looking for a proof for the following theorem which extends the definition of the Chebyshev series of $$f$$ on the complex plane:

Theorem: The Chebyschev series is convergent in the interior of the greatest $$\Gamma_r$$ on wich $$f$$ is analytic. In other words, $$f(z)= \sum_{n=0}^\infty a_n T_n(z), z \in \Omega,$$ where $$\Omega$$ is the interior of interior of the greatest $$\Gamma_r$$ on wich $$f$$ is analytic.

There are different methods to define $$T_n$$ on $$\mathbb{C}.$$ For example, we can extend the definition of $$T_n$$ on the complex plane using the recurrence construction for $$T_n$$, so that $$T_0 = 1,\, \, T_1(z) = z,\,\, T_n(z) = 2z T_{n-1}(z) - T_{n-2}(z), n >1.$$

In many articles, the authors cite this book (Theorem 9.1.1 Page 245) where the above Chebyshev series is a special case ($$\alpha = \beta = -\frac 12$$) of "Jacobi" series, but I didn't found a proof for this general version.

Thank you for any hint...

• Try googling with "ellipse of convergence chebyshev polynomials" – Jean Marie May 8 at 11:15