泛函分析基础：Hahn-Banach定理与巴拿赫空间

需积分: 12 49 浏览量更新于2024-07-18 收藏 600KB PDF 举报

"这篇资源是Jeﬀ Schenker教授在密歇根州立大学春季学期Math920课程的泛函分析讲义。它详细涵盖了泛函分析的基础知识，包括哈恩-班纳赫定理及其应用、巴拿赫空间、希尔伯特空间以及局部凸空间，还涉及弱收敛和弱拓扑的概念。" 在泛函分析中，拓扑线性空间是一个关键概念，它结合了线性空间（具有加法和数乘运算的集合）和拓扑学的元素，确保这些运算在给定的拓扑下是连续的。这一理论起源于对函数空间的研究，其中函数序列的多种收敛性（如逐点收敛、一致收敛、弱收敛）是分析的核心。巴拿赫空间是拓扑线性空间的特殊类型，是那些完备的、能够满足Cauchy序列收敛的线性空间。常见的巴拿赫空间例子包括连续函数空间和可微函数空间。在巴拿赫空间中，对偶空间是研究的重要领域，它是原空间中所有连续线性泛函构成的空间。对偶空间的对偶空间可能不与原空间同构，但是可以通过特定映射从巴拿赫空间到其对偶空间建立关系。微分的概念在巴拿赫空间中得以扩展，微分算子可以作用于所有函数，且函数在某点的微分表现为一个连续线性映射。希尔伯特空间是内积空间的完备化，它提供了一个框架来研究量子力学和波动方程等问题。希尔伯特空间的特性允许我们进行傅里叶分析和泛函分析的许多其他技术。对于有限维希尔伯特空间，线性算子与线性代数中的线性变换相对应，而对于无穷维空间，研究通常集中在可数维度的算子上。希尔伯特空间的一个未解问题是，是否存在一个希尔伯特空间上的算子没有真不变子空间。局部凸空间（LCS）是另一种重要的泛函分析工具，它们是由半范数生成的拓扑线性空间。这类空间特别适用于描述测试函数和分布的空间，这对于解决偏微分方程（PDEs）非常有用。弱收敛和弱拓扑是研究LCS时不可忽视的概念，它们在处理非紧集和非线性问题时尤其重要。这些讲义涵盖了泛函分析的基本理论，并通过具体的例子和应用展示了其在数学和物理领域的广泛影响。通过深入学习，学生可以掌握泛函分析的基本工具，为理解和解决实际问题打下坚实基础。

LECTURE 3

Applications of Hahn-Banach

Reading: §4.1, §4.2 of Lax.

To get an idea what one can do with the Hahn-Banach theorem let’s consider a concrete

application on the linear space X = B(S) of all real valued bounded functions on a set S.

B(S) has a natural partial order, namely x ≤ y if x(s) ≤ y(s) for all s ∈ S. If 0 ≤ x then x

is nonnegative. On B(S) a positive linear functional ` satisﬁes `(y) ≥ 0 for all y ≥ 0.

Theorem 3.1. Let Y be a linear subspace of B(S) that contains y

≥ 1, so y

(s) ≥ 1

for all s ∈ S. If ` is a positive linear functional on Y then ` can be extended to all of B as

a positive linear functional.

This theorem can be formulated in an abstract context as follows. The nonnegative

functions form a cone, where

Definition 3.1. A subset P ⊂ X of a linear space over R is a cone if tx + sy ∈ P

whenever x, y ∈ P and t, s ≥ 0. A linear functional on X is P -nonnegative if `(x) ≥ 0 for

all x ∈ P .

Theorem 3.2. Let P ⊂ X be a cone with an interior point x

. If Y is a subspace

containing x

on which is deﬁned a P ∩Y -positive linear functional `, then ` has an extension

to X which is P -positive.

Proof. Deﬁne a dominating function p as follows

p(x) = inf{`(y) : y −x ∈ P, y ∈ Y }.

Note that y

− tx ∈ P for some t > 0 (since y

is interior to P ), so

− x ∈ P . This

shows that p(x) is well deﬁned. It is clear that p is positive homegeneous. To see that it is

sub-additive, let x

, x

∈ X and let y

, y

be so that y

−x

∈ P . Then y

−(x

) ∈ P ,

p(x

+ x

) ≤ `(y

) + `(y

Minimizing over y

1,2

gives sub-additivity.

Since `(x) = `(x −y) + `(y) ≤ `(y) if x ∈ Y and y −x ∈ P we conclude that `(x) ≤ p(x)

for all x ∈ Y . By Hahn-Banach we may extend ` to all of X so that `(x) ≤ p(x) for all x.

Now let x ∈ P . Then p(−x) ≤ 0 (why?), so −`(x) = `(−x) ≤ 0 which shows that ` is

P -positive. 

The theorem on B(S) follows from the Theorem 3.2 once we observe that the condition

≥ 1 implies that y

is an interior point of the cone of positive functions. The linear

functional that one constructs in this way is monotone:

x ≤ y =⇒ `(x) ≤ `(y),

which is in fact equivalent to positivity. This can allow us to do a little bit of analysis, even

though we haven’t introduced notions of topology or convergence.

3-1

3-2 3. APPLICATIONS OF HAHN-BANACH

To see how this could work, let’s apply the result to the Riemann integral on C[0, 1].

We conclude the existence of a positive linear functional ` : B[0, 1] → R that gives `(f) =

f(x)dx. This gives an “integral” of arbitrary bounded functions, without a measurability

condition. Since the integral is a linear functional, it is ﬁnitely additive. Of course it is

not countably additive since we know from real analysis that such a thing doesn’t exist.

Furthermore, the integral is not uniquely deﬁned.

Nonetheless, the extension is also not arbitrary, and the constraint of positivity actually

pins down `(f) for many functions. For example, consider χ

(a,b)

(the characteristic function

of an open interval). By positivity we know that

f ≤ χ

(a,b)

≤ g, f, g ∈ C[0, 1] =⇒

f(x)dx ≤ `(χ

(a,b)

) ≤

g(x)dx.

Taking the sup over f and inf over g, using properties only of the Riemann integral, we see

that `(χ

(a,b)

) = b − a (hardly a surprising result). Likewise, we can see that `(χ

) = |U| for

any open set, and by ﬁnite additivity `(χ

) = 1 − `(F

) = 1 − |F

| = |F | for any closed set

(| · | is Lebesgue measure). Finally if S ⊂ [0, 1] and

sup {`(F ) : F closed and F ⊂ S} = inf {`(U) : U open and U ⊃ S}

then `(χ

) must be equal to these two numbers. We see that `(χ

) = |S| for any Lebesgue

measurable set. We have just painlessly constructed Lebesgue measure from the Riemann

integral without using any measure theory!

The rigidity of the extension is a bit surprising if we compare with what happens in

ﬁnite dimensions. For instance, consider the linear functional `(x, 0) = x deﬁned on Y =

{(x, 0)} ⊂ R

. Let P be the cone {(x, y) : |y| ≤ αx}. So ` is P ∩ Y -positive. To extend `

to all of R

we need to deﬁne `(0, 1). To keep the extension positive we need only require

y`(0, 1) + 1 ≥ 0

if |y| ≤ α. Thus we must have |`(0, 1)| ≤ 1/α, and any choice in this interval will work.

Only when α = ∞ and the cone degenerates to a half space does the condition pin `(0, 1)

down. So some interesting things happen in ∞ dimensions.

A second example application assigning a limiting value to sequences

a = (a

, a

, . . .).

Let B denote the space of all bounded R-valued sequences and let L denote the subspace of

sequences with a limit. We quickly conclude that there is an positive linear extension of the

positive linear functional lim to all of B. We would like to conclude a little more, however.

After all, for convergent sequences,

lim a

n+k

= lim a

for any k.

To formalize this property we deﬁne a linear map on B by

T (a

, a

, . . .) = (a

, a

, . . .).

(T is the “backwards shift” operator or “left translation.”) Here,

Definition 3.2. Let X

, X

be linear spaces over a ﬁeld F . A linear map T : X

→ X

is a function such that

T (x + ay) = T (x) + aT (y) ∀x, y ∈ X

and a ∈ F.

3. APPLICATIONS OF HAHN-BANACH 3-3

Thus lim T a = lim a for a ∈ L. We would like the extension to have this property. Can

this be done?

The answer is given by the following a simpliﬁed version of the Theorem of Agnew and

Morse given in Chapter 3 of Lax:

Theorem 3.3. Let X be a linear space over R and let T : X → X be a linear map. Let

p : X → R be a positive homogeneous, sub-additive function invariant under T :

p(T x) = p(x) ∀x ∈ X.

Let ` be a linear functional deﬁned on a subspace Y ⊂ X invariant under T ,

T y ∈ Y ∀y ∈ Y.

If ` is invariant under T — `(T y) = `(y) — and dominated by p then ` may be extended to

all of X so that the extension is also invariant under T and dominated by p.

Remark. Applying this to the space B of bounded sequences and letting

p(a) = lim sup a,

we ﬁnd that there is an extension of the linear functional lim from L to all of B such that

for all a = (a

, a

, . . .)

(1) lim

n→∞

n+k

= lim

n→∞

for all k.

(2) lim inf a ≤ lim a ≤ lim sup a . (The ﬁrst inequality follows from the second applied

to −a.)

Proof. Let us deﬁne

q(x) = inf p(Ax),

where the inﬁmum is over all convex combinations of powers of T ,

A =

j=0

(3.1)

where T

= 1, the identity operator on X, 1x = x, and a

is any ﬁnite, non-negative sequence

with

= 1. Clearly q(x) is positive homogeneous and q(x) ≤ p(x). Also, since ` and Y

are invariant under any A of the form (3.1) we have

`(y) ≤ q(y) ∀y ∈ Y.

To apply Hahn-Banach with q as the dominating function we need to show it is sub-

additivie. To this end, let x, y ∈ X and let A, B be of the form (3.1) so that

p(Ax) ≤ q(x) +  and p(By) ≤ q(y) + .

Then A ◦ B = B ◦ A is of the form (3.1) and

q(x + y) ≤ p(AB(x + y)) ≤ p(BAx) + p(ABy) ≤ p(Ax) + p(By) ≤ q(x) + q(y) + 2.

Thus q is sub-additive.

So now we know that an extension exists to all of X with `(x) ≤ q(x) for all x. However,

q(x − T x) ≤ p

j=0

(1 − T )x

p(x − T

n+1

x) ≤

(p(x) + p(−x)) → 0.

So q(x − T x) = 0 for all x and thus `(x) − `(T x) = 0 for all x. 

剩余168页未读，继续阅读

qq_40620959

粉丝: 0
资源: 1

泛函分析基础：Hahn-Banach定理与巴拿赫空间

泛函分析知识总结汇总.doc

泛函分析

一本很好的泛函分析教程

泛函分析1

通信与网络中的瑞典Schenker AB快递公司采用FKI的高速分捡系统

全球汽车供应链管理解决方案市场增长潜力与影响因素分析.pdf

schenker怎么样

泛函分析答案

泛函分析论文

【机器人】将ChatGPT飞书机器人钉钉机器人企业微信机器人公众号部署到vercel及docker_pgj.zip

最新资源