维也纳大学2004-2009年泛函分析教程简介

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《泛函分析基础(外文版的泛函分析基础教程)》是一本由Gerald Teschl撰写的专业教材，旨在为他在维也纳大学开设的泛函分析课程提供教学资料。这本书在2004年和2009年的冬季授课中被使用，并且作者会定期更新以修正错误，确保内容的最新性，读者可以从他的个人网站http://www.mat.univie.ac.at/~gerald/ftp/book-fa/ 获取最新版本。本书主要关注的是基础的泛函分析理论，包括但不限于Banach空间和希尔伯特空间的概念与应用。作者首先引导读者理解线性偏微分方程（在第一章的 §0.1 中）作为引入泛函分析的一个动机，以此展示该领域的实际应用价值。第1章深入探讨了Banach空间和Hilbert空间的基础。首先，通过介绍度量空间和拓扑空间的基本概念（§1.1），帮助读者建立数学背景。然后，作者详细介绍了连续函数的空间C(X)，并强调了其在Banach空间中的地位（§1.2）。接下来，作者通过讨论希尔伯特空间的几何特性（§1.3），如正交基和内积，揭示了这些空间的独特性质。完整性（§1.4）是Hilbert空间的一个关键属性，它确保了空间中序列收敛的特性。章节的后半部分聚焦于Banach空间的运算，如算子（§1.5）和空间的和与商，这些都是泛函分析中的核心概念。这些内容为后续讨论提供了坚实的基础，尤其是在处理函数空间和算子理论时。第二章专攻希尔伯特空间，深入研究了正交基的构造和应用（§2.1），这些对于理解和解析希尔伯特空间上的线性问题至关重要。此外，本章还可能涉及其他高级主题，如希尔伯特空间的对偶性和相关的谱理论。《泛函分析基础》以清晰易懂的方式介绍了基础的泛函分析理论，适合希望入门或进一步学习这个领域的学生和研究人员。它不仅涵盖了理论层面的内容，而且与实际问题紧密结合，使读者能够将所学应用于解决实际问题，尤其是那些涉及线性微分方程和函数空间的问题。

1.1. Warm up: Metric and topological spaces 9

Proof. Let A = {x

}

n∈N

be a dense set in X. The only problem is that

A ∩Y might contain no elements at all. However, some elements of A must

be at least arbitrarily close: Let J ⊆ N

be the set of all pairs (n, m) for

which B

1/m

) ∩ Y 6= ∅ and choose some y

n,m

∈ B

1/m

) ∩ Y for all

(n, m) ∈ J. Then B = {y

n,m

}

(n,m)∈J

⊆ Y is countable. To see that B is

dense, choose y ∈ Y . Then there is some sequence x

with d(x

, y) < 1/k.

Hence (n

, k) ∈ J and d(y

, y) ≤ d(y

, x

) + d(x

, y) ≤ 2/k → 0. 

Next we come to functions f : X → Y , x 7→ f(x). We use the usual

conventions f(U) = {f(x)|x ∈ U } for U ⊆ X and f

−1

(V ) = {x|f(x) ∈ V }

for V ⊆ Y . The set Ran(f) = f(X) is called the range of f and X is called

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 set is again open (equivalently, the inverse image of

every closed set is closed). If the image of every open set is open, then f

is called open. A bijection f is called a homeomorphism if both f and

its inverse f

−1

are continuous. Note that if f is a bijection, then f

−1

continuous if and only if f is open.

In a topological space, (iii) is used as the deﬁnition for continuity. How-

ever, 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.

10 1. A ﬁrst look at Banach and Hilbert spaces

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.

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.

1.1. Warm up: Metric and topological spaces 11

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. 

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, every compact set is closed.

(iv) The product of ﬁnitely many compact sets is 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 (in the induced

topology). Then there are open sets

with O

∩Y and {

}∪{X\Y }

is an open cover for X which has a ﬁnite subcover. This subcover induces a

ﬁnite subcover for Y .

(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 . 

As a consequence we obtain a simple criterion when a continuous func-

tion is a homeomorphism.

Corollary 1.9. Let X and Y be topological spaces with X compact and Y

Hausdorﬀ. Then every continuous bijection f : X → Y is a homeomor-

phism.

12 1. A ﬁrst look at Banach and Hilbert spaces

Proof. It suﬃces to show that f maps closed sets to closed sets. By (ii)

every closed set is compact, by (i) its image is also compact, and by (iii)

also closed. 

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

has a convergent subsequence. In a metric space compact and sequentially

compact are equivalent.

Lemma 1.10. Let X be a metric space. Then a subset is compact if and

only if it is sequentially compact.

Proof. Suppose X is compact and let x

be a sequence which has no conver-

gent subsequence. Then K = {x

} has no limit points and is hence compact

by Lemma 1.8 (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.

Conversely, suppose X 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

∈ 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

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. 

In a metric space, a set is called bounded if it is contained inside some

ball. Note that compact sets are always bounded (show this!). In R

(or

) the converse also holds.

Theorem 1.11 (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 1.10 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

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