The Goal

We are going to show the completeness of $X/N$ where $X$ is a TVS and $N$ a closed subspace. Alongside, a bunch of useful analysis tricks will be demonstrated (and that’s why you may find this blog post a little tedious.). But what’s more important, the theorem proved here will be used in the future.

The main process

To make it clear, we should give a formal definition of $F$-space.

A topological space $X$ is an $F$-space if its topology $\tau$ is induced by a complete invariant metric $d$.

A metric $d$ on a vector space $X$ will be called invariant if for all $x,y,z \in X$, we have

$$d(x+z,y+z)=d(x,y).$$

By complete we mean every Cauchy sequence of $(X,d)$ converges.

Defining the quotient metric $\rho$

The metric can be inherited to the quotient space naturally (we will use this fact latter), that is

If $X$ is a $F$-space, $N$ is a closed subspace of a topological vector space $X$, then $X/N$ is still a $F$-space.

Suppose $d$ is a complete invariant metric compatible with $\tau_X$. The metric on $X/N$ is defined by

$$\boxed{\rho(\pi(x),\pi(y))=\inf_{z \in N}d(x-y,z)}$$

$\rho$ is a metric

Proof. First, if $\pi(x)=\pi(y)$, that is, $x-y \in N$, we see

$$\rho(\pi(x),\pi(y))=\inf_{z \in N}d(x-y,z)=d(x-y,x-y)=0.$$

If $\pi(x) \neq \pi(y)$ however, we shall show that $\rho(\pi(x),\pi(y))>0$. In this case, we have $x-y \notin N$. Since $N$ is closed, $N^c$ is open, and $x-y$ is an interior point of $X-N$. Therefore there exists an open ball $B_r(x-y)$ centered at $x-y$ with radius $r>0$ such that $B_r(x-y) \cap N = \varnothing$. Notice we have $d(x-y,z)>r$ since otherwise $z \in B_r(x-y)$. By putting

$$r_0=\sup\{r:B_r(x-y) \cap N = \varnothing\},$$

we see $d(x-y,z) \geq r_0$ for all $z \in N$ and indeed $r_0=\inf_{z \in N}d(x-y,z)>0$ (the verification can be done by contradiction). In general, $\inf_z d(x-y,z)=0$ if and only if $x-y \in \overline{N}$.

Next, we shall show that $\rho(\pi(x),\pi(y))=\rho(\pi(y),\pi(x))$, and it suffices to assume that $\pi(x) \neq \pi(y)$. Sgince $d$ is translate invariant, we get

$$\begin{aligned} d(x-y,z)&=d(x-y-z,0) \\ &=d(0,y-x+z) \\ &=d(-z,y-x) \\ &=d(y-x,-z). \end{aligned}$$

Therefore the $\inf$ of the left hand is equal to the one of the right hand. The identity is proved.

Finally, we need to verify the triangle inequality. Let $r,s,t \in X$. For any $\varepsilon>0$, there exist some $z_\varepsilon$ and $z_\varepsilon’$ such that

$$d(r-s,z_\varepsilon)<\rho(\pi(r),\pi(s))+\frac{\varepsilon}{2},\quad d(s-t,z'_\varepsilon)<\rho(\pi(s),\pi(t))+\frac{\varepsilon}{2}.$$

Since $d$ is invariant, we see

$$\begin{aligned} d(r-t,z_\varepsilon+z'_\varepsilon)&=d((r-s)+(s-t)-(z_\varepsilon+z'_\varepsilon),0) \\ &=d([(r-s)-z_\varepsilon]+[(s-t)-z'_\varepsilon],0) \\ &=d(r-s-z_\varepsilon,t-s+z'_\varepsilon) \\ &\leq d(r-s-z_\varepsilon,0)+d(t-s+z'_\varepsilon,0) \\ &=d(r-s,z_\varepsilon)+d(s-t,z'_\varepsilon) \end{aligned}$$

(I owe @LeechLattice for the inequality above.)

Therefore

$$\begin{aligned} d(r-t,z_\varepsilon+z'_\varepsilon)&\leq d(r-s,z_\varepsilon)+d(s-t,z'_\varepsilon) \\ &<\rho(\pi(r),\pi(s))+\rho(\pi(s),\pi(t))+\varepsilon. \end{aligned}$$

(Warning: This does not imply that $\rho(\pi(r),\pi(s))+\rho(\pi(s),\pi(t))=\inf_z d(r-t,z)$ since we don’t know whether it is the lower bound or not.)

If $\rho(\pi(r),\pi(s))+\rho(\pi(s),\pi(t))<\rho(\pi(r),\pi(t))$ however, let

$$0<\varepsilon<\rho(\pi(r),\pi(t))-(\rho(\pi(r),\pi(s))+\rho(\pi(s),\pi(t)))$$

then there exists some $z’’_\varepsilon=z_\varepsilon+z’_\varepsilon$ such that

$$d(r-t,z''_\varepsilon)<\rho(\pi(r),\pi(t))$$

which is a contradiction since $\rho(\pi(r),\pi(t)) \leq d(r-t,z)$ for all $z \in N$.

(We are using the $\varepsilon$ definition of $\inf$. See here.)

$\rho$ is translate invariant

Since $\pi$ is surjective, we see if $u \in X/N$, there exists some $a \in X$ such that $\pi(a)=u$. Therefore

$$\begin{aligned} \rho(\pi(x)+u,\pi(y)+u) &=\rho(\pi(x)+\pi(a),\pi(y)+\pi(a)) \\ &=\rho(\pi(x+a),\pi(y+a)) \\ &=\inf_{z \in N}d(x+a-y-a,z) \\ &=\rho(\pi(x),\pi(y)). \end{aligned}$$

$\rho$ is well-defined

If $\pi(x)=\pi(x’)$ and $\pi(y)=\pi(y’)$, we have to show that $\rho(\pi(x),\pi(y))=\rho(\pi(x’),\pi(y’))$. In fact,

$$\begin{aligned} \rho(\pi(x),\pi(y)) &\leq \rho(\pi(x),\pi(x'))+\rho(\pi(x'),\pi(y'))+\rho(\pi(y'),\pi(y)) \\ &=\rho(\pi(x'),\pi(y')) \end{aligned}$$

since $\rho(\pi(x),\pi(x’))=0$ as $\pi(x)=\pi(x’)$. Meanwhile

$$\begin{aligned} \rho(\pi(x'),\pi(y')) &\leq \rho(\pi(x'),\pi(x)) + \rho(\pi(x),\pi(y)) + \rho(\pi(y),\pi(y')) \\ &= \rho(\pi(x),\pi(y)). \end{aligned}$$

therefore $\rho(\pi(x),\pi(y))=\rho(\pi(x’),\pi(y’))$.

$\rho$ is compatible with $\tau_N$

By proving this, we need to show that a set $E \subset X/N$ is open with respect to $\tau_N$ if and only if $E$ is a union of open balls. But we need to show a generalized version:

If $\mathscr{B}$ is a local base for $\tau$, then the collection $\mathscr{B}_N$, which contains all sets $\pi(V)$ where $V \in \mathscr{B}$, forms a local base for $\tau_N$.

Proof. We already know that $\pi$ is continuous, linear and open. Therefore $\pi(V)$ is open for all $V \in \mathscr{B}$. For any open set around $E \subset X/N$ containing $\pi(0)$, we see $\pi^{-1}(E)$ is open, and we have

$$\pi^{-1}(E)=\bigcup_{V\in\mathscr{B}}V$$

and therefore

$$E=\bigcup_{V \in \mathscr{B}}\pi(V).$$

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Now consider the local base $\mathscr{B}$ containing all open balls around $0 \in X$. Since

$$\pi(\{x:d(x,0)<r\})=\{u:\rho(u,\pi(0))<r\}$$

we see $\rho$ determines $\mathscr{B}_N$. But we have already proved that $\rho$ is invariant; hence $\mathscr{B}_N$ determines $\tau_N$.

If $d$ is complete, then $\rho$ is complete.

Once this is proved, we are able to claim that, if $X$ is a $F$-space, then $X/N$ is still a $F$-space, since its topology is induced by a complete invariant metric $\rho$.

Proof. Suppose $(x_n)$ is a Cauchy sequence in $X/N$, relative to $\rho$. There is a subsequence $(x_{n_k})$ with $\rho(x_{n_k},x_{n_{k+1}})<2^{-k}$. Since $\pi$ is surjective, we are able to pick some $z_k \in X$ such that $\pi(z_k) = x_{n_k}$ and such that

$$d(z_{k},z_{k+1})<2^{-k}.$$

(The existence can be verified by contradiction still.) By the inequality above, we see $(z_k)$ is Cauchy (can you see why?). Since $X$ is complete, $z_k \to z$ for some $z \in X$. By the continuity of $\pi$, we also see $x_{n_k} \to \pi(z)$ as $k \to \infty$. Therefore $(x_{n_k})$ converges. Hence $(x_n)$ converges since it has a convergent subsequence. $\rho$ is complete.

Remarks

This fact will be used to prove some corollaries in the open mapping theorem. For instance, for any continuous linear map $\Lambda:X \to Y$, we see $\ker(\Lambda)$ is closed, therefore if $X$ is a $F$-space, then $X/\ker(\Lambda)$ is a $F$-space as well. We will show in the future that $X/\ker(\Lambda)$ and $\Lambda(X)$ are homeomorphic if $\Lambda(X)$ is of the second category.

There are more properties that can be inherited by $X/N$ from $X$. For example, normability, metrizability, local convexity. In particular, if $X$ is Banach, then $X/N$ is Banach as well. To do this, it suffices to define the quotient norm by

$$\lVert \pi(x) \rVert = \inf\{\lVert x-z \rVert:z \in N\}.$$

Continuity

We are restricting ourselves into $\mathbb{R}$ endowed with normal topology. Recall that a function is continuous if and only if for any open set $U \subset \mathbb{R}$, we have

$$\{x:f(x) \in U\}=f^{-1}(U)$$

to be open. One can rewrite this statement using $\varepsilon-\delta$ language. To say a function $f: \mathbb{R} \to \mathbb{R}$ continuous at $f(x)$, we mean for any $\varepsilon>0$, there exists some $\delta>0$ such that for $t \in (x-\delta,x+\delta)$, we have

$$|f(x)-f(t)|<\varepsilon.$$

$f$ is continuous on $\mathbb{R}$ if and only if $f$ is continuous at every point of $\mathbb{R}$.

If $(x-\delta,x+\delta)$ is replaced with $(x-\delta,x)$ or $(x,x+\delta)$, we get left continuous and right continuous, one of which plays an important role in probability theory.

But the problem is, sometimes continuity is too strong for being a restriction, but the ‘direction’ associated with left/right continuous functions are unnecessary as well. For example the function

$$f(x)=\chi_{(0,1)}(x)$$

is neither left nor right continuous (globally), but it is a thing. Left/right continuity is not a perfectly weakened version of continuity. We need something different.

Definition of semicontinuous

Let $f$ be a real (or extended-real) function on $\mathbb{R}$. The semicontinuity of $f$ is defined as follows.

If

$$\{x:f(x)>\alpha\}$$

is open for all real $\alpha$, we say $f$ is lower semicontinuous.

If

$$\{x:f(x)<\alpha\}$$

is open for all real $\alpha$, we say $f$ is upper semicontinuous.

Is it possible to rewrite these definitions à la $\varepsilon-\delta$? The answer is yes if we restrict ourselves in metric space.

$f: \mathbb{R} \to \mathbb{R}$ is upper semicontinuous at $x$ if, for every $\varepsilon>0$, there exists some $\delta>0$ such that for $t \in (x-\delta,x+\delta)$, we have

$$f(t)<f(x)+\varepsilon$$

$f: \mathbb{R} \to \mathbb{R}$ is lower semicontinuous at $x$ if, for every $\varepsilon>0$, there exists some $\delta>0$ such that for $t \in (x-\delta,x+\delta)$, we have

$$f(t)>f(x)-\varepsilon$$

Of course, $f$ is upper/lower semicontinuous on $\mathbb{R}$ if and only if it is so on every point of $\mathbb{R}$. One shall find no difference between the definitions in different styles.

Relation with continuous functions

Here is another way to see it. For the continuity of $f$, we are looking for arbitrary open subsets $V$ of $\mathbb{R}$, and $f^{-1}(V)$ is expected to be open. For the lower/upper semicontinuity of $f$, however, the open sets are restricted to be like $(\alpha,+\infty]$ and $[-\infty,\alpha)$. Since all open sets of $\mathbb{R}$ can be generated by the union or intersection of sets like $[-\infty,\alpha)$ and $(\beta,+\infty]$, we immediately get

$f$ is continuous if and only if $f$ is both upper semicontinuous and lower semicontinuous.

Proof. If $f$ is continuous, then for any $\alpha \in \mathbb{R}$, we see $[-\infty,\alpha)$ is open, and therefore

$$f^{-1}([-\infty,\alpha))$$

has to be open. The upper semicontinuity is proved. The lower semicontinuity of $f$ is proved in the same manner.

If $f$ is both upper and lower semicontinuous, we see

$$f^{-1}((\alpha,\beta))=f^{-1}([-\infty,\beta)) \cap f^{-1}((\alpha,+\infty])$$

is open. Since every open subset of $\mathbb{R}$ can be written as a countable union of segments of the above types, we see for any open subset $V$ of $\mathbb{R}$, $f^{-1}(V)$ is open. (If you have trouble with this part, it is recommended to review the definition of topology.) $\square$

Examples

There are two important examples.

1. If $E \subset \mathbb{R}$ is open, then $\chi_E$ is lower semicontinuous.
2. If $F \subset \mathbb{R}$ is closed, then $\chi_F$ is upper semicontinuous.

We will prove the first one. The second one follows in the same manner of course. For $\alpha<0$, the set $A=\chi_E^{-1}((\alpha,+\infty])$ is equal to $\mathbb{R}$, which is open. For $\alpha \geq 1$, since $\chi_E \leq 1$, we see $A=\varnothing$. For $0 \leq \alpha < 1$ however, the set of $x$ where $\chi_E>\alpha$ has to be $E$, which is still open.

When checking the semicontinuity of a function, we check from bottom to top or top to bottom. The function $\chi_E$ is defined by

$$\chi_E(x)=\begin{cases} 1 \quad x \in E \\ 0 \quad x \notin E \end{cases}.$$

Addition of semicontinuous functions

If $f_1$ and $f_2$ are upper/lower semicontinuous, then so is $f_1+f_2$.

Proof. We are going to prove this using different tools. Suppose now both $f_1$ and $f_2$ are upper semicontinuous. For $\varepsilon>0$, there exists some $\delta_1>0$ and $\delta_2>0$ such that

$$f_1(t) < f_1(x)+\varepsilon/2 \quad t \in (x-\delta_1,x+\delta_1), \\ f_2(t) < f_2(x) + \varepsilon/2 \quad t \in (x-\delta_2,x+\delta_2).$$

Proof. If we pick $\delta=\min(\delta_1,\delta_2)$, then we see for all $t \in (x-\delta,x+\delta)$, we have

$$f_1(t)+f_2(t)<f_1(x)+f_2(x)+\varepsilon.$$

The upper semicontinuity of $f_1+f_2$ is proved by considering all $x \in \mathbb{R}$.

Now suppose both $f_1$ and $f_2$ are lower semicontinuous. We have an identity by

$$\{x:f_1+f_2>\alpha\}=\bigcup_{\beta\in\mathbb{R}}\{x:f_1>\beta\}\cap\{x:f_2>\alpha-\beta\}.$$

The set on the right side is always open. Hence $f_1+f_2$ is lower semicontinuous. $\square$

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However, when there are infinite many semicontinuous functions, things are different.

Let $\{f_n\}$ be a sequence of nonnegative functions on $\mathbb{R}$, then

• If each $f_n$ is lower semicontinuous, then so is $\sum_{1}^{\infty}f_n$.
• If each $f_n$ is upper semicontinuous, then $\sum_{1}^{\infty}f_n$ is not necessarily upper semicontinuous.

Proof. To prove this we are still using the properties of open sets. Put $g_n=\sum_{1}^{n}f_k$. Now suppose all $f_k$ are lower. Since $g_n$ is a finite sum of lower functions, we see each $g_n$ is lower. Let $f=\sum_{n}f_n$. As $f_k$ are non-negative, we see $f(x)>\alpha$ if and only if there exists some $n_0$ such that $g_{n_0}(x)>\alpha$. Therefore

$$\{x:f(x)>\alpha\}=\bigcup_{n \geq n_0}\{x:g_n>\alpha\}.$$

The set on the right hand is open already.

For the upper semicontinuity, it suffices to give a counterexample, but before that, we shall give the motivation.

As said, the characteristic function of a closed set is upper semicontinuous. Suppose $\{E_n\}$ is a sequence of almost disjoint closed set, then $E=\cup_{n\geq 1}E_n$ is not necessarily closed, therefore $\chi_E=\sum\chi_{E_n}$ (a.e.) is not necessarily upper semicontinuous. Now we give a concrete example. Put $f_0=\chi_{[1,+\infty]}$ and $f_n=\chi_{E_n}$ for $n \geq 1$ where

$$E_n=\{x:\frac{1}{1+n} \leq x \leq \frac{1}{n}\}.$$

For $x > 0$, we have $f=\sum_nf_n \geq 1$. Meanwhile, $f^{-1}([-\infty,1))=[-\infty,0]$, which is not open. $\square$

Notice that $f$ can be defined on any topological space here.

Maximum and minimum

There is one fact we already know about continuous functions.

If $X$ is compact, $f: X \to \mathbb{R}$ is continuous, then there exists some $a,b \in X$ such that $f(a)=\min f(X)$, $f(b)=\max f(X)$.

In fact, $f(X)$ is compact still. But for semicontinuous functions, things will be different but reasonable. For upper semicontinuous functions, we have the following fact.

If $X$ is compact and $f: X \to (-\infty,+\infty)$ is upper semicontinuous, then there exists some $a \in X$ such that $f(a)=\max f(X)$.

Notice that $X$ is not assumed to hold any other topological property. It can be Hausdorff or Lindelöf, but we are not asking for restrictions like this. The only property we will be using is that every open cover of $X$ has a finite subcover. Of course, one can replace $X$ with any compact subset of $\mathbb{R}$, for example, $[a,b]$.

Proof. Put $\alpha=\sup f(X)$, and define

$$E_n=\{x:f(x)<\alpha-\frac{1}{n}\}.$$

If $f$ attains no maximum, then for any $x \in X$, there exists some $n \geq 1$ such that $f(x)<\alpha-\frac{1}{n}$. That is, $x \in E_n$ for some $n$. Therefore $\bigcup_{n \geq 1}E_n$ covers $X$. But this cover has no finite subcover of $X$. A contradiction since $X$ is compact. $\square$

Approximating integrable functions

This is a comprehensive application of several properties of semicontinuity.

(Vitali–Carathéodory theorem) Suppose $f \in L^1(\mathbb{R})$, where $f$ is a real-valued function. For $\varepsilon>0$, there exist some functions $u$ and $v$ on $\mathbb{R}$ such that $u \leq f \leq v$, $u$ is an upper semicontinuous function bounded above, and $v$ is lower semicontinuous bounded below, and

$$\boxed{\int_{\mathbb{R}}(v-u)dm<\varepsilon}$$

It suffices to prove this theorem for $f \geq 0$ (of course $f$ is not identically equal to $0$ since this case is trivial). Since $f$ is the pointwise limit of an increasing sequence of simple functions $s_n$, can to write $f$ as

$$f=s_1+\sum_{n=2}^{\infty}(s_n-s_{n-1}).$$

By putting $t_1=s_1$, $t_n=s_n-s_{n-1}$ for $n \geq 2$, we get $f=\sum_n t_n$. We can write $f$ as

$$f=\sum_{k=1}^{\infty}c_k\chi_{E_k}$$

where $E_k$ is measurable for all $k$. Also, we have

$$\int_X f d\mu = \sum_{k=1}^{\infty}c_km(E_k),$$

and the series on the right hand converges (since $f \in L^1$. By the properties of Lebesgue measure, there exists a compact set $F_k$ and an open set $V_k$ such that $F_k \subset E_k \subset V_k$ and $c_km(V_k-F_k)<\frac{\varepsilon}{2^{k+1}}$. Put

$$v=\sum_{k=1}^{\infty}c_k\chi_{V_k},\quad u=\sum_{k=1}^{N}c_k\chi_{F_k}$$

(now you can see $v$ is lower semicontinuous and $u$ is upper semicontinuous). The $N$ is chosen in such a way that

$$\sum_{k=N+1}^{\infty}c_km(E_K)<\frac{\varepsilon}{2}.$$

Since $V_k \supset E_k$, we have $\chi_{V_k} \geq \chi_{E_k}$. Therefore $v \geq f$. Similarly, $f \geq u$. Now we need to check the desired integral inequality. A simple recombination shows that

$$\begin{aligned} v-u&=\sum_{k=1}^{\infty}c_k\chi_{V_k}-\sum_{k=1}^{N}c_k\chi_{F_k} \\ &\leq \sum_{k=1}^{\infty}c_k\chi_{V_k}-\sum_{k=1}^{N}c_k\chi_{F_k}+\sum_{k=N+1}^{\infty}c_k(\chi_{E_k}-\chi_{F_k}) \\ &=\sum_{k=1}^{\infty}c_k(\chi_{V_k}-\chi_{F_k})+\sum_{k=N+1}^{\infty}c_k\chi_{E_k}. \end{aligned}.$$

If we integrate the function above, we get

$$\begin{aligned} \int_{\mathbb{R}}(v-u)dm &\leq \sum_{k=1}^{\infty}c_k\mu(V_k-E_k)+\sum_{k=N+1}^{\infty}c_k\chi_{E_k} \\ &< \sum_{k=1}^{\infty}\frac{\varepsilon}{2^{k+1}}+\frac{\varepsilon}{2} \\ &=\varepsilon. \end{aligned}$$

This proved the case when $f \geq 0$. In the general case, we write $f=f^{+}-f^{-}$. Attach the semicontinuous functions to $f^{+}$ and $f^{-}$ respectively by $u_1 \leq f^{+} \leq v_1$ and $u_2 \leq f^{-} \leq v_2$. Put $u=u_1-v_2$, $v=v_1-u_2$. As we can see, $u$ is upper semicontinuous and $v$ is lower semicontinuous. Also, $u \leq f \leq v$ with the desired property since

$$\int_\mathbb{R}(v-u)dm=\int_\mathbb{R}(v_1-u_1)dm+\int_\mathbb{R}(v_2-u_2)dm<2\varepsilon,$$

and the theorem follows. $\square$

Generalisation

Indeed, the only property about measure used is the existence of $F_k$ and $V_k$. The domain $\mathbb{R}$ here can be replaced with $\mathbb{R}^k$ for $1 \leq k < \infty$, and $m$ be replaced with the respective $m_k$. Much more generally, the domain can be replaced by any locally compact Hausdorff space $X$ and the measure by any measure associated with the Riesz-Markov-Kakutani representation theorem on $C_c(X)$.

Is the reverse approximation always possible?

The answer is no. Consider the fat Cantor set $K$, which has Lebesgue measure $\frac{1}{2}$. We shall show that $\chi_K$ can not be approximated below by a lower semicontinuous function.

If $v$ is a lower semicontinuous function such that $v \leq \chi_K$, then $v \leq 0$.

Proof. Consider the set $V=v^{-1}((0,1])=v^{-1}((0,+\infty))$. Since $v \leq \chi_K$, we have $V \subset K$. We will show that $V$ has to be empty.

Pick $t \in V$. Since $V$ is open, there exists some neighbourhood $U$ containing $t$ such that $U \subset V$. But $U=\varnothing$ since $U \subset K$ and $K$ has an empty interior. Therefore $V = \varnothing$. That is, $v \leq 0$ for all $x$. $\square$

Suppose $u$ is an upper semicontinuous function such that $u \geq f$. For $\varepsilon=\frac{1}{2}$, we have

$$\int_{\mathbb{R}}(u-v)dm \geq \int_\mathbb{R}(f-v)dm \geq \frac{1}{2}.$$

This example shows that there exist some integrable functions that are not able to reversely approximated in the sense of the Vitali–Carathéodory theorem.

I’m assuming the reader has some abstract algebra and functional analysis background. You may have learned this already in your linear algebra class, but we are making our way to functional analysis problems.

Motivation

The trouble with $L^p$ spaces

Fix $p$ with $1 \leq p \leq \infty$. It’s easy to see that $L^p(\mu)$ is a topological vector space. But it is not a metric space if we define

$$d(f,g)=\lVert f-g \rVert_p.$$

The reason is, if $d(f,g)=0$, we can only get $f=g$ a.e., but they are not strictly equal. With that being said, this function $d$ is actually a pseudo metric. This is unnatural. However, the relation $\sim$ by $f \sim g \mathbb{R}ightarrow d(f,g)=0$ is a equivalence relation. This inspires us to take the quotient set into consideration.

Vector spaces are groups anyway

For a vector space $V$, every subspace of $V$ is a normal subgroup. There is no reason to prevent ourselves from considering the quotient group and looking for some interesting properties. Further, a vector space is an abelian group, therefore any subspace is automatically normal.

Definition

Let $N$ be a subspace of a vector space $X$. For every $x \in X$, let $\pi(x)$ be the coset of $N$ that contains $x$, that is

$$\pi(x)=x+N.$$

Trivially, $\pi(x)=\pi(y)$ if and only if $x-y \in N$ (say, $\pi$ is well-defined since $N$ is a vector space). This is a linear function since we also have the addition and multiplication by

$$\pi(x)+\pi(y)=\pi(x+y) \quad \alpha\pi(x)=\pi(\alpha{x}).$$

These cosets are the elements of a vector space $X/N$, which reads, the quotient space of $X$ modulo $N$. The map $\pi$ is called the canonical map as we all know.

Examples

First, we shall treat $\mathbb{R}^2$ as a vector space, and the subspace $\mathbb{R}$, which is graphically represented by $x$-axis, as a subspace (we will write it as $X$). For a vector $v=(2,3)$, which is represented by $AB$, we see the coset $v+X$ has something special. Pick any $u \in X$, for example, $AE$, $AC$, or $AG$. We see $v+u$ has the same $y$ value. The reason is simple since we have $v+u=(2+x,3)$, where the $y$ value remains fixed however $u$ may vary.

With that being said, the set $v+X$, which is not a vector space, can be represented by $\overrightarrow{AD}$. This proceed can be generalized to $\mathbb{R}^n$ with $\mathbb{R}^m$ as a subspace with ease.

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We now consider a fancy example. Consider all rational Cauchy sequences, that is

$$(a_n)=(a_1,a_2,\cdots)$$

where $a_k\in\mathbb{Q}$ for all $k$. In analysis class, we learned two facts.

1. Any Cauchy sequence is bounded.
2. If $(a_n)$ converges, then $(a_n)$ is Cauchy.

However, the reverse of 2 does not hold in $\mathbb{Q}$. For example, if we put $a_k=(1+\frac{1}{k})^k$, we should have the limit to be $e$, but $e \notin \mathbb{Q}$.

If we define the addition and multiplication term by term, namely

$$(a_n)+(b_n)=(a_1+b_1,a_2+b_2,\cdots)$$

and

$$(\alpha a_n)=(\alpha a_1,\alpha a_2,\cdots)$$

where $\alpha \in \mathbb{Q}$, we get a vector space (the verification is easy). The zero vector is defined by

$$(0)=(0,0,\cdots).$$

This vector space is denoted by $\overline{\mathbb{Q}}$. The subspace containing all sequences converges to $0$ will be denoted by $\overline{\mathbb{O}}$. Again, $(a_n)+\overline{\mathbb{O}}=(b_n)+\overline{\mathbb{O}}$ if and only if $(a_n-b_n) \in \overline{\mathbb{O}}$. Using the language of equivalence relation, we also say $(a_n)$ and $(b_n)$ are equivalent if $(a_n-b_n) \in \overline{\mathbb{O}}$. For example, the two following sequences are equivalent:

$$(1,1,1,\cdots,1,\cdots)\quad\quad (0.9,0.99,0.999,\cdots).$$

Actually, we will get $\mathbb{R} \simeq \overline{\mathbb{Q}}/\overline{\mathbb{O}}$ in the end. But to make sure that this quotient space is exactly the one we meet in our analysis class, there are a lot of verifications should be done.

We shall give more definitions for calculation. The multiplication of two Cauchy sequences is defined term by term à la the addition. For $\overline{\mathbb{Q}}/\overline{\mathbb{O}}$ we have

$$((a_n)+\overline{\mathbb{O}})+((b_n)+\overline{\mathbb{O}})=(a_n+b_n) + \overline{\mathbb{O}}$$

and

$$((a_n)+\overline{\mathbb{O}})((b_n)+\overline{\mathbb{O}})=(a_nb_n)+\overline{\mathbb{O}}.$$

As for inequality, a partial order has to be defined. We say $(a_n) > (0)$ if there exists some $N>0$ such that $a_n>0$ for all $n \geq N$. By $(a_n) > (b_n)$ we mean $(a_n-b_n)>(0)$ of course. For cosets, we say $(a_n)+\overline{\mathbb{O}}>\overline{\mathbb{O}}$ if $(x_n) > (0)$ for some $(x_n) \in (a_n)+\overline{\mathbb{O}}$. This is well defined. That is, if $(x_n)>(0)$, then $(y_n)>(0)$ for all $(y_n) \in (a_n)+\overline{\mathbb{O}}$.

With these operations being defined, it can be verified that $\overline{\mathbb{Q}}/\overline{\mathbb{O}}$ has the desired properties, for example, the least-upper-bound property. But this goes too far from the topic, we are not proving it here. If you are interested, you may visit here for more details.

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Finally, we are trying to make $L^p$ a Banach space. Fix $p$ with $1 \leq p < \infty$. There is a seminorm defined for all Lebesgue measurable functions on $[0,1]$ by

$$p(f)=\lVert f \rVert_p=\left\{\int_{0}^{1}|f(t)|^pdt\right\}^{1/p}$$

$L^p$ is a vector space containing all functions $f$ with $p(f)<\infty$. But it’s not a normed space by $p$, since $p(f)=0$ only implies $f=0$ almost everywhere. However, the set $N$ which contains all functions that equal $0$ is also a vector space. Now consider the quotient space by

$$\tilde{p}(\pi(f))=p(f),$$

where $\pi$ is the canonical map of $L^p$ into $L^p/N$. We shall prove that $\tilde{p}$ is well-defined here. If $\pi(f)=\pi(g)$, we have $f-g \in N$, therefore

$$0=p(f-g)\geq |p(f)-p(g)|,$$

which forces $p(f)=p(g)$. Therefore in this case we also have $\tilde{p}(\pi(f))=\tilde{p}(\pi(g))$. This indeed ensures that $\tilde{p}$ is a norm, and $L^p/N$ a Banach space. There are some topological facts required to prove this, we are going to cover a few of them.

Topology of quotient space

Definition

We know if $X$ is a topological vector space with a topology $\tau$, then the addition and scalar multiplication are continuous. Suppose now $N$ is a closed subspace of $X$. Define $\tau_N$ by

$$\tau_N=\{E \subset X/N:\pi^{-1}(E)\in \tau\}.$$

We are expecting $\tau_N$ to be properly-defined. And fortunately, it is. Some interesting techniques will be used in the following section.

$\tau_N$ is a vector topology

There will be two steps to get this done.

$\tau_N$ is a topology.

It is trivial that $\varnothing$ and $X/N$ are elements of $\tau_N$. Other properties are immediate as well since we have

$$\pi^{-1}(A \cap B) = \pi^{-1}(A) \cap \pi^{-1}(B)$$

and

$$\pi^{-1}(\cup A_\alpha)=\cup\pi^{-1}( A_{\alpha}).$$

That said, if we have $A,B\in \tau_N$, then $A \cap B \in \tau_N$ since $\pi^{-1}(A \cap B)=\pi^{-1}(A) \cap \pi^{-1}(B) \in \tau$.

Similarly, if $A_\alpha \in \tau_N$ for all $\alpha$, we have $\cup A_\alpha \in \tau_N$. Also, by definition of $\tau_N$, $\pi$ is continuous.

$\tau_N$ is a vector topology.

First, we show that a point in $X/N$, which can be written as $\pi(x)$, is closed. Notice that $N$ is assumed to be closed, and

$$\pi^{-1}(\pi(x))=x+N$$

therefore has to be closed.

In fact, $F \subset X/N$ is $\tau_N$-closed if and only if $\pi^{-1}(F)$ is $\tau$-closed. To prove this, one needs to notice that $\pi^{-1}(F^c)=(\pi^{-1}(F))^{c}$.

Suppose $V$ is open, then

$$\pi^{-1}(\pi(V))=N+V$$

is open. By definition of $\tau_N$, we have $\pi(V) \in \tau_N$. Therefore $\pi$ is an open mapping.

If now $W$ is a neighbourhood of $0$ in $X/N$, there exists a neighbourhood $V$ of $0$ in $X$ such that

$$V + V \subset \pi^{-1}(W).$$

Hence $\pi(V)+\pi(V) \subset W$. Since $\pi$ is open, $\pi(V)$ is a neighbourhood of $0$ in $X/N$, this shows that the addition is continuous.

The continuity of scalar multiplication will be shown in a direct way (so can the addition, but the proof above is intended to offer some special technique). We already know, the scalar multiplication on $X$ by

$$\begin{aligned} \varphi:\Phi \times X &\to X \\ (\alpha,x) &\mapsto \alpha{x} \end{aligned}$$

is continuous, where $\Phi$ is the scalar field (usually $\mathbb{R}$ or $\mathbb{C}$. Now the scalar multiplication on $X/N$ is by

$$\begin{aligned} \psi: \Phi \times X/N &\to X/N \\ (\alpha,x+N) &\mapsto \alpha{x}+N. \end{aligned}$$

We see $\psi(\alpha,x+N)=\pi(\varphi(\alpha,x))$. But the composition of two continuous functions is continuous, therefore $\psi$ is continuous.

A commutative diagram by quotient space

We are going to talk about a classic commutative diagram that you already see in algebra class.

There are some assumptions.

1. $X$ and $Y$ are topological vector spaces.
2. $\Lambda$ is linear.
3. $\pi$ is the canonical map.
4. $N$ is a closed subspace of $X$ and $N \subset \ker\Lambda$.

Algebraically, there exists a unique map $f: X/N \to Y$ by $x+N \mapsto \Lambda(x)$. Namely, the diagram above is commutative. But now we are interested in some analysis facts.

$f$ is linear.

This is obvious. Since $\pi$ is surjective, for $u,v \in X/N$, we are able to find some $x,y \in X$ such that $\pi(x)=u$ and $\pi(y)=v$. Therefore we have

$$\begin{aligned} f(u+v)=f(\pi(x)+\pi(y))&=f(\pi(x+y)) \\ &=\Lambda(x+y) \\ &=\Lambda(x)+\Lambda(y) \\ &= f(\pi(x))+f(\pi(y)) \\ &=f(u)+f(v) \end{aligned}$$

and

$$\begin{aligned} f(\alpha{u})=f(\alpha\pi(x))&=f(\pi(\alpha{x})) \\ &= \Lambda(\alpha{x}) \\ &= \alpha\Lambda(x) \\ &= \alpha{f(\pi(x))} \\ &= \alpha{f(u)}. \end{aligned}$$

$\Lambda$ is open if and only if $f$ is open.

If $f$ is open, then for any open set $U \subset X$, we have

$$\Lambda(U)=f(\pi(U))$$

to be an open set since $\pi$ is open, and $\pi(U)$ is an open set.

If $f$ is not open, then there exists some $V \subset X/N$ such that $f(V)$ is closed. However, since $\pi$ is continuous, we have $\pi^{-1}(V)$ to be open. In this case, we have

$$f(\pi(\pi^{-1}(V)))=f(V)=\Lambda(\pi^{-1}(V))$$

to be closed. $\Lambda$ is therefore not open. This shows that if $\Lambda$ is open, then $f$ is open.

$\Lambda$ is continuous if and only if $f$ is continuous.

If $f$ is continuous, for any open set $W \subset Y$, we have $\pi^{-1}(f^{-1}(W))=\Lambda^{-1}(W)$ to be open. Therefore $\Lambda$ is continuous.

Conversely, if $\Lambda$ is continuous, for any open set $W \subset Y$, we have $\Lambda^{-1}(W)$ to be open. Therefore $f^{-1}(W)=\pi(\Lambda^{-1}(W))$ has to be open since $\pi$ is open.

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

$$|z_{n(r)}| \geq \sqrt[n(r)]{\prod_{n=1}^{n(r)}|z_n|}\geq r\exp(-Ar^{k-n(r)})$$

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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

$$|z_{n(r)}| \geq \exp(\frac{2A}{\pi}r^{1-n(r)})$$

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

$$\mu_1 \perp \mu_2$$

The Theorem of Lebesgue-Radon-Nikodym

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 $\lambda_{ac}\perp\lambda_s$.)

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.