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I try to prove:

If $f$ is analytic on an open disk $B(0, R) \subseteq \mathbb C$ where $R>0$ then the radius of convergence of its Taylor series is $\ge R$.

But I stuck with my proof. Please can someone help me? Here is what I have so far:

Let $f:B(0,R) \to \mathbb C$ be analytic on $B(0,R)$ and let $T_0(x) = \sum_{n = 0}^\infty c_n z^n$ denote its Taylor series at $0$. The radius of convergence is given by $$ r = {1 \over \lim \sup \sqrt[n]{|c_n|}}$$ where $$ c_n = {1 \over 2 \pi i} \int_\gamma {f(w) \over w^{n+1}}dw$$ where $\gamma$ is a counterclockwise curve around $0$ and contained in $B(0,R)$.

Now how to show $r \ge R$? If $\sqrt[n]{|c_n|} < {1\over R}$ it would be useful but how? Because $f(w)$ can be arbitrary.

blue
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1 Answers1

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To finish on the way you started, let $\gamma$ be a circle with centre $0$ and radius $0 < \rho < R$. Then the standard estimate tells you

$$\lvert c_n\rvert = \frac{1}{2\pi}\left\lvert\int_{\lvert w\rvert = \rho} \frac{f(w)}{w^{n+1}}\,dw\right\rvert \leqslant \frac{1}{2\pi} \int_0^{2\pi} \frac{\lvert f(\rho e^{i\varphi})\rvert}{\rho^n}\,d\varphi \leqslant \rho^{-n}\max \{\lvert f(\zeta)\rvert : \lvert\zeta\rvert = \rho\}.$$

That gives you an upper bound on $\limsup\limits_{n\to\infty} \sqrt[n]{\lvert c_n\rvert}$ in terms of $\rho$. As $0 < \rho < R$ was arbitrary, you get the desired conclusion.

Another way to obtain the result is to use Cauchy's integral formula,

$$f(z) = \int_{\lvert w\rvert = \rho} \frac{f(w)}{w-z}\,dw,$$

for $\lvert z\rvert < \rho$, and expand $\dfrac{1}{w-z}$ into a geometric series, which you know converges uniformly on the circle $\lvert w\rvert= \rho$. Interchanging summation and integration then yields the convergence of the Taylor series in $\lvert z\rvert < \rho$. Let $\rho\to R$.

Daniel Fischer
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  • Thank you! Why does the exponent go from $n+1$ to $n$ in the inequality? ($w^{n+1}$ $\to $ $\rho^n$) I don't understand how it will get the desired conclusion: Let $C=\max |f|$. Then $$ |c_n| \le {C \over \rho^n}$$ but because $\rho < R$ it does not follow that $$ |c_n| < {C \over R^n}$$ Doesn't $c_n$ have to be bounded by $R$? – blue Feb 01 '14 at 15:16
  • Writing $w = \rho e^{i\varphi}$, we have $dw = i \rho e^{i\varphi},d\varphi$, or $d\varphi = \dfrac{dw}{iw}$; "one $w$ is eaten by the $dw$". When you have $$\lvert c_n\rvert \leqslant \frac{C}{\rho^n},$$ then you get $$\sqrt[n]{\lvert c_n\rvert} \leqslant \frac{\sqrt[n]{C}}{\rho}.$$ So $\limsup\limits_{n\to\infty}\sqrt[n]{\lvert c_n\rvert} \leqslant B(\rho)$. But the left hand side doesn't depend on $\rho$, so you can pass to the limit on the right hand side. – Daniel Fischer Feb 01 '14 at 15:22
  • Thank you! And what is $B(\rho)$? Do you write $B(\rho)$ for the limit ${1 \over \rho}$ or does it have a different meaning? So is $\sqrt[n]{C} \to 1$! I did not know. Thank you very much for your patience. – blue Feb 01 '14 at 19:31
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    $B(\rho)$ stands for an expression depending on $\rho$ that bounds $\limsup \sqrt[n]{\lvert c_n\rvert}$. Since you know that $\sqrt[n]{C}\to 1$, you know (that you can choose) $B(\rho) = \dfrac{1}{\rho}$. We don't need the substitution, we can also use $\lvert \int_\gamma g(z),dz\rvert \leqslant \operatorname{length}\gamma \cdot \sup {\lvert g(z)\rvert : z \in \operatorname{Tr}\gamma}$. Since the length of the circle is $2\pi\rho$, that gives the same result. But to prove that inequality, you expand $dz = \gamma'(t),dt$, so in the end it's quite the same. – Daniel Fischer Feb 01 '14 at 19:38