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...

  • 1
    $\begingroup$ Try googling with "ellipse of convergence chebyshev polynomials" $\endgroup$ – Jean Marie May 8 at 11:15

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