泛函分析基础：希尔伯特空间与巴拿赫空间导论

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"Gerald Teschl的泛函分析教材，英文版，包含了基本的希尔伯特空间和巴拿赫空间理论，以及勒贝格空间及其对偶，无需预先了解勒贝格积分。" 《Functional Analysis》是Gerald Teschl撰写的一本关于泛函分析的教材，适用于初学者。书中主要探讨了线性偏微分方程相关的数学理论，并深入介绍了巴拿赫空间和希尔伯特空间的基础知识。 1. 泛函分析基础泛函分析是一门数学分支，主要研究无限维空间中的算子理论和函数空间。该书首先通过引入线性偏微分方程（PDE）来引入这个主题，展示泛函分析在解决实际问题中的应用。 2. 巴拿赫空间与希尔伯特空间 - 巴拿赫空间是完备的赋范向量空间，即任何柯西序列在其内都收敛。第1章深入介绍了巴拿赫空间的基本概念，如度量空间、拓扑空间和连续函数空间，为后续章节打下基础。 - 希尔伯特空间是具有内积的巴拿赫空间，它拥有丰富的几何结构，如正交性和等距性质。第1.3节探讨了希尔伯特空间的几何特性，而第1.4节和1.5节则涉及完备性和有界算子的概念。 3. 勒贝格空间及其对偶勒贝格空间是实值函数的集合，其中的函数满足一定的积分可积性条件。尽管书中假设读者对勒贝格积分没有先验知识，但依然能够讲解这部分内容，使得初学者也能理解。 4. 希尔伯特空间深入第2章专门讨论希尔伯特空间，包括： - 第2.1节介绍了正交基的概念，这是希尔伯特空间中非常重要的一个工具，它在解析问题和信号处理等领域有广泛应用。 - 第2.2节的投影定理和黎斯引理是希尔伯特空间理论的核心部分，它们揭示了希尔伯特空间中线性映射的结构性质，对于理解正交系统的性质极其关键。这本书适合对泛函分析感兴趣的数学学生和研究人员，通过清晰的解释和逐步引导，帮助他们建立起对无限维空间和算子理论的深入理解。作者Gerald Teschl是维也纳大学的数学教授，他的著作以其严谨性和易读性著称，是学习泛函分析的理想参考资料。

1.1. Warm up: Metric and topological spaces 9

(iii) For every neighborhood V of f (x), f

−1

(V ) is a neighborhood of x.

Proof. (i) ⇒ (ii) is obvious. (ii) ⇒ (iii): If (iii) does not hold there is

a neighborhood V of f(x) s uch that B

(x) 6⊆ f

−1

(V ) for e very δ. Hence

we can choose a sequence x

∈ B

1/n

(x) such that f (x

) 6∈ f

−1

(V ). Thus

→ x but f(x

) 6→ f(x). (iii) ⇒ (i): Choose V = B

(f(x)) and observe

that by (iii) B

(x) ⊆ f

−1

(V ) for some δ. 

The last item implies that f is continuous if and only if the inverse image

of every open (closed) set is again open (closed).

Note: In a topological space, (iii) is used as deﬁnition for continuity.

However, in general (ii) and (iii) will no longer be equivalent unless one uses

generalized sequences, so called nets, where the index set N is replaced by

arbitrary directed sets.

If X and X are metric spaces then X × Y together with

d((x

, y

), (x

, y

)) = d

, x

) + d

, y

) (1.9)

is a metric space. A sequence (x

, y

) converges to (x, y) if and only if

→ x and y

→ y. In particular, the projections onto the ﬁrst (x, y) 7→ x

respectively onto the second (x, y) 7→ y coordinate are continuous.

In particular, by

|d(x

, y

) − d(x, y)| ≤ d(x

, x) + d(y

, y) (1.10)

we see that d : X × X → R is continuous.

Example. If we consider R ×R we do not get the Euclidean distance of R

unless we modify (

1.9) as follows:

d((x

, y

), (x

, y

)) =

, x

)

+ d

, y

)

. (1.11)

As noted in our previous example, the topology (and thus also conver-

gence/continuity) is independent of this choice. 

If X and Y are just topological spaces, the product topology is deﬁned

by calling O ⊆ X × Y open if for every point (x, y) ∈ O there are open

neighborhoods U of x and V of y such that U × V ⊆ O. In the case of

metric spaces this clearly agrees with the topology deﬁned via the product

metric (

1.9).

A cover of a set Y ⊆ X is a family of sets {U

} such that Y ⊆

. A

cover is call open if all U

are open. A subset of {U

} is called a subcover.

A subset K ⊂ X is called compact if every open cover has a ﬁnite

subcover.

10 1. A ﬁrst look at Banach and Hilbert spaces

Lemma 1.6. A topological space is compact if and only if it has the ﬁnite

intersection property: The intersection of a family of closed sets is empty

if and only if the intersection of some ﬁnite subfamily is empty.

Proof. By taking complements, to every family of open sets there is a cor-

responding family of closed sets and vice versa. Moreover, the open sets

are a cover if and only if the corresponding closed sets have empty intersec-

tion. 

A subset K ⊂ X is called sequentially compact if every sequence has

a convergent subsequence.

Lemma 1.7. Let X be a topological space.

(i) The continuous image of a compact set is compact.

(ii) Every closed subset of a compact set is compact.

(iii) If X is Hausdorﬀ, any compact set is closed.

(iv) The product of compact sets is compact.

(v) A compact set is also sequentially compact.

Proof. (i) Just observe that if {O

} is an open cover for f(Y ), then {f

−1

)}

is one for Y .

(ii) Let {O

} be an open cover for the closed subset Y . Then {O

} ∪

{X\Y } is an open cover for X.

(iii) Let Y ⊆ X be compact. We show that X\Y is open. Fix x ∈ X\Y

(if Y = X there is nothing to do). By the deﬁnition of Hausdorﬀ, for

every y ∈ Y there are disjoint neighborhoods V (y) of y and U

(x) of x. By

compactness of Y , there are y

, . . . y

such that V (y

) cover Y . But then

U(x) =

j=1

(x) is a neighborhood of x which does not intersect Y .

(iv) Let {O

} be an open cover for X × Y . For every (x, y) ∈ X × Y

there is some α(x, y) such that (x, y) ∈ O

α(x,y)

. By deﬁnition of the product

topology there is some open rectangle U(x, y) × V (x, y) ⊆ O

α(x,y)

. Hence

for ﬁxed x, {V (x, y)}

y∈Y

is an open cover of Y . Hence there are ﬁnitely

many points y

(x) such V (x, y

(x)) cover Y . Set U(x) =

U(x, y

(x)).

Since ﬁnite intersections of open sets are open, {U(x)}

x∈X

is an open cover

and there are ﬁnitely many points x

such U(x

) cover X. By construction,

U(x

) × V (x

, y

)) ⊆ O

α(x

))

cover X × Y .

(v) Let x

be a sequence which has no convergent subsequence. Then

K = {x

} has no limit points and is hence compact by (ii). For every n

there is a ball B

) which contains only ﬁnitely many e leme nts of K.

However, ﬁnitely many suﬃce to cover K, a contradiction. 

In a metric space compact and sequentially compact are equivalent.

1.1. Warm up: Metric and topological spaces 11

Lemma 1.8. Let X be a metric space. Then a subset is compact if and only

if it is sequentially compact.

Proof. First of all note that every cover of open balls with ﬁxed radius

ε > 0 has a ﬁnite subcover. Since if this were false we could construct a

sequence x

∈ X\

n−1

m=1

) such that d(x

, x

) > ε for m < n.

In particular, we are done if we can show that for every open c over

} there is some ε > 0 such that for every x we have B

(x) ⊆ O

for

some α = α(x). Indeed, choosing {x

}

k=1

such that B

) is a cover, we

have that O

α(x

)

is a cover as well.

So it remains to show that there is such an ε. If there were none, for

every ε > 0 there must be an x such that B

(x) 6⊆ O

for every α. Choose

ε =

and pick a c orresponding x

. Since X is sequentially compact, it is no

restriction to assume x

converges (after maybe passing to a subsequence).

Let x = lim x

, then x lies in some O

and hence B

(x) ⊆ O

. But choosing

n so large that

and d(x

, x) <

we have B

1/n

) ⊆ B

(x) ⊆ O

contradicting our assumption. 

Please also recall the Heine-Borel theorem:

Theorem 1.9 (Heine-Borel). In R

(or C

) a set is compact if a nd only if

it is bounded and closed.

Proof. By Lemma 1.7 (ii) and (iii) it suﬃces to show that a closed interval

in I ⊆ R is compact. Moreover, by Lemm a 1.8 it suﬃces to show that

every sequence in I = [a, b] has a c onvergent subsequence. Let x

be our

sequence and divide I = [a,

a+b

] ∪ [

a+b

]. Then at least one of these two

intervals, call it I

, contains inﬁnitely many elements of our sequence. Let

= x

be the ﬁrst one. Subdivide I

and pick y

= x

, with n

> n

before. Proceeding like this we obtain a Cauchy sequence y

(note that by

construction I

n+1

⊆ I

and hence |y

− y

| ≤

b−a

for m ≥ n). 

A topological space is called locally com pact if every point has a com-

pact neighborhood.

Example. R

is locally compact. 

The distance between a point x ∈ X and a subset Y ⊆ X is

dist(x, Y ) = inf

y∈Y

d(x, y). (1.12)

Note that x ∈

Y if and only if dist(x, Y ) = 0.

Lemma 1.10. Let X be a metric space, then

|dist(x, Y ) − dist(z, Y )| ≤ dist(x, z). (1.13)

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