# A Step-by-step of the Analytic Continuation of the Riemann Zeta Function

We compute the analytic continuation of the Riemann Zeta function and after that the reader will realise that asserting $1+2+\dots=-\frac{1}{12}$ without enough caution is not a good idea.

# A Detailed Proof of the Riemann Mapping Theorem

We offer a detailed proof of the Riemann mapping theorem, which states that every proper simply connected region is conformally equivalent to the open unit disc.

# Quasi-analytic Classes

We study the concept of quasi-analytic functions, which are quite close to being analytic.

# The Fourier transform of sinx/x and (sinx/x)^2 and more

In this post we compute the Fourier transform of $\sin{x}/x$ and $(\sin{x}/x)^2$ through contour integration.

# More properties of zeros of an entire function

## What’s going on again

In this post we discussed the topological properties of the zero points of an entire nonzero function, or roughly, how those points look like. The set of zero points contains no limit point, and at most countable (countable or finite). So if it’s finite, then we can find them out one by one. For example, the function $f(z)=z$ has simply one zero point. But what if it’s just countable? How fast the number grows?

Another question. Suppose we have an entire function $f$, and the zeros of $f$, namely $z_1,z_2,\cdots,z_n$, are ordered increasingly by moduli:

Is it possible to get a fine enough estimation of $|z_n|$? Interesting enough, we can get there with the help of Jensen’s formula.

## Jensen’s formula

Suppose $\Omega=D(0;R)$, $f \in H(\Omega)$, $f(0) \neq 0$, $0<r<R$, and $z_1,z_2,\cdots,z_{n(r)}$ are the zeros of $f$ in $\overline{D}(0;R)$, then

There is no need to worry about the assumption $f(0) \neq 0$. Take another look at this proof. Every zero point $a$ has a unique positive number $m$ such that $f(z)=(z-a)^mg(z)$ and $g \in H(\Omega)$ but $g(a) \neq 0$. The number $m$ is called the order of the zero at $a$. Therefore if we have $f(0)=0$ we can simply consider another function, namely $\frac{f}{z^m}$ where $m$ is the order of zero at $0$.

We are not proving this identity at this point. But it can be done by considering the following function

where $m$ is found by ordering $z_j$ in such a way that $z_1,\cdots,z_m \in D(0;r)$ and $|z_{m+1}|=\cdots=|z_{n}|$. One can prove this identity by considering $|g(0)|$ as well as $\log|g(re^{i\theta})|$.

## Several applications

### The number of zeros of $f$ in $\overline{D}(0;r)$

For simplicity we shall assume $f(0)=1$ which has no loss of generality. Let

and $n(r)$ be the number of zeros of $f$ in $\overline{D}(0;r)$. By the maximum modulus theorem, we have

If we insert Jensen’s formula into this inequality and order $|z_n|$ by increasing moduli, we get

Which implies

So $n(r)$ is controlled by $M(2r)$. The second and third inequalities look tricky, which require more explanation.

First we should notice the fact that $z_n \in \overline{D}(0;R)$ for all $R \in \mathbb{R}$. Hence we have $\log\frac{2r}{|z_n|} \geq \log1=0$ for all $z_n \in \overline{D}(0;R)$. Hence the second inequality follows. For the third one, we simply have

So this is it, the rapidity with which $n(r)$ can grow is dominated by $M(r)$. Namely, the number of zeros of $f$ in the closed disc with radius $r$ is controlled by the maximum modulus of $f$ on a circle with bigger radius.

### Examples based on different $M(r)$

Let’s begin with a simple example. Let $f(z)=1$, we have $M(r)=1$ for all $r$, but also we have $n(r)=0$, in which sense this estimation does nothing. Indeed, as long as $M(r)$ is bounded by a constant, which implies $f(z)$ is bounded, then by Liouville’s theorem, $f(z)$ is constant and this estimation is not available.

But if $M(r)$ grows properly, things become interesting. For example, if we have

where $A$ and $k$ are given positive numbers, we have a good enough estimation by

This estimation becomes interesting if we consider the logarithm of $n(r)$ and $r$, that is

If we have $f(z)=1-\exp(z^k)$ where $k$ is a positive integer, we have $n(r) \sim \frac{kr^k}{\pi}$, also

### Lower bound of $|z_{n(r)}|$

We’ll see here, how to evaluate the lower bound of $|z_{n(r)}|$ using Jensen’s formula, provided that $M(r)$, or simply the upper bound of $f(z)$ is properly described. Without loss of generality we shall assume that $f(0)=1$. Also, we assume that the zero points of $f(z)$ are ordered by increasing moduli.

First we still consider

and see what will happen.

By Jensen’s, we have

This gives

By the arrangement of $\{z_n\}$, we have

Another example is when we have

where $\Im{z}$ means the imagine part of $z$.

We shall notice that in this case,

Following Jensen’s formula, we therefore have

# Topological properties of the zeros of a holomorphic function

### What’s going on

If for every $z_0 \in \Omega$ where $\Omega$ is a plane open set, the limit

exists, we say that $f$ is holomorphic (a.k.a. analytic) in $\Omega$. If $f$ is holomorphic in the whole plane, it’s called entire. The class of all holomorphic functions (denoted by $H(\Omega)$) has many interesting properties. For example it does form a ring.

But what happens if we talk about the points where $f$ is equal to $0$? Is it possible to find an entire function $g$ such that $g(z)=0$ if and only if $z$ is on the unit circle? The topological property we will discuss in this post answers this question negatively.

### Zeros

Suppose $\Omega$ is a region, the set

is a at most countable set without limit point, as long as $f$ is not identically equal to $0$ on $\Omega$.

Trivially, if $f(\Omega)=\{0\}$, we have $Z(f)=\Omega$. The set of unit circle is not at most countable and every point is a limit point. Hence if an entire function is equal to $0$ on the unit circle, then the function equals to $0$ on the whole plane.

Note: the connectivity of $\Omega$ is important. For example, for two disjoint open sets $\Omega_0$ and $\Omega_1$, define $f(z)=0$ on $\Omega_0$ and $f(z)=1$ on $\Omega_1$, then everything fails.

### A simple application (Feat. Baire Category Theorem)

Before establishing the proof, let’s see what we can do using this result.

Suppose that $f$ is an entire function, and that in every power series

has at leat one coefficient is $0$, then $f$ is a polynomial.

Clearly we have $n!c_n=f^{(n)}(a)$, thus for every $a \in \mathbb{C}$, we can find a postivie integer $n_0$ such that $f^{(n_0)}(a)=0$. Thus we establish the identity:

Notice the fact that $f^{(n)}$ is entire. So $Z(f^{n})$ is either an at most countable set without limit point, or simply equal to $\mathbb{C}$. If there exists a number $N$ such that $Z(f^{N})=\mathbb{C}$, then naturally $Z(f^{n})=\mathbb{C}$ holds for all $n \geq N$. Whilst we see that $f$’s power series has finitely many nonzero coefficients, thus polynomial.

So the question is, is this $N$ always exist? Being an at most countable set without limit points , $Z(f^{(n)})$ has empty interior (nowhere dense). But according to Baire Category Theorem, $\mathbb{C}$ could not be a countable union of nowhere dense sets (of the first category if you say so). This forces the existence of $N$.

### Proof

The proof will be finished using some basic topology techniques.

Let $A$ be the set of all limit points of $Z(f)$ in $\Omega$. The continuity of $f$ shows that $A \subset Z(f)$. We’ll show that if $A \neq \varnothing$, then $Z(f)=\Omega$.

First we claim that if $a \in A$, then $a \in \bigcap_{n \geq 0}Z(f^{(n)})$. That is, $f^{(k)}(a) = 0$ for all $k \geq 0$. Suppose this fails, then there is a smallest positive integer $m$ such that $c_m \neq 0$ for the power series on the disc $D(a;r)$:

Define

It’s clear that $g \in H(D(a;r))$ since we have

But the continuity shows that $g(a)=0$ while $c_m \neq 0$. A contradiction.

Next fix a point $b \in \Omega$. Choose a curve (continuous mapping) defined $\gamma$ on $[0,1]$ such that $\gamma(0)=a$ and $\gamma(1)=b$. Let

By hypothesis, $0 \in \Gamma$. We shall prove that $1 \in \Gamma$. Let

There exists a sequence $\{t_n\}\subset\Gamma$ such that $t_n \to s$. The continuity of $f^{(k)}$ and $\gamma$ shows that

Hence $s \in \Gamma$. Choose a disc $D(\gamma(s);\delta)\subset\Omega$. On this disc, $f$ is represented by its power series but all coefficients are $0$. It follows that $f(z)=0$ for all $z \in D(\gamma(s);\delta)$. Further, $f^{(k)}(z)=0$ for all $z \subset D(\gamma(s);\delta)$ for all $k \geq 0$. Therefore by the continuity of $\gamma$, there exists $\varepsilon>0$ such that $\gamma(s-\varepsilon,s+\varepsilon)\subset D(\gamma(s);\delta)$, which implies that $(s-\varepsilon, s+\varepsilon)\cap[0,1]\subset\Gamma$. Since $s=\sup\Gamma$, we have $s=1$, therefore $1 \in \Gamma$.

So far we showed that $\Omega = \bigcap_{n \geq 0}Z(f^{(n)})$, which forces $Z(f)=\Omega$. This happens when $Z(f)$ contains limit points, which is equivalent to what we shall prove.

When $Z(f)$ contains no limit point, all points of $Z(f)$ are isolated points; hence in each compact subset of $\Omega$, there are at most finitely many points in $Z(f)$. Since $\Omega$ is $\sigma$-compact, $Z(f)$ is at most countable. $Z(f)$ is also called a discrete set in this situation.