The completeness of the quotient space (topological vector space)

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 τ 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,zX, we have

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

Defining the quotient metric ρ

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 τX. The metric on X/N is defined by

ρ is a metric

Proof. First, if π(x)=π(y), that is, xyN, we see

If π(x)π(y) however, we shall show that ρ(π(x),π(y))>0. In this case, we have xyN. Since N is closed, Nc is open, and xy is an interior point of XN. Therefore there exists an open ball Br(xy) centered at xy with radius r>0 such that Br(xy)N=. Notice we have d(xy,z)>r since otherwise zBr(xy). By putting

we see d(xy,z)r0 for all zN and indeed r0=infzNd(xy,z)>0 (the verification can be done by contradiction). In general, infzd(xy,z)=0 if and only if xyN.

Next, we shall show that ρ(π(x),π(y))=ρ(π(y),π(x)), and it suffices to assume that π(x)π(y). Sgince d is translate invariant, we get

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,tX. For any ε>0, there exist some zε and zε such that

Since d is invariant, we see

(I owe @LeechLattice for the inequality above.)

Therefore

(Warning: This does not imply that ρ(π(r),π(s))+ρ(π(s),π(t))=infzd(rt,z) since we don’t know whether it is the lower bound or not.)

If ρ(π(r),π(s))+ρ(π(s),π(t))<ρ(π(r),π(t)) however, let

then there exists some zε=zε+zε such that

which is a contradiction since ρ(π(r),π(t))d(rt,z) for all zN.

(We are using the ε definition of inf. See here.)

ρ is translate invariant

Since π is surjective, we see if uX/N, there exists some aX such that π(a)=u. Therefore

ρ is well-defined

If π(x)=π(x) and π(y)=π(y), we have to show that ρ(π(x),π(y))=ρ(π(x),π(y)). In fact,

since ρ(π(x),π(x))=0 as π(x)=π(x). Meanwhile

therefore ρ(π(x),π(y))=ρ(π(x),π(y)).

ρ is compatible with τN

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

If B is a local base for τ, then the collection BN, which contains all sets π(V) where VB, forms a local base for τN.

Proof. We already know that π is continuous, linear and open. Therefore π(V) is open for all VB. For any open set around EX/N containing π(0), we see π1(E) is open, and we have

and therefore


Now consider the local base B containing all open balls around 0X. Since

we see ρ determines BN. But we have already proved that ρ is invariant; hence BN determines τN.

If d is complete, then ρ 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 ρ.

Proof. Suppose (xn) is a Cauchy sequence in X/N, relative to ρ. There is a subsequence (xnk) with ρ(xnk,xnk+1)<2k. Since π is surjective, we are able to pick some zkX such that π(zk)=xnk and such that

(The existence can be verified by contradiction still.) By the inequality above, we see (zk) is Cauchy (can you see why?). Since X is complete, zkz for some zX. By the continuity of π, we also see xnkπ(z) as k. Therefore (xnk) converges. Hence (xn) converges since it has a convergent subsequence. ρ is complete.

Remarks

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

Basic Facts of Semicontinuous Functions

Continuity

We are restricting ourselves into R endowed with normal topology. Recall that a function is continuous if and only if for any open set UR, we have

to be open. One can rewrite this statement using εδ language. To say a function f:RR continuous at f(x), we mean for any ε>0, there exists some δ>0 such that for t(xδ,x+δ), we have

f is continuous on R if and only if f is continuous at every point of R.

If (xδ,x+δ) is replaced with (xδ,x) or (x,x+δ), 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

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 R. The semicontinuity of f is defined as follows.

If

is open for all real α, we say f is lower semicontinuous.

If

is open for all real α, we say f is upper semicontinuous.

Is it possible to rewrite these definitions à la εδ? The answer is yes if we restrict ourselves in metric space.

f:RR is upper semicontinuous at x if, for every ε>0, there exists some δ>0 such that for t(xδ,x+δ), we have

f:RR is lower semicontinuous at x if, for every ε>0, there exists some δ>0 such that for t(xδ,x+δ), we have

Of course, f is upper/lower semicontinuous on R if and only if it is so on every point of 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 R, and f1(V) is expected to be open. For the lower/upper semicontinuity of f, however, the open sets are restricted to be like (α,+] and [,α). Since all open sets of R can be generated by the union or intersection of sets like [,α) and (β,+], 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 αR, we see [,α) is open, and therefore

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

is open. Since every open subset of R can be written as a countable union of segments of the above types, we see for any open subset V of R, f1(V) is open. (If you have trouble with this part, it is recommended to review the definition of topology.)

Examples

There are two important examples.

  1. If ER is open, then χE is lower semicontinuous.
  2. If FR is closed, then χF is upper semicontinuous.

We will prove the first one. The second one follows in the same manner of course. For α<0, the set A=χE1((α,+]) is equal to R, which is open. For α1, since χE1, we see A=. For 0α<1 however, the set of x where χE>α 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 χE is defined by

Addition of semicontinuous functions

If f1 and f2 are upper/lower semicontinuous, then so is f1+f2.

Proof. We are going to prove this using different tools. Suppose now both f1 and f2 are upper semicontinuous. For ε>0, there exists some δ1>0 and δ2>0 such that

Proof. If we pick δ=min(δ1,δ2), then we see for all t(xδ,x+δ), we have

The upper semicontinuity of f1+f2 is proved by considering all xR.

Now suppose both f1 and f2 are lower semicontinuous. We have an identity by

The set on the right side is always open. Hence f1+f2 is lower semicontinuous.


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

Let {fn} be a sequence of nonnegative functions on R, then

  • If each fn is lower semicontinuous, then so is 1fn.
  • If each fn is upper semicontinuous, then 1fn is not necessarily upper semicontinuous.

Proof. To prove this we are still using the properties of open sets. Put gn=1nfk. Now suppose all fk are lower. Since gn is a finite sum of lower functions, we see each gn is lower. Let f=nfn. As fk are non-negative, we see f(x)>α if and only if there exists some n0 such that gn0(x)>α. Therefore

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 {En} is a sequence of almost disjoint closed set, then E=n1En is not necessarily closed, therefore χE=χEn (a.e.) is not necessarily upper semicontinuous. Now we give a concrete example. Put f0=χ[1,+] and fn=χEn for n1 where

For x>0, we have f=nfn1. Meanwhile, f1([,1))=[,0], which is not open.

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:XR is continuous, then there exists some a,bX such that f(a)=minf(X), f(b)=maxf(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(,+) is upper semicontinuous, then there exists some aX such that f(a)=maxf(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 R, for example, [a,b].

Proof. Put α=supf(X), and define

If f attains no maximum, then for any xX, there exists some n1 such that f(x)<α1n. That is, xEn for some n. Therefore n1En covers X. But this cover has no finite subcover of X. A contradiction since X is compact.

Approximating integrable functions

This is a comprehensive application of several properties of semicontinuity.

(Vitali–Carathéodory theorem) Suppose fL1(R), where f is a real-valued function. For ε>0, there exist some functions u and v on R such that ufv, u is an upper semicontinuous function bounded above, and v is lower semicontinuous bounded below, and

It suffices to prove this theorem for f0 (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 sn, can to write f as

By putting t1=s1, tn=snsn1 for n2, we get f=ntn. We can write f as

where Ek is measurable for all k. Also, we have

and the series on the right hand converges (since fL1. By the properties of Lebesgue measure, there exists a compact set Fk and an open set Vk such that FkEkVk and ckm(VkFk)<ε2k+1. Put

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

Since VkEk, we have χVkχEk. Therefore vf. Similarly, fu. Now we need to check the desired integral inequality. A simple recombination shows that

If we integrate the function above, we get

This proved the case when f0. In the general case, we write f=f+f. Attach the semicontinuous functions to f+ and f respectively by u1f+v1 and u2fv2. Put u=u1v2, v=v1u2. As we can see, u is upper semicontinuous and v is lower semicontinuous. Also, ufv with the desired property since

and the theorem follows.

Generalisation

Indeed, the only property about measure used is the existence of Fk and Vk. The domain R here can be replaced with Rk for 1k<, and m be replaced with the respective mk. 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 Cc(X).

Is the reverse approximation always possible?

The answer is no. Consider the fat Cantor set K, which has Lebesgue measure 12. We shall show that χK can not be approximated below by a lower semicontinuous function.

If v is a lower semicontinuous function such that vχK, then v0.

Proof. Consider the set V=v1((0,1])=v1((0,+)). Since vχK, we have VK. We will show that V has to be empty.

Pick tV. Since V is open, there exists some neighbourhood U containing t such that UV. But U= since UK and K has an empty interior. Therefore V=. That is, v0 for all x.

Suppose u is an upper semicontinuous function such that uf. For ε=12, we have

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

An Introduction to Quotient Space

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 Lp spaces

Fix p with 1p. It’s easy to see that Lp(μ) is a topological vector space. But it is not a metric space if we define

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 by fgRightarrowd(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 xX, let π(x) be the coset of N that contains x, that is

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

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

Examples

R^2-quotient

R^2-quotient

First, we shall treat R2 as a vector space, and the subspace 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 uX, 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 AD. This proceed can be generalized to Rn with Rm as a subspace with ease.


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

where akQ for all k. In analysis class, we learned two facts.

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

However, the reverse of 2 does not hold in Q. For example, if we put ak=(1+1k)k, we should have the limit to be e, but eQ.

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

and

where αQ, we get a vector space (the verification is easy). The zero vector is defined by

This vector space is denoted by Q. The subspace containing all sequences converges to 0 will be denoted by O. Again, (an)+O=(bn)+O if and only if (anbn)O. Using the language of equivalence relation, we also say (an) and (bn) are equivalent if (anbn)O. For example, the two following sequences are equivalent:

Actually, we will get RQ/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 Q/O we have

and

As for inequality, a partial order has to be defined. We say (an)>(0) if there exists some N>0 such that an>0 for all nN. By (an)>(bn) we mean (anbn)>(0) of course. For cosets, we say (an)+O>O if (xn)>(0) for some (xn)(an)+O. This is well defined. That is, if (xn)>(0), then (yn)>(0) for all (yn)(an)+O.

With these operations being defined, it can be verified that Q/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.


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

Lp is a vector space containing all functions f with p(f)<. 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

where π is the canonical map of Lp into Lp/N. We shall prove that p~ is well-defined here. If π(f)=π(g), we have fgN, therefore

which forces p(f)=p(g). Therefore in this case we also have p~(π(f))=p~(π(g)). This indeed ensures that p~ is a norm, and Lp/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 τ, then the addition and scalar multiplication are continuous. Suppose now N is a closed subspace of X. Define τN by

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

τN is a vector topology

There will be two steps to get this done.

τN is a topology.

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

and

That said, if we have A,BτN, then ABτN since π1(AB)=π1(A)π1(B)τ.

Similarly, if AατN for all α, we have AατN. Also, by definition of τN, π is continuous.

τN is a vector topology.

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

therefore has to be closed.

In fact, FX/N is τN-closed if and only if π1(F) is τ-closed. To prove this, one needs to notice that π1(Fc)=(π1(F))c.

Suppose V is open, then

is open. By definition of τN, we have π(V)τN. Therefore π 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

Hence π(V)+π(V)W. Since π is open, π(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

is continuous, where Φ is the scalar field (usually R or C. Now the scalar multiplication on X/N is by

We see ψ(α,x+N)=π(φ(α,x)). But the composition of two continuous functions is continuous, therefore ψ 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.

diagram-000001

diagram-000001

There are some assumptions.

  1. X and Y are topological vector spaces.
  2. Λ is linear.
  3. π is the canonical map.
  4. N is a closed subspace of X and NkerΛ.

Algebraically, there exists a unique map f:X/NY by x+NΛ(x). Namely, the diagram above is commutative. But now we are interested in some analysis facts.

f is linear.

This is obvious. Since π is surjective, for u,vX/N, we are able to find some x,yX such that π(x)=u and π(y)=v. Therefore we have

and

Λ is open if and only if f is open.

If f is open, then for any open set UX, we have

to be an open set since π is open, and π(U) is an open set.

If f is not open, then there exists some VX/N such that f(V) is closed. However, since π is continuous, we have π1(V) to be open. In this case, we have

to be closed. Λ is therefore not open. This shows that if Λ is open, then f is open.

Λ is continuous if and only if f is continuous.

If f is continuous, for any open set WY, we have π1(f1(W))=Λ1(W) to be open. Therefore Λ is continuous.

Conversely, if Λ is continuous, for any open set WY, we have Λ1(W) to be open. Therefore f1(W)=π(Λ1(W)) has to be open since π is open.

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 z1,z2,,zn, are ordered increasingly by moduli:

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

Jensen’s formula

Suppose Ω=D(0;R), fH(Ω), f(0)0, 0<r<R, and z1,z2,,zn(r) are the zeros of f in D(0;R), then

There is no need to worry about the assumption f(0)0. Take another look at this proof. Every zero point a has a unique positive number m such that f(z)=(za)mg(z) and gH(Ω) but g(a)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 fzm 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 zj in such a way that z1,,zmD(0;r) and |zm+1|==|zn|. One can prove this identity by considering |g(0)| as well as log|g(reiθ)|.

Several applications

The number of zeros of f in 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 D(0;r). By the maximum modulus theorem, we have

If we insert Jensen’s formula into this inequality and order |zn| 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 znD(0;R) for all RR. Hence we have log2r|zn|log1=0 for all znD(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)=1exp(zk) where k is a positive integer, we have n(r)krkπ, also

Lower bound of |zn(r)|

We’ll see here, how to evaluate the lower bound of |zn(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 {zn}, we have


Another example is when we have

where z means the imagine part of z.

We shall notice that in this case,

Following Jensen’s formula, we therefore have

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

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

For measurable sets EM, it can be seen that if m(E)=0, we have m(g(E))=0. Moreover, g(E)M, and g is one-to-one. Therefore we can define a measure like

If we have a relation

(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

which trivially implies

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

one may write as

or, more generally, a measure μ with respect to another measure λ? Does this μ exist with respect to λ? Does this h exist? Lot of questions. Luckily the Lebesgue decomposition and Radon-Nikodym theorem make it possible.

Notations

Let μ be a positive measure on a σ-algebra M, let λ be any arbitrary measure (positive or complex) defined on M.

We write

if λ(E)=0 for every EM for which μ(E)=0. (You may write μm in the previous section.) We say λ is absolutely continuous with respect to μ.

Another relation between measures worth consideration is being mutually singular. If we have λ(E)=λ(AE) for every EM, we say λ is concentrated on A.

If we now have two measures μ1 and μ2, two disjoint sets A and B such that μ1 is concentrated on A, μ2 is concentrated on B, we say μ1 and μ2 are mutually singular, and write

The Theorem of Lebesgue-Radon-Nikodym

Let μ be a positive σ-finite measure on M, and λ a complex measure on M.

  • There exists a unique pair of complex measures λac and λs on M such that
  • There is a unique hL1(μ) such that

for every EM.

The unique pair (λac,λs) 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 dλac=hdμ or dλacdμ=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 λμ, then λs=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 σ-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 λ is clear since it’s complex, but μ is only assumed to be σ-finite. Therefore we need some adjustment onto μ.

1.1 Replacing μ with a finite measure

If μ is a positive σ-finite measure on a σ-algebra M in a set X, then there is a function w such that wL1(μ) and 0<w(x)<1 for every xX.

The σ-finiteness of μ denotes that, there exist some sets En such that

and that μ(En)< for all n.

Define

(you can also say that wn=12n(1+μ(En))χEn), then we have

satisfies 0<w<1 for all x. With w, we are able to define a new measure, namely

The fact that μ~(E) is a measure can be validated by considering Ewdμ=XχEwdμ. It’s more important that μ~(E) is bounded and μ~(E)=0 if and only if μ(E)=0. The second one comes from the strict positivity of w. For the first one, notice that

1.2 A bounded linear functional associated with λ

Since λ is complex, without loss of generality, we are able to assume that λ is a positive bounded measure on M. By 1.1, we are able to obtain a positive bounded measure by

Following the construction of Lebesgue measure, we have

for all nonnegative measurable function f. Also, notice that λφ, we have

for fL2(φ) by Schwarz inequality.

Since φ(X)<, we have

to be a bounded linear functional on L2(φ).

Step 2 - Find the associated function with respect to λ

Since L2(φ) 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 L2(φ), there exists a function g such that

The properties of L2 space shows that g is determined almost everywhere with respect to φ.

For EM, we got

which implies 0g1 for almost every x with respect to φ. Therefore we are able to assume that 0g1 without ruining the identity. The proof is in the bag once we define A to be the set where 0g<1 and B the set where g=1.

Step 3 - Generate λac and λs and the Radon-Nikodym derivative at one stroke

We claim that λ(AE) and λ(BE) form the decomposition we are looking for, λac and λs, respectively. Namely, λac=λ(AE), λs=λ(BE).

Proving λsμ

If we combine Λf=(f,g) and φ=λ+μ~ together, we have

Put f=χB, we have

Since w is strictly positive, we see that μ(B)=0. Notice that AB= and AB=X. For EM, we write E=EAEB, where EAA and EBB. Therefore

Therefore μ is concentrated on A.

For λs, observe that

Hence λs is concentrated on B. This observation shows that λsμ.

Proving λacμ by the Radon-Nikodym derivative

The relation that λacμ will be showed by the existence of the Radon-Nikodym derivative.

If we replace f by

where EM, we have

Notice that

Define hn=g(1+g+g2++gn)w, we see that on A, hn converges monotonically to

By monotone convergence theorem, we got

for every EM.

The measurable function h is the desired Radon-Nikodym derivative once we show that hL1(μ). Replacing E with X, we see that

Clearly, if μ(E)=0, we have

which shows that

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

where λ1 and λ2 are real.

Step 4 - Uniqueness of the decomposition

If we have two Lebesgue decompositions of the same measure, namely (λac,λs) and (λac,λs), we shall show that

By the definition of the decomposition we got

with λacλacμ and λsλsμ. This implies that λsλsμ as well.

Since λsλsμ, there exists a set with μ(A)=0 on which λsλs is concentrated; the absolute continuity shows that λs(E)λs(E)=0 for all EA. Hence λsλs is concentrated on XA. Therefore we got (λsλs)(λsλs), which forces λsλs=0. The uniqueness is proved.

(Following the same process one can also show that λacλs.)

Topological properties of the zeros of a holomorphic function

What’s going on

If for every z0Ω where Ω is a plane open set, the limit

exists, we say that f is holomorphic (a.k.a. analytic) in Ω. If f is holomorphic in the whole plane, it’s called entire. The class of all holomorphic functions (denoted by H(Ω)) 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 Ω 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 Ω.

Trivially, if f(Ω)={0}, we have Z(f)=Ω. 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 Ω is important. For example, for two disjoint open sets Ω0 and Ω1, define f(z)=0 on Ω0 and f(z)=1 on Ω1, 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!cn=f(n)(a), thus for every aC, we can find a postivie integer n0 such that f(n0)(a)=0. Thus we establish the identity:

Notice the fact that f(n) is entire. So Z(fn) is either an at most countable set without limit point, or simply equal to C. If there exists a number N such that Z(fN)=C, then naturally Z(fn)=C holds for all nN. 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, 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 Ω. The continuity of f shows that AZ(f). We’ll show that if A, then Z(f)=Ω.

First we claim that if aA, then an0Z(f(n)). That is, f(k)(a)=0 for all k0. Suppose this fails, then there is a smallest positive integer m such that cm0 for the power series on the disc D(a;r):

Define

It’s clear that gH(D(a;r)) since we have

But the continuity shows that g(a)=0 while cm0. A contradiction.

Next fix a point bΩ. Choose a curve (continuous mapping) defined γ on [0,1] such that γ(0)=a and γ(1)=b. Let

By hypothesis, 0Γ. We shall prove that 1Γ. Let

There exists a sequence {tn}Γ such that tns. The continuity of f(k) and γ shows that

Hence sΓ. Choose a disc D(γ(s);δ)Ω. On this disc, f is represented by its power series but all coefficients are 0. It follows that f(z)=0 for all zD(γ(s);δ). Further, f(k)(z)=0 for all zD(γ(s);δ) for all k0. Therefore by the continuity of γ, there exists ε>0 such that γ(sε,s+ε)D(γ(s);δ), which implies that (sε,s+ε)[0,1]Γ. Since s=supΓ, we have s=1, therefore 1Γ.

So far we showed that Ω=n0Z(f(n)), which forces Z(f)=Ω. 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 Ω, there are at most finitely many points in Z(f). Since Ω is σ-compact, Z(f) is at most countable. Z(f) is also called a discrete set in this situation.