泛函分析基础：从Banach到Hilbert空间

Functional

Analysis

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"这是一本由Gerald Teschl编写的英文原版《Functional Analysis》教材，涵盖了泛函分析的基础知识，包括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空间（假设读者对勒贝格积分没有预先知识）。书中详细讨论了线性偏微分方程、度量空间、拓扑空间、连续函数的巴拿赫空间、希尔伯特空间的几何性质、完备性、有界算子、正交基、投影定理、瑞利引理、紧算子、谱定理、斯托克斯-韦尔斯特拉斯定理、弗雷德霍姆理论、勒贝格测度、可测函数、积分、Lp空间、弱收敛、Lp空间的对偶、巴拿赫代数、算子C*代数、谱定理和斯通-魏尔斯特拉斯定理等主题。该书适合对泛函分析感兴趣的读者，特别是那些希望深入了解数学分析的学者和学生。" 《Functional Analysis》是学习和理解泛函分析的重要参考资料。书中首先介绍了线性偏微分方程，这是许多物理和工程问题的核心工具。接着，通过度量空间和拓扑空间的概念，引入了巴拿赫空间，特别是连续函数构成的巴拿赫空间。希尔伯特空间的几何特性被详细阐述，强调了其在量子力学和其他物理理论中的重要性。完备性和有界算子的概念帮助读者理解这些空间的结构。希尔伯特空间章节深入探讨了正交基、投影定理和瑞利引理，这些都是理解无限维空间中算子理论的关键。紧算子章节涵盖了谱定理，对于理解和解决特定类型的偏微分方程，如斯托克斯-黎欧维尔算子的问题非常有用。此外，书中还讨论了弗雷德霍姆理论，这对于理解和处理边值问题至关重要。在勒贝格积分部分，书中不仅介绍了Borel测度和可测函数，还详细讲解了勒贝格积分的构造和性质，包括积测度。Lp空间的章节则阐述了几乎处处有定义的函数、不等式理论（如Jensen不等式、Holder不等式和Minkowski不等式）以及Lp空间的完备性，这些都是泛函分析中的基本工具。关于巴拿赫空间的主要定理，如巴拿赫定理和汉恩-巴拿赫定理，以及弱收敛的概念，提供了深入的理解，这对于研究函数空间的结构非常重要。Lp空间对偶性的讨论，特别是L1空间的对偶和Riesz表示定理，揭示了积分与测度之间的深刻联系。最后，书中介绍了巴拿赫代数和算子C*代数，以及斯通-魏尔斯特拉斯定理，这为理解算子理论和抽象代数在分析中的应用奠定了基础。整体而言，《Functional Analysis》是一本全面而深入的教材，适合希望掌握泛函分析核心概念的读者。

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1.1. Warm up: Metric and topological spaces 9

the domain of f. A function f is called injective if for each y ∈ Y there

is at most one x ∈ X with f (x) = y (i.e., f

−1

({y}) contains at most one

point) and surjective or onto if Ran(f) = Y . A function f which is both

injective and surjective is called bijective.

A function f between metric spaces X and Y is called continuous at a

point x ∈ X if for every ε > 0 we can ﬁnd a δ > 0 such that

(f(x), f(y)) ≤ ε if d

(x, y) < δ. (1.9)

If f is continuous at every point, it is called continuous.

Lemma 1.5. Let X be a metric space. The following are equivalent:

(i) f is continuous at x (i.e, (1.9) holds).

(ii) f(x

) → f(x) whenever x

→ x.

(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) such that B

(x) 6⊆ f

−1

(V ) for every 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 the 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.

The support of a function f : X → C

is the closure of all points x for

which f(x) does not vanish; that is,

supp(f) = {x ∈ X|f (x) 6= 0}. (1.10)

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

d((x

, y

), (x

, y

)) = d

, x

) + d

, y

) (1.11)

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. More-

over, if X and Y are complete, so is X × Y .

In particular, by the inverse triangle inequality (1.1),

|d(x

, y

) − d(x, y)| ≤ d(x

, x) + d(y

, y), (1.12)

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

10 1. A ﬁrst look at Banach and Hilbert spaces

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

unless we modify (1.11) as follows:

d((x

, y

), (x

, y

)) =

, x

)

+ d

, y

)

. (1.13)

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 (1.11).

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

} such that Y ⊆

A cover is called open if all U

are open. Any subset of {U

} which still

covers Y is called a subcover.

Lemma 1.6 (Lindel¨of). If X is second countable, then every open cover

has a countable subcover.

Proof. Let {U

} be an open cover for Y and let B be a countable base.

Since every U

can be written as a union of elements from B, the set of all

B ∈ B which satisfy B ⊆ U

for some α form a countable open cover for Y .

Moreover, for every B

in this set we can ﬁnd an α

such that B

⊆ U

By construction {U

} is a countable subcover. 

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

subcover.

Lemma 1.7. 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 1.8. 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.

1.1. Warm up: Metric and topological spaces 11

(iv) The product of ﬁnitely many compact sets is compact.

(v) A compact set is also sequentially compact.

Proof. (i) 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 the 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 that the 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 that the U (x

) cover X. By

construction, the 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 elements of K.

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

In a metric space compact and sequentially compact are equivalent.

Lemma 1.9. Let X be a metric space. Then a subset is compact if and only

if it is sequentially compact.

Proof. By item (v) of the previous lemma it suﬃces to show that X is

compact if it is sequentially compact.

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\

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 cover

} 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

12 1. A ﬁrst look at Banach and Hilbert spaces

ε =

and pick a corresponding 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 1.10 (Heine–Borel). In R

(or C

) a set is compact if and only

if it is bounded and closed.

Proof. By Lemma 1.8 (ii) and (iii) it suﬃces to show that a closed interval

in I ⊆ R is compact. Moreover, by Lemma 1.9 it suﬃces to show that

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

be our

sequence and divide I = [a,

a+b

] ∪ [

a+b

, 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 compact 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). (1.14)

Note that x is a limit point of Y if and only if dist(x, Y ) = 0.

Lemma 1.11. Let X be a metric space. Then

|dist(x, Y ) − dist(z, Y )| ≤ d(x, z). (1.15)

In particular, x 7→ dist(x, Y ) is continuous.

Proof. Taking the inﬁmum in the triangle inequality d(x, y) ≤ d(x, z) +

d(z, y) shows dist(x, Y ) ≤ d(x, z)+dist(z, Y ). Hence dist(x, Y )−dist(z, Y ) ≤

d(x, z). Interchanging x and z shows dist(z, Y ) − dist(x, Y ) ≤ d(x, z). 

Lemma 1.12 (Urysohn). Suppose C

and C

are disjoint closed subsets of

a metric space X. Then there is a continuous function f : X → [0, 1] such

that f is zero on C

and one on C

If X is locally compact and C

is compact, one can choose f with compact

support.

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