Let \[ f(z)=\sum_{n=0}^\infty a_nz^n, \] where every \(a_n\) is a nonnegative real number and the series has radius of convergence \(1\). Prove that \(f\) cannot be analytically continued to a function analytic in a neighborhood of \(z=1\).
Solution
Suppose, for contradiction, that \(f\) admits an analytic continuation to a neighborhood of \(1\). Write \[ D(c,r)=\{z\in\CC:\abs{z-c}<r\}. \]
1. There is a number \(\rho>0\) and a function \(F\) analytic on\[ D(0,1)\cup D(1,\rho) \]such that \(F=f\) on \(D(0,1)\).
By assumption, there is a function analytic on some disk \(D(1,\rho)\) that agrees with \(f\) on the nonempty overlap with the unit disk. The two analytic functions therefore glue to an analytic function \(F\) on the displayed union.
2. Choose\[ 0<\delta<\min\left\{\frac12,\frac\rho4\right\}, \qquad t=1-\delta. \]Then\[ D(t,2\delta) \subset D(0,1)\cup D(1,\rho). \]
Let \(z\in D(t,2\delta)\). If \(\abs{z}<1\), then \(z\in D(0,1)\). If \(\abs{z}\ge1\), then \[ \abs{z-1} \le \abs{z-t}+\abs{t-1} < 2\delta+\delta < \rho. \] Thus \(z\in D(1,\rho)\). This proves the inclusion.
3. For every integer \(k\ge0\),\[ \frac{F^{(k)}(t)}{k!} = \sum_{n=k}^\infty a_n\binom{n}{k}t^{n-k}, \]and this number is nonnegative.
Since \(0<t<1\), the original power series converges in a neighborhood of \(t\) and may be differentiated term by term. Hence \[ F^{(k)}(t) = \sum_{n=k}^\infty a_n\frac{n!}{(n-k)!}t^{n-k}. \] Dividing by \(k!\) gives the displayed formula. Every term is nonnegative because \(a_n\ge0\) and \(t>0\).
4. Set\[ h=\frac{3\delta}{2}. \]Then\[ t+h=1+\frac\delta2>1, \]and the Taylor series of \(F\) about \(t\) converges at \(t+h\).
By step 2, \(F\) is analytic on the disk \(D(t,2\delta)\), so its Taylor series about \(t\) converges for \(\abs{z-t}<2\delta\). Since \(h=3\delta/2<2\delta\), it converges at \(z=t+h\).
5. One has\[ \sum_{n=0}^\infty a_n(t+h)^n<\infty. \]
By step 4, \[ F(t+h) = \sum_{k=0}^\infty \frac{F^{(k)}(t)}{k!}h^k. \] Substituting step 3 gives \[ F(t+h) = \sum_{k=0}^\infty \sum_{n=k}^\infty a_n\binom{n}{k}t^{n-k}h^k. \] Every term in this double series is nonnegative. Hence its order of summation may be interchanged, giving \[ \begin{aligned} F(t+h) &= \sum_{n=0}^\infty a_n \sum_{k=0}^n \binom{n}{k}t^{n-k}h^k\\ &= \sum_{n=0}^\infty a_n(t+h)^n. \end{aligned} \] The left-hand side is finite, so the displayed series converges.
6. The convergence in step 5 contradicts the assumption thatthe radius of convergence of\[ \sum_{n=0}^\infty a_nz^n \]is \(1\).
Step 4 gives \(t+h>1\). A power series that converges at the point \(z=t+h\) has radius of convergence at least \(t+h>1\), contradicting the stated radius \(1\).
7. Therefore \(f\) cannot be analytically continued to a functionanalytic in a neighborhood of \(1\).
The assumption of such a continuation led to the contradiction in step 6.
8. Q.E.D.
Step 7 is the required conclusion.