Proof that Radius of Convergence Extend to Nearest Singularity

Question

Can someone provide a proof for the fact that the radius of convergence of the power series of an analytic function is the distance to the nearest singularity? I've read the identity theorem, but I don't see how it implies that the two functions must be equal everywhere.

Answer 1

http://en.wikipedia.org/wiki/Analyticity_of_holomorphic_functions

I wrote the initial draft of the article linked above in February 2004, and mostly it's still as I wrote it, although others have contributed.

Postscript:

Let $C$ be a positively oriented circle centered at $a$ that encloses a point $z$ that is closer to $a$ then is any place where $f$ blows up, and that does not enclose, nor pass through, any point where $f$ blows up.

\begin{align}f(z) &{}= {1 \over 2\pi i}\int_C {f(w) \over w-z}\,dw \tag1 \\[10pt] &{}= {1 \over 2\pi i}\int_C {f(w) \over (w-a)-(z-a)} \,dw \tag2 \\[10pt] &{}={1 \over 2\pi i}\int_C {1 \over w-a}\cdot{1 \over 1-{z-a \over w-a}}f(w)\,dw\tag3 \\[10pt] &{}={1 \over 2\pi i}\int_C {1 \over w-a}\cdot{\sum_{n=0}^\infty\left({z-a \over w-a}\right)^n} f(w)\,dw\tag4 \\[10pt] &{}=\sum_{n=0}^\infty{1 \over 2\pi i}\int_C {(z-a)^n \over (w-a)^{n+1}} f(w)\,dw\tag5 \\[10pt] & = \sum_{n=0}^\infty (z-a)^n \underbrace{{1 \over 2\pi i}\int_C {f(w) \over (w-a)^{n+1}} \,dw}_{\text{No $z$ appears here!}}.\tag6 \end{align}

Step $(1)$ above is Cauchy's formula.

Step $(4)$ is summing a geometric series.

Step $(6)$ can be done because "$(z-a)^n$" has no $w$ in it; thus does not change as $w$ goes around the circle $C$.

The fact that no "$z$" appears in the expression in $(6)$ where that is noted, means that the last expression is a power series in $z-a$.

Answer 2

The answer to this question is a bit hard to find, because there is no general agreement on the name of the relevant theorem. Ahlfors, for example, just calls it « Theorem 3 » in the standard work on Complex Analysis that he wrote (2nd edition, p. 177). The point to remember is that it comes under the heading of « Taylor Series ». You need to take a look at how Cauchy's formula is applied to write down an integral expression for the remainder term of the finite Taylor series, which is expressed as a complex line integral that follows a circle close to the edge of the largest disk where the function f is holomorphic. From the obtained expression for the remainder term, you will see that it approaches zero as the number of terms in the series is increased, which in its turn means that the series is convergent, and that it converges to the value of f at the centre of the circle. And voila! There you have the theorem. Henri Cartan has a very clear statement and proof of the theorem (Théorie élementaire, Paris 1961, p. 73), and you will also find it in Ruel V. Churchill's book.

Answer 3

Suppose $f$ is holomorphic in an region $\Omega\subset \mathbb{R}$. Then, for any $a\in \Omega$ $f$ admits an expansion as a power series within a neighborhood of $a$. To be more precise, let $R:=d(a,\mathbb{C}\setminus\Omega)=\{|w-a|:w\in \mathbb{C}\setminus \Omega\}$. For any $0<r<R$, define $B_r(a)=\{z\in\mathbb{C}:||z-a|<r\}$. Then, for $w\in \partial B_r(a)=\{w\in\mathbb{C}: |w-a|=r\}$ and $z\in B_r(a)$ $$\Big|\frac{z-a}{w-a}\Big|<\frac{|z-a|}{r}<1$$ Hence $$\frac{f(w)}{w-z}=\frac{f(w)}{w-a}\frac{1}{1- \frac{z-a}{w-a}}=\sum^\infty_{n=0}\frac{f(w)}{(w-a)^{n+1}}(z-a)^n$$ and the convergence is uniform and absolute on any compact subset of $B_r(a)$. Consequently, integrating along the curve $\gamma_r(t)=a+re^{it}$, $0\leq t\leq 2\pi$, can be done by exchanging the order of summation and integration. Moreover, by Cauchy's theorem we have that:

1. For $z\in B_r(a)$ \begin{align} f(z)=\frac{1}{2\pi i}\int_\gamma\frac{f(w)}{w-z}\,dw=\sum^\infty_{n=0}c_n(z-a)^n,\tag{0}\label{0} \end{align} where $c_n=\frac{1}{2\pi I}\int_{\gamma_r}\frac{f(w)}{(w-a)^{n+1}}\,dw$.
2. The integral expression for each $c_n$ is independent of $r$, that is $$\int_{\gamma_r}\frac{f(w)}{w-a}\,dw= \int_{\gamma_s}\frac{f(w)}{w-a}\,dw$$ for all $0<r,s<R$.

The power series in the RHS of \eqref{0} converges uniformly and absolutely inside a ball of radius $$\rho=\frac{1}{\limsup_n\sqrt[n]{|c_n|}}$$ that is,

3. the power series in the RHS of \eqref{0} converges for all $z\in B_\rho(a)$, and defines a Holomorphic function on $B_\rho(a)$.

4. The power series diverges for all $z$ with $|z-a|>\rho$.

5. Consequently, $f$ is holomorphic (or rather can be extended as an holomorphic function) on $\Omega\cup B_\rho(a)$. Furthermore, by (1.) and (2.), $ r\leq \rho$ for all $0<r<R$ and so, $R\leq \rho$.

6. It follows that on the boundary of convergence $S_\rho(a):=\partial B_\rho(a)=\{w\in\mathbb{C}:|w-a|=\rho\}$, there must be a point $p$ at which $f$ has no analytic continuation, that is, there is no neighborhood $V$ around $p$ and an no function $g\in H(V)$ such that $f=g$ on $B_r(a)\cap V$. Such a point is called singular point. If that were not the case, then it would be possible to enlarge to convergence of the power series \eqref{0} describing $f$ to a larger ball $B_{\rho+\varepsilon}(a)$, $\varepsilon>0$. This is a standard compactness argument and the fact that two holomorphic $f$ and $g$ functions defined on an open connected set $D$ that coincide in a set $A\subset D$ having limit points in $D$ are in fact equal on $D$. This extension would lead to a contradiction since the power series diverges for all $|z-a|>\rho$. Hence $\rho$ is indeed the distance to from $a$ to the nearest singular point.

7. A singular point $p\in \partial S_r(a)$ does not have to be pole, or an essential singularity or a branch point a branch point. Furthermore, it might also be that all points on $S_r(a)$ are singular. In such a case the circle $S_r(a)$ is called a natural boundary. This is true in particular for Hadamard lacunary power series.

Examples

8. $f(z)=\frac{1}{1-z}$ is holomorphic on $\mathbb{C}\setminus\{1\}$. $f$ has a power series representation $f(z)=\sum^\infty_{n=0}z^n$ for all $|z|<1$. The radius of converges of this series is $1$. The only singular point on the boundary of convergence $\mathbb{S}^1$ occurs at $z=1$ which happens to be a pole of order $1$.

9. If $\log$ is the priuncipal branch of logarithm, $f(z)=\log (1-z)$ is holomphic on $\mathbb{C}\setminus[1,\infty)\times\{0\}$. Around $a=0$, $f$ has the power expansion $f(z)=-\sum^\infty_{n=0}\frac{z^n}{n}$ and this power series has radius of convergence $1$. The point $p=1\in\mathbb{S}^1$ is a singular point; a branch point type of singularity.

10. The dilogarithm function $f(z)=\sum^\infty_{n=0}\frac{z^n}{n^2}$ is holomorphic on $B_1(0)$. The power series has radius of convergence $1$. The function $f$ can be extended continuously to the closed ball $\overline{B_1(0)}$. It can be extended analytically to $\mathbb{C}\setminus[1,\infty)\times\{0\}$. $p=1\in\mathbb{S}^1$ is a singular point; a branch singularity.

11. The function $f(z)=\frac{z}{(z-1)^2}$ is holomorphic on $\mathbb{C}\setminus\{1\}$. It has the power series repersentation $f(z)=\sum^\infty_{n=0}nz^n=$ around $a=0$ and has radius of convergence $1$. $z=1$ is a pole of order $2$ for $f$, the power series diverges at every point on the boundary of convergence $\mathcal{S}^1$.

12. The function $f(z)=e^z$ is entire, it admits power series $f(z)=\sum^\infty_{n=0}\frac{z^n}{n!}$ around $a=0$. The radius of convergence of this series of of course $\infty$. $f$ has no singular points on $\mathbb{C}$ (the boundary of convergence is $\emptyset$.)

13. If a power series $f(z)=\sum^\infty_{n=0}c_n z^{p_n}$, is such that $1=\limsup_n\sqrt[n]{|c_n|}$ and $p_{n+1}>\Big(1+\frac1\lambda\big)p_n$, $k\in\mathbb{Z}_+$, for some integere $\lambda>0$, then every point in the boundary of convergence $\mathbb{S}^1$ is a singular point. This is a Theorem by Hadamard.

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Details to all facts stated in this posting are covered in many books on Complex Analysis, for example Rudin W., Real and Complex Analysis, third Edition, McGraw Hill, 1986., or Simon, B., Basic Complex Analysis 2A, AMS, 2015.

Answer 4

Here we use as mentioned in OPs post the Identity Theorem to show the existence of a singularity on the boundary of the disc of convergenc of an analytic function. We start with a sketch of the proof.

Sketch of the proof: The proof is done by contradiction.

• We assume there is no singularity on the boundary of the disc $B_R(c)$ of convergence at $c$ with radius $R$ of an analytic function $f$.

• We can then find for every point of the boundary a small disc and an analytic function defined there which is identical with $f$ when restricted on their common domain.

• Since the boundary $\partial B_R(c)$ is compact, we can choose a finite number of small discs and corr. analytic functions which cover the boundary of the disc $B_R(c)$.

• With the help of these finite number of small discs together with $B_R(c)$ we obtain an enlarged disc $\tilde{B}_R(c)$ of convergence on which a function can be defined which coincides with $f$ due to the Identity Theorem.

• We so have enlarged the radius of convergence of $f$. This is a contradition invalidating the assumption there is no singularity.

Since we closely follow section 8.5 Existence of singular points in Theory of Complex Functions by R. Remmert, we also cite the Identity Theorem stated there.

Identity Theorem: The following statements about a pair $f,g$ of holomorphic functions in a region $G\subset \mathbb{C}$ are equivalent:

• i) $f=g$.

• ii) The coincidence set $\{w\in G:f(w)=g(w)\}$ has a cluster point in $G$.

• iii) There is a point $c \in G$ such that $f^{(n)}(c)=g^{(n)}(c)$ for all $n\in\mathbb{N}$.

In the following we denote with $\mathcal{O}(G)$ the set of holomorphic functions defined on a region $G\subset\mathbb{C}$.

Existence of singular points: On the boundary of the disc of convergence of a power series $f(z)=\sum a_{\nu}(z-c)^{\nu}$ there is always at least one singular point of $f$.

Proof: Starting with a proof by contradiction we assume there is an $f$ which has no singular point on $B:=B_{R}(c)$.

• Then for every $w\in\partial B$ there is a disc $B_r(w), r=r(w)>0$ and a function $g\in\mathcal{O}\left(B_r(w)\right)$ such that \begin{align*} f\big|_{B\cap B_r(w)}=g\big|_{B\cap B_r(w)} \end{align*}

• Since $\partial B$ is compact we can choose a final number $1\leq j\leq l$ of discs, say $K_j$ which cover $\partial B$. We can choose $g_j, 1\leq j\leq l$ such that \begin{align*} f\big|_{B\cap K_j}=g_j\big|_{B\cap K_j}\qquad 1\leq j\leq l\tag{2} \end{align*}

• With (2) in mind we take a radius $\tilde{R}>R$ such that $\tilde{B}:=B_{\tilde{R}}(c)\subset B\cup K_1\cup\cdots\cup K_l$. We now define a function $\tilde{f}$ in $\tilde{B}$ as follows \begin{align*} \tilde{f}(z):= \begin{cases} f(z)&\qquad z\in B\\ g_j(z)&\qquad z\in (\tilde{B}\setminus B) \mathrm{\ whereby\ } z\in K_j\tag{3} \end{cases} \end{align*} Note, the extension $\tilde{f}(z)$ for $z\in \tilde{B}\setminus B$ is sound. If $z$ is in more than one of the discs $K_j, 1\leq j\leq l$, say $z\in K_{j_1}\cap K_{j_2}$, then \begin{align*} z\in K_{j_1}\cap K_{j_2}\cap B\ne \emptyset\tag{4} \end{align*} and in this open set (4) the functions $g_{j_1}$ and $g_{j_2}$ each coincide with $f$. From this and the Identity Theorem it follows that $g_{j_1}$ and $g_{j_2}$ coincide throughout the region $K_{j_1}\cap K_{j_2}$ and in particular at the point $z$.

Conclusion: The function $\tilde{f}$ is well-defined and holomorphic in $\tilde{B}$. It can therefore be represented by a power series with center $c$ and is convergent in $\tilde{B}$. This power series is the same which represents $f$. So, the smaller disc $B$ is not the disc of convergence of $f$ and we have reached a contradiction.

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