# Continuity of analytic function implies convergence of power series?

Suppose that $f(z)$ is analytic in the unit disk $\Delta:\,|z|<1$. Then $f(z)$ has a Taylor series $\sum=\sum a_nz^n$ in the unit disk. One may assume that $\sum$ has $R=1$ as its radius of convergence for convenience.

Question 1 (local version): If $f(z)$ is continous at some point, say $z=1$, on the unit circle $S^1$, then its Taylor series $\sum$ is covergent at $z=1$?

Question 2 (global version): Suppose that $f(z)$ is continous on the closed unit disk $\overline\Delta$. Is the Taylor series $\sum$ covergent at every poin of $S^1$?

Of course, if the answer to the local version is "yes", then so is the answer to the global version.

EDIT. It seems that Question 1 obviously has a negative answer. So I modify it into a new question.

Question 3 (New local version): Suppose that $f(z)$ is continous at some point, say $z=1$, on the unit circle $S^1$ and the coefficients of $\sum$ satisfies $a_n\to 0$. Is the Taylor series $\sum$ covergent at $z=1$?

Note. If $f$ is analytic at $z=1$, then the answer to Question 3 is yes. This is Fatou's Theorem.

• For Question 3, $a_n=O(1/n)$ would suffice and my impression is that the order of magnitude cannot be improved, but I don't remember ever seeing an example to this effect. – user53153 Jan 24 '13 at 23:16
• Congratulations on being the 100,000th question on MSE :P – Ben Millwood Jan 26 '13 at 15:31
• At the time the 100,000th question was posted, a sreenshot was taken: i.stack.imgur.com/sslH5.jpg – Andreas Oct 16 '18 at 11:47

## 2 Answers

The answer to Question 2 is no. The essential reason is the following: Du Bois - Reymond constructed a continuous function $f$ on the unit circle whose Fourier series, $\sum_n a_n e^{i n \theta}$, does not converge at $\theta=0$. If you look at the construction, you'll see that it is easy to arrange that $a_n=0$ for $n<0$. Then $\sum_{n=0}^{\infty} a_n z^n$ defines an analytic function inside the unit disc, which extends continuously to the boundary, but such that $\sum a_n$ is divergent.

Below, I give the details. I used Pinsky's book as my reference for the construction of du Bois - Reymond.

Define $S_M(z) = \sum_{r=1}^M \frac{z^r-z^{-r}}{r}$. Our function will be of the form $$f(z) := \sum_{k=1}^{\infty} \frac{z^{N_k}}{k^2} S_{M_k}(z) \quad (1)$$ where $M_k$ and $N_k$ are sequences of positive integers chosen such that $$0 < N_1 - M_1 < N_1 + M_1 < N_2 - M_2 < N_2 + M_2 < N_3 - M_3 < N_3+ M_3 < \cdots \quad (2)$$ and $$\frac{\log M_k}{k^2} \to \infty \ \mbox{as} \ k \to \infty. \quad (3)$$

We will show below that the sum (1) is uniformly convergent in the closed unit disc. Hence, it defines a continuous function on the closed disc and an analytic function in the interior. Condition (2) forces that the polynomials $z^{N_k} s_{M_k}(z)$ have no overlapping terms, so the Taylor series of $f$ just looks like blocks of $-1/M_k$, $-1/(M_k-1)$, ..., $-1$, $0$, $1$, $1/2$, ..., $1/M_k$, separated by long blocks of zeroes. Define $a_n$ to be the coefficients of $f(z) = \sum_{n=0}^{\infty} a_n z^n$.

We now check that $\sum a_n$ is divergent. We have $$\sum_{n=0}^{N_k} a_n = \sum_{j=1}^{k-1} \frac{1}{j^2} \left( \frac{-1}{M_j} + \cdots + \frac{-1}{1} + \frac{1}{1} + \cdots + \frac{1}{M_j} \right) - \frac{1}{k^2} \left( \frac{1}{M_k} + \frac{1}{M_k -1} + \cdots + \frac{1}{2} + \frac{1}{1} \right)$$ $$=0+0+\cdots + 0 - \frac{1}{k^2} \left( \frac{1}{M_k} + \frac{1}{M_k -1} + \cdots + \frac{1}{2} + \frac{1}{1} \right) \approx - \frac{\log M_k}{k^2}.$$ Using condition (3), this goes to $- \infty$. So there is a subsequence of partial sums of $\sum a_n$ which goes to $- \infty$ and $\sum a_n$ diverges.

We now must prove that (1) is uniformly convergent. We need

Lemma There is an absolute constant $C$ so that, for any real angle $\theta$, and any positive integer $M$, we have $$\left| \sum_{r=1}^M \frac{\sin (r \theta)}{r} \right| \leq C.$$

Given the lemma, uniform convergence is easy. By the maximum modulus principle, $\left| z^{N_k} s_{M_k}(z) \right|$ is maximized on the boundary of the unit disc. On that boundary, $|e^{i N_k \theta} s_{M_k}(e^{i \theta} )| = 2 \left| \sum_{r=1}^{M_k} \frac{\sin (r \theta)}{r} \right|$ and so, by the lemma, is bounded independently of $\theta$. The $\frac{1}{k^2}$ factor in front then forces uniform convergence.

We now prove the lemma. This is the only part I'm not adapting from Pinsky, because he treats this as obvious.

Proof of lemma: Since the sum of sines is clearly periodic modulo $2 \pi$, and is clearly odd, we may assume that $\theta \in (0,\pi)$. We break the sum at $r=K$, for a parameter $K$ to be chosen later. For the first part of the sum, $$\left| \sum_{r=1}^{K} \frac{\sin (r \theta)}{r} \right| \leq \sum_{r=1}^{K} \frac{r \theta}{r} = K \theta.$$

For the second part of the sum, we start by noting $$\sin \frac{\theta}{2} \cdot \sum_{r=K+1}^M \frac{\sin (r \theta)}{r} =$$ $$\frac{\cos ((K+1/2) \theta)}{K+1} + \sum_{r=K+2}^M \cos ((r-1/2) \theta)\left( \frac{1}{r-1} - \frac{1}{r} \right)+ \frac{\cos((M+1/2) \theta)}{M}$$ so $$\left| \sum_{r=K+1}^M \frac{\sin (r \theta)}{r} \right| \leq \frac{1}{\sin (\theta/2)} \left( \frac{1}{K+1} + \sum_{r=K+2}^M \left( \frac{1}{r-1} - \frac{1}{r} \right)+ \frac{1}{M}\right)$$ $$= \frac{2}{\sin (\theta/2) (K+1)} \leq \frac{2}{(K+1)(\theta/\pi)} = \frac{2 \pi}{(K+1)\theta}.$$ We have used the bound $\sin (\theta/2) \geq \theta/\pi$ for $\theta \in (0 , \pi)$, which is true because $\sin$ is concave.

In short, $$\left| \sum_{r=1}^M \frac{\sin (r \theta)}{r} \right| \leq K \theta + \frac{2 \pi}{(K+1) \theta}.$$ Choose $K$ such that $K \theta$ is neither near $0$ nor $\infty$, and this quantity will be bounded, so we have proved the lemma. $\square$.

• The Lemma was an exercise in the 2nd edition of Rudin's Principles, but he took it out in the 3rd, while trimming the Fourier series part. Too bad. // His approach was: (i) the Fourier series of any function of bounded variation has uniformly bounded partial sums; (ii) $\sum \sin(r\theta)/r$ is the Fourier series of $\pi-t$ on $(0,\pi)$, extended to $(-\pi, \pi)$ as an odd function. // The idea of (i): BV gives $c_n=O(1/n)$ for the coefficients, which is enough to bound the difference between Fejer and Dirichlet sums. (And Fejer sums are trivially bounded). – user53153 Jan 27 '13 at 3:14
• @DavidSpeyer : Could you explain why $\sum_{n=0}^{\infty}a_nz^n$ extends continuously to the boundary? – Malik Younsi Jan 27 '13 at 14:12
• @MalikYounsi Because the sum $\sum_k \frac{z^{N_k}}{k^2} S_{M_k}(z)$ is uniformly convergent on the closed disc, so it converges to a continuous function there. – David E Speyer Jan 27 '13 at 16:44
• To emphasize: The sum $\sum_n a_n z^n$ doesn't converge at $1$, but if we group it together into chunks as $\sum_k \frac{z^{N_k}}{k^2} s_{M_k}(z)$, then it converges everywhere on the closed disc. – David E Speyer Jan 27 '13 at 16:45
• (David: It is only one person, Paul du Bois-Reymond) – Andrés E. Caicedo Jan 27 '13 at 17:14

Question 1

In general, no. Consider, for example, $f(z) := \frac{1}{1+z}$. This function has a Taylor series, centred at $z=0$, with radius of convergence $\rho = 1$. We see that $f(z) \equiv 1 - z + z^2 - \cdots + (-1)^kz^k + \cdots$ for all $|z| < 1$. The function $f$ is continuous for all $z \in \mathbb{C}$ except $z=-1$. (In fact, $f$ is holomorphic for all $z \in \mathbb{C}$ except $z=-1$, which is much stronger than just continuous.)

Now, consider the point $z=1$. The function is holomorphic at $z=1$ and $f(1)=1/2$. However, trying to apply the power series at $z=1$ gives the divergent series $1-1+1-1+1-1+\cdots$.

Question 2

I know that this isn't what you asked, but there exist power series which converge for all $|z| \le 1$ and yet are discontinuous on the circle $|z|=1{}$. This is a result by Sierpinski (1916).

• Do you mean that Sierpinski's example is nowhere continous on the unit circle? Can you tell me in which article his example is constructed. – Wall Great Jan 24 '13 at 21:21
• @WallGreat Sierpinski's example was thoroughly discussed on MO – user53153 Jan 24 '13 at 21:43