# The Big Three Pt. 2 - The Banach-Steinhaus Theorem

People call the Banach-Steinhaus theorem the first of the big three, which sits at the foundation of linear functional analysis. None of them can go without the Baire's category theorem.

This blog post offers the Banach-Steinhaus theorem on different abstract levels. Recall that we have $\text{TVS} \supset \text{Metrizable TVS} \supset \text{F-space} \supset \text{Fréchet space}\supset\text{Banach space} \supset \text{Hilbert space}$ First, there will be a simple version for Banach spaces, which may be more frequently used, and you will realize why it's referred to as the uniform boundedness principle. After that, there will be a much more generalized version for TVS. Typically, the metrization of the space will not be considered.

Also, it will be a good chance to get a better view of the first and second space by Baire.

## Equicontinuity

For metric spaces, equicontinuity is defined as follows. Let $(X,d_X)$ and $(Y,d_Y)$ be two metric spaces.

Let $\Lambda$ be a collection of functions from $X$ to $Y$. We have three different levels of equicontinuity.

1. Equicontinuous at a point. For $x_0 \in X$, if for every $\varepsilon>0$, there exists a $\delta>0$ such that $d_Y(Lx_0,Lx)<\varepsilon$ for all $L \in \Lambda$ and $d_X(x_0,x)<\delta$ (that is, the continuity holds for all $L$ in a ball centered at $x_0$ with radius $r$).
2. Pointwise equicontinuous. $\Lambda$ is equicontinuous at each point of $X$.
3. Uniformly equicontinuous. For every $\varepsilon>0$, there exists a $\delta>0$ such that $d_Y(Lx,Ly)<\varepsilon$ for all $x \in \Lambda$ and $x,y \in X$ such that $d_X(x,y) < \delta$.

Indeed, if $\Lambda$ contains only one element, namely $L$, then everything goes with the continuity and uniform continuity.

But for Banach-Steinhaus theorem, we need a little more restrictions. In fact, $X$ and $Y$ should be considered Banach spaces, and $\Lambda$ contains linear functions only. In this sense, for $L \in \Lambda$, we have the following three conditions equivalent.

1. $L$ is bounded.
2. $L$ is continuous.
3. $L$ is continuous at one point of $X$.

For topological vector spaces, where only topology and linear structure are taken into consideration, things get different. Since no metrization is considered, we have to state it in the language of topology.

Suppose $X$ and $Y$ are TVS and $\Lambda$ is a collection of linear functions from $X$ to $Y$. $\Lambda$ is equicontinuous if for every neighborhood $N$ of $0$ in $Y$, there corresponds a neighborhood $V$ of $0$ in $X$ such that $L(V) \subset N$ for all $L \in \Lambda$.

Indeed, for TVS, $L \in \Lambda$ has the three conditions equivalent as well. With that being said, equicontinuous collection has the boundedness property in a uniform manner. That's why the Banach-Steinhaus theorem is always referred to as the uniform boundedness principle.

## The Banach-Steinhaus theorem, a sufficient condition for being equicontinuous

### Banach space version

Suppose $X$ is a Banach space, $Y$ is a normed linear space, and ${F}$ is a collection of bounded linear transformation of $X$ into $Y$, we have two equivalent statements: 1. (The Resonance Theorem) If $\sup\limits_{L \in \Lambda}\left\Vert{L}\right\Vert=\infty$, then there exists some $x \in X$ such that $\sup\limits_{L \in {L}}\left\Vert{Lx}\right\Vert=\infty$. (In fact, these $x$ form a dense $G_\delta$.)

1. (The Uniform Boundedness Principle) If $\sup\limits_{L \in {\Lambda}}\left\Vert{Lx}\right\Vert<\infty$ for all $x \in X$, then we have $L M$ for all $L \in {\Lambda}$ and some $M<\infty$.
2. (A summary of 1 and 2) Either there exists an $M<\infty$ such that $\lVert L \rVert \leq M$ for all $L \in {L}$, or $\sup\lVert Lx \rVert = \infty$ for all $x$ belonging to some dense $G_\delta$ in $X$.

#### Proof

Though it would be easier if we finish the TVS version proof, it's still a good idea to leave the formal proof without the help of TVS here. The equicontinuity of $\Lambda$ will be shown in the next section.

##### An elementary proof of the Resonance theorem

First, we offer an elementary proof in which the hardest part is the Cauchy sequence.

(Lemma) For any $x \in X$ and $r >0$, we have $\sup_{y\in B(x,r)}\lVert Lx \rVert \geq \lVert L \rVert r$ where $B(x,r)=\{y \in X:\lVert x-y \rVert < r\}$.

(Proof of the lemma)

For $t \in X$ we have a simple relation \begin{aligned} \max(\lVert{L(x+t)}\rVert,\lVert{L(x-t)}\rVert)&=\frac{1}{2}(\lVert{L(x+t)}\rVert+\lVert{L(x-t)}\rVert)+\frac{1}{2}\left\vert\lVert{L(x+t)}\rVert-\lVert{L(x-t)}\rVert\right\vert \\ &\geq \frac{1}{2}(\lVert{L(x+t)}\rVert+\lVert{L(x-t)}\rVert) \\ &\geq \frac{1}{2}\lVert{L(2t)}\rVert=\lVert Lt \rVert \end{aligned} If we have $t \in B(0,r)$, then $x+t,x-t\in{B(x,r)}$. And the desired inequality follows by taking the supremum over $t \in B(0,r)$. (If you find trouble understanding this, take a look at the definition of $\lVert L \rVert$.)

Suppose now $\sup\limits_{L \in \Lambda}\left\Vert{L}\right\Vert=\infty$. Pick a sequence of linear transformation in $\Lambda$, say $(L_n)_{n=1}^{\infty}$, such that $\lVert L_n \rVert \geq 4^n$. Pick $x_0 \in X$, and for $n \geq 1$, we pick $x_n$ inductively.

Set $r_n=3^{-n}$. With $x_{n-1}$ being picked, $x_n \in B(x_{n-1},r_n)$ is picked in such a way that $\lVert L_n x_n \rVert \geq \frac{2}{3}\lVert L_n \rVert r_n$ (It's easy to validate this inequality by reaching a contradiction.) Also, it's easy to check that $(x_n)_{n=1}^{\infty}$ is Cauchy. Since $X$ is complete, $(x_n)$ converges to some $x \in X$. Further we have \begin{aligned} \lVert x-x_n \rVert &\leq \sum_{k=n}^{\infty}\lVert x_k - x_{k+1}\rVert \\ &=\frac{1}{2\cdot 3^n} \end{aligned} Therefore we have \begin{aligned} \lVert L_n x \rVert &=\lVert L_n[x_n-(x_n-x)] \rVert \\ &\geq \lVert L_nx_n \rVert - \lVert L_n(x_n-x) \rVert \\ &\geq \frac{2}{3}\lVert{L_n}\rVert{3}^{-n}-\lVert{L_n}\rVert\lVert{x_n-x}\rVert\\ &\geq \frac{1}{6}\lVert{L_n}\rVert{3}^{-n} \\ & \geq \frac{1}{6}\left(\frac{4}{3}\right)^n \to\infty \end{aligned}

##### A topology-based proof

The previous proof is easy to understand but it's not easy to see the topological properties of the set formed by such $x$. Thus we are offering a topology-based proof which enables us to get a topology view.

Put $\varphi(x)=\sup_{L \in \Lambda}\lVert Lx \rVert$ and let $V_n=\{x:\varphi(x)>n\}$ we claim that each $V_n$ is open. Indeed, we have to show that $x \mapsto \lVert Lx \rVert$ is continuous. It suffice to show that $\lVert\cdot\rVert$ defined in $Y$ is continuous. This follows immediately from triangle inequality since for $x,y \in Y$ we have $\lVert x \rVert \leq \lVert x-y \rVert + \lVert y \rVert$ which implies $\lVert x \rVert - \lVert y \rVert \leq \lVert x-y \rVert$ by interchanging $x$ and $y$, we get $|\lVert x \rVert - \lVert y \rVert | \leq \lVert x-y \rVert$ Thus $x \mapsto \lVert Lx \rVert$ is continuous since it's a composition of $\lVert\cdot\rVert$ and $L$. Hence $\varphi$, by the definition, is lower semicontinuous, which forces $V_n$ to be open.

If every $V_n$ is dense in $X$ (consider $\sup\lVert L \rVert=\infty$), then by BCT, $B=\bigcap_{n=1}^{\infty} V_n$ is dense in $X$. Since each $V_n$ is open, $B$ is a dense $G_\delta$. Again by the definition of $B$, we have $\varphi(x)=\infty$ for all $x \in B$.

If one of these sets, namely $V_N$, fails to be dense in $X$, then there exist an $x_0 \in X - V_N$ and an $r>0$ such that for $x \in B(0,r)$ we have $x_0+x \notin V_N$, which is equivalent to $\varphi(x+x_0) \leq N$ considering the definition of $\varphi$, we also have $\lVert L(x+x_0) \rVert \leq N$ for all $L \in \Lambda$. Since $x=(x+x_0)-x_0$, we also have $\lVert Lx \rVert \leq \lVert L(x+x_0) \rVert+\lVert Lx_0 \rVert \leq 2N$ Dividing $r$ on two sides, we got $\lVert L\frac{x}{r}\rVert \leq \frac{2N}{r}$ therefore $\lVert L \rVert \leq M=\frac{2N}{r}$ as is to be shown. Again, this follows from the definition of $\lVert L \rVert$.

### Topological vector space version

Suppose $X$ and $Y$ are topological vector spaces, $\Lambda$ is a collection of continuous linear mapping from $X$ into $Y$, and $B$ is the set of all $x \in X$ whose orbits $\Lambda(x)=\{Lx:L\in\Lambda\}$ are bounded in $Y$. For this $B$, we have:

• If $B$ is of the second category, then $\Lambda$ is equicontinuous.
##### A proof using properties of TVS

Pick balanced neighborhoods $W$ and $U$ of the origin in $Y$ such that $\overline{U} + \overline{U} \subset W$. The balanced neighborhood exists since every neighborhood of $0$ contains a balanced one.

Put $E=\bigcap_{L \in \Lambda}L^{-1}(\overline{U}).$ If $x \in B$, then $\Lambda(x)$ is bounded, which means that to $U$, there exists some $n$ such that $\Lambda(x) \subset nU$ (Be aware, no metric is introduced, this is the definition of boundedness in topological space). Therefore we have $x \in nE$. Consequently, $B\subset \bigcup_{n=1}^{\infty}nE.$ If no $nE$ is of the second category, then $B$ is of the first category. Therefore, there exists at least one $n$ such that $nE$ is of the second category. Since $x \mapsto nx$ is a homeomorphism of $X$ onto $X$, $E$ is of the second category as well. But $E$ is closed since each $L$ is continuous. Therefore $E$ has an interior point $x$. In this case, $x-E$ contains a neighborhood $V$ of $0$ in $X$, and $L(V) \subset Lx-L(E) \subset \overline{U} - \overline{U} \subset W$ This proves that $\Lambda$ is equicontinuous.

##### Equicontinuity and uniform boundedness

We'll show that $B=X$. But before that, we need another lemma, which states the connection between equicontinuity and uniform boundedness

(Lemma) Suppose $X$ and $Y$ are TVS, $\Gamma$ is an equicontinuous collection of linear mappings from $X$ to $Y$, and $E$ is a bounded subset of $X$. Then $Y$ has a bounded subset $F$ such that $T(E) \subset F$ for every $T \in \Gamma$.

(Proof of the lemma) We'll show that, the set $F=\bigcup_{T \in \Gamma}T(E)$ is bounded. By the definition of equicontinuity, there is an neighborhood $V$ of the origin in $X$ such that $T(V) \subset W$ for all $T \in \Gamma$. Since $E$ is bounded, there exists some $t$ such that $E \subset tV$. For these $t$, by the definition of linear functions, we have $T(E) \subset T(tV)=tT(V) \subset tW$ Therefore $F \subset tW$. $F$ is bounded.

Thus $\Lambda$ is uniformly bounded. Picking $E=\{x\}$ in the lemma, we also see $\Lambda(x)$ is bounded in $Y$ for every $x$. Thus $B=X$.

#### A special case when $X$ is a $F$-space or Banach space

$X$ is a $F$-space if its topology $\tau$ is induced by a complete invariant metric $d$. By BCT, $X$ is of the second category. If we already have $B=X$, in which case $B$ is of the second category, then by Banach-Steinhaus theorem, $\Lambda$ is equicontinuous. Formally speaking, we have:

If $\Lambda$ is a collection of continuous linear mappings from an $F$-space $X$ into a topological vector space $Y$, and if the sets $\Lambda(x)=\{Lx:L\in\Lambda\}$ are bounded in $Y$ for every $x \in X$, then $\Lambda$ is equicontinuous.

Notice that all Banach spaces are $F$-spaces. Therefore we can restate the Uniform Boundedness Principle in Banach space with equicontinuity.

Suppose $X$ is a Banach space, $Y$ is a normed linear space, and ${F}$ is a collection of bounded linear transformation of $X$ into $Y$, we have:

• (The Uniform Boundedness Principle) If $\sup\limits_{L \in {\Lambda}}\left\Vert{Lx}\right\Vert<\infty$ for all $x \in X$, then we have $\|L\| \le M$ for all $L \in {\Lambda}$ and some $M<\infty$. Further, $\Lambda$ is equicontinuous.

## Application

Surprisingly enough, the Banach-Steinhaus theorem can be used to do Fourier analysis. An important example follows.

There is a periodic continuous function $f$ on $[0,1]$ such that the Fourier series $\sum_{n\in\mathbb{Z}}\hat{f}(n)e^{2\pi inx}$ of $f$ diverges at $0$. $\hat{f}(n)$ is defined by $\hat{f}(n)=\int_{0}^{1}e^{-2\pi inx}f(x)dx$

Notice that $f \mapsto \hat{f}$ is linear, and the divergence of the series at $0$ can be considered by $\sum_{n\in\mathbb{Z}}\hat{f}(n)e^{2\pi in\cdot0}=\sum_{n\in\mathbb{Z}}\hat{f}(n)$ To invoke Banach-Steinhaus theorem, the family of linear functionals are defined by $\lambda_N(f)=\sum_{|n| \leq N}\hat{f}(n)$ It can be proved that $\lVert \lambda_N \rVert=\int_0^1\left\vert\sum_{|n| \leq N}e^{-2\pi inx}\right\vert dx$ which goes to infinity as $N \to \infty$. The existence of such $f$ that $\sup_{N}|\lambda_N(f)|=+\infty$ follows from the resonance theorem. Further, we also know that these $f$ are in a dense $G_\delta$ subset of the vector space generated by all periodic continuous functions on $[0,1]$.

## The series

Since there is no strong reason to write more posts on this topic, i.e. the three fundamental theorems of linear functional analysis, I think it's time to make a list of the series. It's been around half a year.

1. arXiv:1005.1585v2
2. W. Rudin, Real and Complex Analysis
3. W. Rudin, Functional Analysiss
4. Applications to Fourier series

# The Big Three Pt. 1 - Baire Category Theorem Explained

There are three theorems about Banach spaces that occur frequently in the crux of functional analysis, which are called the 'big three':

1. The Hahn-Banach Theorem
2. The Banach-Steinhaus Theorem
3. The Open Mapping Theorem

The incoming series of blog posts is intended to offer a self-read friendly explanation with richer details. Some basic analysis and topology backgrounds are required.

## First and second category

The term 'category' is due to Baire, who developed the category theorem afterwards. Let $X$ be a topological space. A set $E \subset X$ is said to be nowhere dense if $\overline{E}$ has empty interior, i.e. $\text{int}(\overline{E})= \varnothing$.

There are some easy examples of nowhere dense sets. For example, suppose $X=\mathbb{R}$, equipped with the usual topology. Then $\mathbb{N}$ is nowhere dense in $\mathbb{R}$ while $\mathbb{Q}$ is not. It's trivial since $\overline{\mathbb{N}}=\mathbb{N}$, which has empty interior. Meanwhile $\overline{\mathbb{Q}}=\mathbb{R}$. But $\mathbb{R}$ is open, whose interior is itself. The category is defined using nowhere dense set. In fact,

• A set $S$ is of the first category if $S$ is a countable union of nowhere dense sets.
• A set $T$ is of the second category if $T$ is not of the first category.

## Baire category theorem (BCT)

In this blog post, we consider two cases: BCT in complete metric space and in locally compact Hausdorff space. These two cases have nontrivial intersection but they are not equal. There are some complete metric spaces that are not locally compact Hausdorff.

There are some classic topological spaces, for example $\mathbb{R}^n$, are both complete metric space and locally compact Hausdorff. If a locally compact Hausdorff space happens to be a topological vector space, then this space has finite dimension. Also, a topological vector space has to be Hausdorff.

By a Baire space we mean a topological space $X$ such that the intersection of every countable collection of dense open subsets of $X$ is also dense in $X$.

Baire category states that

(BCT 1) Every complete metric space is a Baire space.

(BCT 2) Every locally compact Hausdorff space is a Baire space.

By taking the complement of the definition, we can see that, every Baire space is not of the first category.

Suppose we have a sequence of sets $\{X_n\}$ where $X_n$ is dense in $X$ for all $n>0$, then $X_0=\cap_n X_n$ is also dense in $X$. Notice then $X_0^{c} = \cup_n X_n^c$, a nowhere dense set and a countable union of nowhere dense sets, i.e. of the first category.

### Proving BCT 1 and BCT 2 via Choquet game

Let $X$ be the given complete metric space or locally Hausdorff space, and $\{X_n\}$ a countable collection of open subsets of $X$. Pick an arbitrary open subsets of $X$, namely $A_0$ (this is possible due to the topology defined on $X$). To prove that $\cap_n V_n$ is dense, we have to show that $A_0 \cap \left(\cap_n V_n\right) \neq \varnothing$. This follows the definition of denseness. Typically we have

A subset $A$ of $X$ is dense if and only if $A \cap U \neq \varnothing$ for all nonempty open subsets $U$ of $X$.

We pick a sequence of nonempty open sets $\{A_n\}$ inductively. With $A_{n-1}$ being picked, and since $V_n$ is open and dense in $X$, the intersection $V_n \cap A_{n-1}$ is nonempty and open. $A_n$ can be chosen such that $\overline{A}_n \subset V_n \cap A_{n-1}$ For BCT 1, $A_n$ can be chosen to be open balls with radius $< \frac{1}{n}$; for BCT 2, $A_n$ can be chosen such that the closure is compact. Define $C = \bigcap_{n=1}^{\infty}\overline{A}_n$ Now, if $X$ is a locally compact Hausdorff space, then due to the compactness, $C$ is not empty, therefore we have $\begin{cases} K \subset A_0 \\ K \subset V_n \quad(n \in \mathbb{N}) \end{cases}$ which shows that $A_0 \cap V_n \neq \varnothing$. BCT 2 is proved.

For BCT 1, we cannot follow this since it's not ensured that $X$ has the Heine-Borel property, for example when $X$ is the Hilbert space (this is also a reason why BCT 1 and BCT 2 are not equivalent). The only tool remaining is Cauchy sequence. But how and where?

For any $\varepsilon > 0$, we have some $N$ such that $\frac{1}{N} < \varepsilon$. For all $m>n>N$, we have $A_m \subset A_n\subset A_N$, therefore the centers of $\{A_n\}$ form a Cauchy sequence, converging to some point of $K$, which implies that $K \neq \varnothing$. BCT 1 follows.

## Applications of BCT

BCT will be used directly in the big three. It can be considered as the origin of them. But there are many other applications in different branches of mathematics. The applications shown below are in the same pattern: if it does not hold, then we have a Baire space of the first category, which is not possible.

$\mathbb{R}$ is uncountable

Suppose $\mathbb{R}$ is countable, then we have $\mathbb{R}=\bigcup_{n=1}^{\infty}\{x_n\}$ where $x_n$ is a real number. But $\{x_n\}$ is nowhere dense, therefore $\mathbb{R}$ is of the first category. A contradiction.

Suppose that $f$ is an entire function, and that in every power series $f(z)=\sum_{n=1}^{\infty}c_n(z-a)^n$ has at least one coefficient is $0$, then $f$ is a polynomial (there exists a $N$ such that $c_n=0$ for all $n>N$).

You can find the proof here. We are using the fact that $\mathbb{C}$ is complete.

An infinite dimensional Banach space $B$ has no countable basis

Assume that $B$ has a countable basis $\{x_1,x_2,\cdots\}$ and define $B_n=\text{span}\{x_1,x_2,\cdots,x_n\}$ It can be easily shown that $B_n$ is nowhere dense. In this sense, $B=\cup_n B_n$. A contradiction since $B$ is a complete metric space.

## The series

Since there is no strong reason to write more posts on this topic, i.e. the three fundamental theorems of linear functional analysis, I think it's time to make a list of the series. It's been around half a year.

# 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: $|z_1| \leq |z_2| \leq \cdots \leq |z_n| \leq \cdots$ 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 $|f(0)|\prod_{n=1}^{n(r)}\frac{r}{|z_n|}=\exp\left[\frac{1}{2\pi}\int_{-\pi}^{\pi}\log|f(re^{i\theta})|d\theta\right]$

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 $g(z)=f(z)\prod_{n=1}^{m}\frac{r^2-\overline{z}_nz}{r(z_n-z)}\prod_{n=m+1}^{n(r)}\frac{z_n}{z_n-z}$ 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 $M(r)=\sup_{\theta}|f(re^{i\theta})|\quad 0<r<\infty$ and $n(r)$ be the number of zeros of $f$ in $\overline{D}(0;r)$. By the maximum modulus theorem, we have $\log|f(2re^{i\theta})| \leq |f(2re^{i\theta})| \leq M(2r)$ If we insert Jensen's formula into this inequality and order $|z_n|$ by increasing moduli, we get $\log M(2r) \geq \frac{1}{2\pi}\int_{-\pi}^{\pi}\log|f(2re^{i\theta})|d\theta=\sum_{n=1}^{n(2r)}\log\frac{2r}{|z_n|}\geq\sum_{n=1}^{n(r)}\log\frac{2r}{|z_n|}\geq n(r)\log2$ Which implies $n(r)\leq\log_2M(2r)$ 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 $\sum_{n=1}^{n(r)}\log\frac{2r}{|z_n|}=\sum_{n=1}^{n(r)}(\log2+\log\frac{r}{|z_n|}) \geq n(r)\log2.$ 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 $M(r) \leq \exp(Ar^k)$ where $A$ and $k$ are given positive numbers, we have a good enough estimation by $n(r) \leq \frac{A+(2r)^k}{\log2}$ This estimation becomes interesting if we consider the logarithm of $n(r)$ and $r$, that is \begin{aligned} \limsup_{r\to\infty}\frac{\log{n(r)}}{\log{r}} &\leq \lim_{r\to\infty} \frac{\log(A+(2r)^k)-\log{2}}{\log{r}} \\ & =k \end{aligned} 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 $\lim_{r\to\infty}\frac{\log{n(r)}}{\log r}=k$

### 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 $M(r) \leq \exp(Ar^k)$ and see what will happen.

By Jensen's, we have $\prod_{n=1}^{n(r)}\frac{r}{|z_n|}=\exp\left[\frac{1}{2\pi}\int_{-\pi}^{\pi}\log|f(re^{i\theta})|d\theta\right] \leq \exp{Ar^k}$ This gives $\prod_{n=1}^{n(r)}|z_n| \geq r^{n(r)}\exp(-Ar^k)$ By the arrangement of $\{z_n\}$, we have $|z_{n(r)}| \geq \sqrt[n(r)]{\prod_{n=1}^{n(r)}|z_n|}\geq r\exp(-Ar^{k-n(r)})$

Another example is when we have $|f(z)| \leq \exp(A|\Im{z}|)$ where $\Im{z}$ means the imagine part of $z$.

We shall notice that in this case, \begin{aligned} \frac{1}{2\pi}\int_{-\pi}^{\pi}\log|f(re^{i\theta})|d\theta &\leq \frac{1}{2\pi}\int_{-\pi}^{\pi}A|r\sin\theta|d\theta=\frac{2Ar}{\pi} \end{aligned} Following Jensen's formula, we therefore have $|z_{n(r)}| \geq \exp(\frac{2A}{\pi}r^{1-n(r)})$

# The Lebesgue-Radon-Nikodym theorem and how von Neumann proved it

## An introduction

If one wants to learn the fundamental theorem of Calculus in the sense of Lebesgue integral, properties of measures have to be taken into account. In elementary calculus, one may consider something like $df(x)=f'(x)dx$ where $f$ is differentiable, say, everywhere on an interval. Now we restrict $f$ to be a differentiable and nondecreasing real function defined on $I=[a,b]$. There we got a one-to-one function defined by $g(x)=x+f(x)$

For measurable sets $E\in\mathfrak{M}$, it can be seen that if $m(E)=0$, we have $m(g(E))=0$. Moreover, $g(E) \in \mathfrak{M}$, and $g$ is one-to-one. Therefore we can define a measure like $\mu(E)=m(g(E))$ If we have a relation $\mu(E)=\int_{E}hdm$ (in fact, this is the Radon-Nikodym theorem we will prove later), the fundamental theorem of calculus for $f$ becomes somewhat clear since if $E=[a,x]$, we got $g(E)=[a+f(a),x+f(x)]$, thus we got \begin{aligned} \mu(E)=m(g(E))&=g(x)-g(a)\\ &=f(x)-f(a)+\int_a^xdt \\ &=\int_a^xh(t)dt \end{aligned} which trivially implies $f(x)-f(a)=\int_a^x[h(t)-1]dt$ the function $h$ looks like to be $g'=f'+1$.

We are not proving the fundamental theorem here. But this gives rise to a question. Is it possible to find a function such that $\mu(E)=\int_{E}hdm$ one may write as $d\mu=hdm$ or, more generally, a measure $\mu$ with respect to another measure $\lambda$? Does this $\mu$ exist with respect to $\lambda$? Does this $h$ exist? Lot of questions. Luckily the Lebesgue decomposition and Radon-Nikodym theorem make it possible.

### Notations

Let $\mu$ be a positive measure on a $\sigma$-algebra $\mathfrak{M}$, let $\lambda$ be any arbitrary measure (positive or complex) defined on $\mathfrak{M}$.

We write $\lambda \ll \mu$ if $\lambda(E)=0$ for every $E\in\mathfrak{M}$ for which $\mu(E)=0$. (You may write $\mu \ll m$ in the previous section.) We say $\lambda$ is absolutely continuous with respect to $\mu$.

Another relation between measures worth consideration is being mutually singular. If we have $\lambda(E)=\lambda(A \cap E)$ for every $E \in \mathfrak{M}$, we say $\lambda$ is concentrated on $A$.

If we now have two measures $\mu_1$ and $\mu_2$, two disjoint sets $A$ and $B$ such that $\mu_1$ is concentrated on $A$, $\mu_2$ is concentrated on $B$, we say $\mu_1$ and $\mu_2$ are mutually singular, and write $\mu_1 \perp \mu_2$

Let $\mu$ be a positive $\sigma$-finite measure on $\mathfrak{M}$, and $\lambda$ a complex measure on $\mathfrak{M}$.

• There exists a unique pair of complex measures $\lambda_{ac}$ and $\lambda_{s}$ on $\mathfrak{M}$ such that

$\lambda = \lambda_{ac}+\lambda_s \quad \lambda_{ac}\ll\mu\quad \lambda_s \perp \mu$

• There is a unique $h \in L^1(\mu)$ such that

$\lambda_{ac}(E)=\int_{E}hd\mu$

for every $E \in \mathfrak{M}$.

The unique pair $(\lambda_{ac},\lambda_s)$ is called the Lebesgue decomposition; the existence of $h$ is called the Radon-Nikodym theorem, and $h$ is called the Radon-Nikodym derivative. One also writes $d\lambda_{ac}=hd\mu$ or $\frac{d\lambda_{ac}}{d\mu}=h$ in this situation.

These are two separate theorems, but von Neumann gave the idea to prove these two at one stroke.

If we already have $\lambda \ll \mu$, then $\lambda_s=0$ and the Radon-Nikodym derivative shows up in the natural of things.

Also, one cannot ignore the fact that $m$ the Lebesgue measure is $\sigma$-finite.

## Proof explained

### Step 1 - Construct a bounded functional

We are going to employ Hilbert space technique in this proof. Precisely speaking, we are going to construct a bounded linear functional to find another function, namely $g$, which is the epicentre of this proof.

The boundedness of $\lambda$ is clear since it's complex, but $\mu$ is only assumed to be $\sigma$-finite. Therefore we need some adjustment onto $\mu$.

#### 1.1 Replacing $\mu$ with a finite measure

If $\mu$ is a positive $\sigma$-finite measure on a $\sigma$-algebra $\mathfrak{M}$ in a set $X$, then there is a function $w$ such that $w \in L^1(\mu)$ and $0<w(x)<1$ for every $x \in X$.

The $\sigma$-finiteness of $\mu$ denotes that, there exist some sets $E_n$ such that $X=\bigcup_{n=1}^{\infty}E_n$ and that $\mu(E_n)<\infty$ for all $n$.

Define w_n(x)= \begin{aligned} \begin{cases} \frac{1}{2^n(1+\mu(E_n))}\quad &x \in E_n \\ 0 \quad &x\notin E_n \end{cases} \end{aligned} (you can also say that $w_n=\frac{1}{2^n(1+\mu(E_n))}\chi_{E_n}$), then we have \begin{aligned} w &= \sum_{n=1}^{\infty}w_n \\ \end{aligned} satisfies $0<w<1$ for all $x$. With $w$, we are able to define a new measure, namely $\tilde{\mu}(E)=\int_{E}wd\mu.$ The fact that $\tilde{\mu}(E)$ is a measure can be validated by considering $\int_{E}wd\mu=\int_{X}\chi_{E}wd\mu$. It's more important that $\tilde{\mu}(E)$ is bounded and $\tilde{\mu}(E)=0$ if and only if $\mu(E)=0$. The second one comes from the strict positivity of $w$. For the first one, notice that \begin{aligned} \tilde{\mu}(X) &\leq \sum_{n=1}^{\infty}\tilde{\mu}(E_n) \\ &= \sum_{n=1}^{\infty}\frac{1}{2^n(1+\mu(E_n))} \\ &\leq \sum_{n=1}^{\infty}\frac{1}{2^n} \end{aligned}

#### 1.2 A bounded linear functional associated with $\lambda$

Since $\lambda$ is complex, without loss of generality, we are able to assume that $\lambda$ is a positive bounded measure on $\mathfrak{M}$. By 1.1, we are able to obtain a positive bounded measure by $\varphi=\lambda+\tilde{\mu}$ Following the construction of Lebesgue measure, we have $\int_{X}fd\varphi=\int_{X}fd\lambda+\int_{X}fwd\mu$ for all nonnegative measurable function $f$. Also, notice that $\lambda \leq \varphi$, we have $\left\vert \int_{X}fd\lambda \right\vert \leq \int_{X}|f|d\lambda \leq \int_{X}|f|d\varphi \leq \sqrt{\varphi(X)}\left\Vert f \right\Vert_2$ for $f \in L^2(\varphi)$ by Schwarz inequality.

Since $\varphi(X)<\infty$, we have $\Lambda{f}=\int_{X}fd\lambda$ to be a bounded linear functional on $L^2(\varphi)$.

### Step 2 - Find the associated function with respect to $\lambda$

Since $L^2(\varphi)$ is a Hilbert space, every bounded linear functional on a Hilbert space $H$ is given by an inner product with an element in $H$. That is, by the completeness of $L^2(\varphi)$, there exists a function $g$ such that $\Lambda{f}=\int_{X}fd\lambda=\int_{X}fgd\varphi=(f,g).$ The properties of $L^2$ space shows that $g$ is determined almost everywhere with respect to $\varphi$.

For $E \in \mathfrak{M}$, we got $0 \leq (\chi_{E},g)=\int_{E}gd\varphi=\int_{E}d\lambda=\lambda(E)\leq\varphi(E)$ which implies $0 \leq g \leq 1$ for almost every $x$ with respect to $\varphi$. Therefore we are able to assume that $0 \leq g \leq 1$ without ruining the identity. The proof is in the bag once we define $A$ to be the set where $0 \leq g < 1$ and $B$ the set where $g=1$.

### Step 3 - Generate $\lambda_{ac}$ and $\lambda_{s}$ and the Radon-Nikodym derivative at one stroke

We claim that $\lambda(A \cap E)$ and $\lambda(B \cap E)$ form the decomposition we are looking for, $\lambda_{ac}$ and $\lambda_s$, respectively. Namely, $\lambda_{ac}=\lambda(A \cap E)$, $\lambda_s=\lambda(B \cap E)$.

#### Proving $\lambda_s \perp \mu$

If we combine $\Lambda{f}=(f,g)$ and $\varphi=\lambda+\tilde{\mu}$ together, we have $\int_{X}(1-g)fd\lambda=\int_{X}fgwd\mu.$ Put $f=\chi_{B}$, we have $\int_{B}wd\mu=0.$ Since $w$ is strictly positive, we see that $\mu(B)=0$. Notice that $A \cap B = \varnothing$ and $A \cup B=X$. For $E \in \mathfrak{M}$, we write $E=E_A \cup E_B$, where $E_A \subset A$ and $E_B \subset B$. Therefore $\mu(E)=\mu(E_A)+\mu(E_B)=\mu(E \cap A)+\mu(E \cap B)=\mu(E \cap A).$ Therefore $\mu$ is concentrated on $A$.

For $\lambda_s$, observe that $\lambda_s(E)=\lambda(E \cap B)=\lambda((E \cap B) \cap B)=\lambda_s(E \cap B).$ Hence $\lambda_s$ is concentrated on $B$. This observation shows that $\lambda_s \perp \mu$.

#### Proving $\lambda_{ac} \ll \mu$ by the Radon-Nikodym derivative

The relation that $\lambda_{ac} \ll \mu$ will be showed by the existence of the Radon-Nikodym derivative.

If we replace $f$ by $(1+g+\cdots+g^n)\chi_E,$ where $E \in \mathfrak{M}$, we have $\int_X(1-g)fd\lambda=\int_E(1-g^{n+1})d\lambda=\int_Eg(1+g+\cdots+g^n)wd\mu.$ Notice that \begin{aligned} \int_{E}(1-g^{n+1})d\lambda &=\int\limits_{E \cap A}(1-g^{n+1})d\lambda + \int\limits_{E \cap B}(1-g^{n+1})d\lambda \\ &=\int\limits_{E \cap A}(1-g^{n+1})d\lambda \\ &\to\lambda(E \cap A) = \lambda_{ac}(E)\quad(n\to\infty) \end{aligned} Define $h_n=g(1+g+g^2+\cdots+g^n)w$, we see that on $A$, $h_n$ converges monotonically to h= \begin{aligned} \begin{cases} \frac{gw}{1-g} \quad &x\in{A}\\ 0 \quad &x\in{B} \end{cases} \end{aligned} By monotone convergence theorem, we got $\lim_{n\to\infty}\int_{E}h_nd\mu = \int_{E}hd\mu=\lambda_{ac}(E).$ for every $E\in\mathfrak{M}$.

The measurable function $h$ is the desired Radon-Nikodym derivative once we show that $h \in L^1(\mu)$. Replacing $E$ with $X$, we see that $\int_{X}|h|d\mu=\int_{X}hd\mu=\lambda_{ac}(X)\leq\lambda(X)<\infty.$ Clearly, if $\mu(E)=0$, we have $\lambda_{ac}(E)=\int_{E}hd\mu=0$ which shows that $\lambda_{ac}\ll\mu$ as desired.

### Step 3 - Generalization onto complex measures

By far we have proved this theorem for positive bounded measure. For real bounded measure, we can apply the proceeding case to the positive and negative part of it. For all complex measures, we have $\lambda=\lambda_1+i\lambda_2$ where $\lambda_1$ and $\lambda_2$ are real.

### Step 4 - Uniqueness of the decomposition

If we have two Lebesgue decompositions of the same measure, namely $(\lambda_{ac},\lambda_s)$ and $(\lambda'_{ac},\lambda'_s)$, we shall show that $\lambda_{ac}-\lambda_{ac}'=\lambda_s'-\lambda_s=0$ By the definition of the decomposition we got $\lambda_{ac}-\lambda'_{ac}=\lambda'_s-\lambda_s$ with $\lambda_{ac}-\lambda_{ac}' \ll \mu$ and $\lambda_{s}'-\lambda_{s}\perp\mu$. This implies that $\lambda'_{s}-\lambda_{s} \ll \mu$ as well.

Since $\lambda'_s-\lambda_s\perp\mu$, there exists a set with $\mu(A)=0$ on which $\lambda'_s-\lambda_s$ is concentrated; the absolute continuity shows that $\lambda'_s(E)-\lambda_s(E)=0$ for all $E \subset A$. Hence $\lambda_s'-\lambda_s$ is concentrated on $X-A$. Therefore we got $(\lambda'_s-\lambda_s)\perp(\lambda'_s-\lambda_s)$, which forces $\lambda'_s-\lambda_s=0$. The uniqueness is proved.

(Following the same process one can also show that $\lambda_{ac}\perp\lambda_s$.)

# 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 $\lim_{z \to z_0}\frac{f(z)-f(z_0)}{z-z_0}$ 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 $Z(f)=\{z_0\in\Omega:f(z_0)=0\}$ 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 $f(z)=\sum c_n(z-a)^n$ 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: $\bigcup_{n=0}^{\infty} Z(f^{(n)})=\mathbb{C}$ 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 $D(a;r)$: $f(z)=\sum_{n=1}^{\infty}c_n(z-a)^{n}.$

Define

\begin{aligned} ​ g(z)=\begin{cases} ​ (z-a)^{-m}f(z)\quad&(z\in\Omega-\{a\}) \\\ ​ c_m\quad&(z=a) ​ \end{cases} \end{aligned}

It's clear that $g \in H(D(a;r))$ since we have $g(z)=\sum_{n=1}^{\infty}c_{m+n}(z-a)^{n}\quad(z\in D(a;r))$

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

$\Gamma=\{t\in[0,1]:\gamma(t)\in\bigcap_{n \geq 0}Z(f^{(n)})\}$ By hypothesis, $0 \in \Gamma$. We shall prove that $1 \in \Gamma$. Let $s = \sup\Gamma$ There exists a sequence $\{t_n\}\subset\Gamma$ such that $t_n \to s$. The continuity of $f^{(k)}$ and $\gamma$ shows that $f^{(k)}(\gamma(s))=0$

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.

# 洛必达法则的几种不同的证明

### 证明1:线性近似

#### 线性近似的严格证明

$h\to0$时，$r(h)\to0$$s(h)\to0$，故得到了结论。

### 证明2:中值定理

#### 情况1: $-\infty\leq{A}<+\infty$

$a<x<y<c$，由GMVT可知，存在$t\in(x,y)$使得不等式(A)成立: $\frac{f(x)-f(y)}{g(x)-g(y)}=\frac{f’(t)}{g’(t)}<A+\varepsilon$ 最后一个不等式成立是因为$t\in(x,y)\subset(a,b)$，而$(a,b)$中这个不等式成立。

##### 情况1.1: $g(x)\to0$

$x\to{a}$，此时关于$x$$y$的不等式会有$\frac{f(y)}{g(y)}\leq{A+\varepsilon}<q\quad(a<y<a+\delta)$

(注意:这个地方并没有用$\varepsilon-\delta$证明了这个情况下的收敛)

##### 情况1.2: $g(x)\to+\infty$

$r=A+\varepsilon$。固定不等式(A)中的$y$，因为$g(x)\to+\infty$，能找到一个值$c\in(a,b)$使得$g(x)>g(y)$$g(x)>0$对所有$x\in(a,c)$同时成立。那么不等式(A)两边同时乘以$[g(x)-g(y)]/g(x)$，能得到不等式(C) $\frac{f(x)}{g(x)}<r-r\frac{g(y)}{g(x)}+\frac{f(y)}{g(x)}\quad(a<x<c)$

$x\to{a}$，因为$g(x)\to+\infty$，有点$c_1\in(a,c)$使得不等式(D)成立: $\frac{f(x)}{g(x)}<q\quad(a<x<c_1)$