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2. Banach spaces
Definition. Let K be one of the fields R or C. A Banach space over K is a
normed K-vector space (X, k . k), which is complete with respect to the metric
d(x, y) = kx − yk, x, y ∈ X.
Remark 2.1. Completeness for a normed vector space is a purely topological
property. This means that, if k . k is a norm on X, such that (X, k . k) is a Banch
space, then so is (X, k . k0 ), where k . k0 is any norm equivalent to k . k. This is due
to the fact that, if k . k0 is equivalent to k . k, then one has
Ckxk ≤ kxk0 ≤ Dkxk, ∀ x ∈ X,
for some constants C, D > 0. This clearly gives the fact that a sequence (xn )∞
n=1 ⊂
X is Cauchy with respect to k . k, if and only if it is Cauchy with respect to k . k0 .
Example 2.1. The field K, equipped with the absolute value norm, is a Banach
space. More generally, the vector space Kn , equipped with the norm
k(λ1 , . . . , λn )k∞ = max{|λ1 |, . . . , |λn |},
is a Banach space.
Remark 2.2. Since any two norms on a finite dimensional space are equivalent,
by Proposition 1.4, it follows that any finite dimensional normed vector space is a
Banach space.
Below is an interesting application of this fact.
Proposition 2.1. Let X be a normed vector space, and let Y ⊂ X be a finite
dimensional linear subspace.
(i) Y is closed.
(ii) More generally, if Z ⊂ X is a closed subspace, then the linear subspace
Y + Z is also closed.
Proof. (i). Start with some sequence (yn )∞
n=1 ⊂ Y, which is convergent to
some point x ∈ X, and let us show that x ∈ Y. By the above Remark, when
equipped with the norm coming from X, the normed vector space Y is complete.
Since obviously (yn )∞
n=1 is Cauchy, it will converge in Y to some vector y ∈ Y. Of
course, this forces x = y, and we are done.
(ii). Let us consider the quotient space X/Z, equipped with the quotient norm
k . kX/Z , and the quotient map P : X → X/Z. By Proposition 1.5 we know that
P is continuous. Since the linear subspace V = P (Y) ⊂ X/Z is finite dimensional,
by part (i) it follows that V is closed in X/Z. By continuity, its preimage P −1 (V)
is closed in X. Now we are done since we obviously have the equality P −1 (V) =
Y + Z.
Remark 2.3. Using the facts from the general theory of metric spaces, we
know that for a normed vector space (X, k . k), the following are equivalent:
(i) X is a Banach space;
P∞
(ii) given any sequence (xn )n≥1 ⊂ X with Pn=1 kxn k < ∞, the sequence
n
(yn )n≥1 of partial sums, defined by yn = k=1 xk , is convergent;
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CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS
(iii) every Cauchy sequence in X has a convergent subsequence.
This is pretty obvious, since the sequence of partial sums has the property that
d(yn+1 , yn ) = kyn+1 − yn k = kxn+1 k, ∀ n ≥ 1.
There are several techniques for constructing new Banach space out of old ones.
The result below is one example.
Proposition 2.2. Let X be a Banach space, let Y be a normed vector space,
and let T : X → Y be a surjective linear continuous map. Assume there exists some
constant C > 0, such that
• for every y ∈ Y, there exists x ∈ X with T x = y, and kxk ≤ Ckyk.
Then Y is a Banach space.
Proof. We are going to
use the above characterization. Start with some
P∞
∞
sequence (yn )∞
⊂
Y,
with
n=1 P
n=1 kyn k < ∞. Define the sequence (wn )n=1 ⊂ Y of
n
partial sums wn = k=1 yn , and let us prove that (wn )∞
is
convergent
to some
n=1
element in Y. Use the hypothesis to find, for each n ≥ 1, an element
x
P∞ n ∈ X,
with
T
x
=
y
,
and
kx
k
≤
Cky
k.
In
particular,
we
clearly
have
n
n
n
n
n=1 kxn k ≤
P∞
Cky
k
<
∞.
Using
the
fact
that
X
is
a
Banach
space,
it
follows
that the
n
n=1
Pn
sequence (zn )∞
defined
by
z
=
x
is
convergent
to
some
z
∈
X.
Since
we
n
n=1
k=1 n
have T zn = wn , ∀ n ≥ 1, by the continuity of T we get limn→∞ wn = T z.
Corollary 2.1. Let X be a Banach space, and let Y be a closed linear subspace
of X. When equipped with the quotient norm, the quotient space X/Y is a Banach
space.
Proof. Use the notations from Section 1. Consider the quotient map
P : X 3 x 7−→ [x] ∈ X/Y.
We know that P is linear, continnuous, and surjective. Let us check that P satisfies
the hypothesis in Proposition 2.2, with C = 2. Start with some vector v ∈ X/Y,
and let us show that there exists x ∈ X with P x = v (i.e. x ∈ v), such that
kxk ≤ 2kvkX/Y . If v = 0, there is nothing to prove, because we can take x = 0. If
v 6= 0, then we use the definition of the quotient norm
kvkX/Y = inf kxk,
x∈v
combined with 2kvkX/Y > kvkX/Y .
Exercise 1. Let X and Y be normed vector spaces. Consider the product X×Y,
equipped with the natural vector space structure.
(i) Prove that k(x, y)k = kxk + kyk, (x, y) ∈ X × Y defines a norm on X × Y.
(ii) Prove that, when equipped with the above norm, X×Y is a Banach space,
if and only if both X and Y are Banach spaces.
Proposition 2.3. Let X be a normed vector space, and let Y be a Banach
space. Then L(X, Y) is a Banach space, when equipped with the operator norm.
Proof. Start with a Cauchy sequence (Tn )n≥1 ⊂ L(X, Y). This means that
for every ε > 0, there exists some Nε such that
(1)
kTm − Tn k < ε, ∀ m, n ≥ Nε .
§2. Banach spaces
81
Notice that, if one takes for example ε = 1, and we define
C = 1 + max{kT1 k, kT2 k, . . . , kTN1 k},
then we clearly have
kTn k ≤ C, ∀ n ≥ 1.
(2)
Notice that, using (1), we have
(3)
kTm x − Tn xk ≤ εkxk, ∀ m, n ≥ Nε , x ∈ X,
which proves that
• for every x ∈ X, the sequence (Tn x)n≥1 ⊂ Y is Cauchy.
Since Y is a Banach space, for each x ∈ X, the sequence (Tn )n≥1 will be convergent.
We define the map T : X → Y by
T x = lim Tn x, x ∈ X.
n→∞
Using (2) we immediately get
kT xk ≤ Ckxk, ∀ x ∈ X.
Since T is obviously linear, this prove that T is continuous. Finally, if we fix n ≥ Nε
and we take limm→∞ in (3), we get
kTn x − T xk ≤ εkxk, ∀ n ≥ Nε , x ∈ X,
which proves precisely that we have the inequality
kTn − T k ≤ ε, ∀ n ≥ Nε ,
hence (Tn )n≥1 is convergent to T in the norm topology.
Corollary 2.2. If X is a normed vector space, then its topological dual X∗ =
L(X, K) is a Banach space.
Proof. Immediate from the fact that K is a Banach space.
Example 2.2. Let K be either R or C, and let J be a non-empty set. For
every p ∈ [1, ∞], the space `pK (J), introduced in Section 1, is a Banach space. This
follows from the isometric linear isomorphisms `1 ' (c0 )∗ , and `p ' (`q )∗ , where q
is Hölder conjugate to p.
Proposition 2.4. Let X be a Banach space, and let Z ⊂ X be a linear subspace.
The following are equivalent:
(i) Z is a Banach space, ehen equipped with the norm from X;
(ii) Z is closed in X, in the norm topology.
Proof. This is a particular case of a general result from the theory of complete
metric spaces.
Example 2.3. Let J be a non-empty set, and let K be one of the fields R or
∞
C. Then cK
0 (J) is a Banach space, since it is a closed linear subspace in `K (J).
The following results give examples of Banach spaces coming from topology.
Notation. Let K be one of the fields R or C, and let Ω be a topological space.
We define
CbK (Ω) = {f : Ω → K : f bounded and continuous}.
In the case when K = C we use the notation Cb (Ω).
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CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS
Proposition 2.5. With the notations above, if we define
kf k = sup |f (p)|, ∀ f ∈ CbK (Ω),
p∈Ω
then
CbK (Ω)
is a Banach space.
Proof. It is obvious that CbK (Ω) is a linear subspace of `∞
K (Ω), and the norm
is precisely the one coming from `∞
(Ω).
Therefore,
it
suffices
to
prove that CbK (Ω)
K
is closed in `∞
(Ω).
K
Start with some sequence (fn )n≥1 ⊂ CbK (Ω), which convergens in norm to
some f ∈ `∞
K (Ω), and let us prove that f : Ω → K is continuous (the fact that f is
bounded is automatic).
Fix some point p0 ∈ Ω, and some ε > 0. We need to find some neighborhood
V of p0 , such that
|f (p) − f (p0 )| < ε, ∀ p ∈ V.
Start by choosing n such that kfn − f k < 3ε . Use the fact that fn is continuous, to
find a neighborhood V of p0 , such that
ε
|fn (p) − fn (p0 )| < , ∀ Ω ∈ V.
3
Suppose now Ω ∈ V . We have
|f (p) − f (p0 )| ≤ |fn (p) − f (p)| + |fn (p) − fn (p0 )| + |fn (p0 ) − f (p0 )| ≤
ε ε
|fn (p) − fn (p0 )| + 2 sup |fn (q) − f (q)| < 2 + = ε.
3 3
q∈Ω
Notation. If X is a compact Hausdorff space, then every continuous function
f : X → K is automatically bounded. In this case, the Banach space CbK (X) will
simply be denoted by C K (X), again with the convention that when K = C, the
superscript will be ommitted.
Example 2.4. Let K be either R or C, and let Ω be a locally compact space,
which is not compact. We define (see I.5) the space
C0K (Ω) = f : X → K : f continuous, with limit 0 at ∞ .
In other words, for a continuous function f : Ω → K, the condition f ∈ C0K (Ω) is
equivalent to
• for every ε > 0, there exists some compact set Kε ⊂ Ω, such that
(4)
|f (x)| < ε, ∀ x ∈ Ω r Kε .
As before, when K = C, the superscript will be ommitted from the notation.
Proposition 2.6. Let K be either R or C, and let Ω be a locally compact
space, which is not compact. Then C0K (Ω) is a closed linear subspace of CbK (Ω). In
particular, when equipped with the supremum norm, C0K (Ω) is a Banach space.
K
Proof. Let (fn )∞
n=1 ⊂ C0 (Ω) be a sequence, which converges in norm to some
K
f ∈ Cb (Ω), and let us show that f ∈ C0K (Ω). Fix some ε > 0, and let us indicate
how to construct a compact set Kε , with the property (4). Start off by choosing
n ≥ 1, such that kfn − f k < ε/2, and then choose the compact set Kε ⊂ Ω, such
that
(5)
|fn (x)| < ε, ∀ x ∈ Ω r Kε .
§2. Banach spaces
83
Notice now that, if x ∈ Ω r Kε , then
|f (x)| ≤ kf (x) − fn (x)| + |fn (x)| ≤ kfn − f k + |fn (x)|,
and then using the choice of n, combined with (5), the desired property (4) immediately follows.
The Banach space C0K (Ω) has several other equivalent descriptions. One of
them is based on the following notion.
Definitions. Let Ω be a locally compact space. If K is one of the fields R or
C, and f : Ω → K is a continuous function, we define the support of f by
supp f = {ω ∈ Ω : f (ω) 6= 0}.
We define the space
CcK (Ω) = f : Ω → K : f continuous, with compact support .
When K = C, this space will be denoted simply by Cc (Ω). Remark that, when
equipped with pointwise addition and multiplication, the space CcK (Ω) becomes a
K-algebra. One has obviously the inclusion CcK (Ω) ⊂ CbK (Ω).
Proposition 2.7. With the notations above, the Banach space C0K (Ω) is the
closure of CcK (Ω) in CbK (Ω).
Proof. Start with some function f ∈ C0K (Ω), and let us construct a sequence
K
(fn )∞
n=1 ⊂ Cc (Ω), which converges to f in the norm topology. For every integer
n ≥ 1 we choose some compact set Kn ⊂ Ω, such that
1
(6)
|f (x)| < , ∀ x ∈ Ω r Kn .
n
For each n ≥ 1, we also choose some open set Dn with Dn ⊃ Kn , and Dn compact,
and we choose (use Urysohn Lemma for locally compact spaces;
see I.7) some
continuous function hn : Ω → [0, 1], such that hn Kn = 1 and hn ΩrDn = 0. Put
fn = f hn , ∀ n ≥ 1. It is obvious that, since hn ΩrDn = 0, we have supp hn ⊂ Dn ,
and consequently we also have supp fn ⊂ Dn , so we have fn ∈ CcK (Ω), ∀ n ≥ 1. Let
us now estimate the norms kf − fn k, n ≥ 1. Fix for the moment n. On the one
hand, we have 0 ≤ 1 − hn (x) ≤ 1, ∀ x ∈ Ω, which combined with (6), yields
1
|f (x) − fn (x)| = |f (x)| · |1 − hn (x)| ≤ , ∀ x ∈ Ω r Kn .
n
On the other hand, we have f (x) = fn (x), ∀ x ∈ Kn , so the above estimate actually
gives
1
kf − fn k = sup |f (x) − fn (x)| ≤ , ∀ n ≥ 1,
n
x∈Ω
which in particular gives the fact that limn→∞ fn = f , in the norm topology.
At this point, based on the above result we shall adopt the following.
Convention. If Ω is a compact Hausdorff space, the notation C0K (Ω) designates the space C K (Ω). The reason for adopting this convention is the (trvial)
fact that, when Ω is compact, every continuous function f : Ω → K has compact
support. In this spirit, if we adopt the above characterization as the “working
definition” of C0K (Ω), then our convention is legitimate.
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CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS
Exercise 2*. Let X be an infinite dimensional Banach space, and let B be a
linear basis for X. Prove that B is uncountable.
Hint: If B is countable, say B = {bn : n ∈ N}, then
X=
∞
[
Fn ,
n=1
where Fn = Span(b1 , b2 , . . . , bn }. Since the Fn ’s are finite dimensional linear subspaces, they will
be closed. Use Baire’s Theorem to get a contradiction.
Comments. A third method of constructing Banach spaces is the completion.
If we start with a normed K-vector space X, when we regard X as a metric space,
its completion X̃ is constructed as follows. One defines
cs(X) = x = (xn )n≥1 : (xn )n≥1 Cauchy sequence in X .
Two Cauchy sequences x = (xn )n≥1 and x0 = (x0n )n≥1 are said to be equivalent, if
limn→∞ kxn − x0n k = 0. In this case one writes x ∼ x0 . The completion X̃ is then
defined as the space
X̃ = cs(X)/ ∼
of equivalence classes. For x ∈ cs(X), one denotes by x̃ its equivalence class in X̃.
Finally for an element x ∈ X one denotes by hxi ∈ X̃ the equivalence class of the
constant sequence x.
We know from general theory that X̃ is a complete metric space, with the
distance d˜ (correctly) defined by
˜ x̃0 ) = lim kxn − x0 k,
d(x̃,
n
n→∞
for any two Cauchy sequences x = (xn )n≥1 and x0 = (x0n )n≥1 .
It turns out that, in our situation, the space cs(X) carries a natural vector
space structure, defined by pointwise addition and scalar multiplication. Moreover,
the space X̃ is identified as a quotient vector space
X̃ = cs(X)/ns(X),
where
ns(X) = x = (xn )n≥1 : (xn )n≥1 sequence in X with lim xn = 0
n→∞
is the linear subspace of null sequences. It then follows that X̃ carries a natural
vector space structure. More explicitly, if we start with a scalar λ ∈ K, and with
two elements p, q ∈ X̃, which are represented as p = x̃ and q = ỹ, for two Cauchy
sequences x = (xn )n≥1 and y = (yn )n≥1 in X, then the sequence
w = (λxn + yn )n≥1
is Cauchy in X, and the element λp + q ∈ X̃ is then defined as λp + q = w̃.
Finally, there is a natural norm on X̃, (correctly) defined by
˜ h0i) = lim kxn k,
kx̃k = d(x̃,
n→∞
for all Cauchy sequences x = (xn )n≥1 . These considerations then prove that X̃ is
a Banach space, and the map
X 3 x 7−→ hxi ∈ X̃
§2. Banach spaces
85
is linear and isometric, in the sense that
khxik = kxk, ∀ x ∈ X.
In the context of normed vector spaces, the universality property of the completion is stated as follows:
Proposition 2.8. Let X be a normed vector space, let X̃ denote its completion,
and let Y be a Banach space. For every linear continuous map T : X → Y, there
exists a unique linear continuous map T̃ : X̃ → Y, such that
T̃ hxi = T x, ∀ x ∈ X.
Moreover the map
L(X, Y) 3 T 7−→ T̃ ∈ L(X̃, Y)
is an isometric linear isomorphism.
Proof. If T : X → Y is linear an continuous, then T is a Lipschitz map with
Lipschitz constant kT k, because
kT x − T x0 k ≤ kT k · kx − x0 k, ∀ x, x0 ∈ X.
We know, from the theory of metric spaces, that there exists a unique continuous
map T̃ : X̃ → Y, such that
T̃ hxi = T x, ∀ x ∈ X.
We also know that T̃ is Lipschitz, with Lipschitz constant kT k. The only thing we
need to prove is the fact that T̃ is linear. Start with two points p, q ∈ X̃, represented
as p = x̃ and q = z̃, for some Cauchy sequences x = (xn )n≥1 and z = (zn )n≥1 in
X. If λ ∈ K, then λp + q = w̃, where w = (λxn + zn )n≥1 . We then have
T̃ (λp + q) = lim T (λxn + z + n) = λ · lim T xn + lim T zn = λT̃ p + T̃ q.
n→∞
n→∞
n→∞
Let us prove now that kT̃ k = kT k. Since T̃ is Lipschitz, with Lipschitz constant
kT k, we will have kT̃ k ≤ kT k. To prove the other inequality, let us consider the
sets
B0 = {p ∈ X̃ : kpk ≤ 1},
B1 = {hxi : x ∈ X, kxk ≤ 1}.
By definition, we have
kT̃ k = sup kT̃ pk.
p∈B0
Since we clearly have B0 ⊃ B1 , we get
kT̃ k = sup kT̃ pk ≥ sup kT̃ hxik : x ∈ X kxk ≤ 1 =
p∈B1
= sup kT xk : x ∈ X kxk ≤ 1 = kT k.
The fact that the map L(X, Y) 3 T 7−→ T̃ ∈ L(X̃, Y) is linear is obvious.
To prove the surjectivity, start with some S ∈ L(X̃, Y). Consider the map
ι : X 3 x 7−→ hxi ∈ X̃.
Since ι is linear and isometric, in particular it is continuous, so the composition
T = S ◦ ι is linear and continuous. Notice that
Shxi = S ι(x) = (S ◦ ι)x = T x, ∀ x ∈ X,
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CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS
so by uniqueness we have S = T̃ .
Corollary 2.3. Let X be a normed space, let Y be a Banach space, and let
T : X → Y be an isometric linear map.
(i) Let T̃ : X̃ → Y be the linear continuous map defined in the previous result.
Then T̃ is linear, isometric, and T̃ (X̃) = T (X).
(ii) X is complete, if and only of T (X) is closed in Y.
Proof. (i). The fact that T̃ is isometric, and has the range equal to T (X) is
true in general (i.e. for X metric space, and Y complete metric space). The linearity
follows from the previous result.
(ii). This is obvious.
Example 2.5. Let X be a normed vector space. For every x ∈ X define the
map x : X∗ → K by
x (φ) = φ(x), ∀ φ ∈ X∗ .
Then x is a linear and continuous. This is an immediate consequence of the
inequality
|x (φ)| = |φ(x)| ≤ kxk · kφk, ∀ φ ∈ X∗ .
Notice that this also proves
kx k ≤ kxk, ∀ x ∈ X.
Interestingly enough, we actually have
(7)
kx k = kxk, ∀ x ∈ X.
To prove this fact, we start with an arbitrary x ∈ X, and we consider the linear
subspace
Y = Kx = {λx : λ ∈ K}.
If we define φ0 : Y → K, by
φ0 (λx) = λkxk, ∀ λ ∈ K,
then it is clear that φ0 (x) = kxk, and
|φ0 (y)| ≤ kyk, ∀ y ∈ Y.
Use then the Hahn-Banach Theorem to find φ : X → K such that φY = φ0 , and
|φ(z)| ≤ kzk, ∀ z ∈ X.
This will clearly imply kφk ≤ 1, while the first condition will give φ(x) = φ0 (x) =
kxk. In particular, we will have
kxk = |φ(x)| = |x (φ)| ≤ kx k · kφk ≤ kx k.
Having proven (7), we now have a linear isometric map
E : X 3 x 7−→ x ∈ X∗∗ .
Since X∗∗ is a Banach space, we now see that Ẽ : X̃ → E(X) is an isometric linear
isomorphism. In particular, X is Banach, if and only if E(X) is closed in X∗∗ .
We continue with a series of results, which are often regarded as the “principles
of Banach space theory.” These results are consequences of Baire Theorem.
Theorem 2.1 (Uniform Boundedness Principle). Let X be a Banach space, let
Y be normed vector space, and let M ⊂ L(X, Y). The following are equivalent
§2. Banach spaces
87
(i) sup kT k : T ∈ M < ∞;
(ii) sup kT xk : T ∈ M < ∞, ∀ x ∈ X.
Proof. The implication (i) ⇒ (ii) is trivial, because if we define
M = sup kT k : T ∈ M ,
then by the definition of the norm, we clearly have
sup kT xk : T ∈ M ≤ M kxk, ∀ x ∈ X.
(ii) ⇒ (i). Assume M satisfies condition (ii). For each integer n ≥ 1, let us
define the set
Fn = x ∈ X : kT xk ≤ n, ∀ T ∈ M .
It is obvious S
that Fn is a closed subset of X, for each n ≥ 1. Moreover, by (ii) we
∞
clearly have n=1 Fn = X. Using Baire’s Theorem, there exists some n ≥ 1, such
that Int(Fn ) 6= ∅. This means that there exists some x0 ∈ X and some r > 0, such
that
Fn ⊃ B̄r (x0 ) = {y ∈ X : kx − x0 k ≤ r}.
Put M0 = sup kT x0 k : T ∈ M . Fix for the moment some arbitrary x ∈ X,
with kxk ≤ 1, and some arbitrary element T ∈ M. The vector y = x0 + rx clearly
belongs to B̄r (x0 ), so we have kT yk ≤ n. We then get
kT xk = T 1 (y − x0 ) = 1 kT y − T x0 k ≤ 1 kT yk + kT x0 k ≤ 1 (n + M0 ).
r
r
r
r
Keep T fixed, and use the above estimate, which gives
n + M0
,
sup kT xk : x ∈ X, kxk ≤ 1 ≤
r
0
to conclude that kT k ≤ n+M
. Since T ∈ M is arbitrary, we finally get
r
n + M0
sup kT k : T ∈ M ≤
< ∞. r
Theorem 2.2 (Inverse Mapping Theorem). Let X and Y be Banach spaces,
and let let T : X → Y be a bijective linear continuous map. Then the linear map
T −1 : Y → X is also continuous.
Proof. Let us denote by A the open unit ball in X centered at the origin, i.e.
A = x ∈ X : kxk < 1 .
The first step in the proof is contained in the following.
Claim 1: The closure T (A) is a neighborhood of 0 in Y.
∞
Consider the sequence of closed sets kT (A) k=1 . (Here we use the notation kM =
{kv : v ∈ M}.) Since the map v 7−→ kv is a homeomorphism, one has the equalities
kT (A) = kT (A) = T (kA), ∀ k ≥ 1.
In particular, we have
∞
∞
∞
[
[
[
[
kT (A) =
T (kA) ⊃
T (kA) = T
[kA] .
k=1
k=1
k=1
k=1
S∞
Since we obviously have k=1 [kA] = X, and T is surjective, the above equality
S∞
shows that
k=1kT (A) = Y. Using Baire’s Theorem, there exists some k ≥ 1, such
that Int kT (A) 6= ∅. Again using the fact that v 7−→ kv is a homeomorphism,
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CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS
this gives Int T (A) 6= ∅. Fix now some point y ∈ Int T (A) , and some r > 0,
such that T (A) contains the open ball
(8)
Br (y) = z ∈ Y : kz − yk < r .
The proof of the Claim is then finished, once we prove the inclusion
T (A) ⊃ B r2 (0).
To prove this inclusion, start with some arbitrary v ∈ B r2 (0), i.e. v ∈ Y and kvk <
r
2 . Since k(2v+y)−yk = 2kvk < r, using (8) it follows that 2v+y ∈ T (A). i.e. there
exists a sequence (xn )∞
n=1 ⊂ X with kxn k < 1, ∀ n ≥ 1, and 2v + y = limn→∞ T xn .
Since y itself belongs to T (A), there also exists some sequence (zn )∞
n=1 ⊂ X, with
kzn k < 1, ∀ n ≥ 1, and y = limn→∞ T zn . On the one hand, if we consider the
1
sequence (un )∞
n=1 ⊂ X given by un = 2 (xn − zn ), then it is clear that
kun k ≤ 12 kxn k + kzn k < 1, ∀ n ≥ 1,
i.e. (un )∞
n=1 ⊂ A. On the othe hand, we have
1
n→∞ 2
lim T un = lim
n→∞
T xn − T zn ) = 12 (2v + y − y) = v,
so v indeed belongs to T (A).
The next step is a slight (but crucial) improvement of Claim 1.
Claim 2: T (A) is a neighborhood of 0.
Start off by choosing ε > 0, such that
(9)
T (A) ⊃ Bε (0).
The Claim will follow, once we prove the inclusion
(10)
T (A) ⊃ B 2ε (0).
To prove this inclusion, we start with some arbitrary y ∈ Bε (0). We want to
construct a sequence of vectors (xn )∞
n=1 ⊂ A, such that, for every n ≥ 1, we have
the inequality
n
X
ε
1
(11)
T ( 2k xk )
y −
≤ 2n+1 .
k=1
This sequence is constructed inductively as follows. We start by using (9), and we
pick x1 ∈ A such that k2y − T x1 k < 2ε . Once x1 , . . . , xp are constructed, such that
(11) holds with n = p, we consider the vector
p
X
z = 2p+1 y −
T ( 21k T xk ) ∈ Bε (0),
k=1
and we use again (9) to find xp+1 ∈ A, such that kz − T xp+1 k ≤ 2ε . We then claerly
have
p+1
X
z − T xp+1 ε
1
y −
T 2k xk =
≤ p+2 ,
2p+1
2
k=1
P∞
Consider now the series k=1 21k xk . Since kxk k < 1, ∀ k ≥ 1, and X is a Banacch
space, by Remark 3.1, the sequence of (wn )∞
n=1 ⊂ X of partial sums
n
X
1
wn =
x , n ≥ 1,
2k k
k=1
§2. Banach spaces
89
is convergent to some point x ∈ X. Moreover, since we have
n
∞
X
kxk k X kxk k
kwn k ≤
≤
, ∀ n ≥ 1,
2k
2k
k=1
k=1
we get the inequality
kxk ≤
∞
X
kxk k
k=1
2k
< 1,
which means that x ∈ A. Note also that using these partial sums, the inequality
(11) reads
ε
ky − T wn k ≤ n+2 , ∀ n ≥ 1,
2
so by the continuity of T , we have y = T x ∈ T (A).
Let us show now that T −1 is continuous. Use Claim 2, to find some r > 0 such
that
(12)
T (A) ⊃ Br (0),
and let y ∈ Y be an arbitrary vector with kyk ≤ 1. Consider the vector v = 2r y,
which has kvk ≤ 2r < r. By (12), there exists x ∈ A, such that T x = v, which
means that T −1 y = 2r x. This forces kT −1 yk ≤ 2r . This argument shows that
2
sup kT −1 yk : y ∈ Y, kyk ≤ 1 ≤ < ∞,
r
−1
and the continuity of T
follows from Proposition 2.5.
The following two exercises deal with two more “principles of Banach space
theory.”
Exercise 3 ♦ . (Closed Graph Theorem). Let X and Y be Banach spaces, and
let T : X → Y be a linear map. Prove that the following are equivalent:
(i) T is continuous.
(ii) The graph of T
GT = (x, T x) : x ∈ X
is a closed subset of X × Y, in the product topology.
Hint: For the implication (ii) ⇒ (i), use Exercise 3, to get the fact that GT is a Banach space.
Then T is exactly the inverse of πX G , where πX : X × Y → X is the projection onto the first
T
coordinate. Use Theorem 3.2.
Exercise 4 ♦ . (Open Mapping Theorem). Let X and Y be Banach spaces, and
let T : X → Y be a surjective linear continuous map. Prove that T is an open map,
in the sense that
• whenver D ⊂ X is open, it follows that T (D) is open in Y.
Hints: There are two possible proofs available.
First proof. Consider the linear map
S : X × Y 3 (x, y) 7−→ (x, T x + y) ∈ X × Y.
Prove that S is linear, continuous, bijective, hence by Theorem 3.2, it is a homeomorphism. Use
this fact to prove that for every open set D ⊂ X, there exists some open set E ⊂ X × Y, such that
T (D) = PY (E), where PY : X × Y → Y is the projection onto the second coordinate. This reduces
the problem to proving the fact that PY is an open map.
Second proof. Take N = Ker T , and use the Factorization Theorem (Proposition 1.5) to write
T = T̂ ◦ Q, where Q : X → X/N is the quotient map, and T̂ : X/N → Y is some linear continuous
90
CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS
map. Remark that T̂ is bijective (so we can apply the Theorem 3.2 to it). Use the fact that Q is
open (Section 1, Exercise 12).
We conclude with two useful results concerning Banach spaces of the form
C K (X), with X compact Hausdorff space. Additional results for these Banach
spaces will be given in Section 6.
The following is an interesting continuity result.
Theorem 2.3. Let T be a topological space, let X be compact Hausdorff spaces,
and let K be one of the fields R or C. For a map F : T × X → K, the following are
equivalent.
(i) F is continuous.
(ii) For every t ∈ T , the map φt : X 3 p 7−→ F (t, p) ∈ K is continuous, and
moreover, the map
Φ : T 3 t 7−→ φt ∈ C K (X)
is continuous.
Proof. (i) ⇒ (ii). Suppose F is continuous. The first assertion is clear, since,
for t ∈ T , we can write φt = F ◦ ιt , where ιt is the map
ιt : X 3 p 7−→ (t, p) ∈ T × X,
which is obviously continuous.
Next, we show that Φ is continuous, at every point t ∈ T . Fix t ∈ T . We need
to prove that, for every ε > 0, there exists some neighborhood N of t in T , such
that
kφv − φt k < ε, ∀ v ∈ N.
By definition, the above condition reads:
(13)
max F (v, p) − F (t, p)| < ε, ∀ v ∈ N.
p∈X
The neighborhood N is constructed as follows. Fix for the moment p ∈ X, and we
use the continuity of F : T × X → K at the point (t, p), to find a neighborhood Ep
of (t, p) in T × X, such that
ε
|F (v, q) − F (t, p)| < , ∀ (v, q) ∈ Ep .
2
The using the definition of the product topology, there exist open sets Vp ⊂ T and
Dp ⊂ X, such that (t, p) ⊂ Vp × Wp ⊂ Ep , so we have
ε
|F (v, q) − F (t, p)| <
∀ v ∈ Vp , q ∈ Wp .
2
Use now compactness to produce a finite sequence of points p1 , . . . , pn ∈ X, such
that X = Wp1 ∪ · · · ∪ Wpn , and take N = Vp1 ∩ · · · ∩ Vpn . To check the desired
property, we start with some arbitrary v ∈ N . For each p ∈ X, there exists some
k ∈ {1, . . . , n} with p ∈ Wpk , and then we have N × Wpk ⊂ Vpk × Wpk ⊂ Epk . In
particular, both (v, p) and (x, p) belong to Epk , so we get
ε
ε
|F (v, p) − F (t, pk )| < and |F (t, p) − F (t, pk )| < ,
2
2
so we immediately get
|F (v, p) − F (t, p)| ≤ |F (v, p) − F (t, pk )| + |F (t, p) − F (t, pk )| < ε.
Since the above inequality holds for all p ∈ X, it will force (13).
§2. Banach spaces
91
(ii) ⇒ (i). Assume condition (ii), and let us prove that F is continuous at
every point (t, p) ∈ T × X. Fix (t, p) ∈ T × X, as well as some ε > 0. We need to
find two open sets V ⊂ T and W ⊂ X, with (t, p) ∈ V × W , such that
(14)
|F (v, q) − F (t, q)| < ε, ∀ (v, q) ∈ V × W.
Using (ii), we find V such that
max |F (v, q) − F (t, q)| <
q∈X
ε
, ∀ v ∈ V,
2
so we get
ε
+|φt (q)−φt (p)|.
2
Using the continuity of φt : X → K, we can then find an open neighborhood W of
p in X with
ε
|φt (q) − φt (p)| < , ∀ q ∈ W,
2
and then (15) forces (14).
(15) |F (v, q)−F (t, p)| ≤ |F (v, q)−F (t, q)|+|F (t, q)−F (t, p)| <
Exercise 5 ♦ . Let X be a compact Hausdorff space. Prove that, when we equip
X × C K (X) with the product topology, the map
X × C K (X) 3 (p, f ) 7−→ f (p) ∈ K
is continuous. .
The following is a technical result which illustrates perfectly the advantage of
working with Banach spaces.
Theorem 2.4 (Arzela-Ascoli Compactness Theorem). Let X be a compact
Hausdorff space, let K be one of the fields R or C. For a subset A ⊂ C K (X), the
following are equivalent.
(i) The closure A is compact, in the norm topology.
(ii) A is bounded and equicontinuous, in the sense that it satifies the condition:
(ec) for every p ∈ X, and every ε > 0, there exists a neighborhood N of p
in X, such that
|f (q) − f (p)| < ε, ∀ q ∈ N, f ∈ A.
Proof. (i) ⇒ (ii). Assume (ii). In order to prove (i), we are going to employ
Corollary I.6.?? (the fact that we work in C K (X), which is a Banach space, is
crucial here) which states that condition (i) is equivalent to:
(i’) For every ε > 0, all ε-discrete subsets of A are finite.
Recall that a set F ⊂ C K (X) is ε-discrete, if
kf − gk ≥ ε, ∀ f, g ∈ F, f 6= g.
Fix ε > 0, as well as a ε-discrete set F ⊂ A. Start off by using the equicontinuity
condition (ec), to construct, for each p ∈ X, an open neighborhood D(p) of p in
X, such that
ε
(16)
|f (q) − f (p)| < , ∀ q ∈ D(p), f ∈ A.
3
Use compactness of X to find points p1 , . . . , pn , such that X = D(p1 ) ∪ · · · ∪ D(pn ).
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CHAPTER II: ELEMENTS OF FUNCTIONAL ANALYSIS
Consider the finite dimensional Banach space Kn , equipped with the norm
k . k∞ , and the map
T : C K (X) 3 f 7−→ f (p1 ), . . . , f (pn ) ∈ Kn .
It is obvious that T is linear, continuous,
with kT k ≤ 1. Remark also that, if
we put M = sup kf k : f ∈ A , then we have the inclusion T (A) ⊂ B n , where
B = {λ ∈ K : |λ| ≤ M }.
Claim: One has
ε
kT f − T gk∞ ≥ , ∀ f, g ∈ F, f 6= g.
3
Start with f, g ∈ F, with f 6= g. Since F is ε-discrete, we have kf − gk ≥ ε, which
means that there exists q ∈ X, with |f (q) − g(q)| ≥ ε. If we choose k ∈ {1, . . . , n},
such that D(pk ) 3 p, then using (16), we have
ε
ε
|f (q) − f (pk )| < and |g(q) − g(pk )| < .
3
3
We now have
2ε
ε ≤ |f (q)−g(q)| ≤ |f (q)−f (pk )|+|f (pk )−g(pk )|+|g(q)−g(pk )| ≤ |f (pk )−g(pk )|+ ,
3
which forces |f (pk ) − g(pk )| ≥ 3ε , and we are done.
Having proven the Claim, let us observe that now we have the following features:
• the map T F : F → Kn is injective;
• the set T (F) is 3ε -discrete in Kn .
Now we are done, since the fact that B n is compact in Kn , forces T (F) to be
finite.
(i) ⇒ (ii). Assume A is compact. Clearly this forces A to be bounded, and
so will be A. Let us show that A is equicontinuous. Replacing A with A, we can
assume that A itself is compact.
By Exercise 6, the map
X × A 3 (p, f ) 7−→ f (p) ∈ K
is continuous, so by Theorem 2.3, for each p ∈ X, the map
φp : A 3 f 7−→ f (p) ∈ K
is continuous, and moreover, the map
Φ : X 3 p 7−→ φp ∈ C K (A)
is also continuous.
Fix now p ∈ X, and ε > 0. By the continuity of Φ, there exists some open set
U ⊂ X, such that
kΦ(q) − Φ(p)k < ε, ∀ q ∈ U.
This reads
sup |f (q) − f (p)| < ε, ∀ q ∈ U,
f ∈A
which is precisely the equicontinuity condition.