Show F cannot be continued analytically past the unit disc.

Question

This question is from Stein's complex analysis. This question has 4 parts, I don't have question about the first part and the second part, but I have no idea about how to solve the third part. I think there is some question like this posting here, but I forgot how to get to that post, and that post did not help me when I read it.

Let $F(z)=\sum_{n=1}^{\infty}d(n)z^{n}$, where $d(n)$ denotes the number of divisors of n. Observe that the radius of convergence of this series is 1.

Part 1: Verify the identity $F(z)=\sum_{n=1}^{\infty}d(n)z^{n}=\sum_{n=1}^{\infty}\frac{z^{n}}{1-z^{n}}$.

Part 2: Using this identity, show that if $z=r$ with $0<r<1$, then $\mid$$F(r)$$\mid$ $\geq$ $c$$\frac{1}{1-r}log(1/(1-r))$ as $r\rightarrow 1$.

Part 3: Similarly, if $\theta=2\pi p/q$ where $p$ and $q$ are positive integers and $z=re^{i\theta}$, then $\mid$$F(re^{i\theta})$$\mid$ $\geq$ $c_{p/q}$$\frac{1}{1-r}$$log(1/(1-r))$.

Part 4: Conclude that F cannot be continued analytically past the unit disc.

Since I have solved part 1 and part 2, and part 3 told me it is similar, I have done something, but I cannot proceed further.

$\mid F(z)\mid=\mid$$\sum_{n=1}^{\infty}\frac{z^{n}}{1-z^{n}}\mid$$=$$\mid$$\frac{1}{1-z}\sum_{n=1}^{\infty}\frac{z^{n}}{1+\cdots +z^{n-1}}\mid$$=$$\mid$$\frac{1}{1-z}$$\mid$$\mid$$z+\frac{z^{2}}{1+z}+\cdots +\frac{z^{n}}{1+\cdots +z^{n-1}}+\cdots$$\mid$

If this is part 2, I can expand $log(1/1-r)$ in power series and compare term by term, and will get some inequalities, but in this part I don't know how to proceed.

Moreover, how to relate the constant C with $p/q$? I think that it may be related to some period in the unit circle, but I cannot write down anything about it.

Part 4 follows part 3 immediately and I don't have problem with it.

Any hints and detailed explanations are really really appreciated!!!!

Answer 1

Okay, I think I have solved this. Note that the idea of my answer is based on the post Lower bound for $ F(z) = \sum_{n=1}^\infty d(n)z^n $ near radius of convergence., and he was really closed to the correct answer, and he was actually mostly correct. I only added more details in it and presented it here.

Let $z=re^{i\theta}$ with $\theta=2\pi\frac{p}{q}$.

Note that it suffices to show it in the case of $gcd(p,q)=1$, since if $gcd(p,q)\neq1$, we will then be able to get $p'$ and $q'$ such that $gcd(p',q')=1$ and $2\pi\frac{p}{q}=2\pi\frac{p'}{q'}$.

Thus, suppose $gcd(p,q)=1$. Then $z^{n}=r^{n}e^{i2\pi n\frac{p}{q}}$.

If $q\mid n$, then we can let $n=qm$, then we have $z^{n}=z^{qm}$. If $q$ does not divide $n$, then we let $N_{q}=$ {$n\in\mathbb{N}$$\mid$$q$ does not divide $n$}.

Then, we can split the sum $\sum_{n=1}^{\infty}\frac{z^{n}}{1-z^{n}}$ into two parts:

that is: $\mid$$F(z)$$\mid$ $=$ $\mid$$\sum_{n=1}^{\infty}\frac{z^{n}}{1-z^{n}}$$\mid$ $=$ $\mid$$\sum_{m=1}^{\infty}\frac{z^{mq}}{1-z^{mq}}+\sum_{n\in N_{q}}\frac{z^{n}}{1-z^{n}}$$\mid$. Let the first sum be $A$ and the second sum be $B$.

Now, in $A$, $z^{qm}$ is real, so $z=r$ and then be part 2 of this question, we have $A\geq C_{A}\frac{1}{1-r}log(\frac{1}{1-r})$, and $A\geq 0$

Then, in $B$, since $\theta=\frac{p}{q}2\pi$, which implies $q\theta=p2\pi$, we then can observe that $e^{i(p+q)\theta}=e^{ik\theta}e^{iq\theta}=e^{ik\theta}$ for all $k$. Thus, we only have $(q-1)$ direction to worry about along the unit circle.

Now let $k'$ be defined such that $e^{ik'\theta}$ is the closest to $1$ when following the unit circle perimeter.

Let $C_{B}=min${$1$, $inf$$\mid$$1-re^{ik'\theta}$$\mid$:$0\leq r\leq 1$}, so $C_{B}> 0$

Then, we have: $\mid$$B$$\mid$ $\leq$ $\sum_{n\in N_{q}}\frac{\mid z\mid^{n}}{\mid 1-z^{n}\mid}$ $\leq$ $\sum_{n\in N_{q}}\frac{\mid z\mid^{n}}{\mid C_{B}\mid}$ $=$ $C_{B}^{-1}$$\sum_{n\in N_{q}}$$\mid$$z\mid^{n}$ $\leq$ $C_{B}^{-1}$$\sum_{n=1}^{\infty}$$\mid$$z$$\mid^{n}$ $=$ $C_{B}^{-1}$$\frac{1}{1-\mid z\mid}$ $=$ $C_{B}^{-1}$$\frac{1}{1-r}$

Therefore, we have $\mid$$B$$\mid$ $\leq$ $\frac{1}{1-r}C_{B}^{-1}$, and thus $-$$\mid$$B$$\mid$ $\geq$ $-$$\frac{1}{1-r}C_{B}^{-1}$.

Since $A\geq 0$, we have $\mid A+B$$\mid\geq A-\mid$$B\mid$, and thus we have:

$\mid F(z)\mid$ $\geq$ $A-\mid$$B\mid$ $\geq$ $C_{A}\frac{1}{1-r}log(\frac{1}{1-r})$ $-$ $\mid$$B\mid$ $\geq$ $C_{A}\frac{1}{1-r}log(\frac{1}{1-r})$ $-$ $\frac{1}{1-r}C_{B}^{-1}$ $=$ $C_{A}\frac{1}{1-r}(log(\frac{1}{1-r})-C_{B}^{-1})$.

Since, $log(\frac{1}{1-r})$ is far much bigger than $C_{B}^{-1}$ as $r\rightarrow 1$.

We will have finally: $\mid$$F(z)$$\mid$ $\geq$ $C_{\frac{p}{q}}$$\frac{1}{1-r}log(\frac{1}{1-r})$ [QED]

I appreciate and welcome any comments or corrections!

Answer 2

This is exactly the same solution as JacobsonRadical, but I'm writing it since I couldn't solve this and had to consider the post above for some hints but I'm not sure if I now understand his post properly.

Here's the crucial idea: If $S$ is a finite set of points in the unit circle, then $$\min_{r \in \mathbb{R}, s \in S} {|1-rs|}$$ exists and is nonzero (proof: The locus of the points $1-rs$ is a set of straight lines. Now the desired value is the smallest perpendicular distance from one). Using this and the triangle inequality churns out the answer.

Let $\omega = e^{i\frac{2 \pi p}{q}}$ and $z = r \omega$ with $r \in \mathbb{R}$. Let $c =\min_{r \in \mathbb{R}, 0 < n < q}|1-r\omega^n|$, so we get \begin{align} \left| \sum_{q \not \div n} \frac{r^n \omega^n}{1-r^n \omega^n} \right| &< \sum_{q \not \div n} \left|\frac{r^n \omega^n}{1-r^n \omega^n}\right| \\ &< \sum_{q \not \div n} \frac{|r^n \omega^n|}{|1-r^n \omega^n|} \\ &= \frac{1}{c} \cdot \sum_{q \not \div n} r^n \\ &< \frac{1}{c} \sum_{0 \geq n} r^n \\ &= \frac{1}{c(1-r)}. \end{align}

We assume the result of part (2), i.e $$ \sum_{0 \leq n} \frac{r^n}{1-r^n} \geq -\frac{\log(1-r)}{1-r} $$ for all $0 < r < 1$. Now, \begin{align} \left| \sum_{0 \leq n} \frac{r^n \omega^n}{1-r^n \omega^n} \right| &=\left| \sum_{q | n} \frac{r^{n}}{1-r^{n}} + \sum_{q \not \div n} \frac{r^n \omega^n}{1-r^n\omega^n} \right| \\ &= \left| \sum_{0 \geq n} \frac{r^q}{1-r^q} \right| + \left| \sum_{q \not \div| n} \frac{r^n \omega^n}{1-r^n\omega^n} \right| \\ &> - \frac{\log(1-r^q)}{(1-r^q)} + \frac{1}{c(1-r)} \\ &> -\frac{\log(1-r) + \log(q)}{q(1-r)} + \frac{1}{c(1-r)} \\ & \sim \frac{ \log(1-r)}{1-r}, \end{align} as desired.