Download Topological Vector Spaces IV: Completeness and Metrizability

c
Gabriel
Nagy
TVS IV
Topological Vector Spaces IV:
Completeness and Metrizability
Notes from the Functional Analysis Course (Fall 07 - Spring 08)
In this section we isolate two important features of topological vector spaces, which, when
present, are very useful. With few exceptions, the material from sub-section A is is optional.
Convention. Throughout this note K will be one of the fields R or C, and all vector
spaces are over K.
A. Completeness for Topological Vector Spaces
This sub-section is aimed at treating completeness for most general types of topological
vector spaces, beyond the traditional metric framework. Our approach here is essentially
and adaptation of the the theory of uniform spaces1 . As it turns out, a topological vector
space is always a uniform space.
Definitions. Suppose (X , T) is a topological vector space.
A. A net (xλ )λ∈Λ ⊂ X is said to be T-Cauchy, if:
(c) for every T- neighborhood V of 0, there exists λV ∈ Λ, such that: xλ − xµ ∈ V,
∀ λ, µ λV .
This condition is equivalent to the fact that the double net (xλ −xµ )(λ,µ)∈Λ×Λ converges
to 0.
B. The linear topology T is said to be complete, if every T-Cauchy net in X is convergent.
Example 1. Let I be an arbitrary non-empty
Q set. If (Xi )i∈I is a collection of topological
vector spaces, then the product space X = i∈I Xi is complete, when equipped with the
product topology, if and only if all the Xi , i ∈ I, are complete. This follows from the fact
that a net (xλ )λ∈Λ in X , written in coordinates as xλ = (xiλ )i∈I , ∀ λ, is Cauchy in X , if and
only if, for every i ∈ I, the net (xiλ )λ∈Λ is Cauchy in Xi .
In particular, all finite dimensional topological vector spaces are complete.
1
A uniform space is a pair (X, U) consisting
of a non-empty set X and a filter U of subsets U ⊂ X × X,
T
subject to the following conditions: (i) U ∈U U = {(x, x) : x ∈ X} – the diagonal of X; (ii) if U belongs to
U, then its inverse (as a relation) U −1 = {(y, x) : (x, y) ∈ U } also belongs to U; (iii) for every U ∈ U there
exists V ∈ U, such that the composition (as relations) V ◦ V is contained in U . The filter U is then called a
uniformity on X.
1
Exercise 1. Prove that a topological vector space X is complete, if and only if every
Cauchy net has a convergent sub-net.
Comment. If one attempts to construct the completion of a topological vector space in
a manner similar to the one used in the metric framework, then Definition B above is a bit
useless, since there is no such thing as “the set of all nets.” Therefore, one would like to
replace (for now) nets with filters2 , thus the following definition is natural.
Definition. Suppose (X , T) is a topological vector space. A filter F in X is called
Cauchy, if for every T- neighborhood V of 0, there exists F ∈ F, such that: x − y ∈ V,
∀ x, y ∈ F.
With all the preparations in place, the exercises below will guide the reader towards building the completion of a topological vector space (X , T). Except for Notations, Definitions,
and Lemma 1, the remainder of this sub-section is optional.
Exercises 2-8. These three exercises clarify the connections between Cauchy filters and
Cauchy nets. Fix a topological vector space X .
2. Prove that, if a filter F is convergent, that is, there exists x ∈ X , such that, for every
neighborhood V of x, there exists F ∈ F with F ⊂ V, then F is Cauchy.
3. Prove that, for a net (xλ )λ∈Λ in X , the following are equivalent:
(i) (xλ )λ∈Λ is Cauchy;
(ii) the tail filter F = {Fλ }λ∈Λ , defined by Fλ = {xµ : µ λ}, is Cauchy.
4*. Prove that, for a filter F on X , the following are equivalent:
(i) F is Cauchy;
(ii) all F-section nets3 are Cauchy.
(Hint: For (ii) ⇒ (i), equip each F ∈ F with a well ordering F , and consider the set
Λ = {(F, z) : F ∈ F, z ∈ F}, equipped with the following order: (F1 , z1 ) (F2 , z2 ),
if either F1 ( F2 , or F1 = F2 and z1 F1 z2 . Show that Λ is directed. Consider
the obvious map Λ 3 (F, z) 7−→ z ∈ X as a net (xλ )λ∈Λ . Show that this net is an
F-section, and prove that by condition (ii) the filter F is Cauchy.)
5. Prove that if F is a Cauchy filter, then so is any filter F0 ⊃ F. In particular, all
ultrafilters that contain F are Cauchy.
6. Prove that the following conditions are equivalent:
(i) X is complete;
(ii) every Cauchy filter in X is convergent;
2
Recall that, given some non-empty set X, a filter in X is a collection F of non-empty subsets, which
is directed with respect to the reverse inclusion, that is, for every F1 , F2 ∈ F, there exists F ∈ F, with
F ⊂ F1 ∩ F2 .
3
Recall that a net (xλ )λ∈Λ is called an F-section, if for every F ∈ F, there exists λF ∈ Λ, such that
xλ ∈ F, ∀ λ λF .
2
(iii) every Cauchy ultrafilter in X is convergent.
(Hints: Prove the implications (i) ⇒ (ii) ⇒ (iii) ⇒ (i). For (i) ⇒ (ii) use Exercise 4.
For (iii) ⇒ (i), start with a Cauchy net (xλ ) and choose an ultrafilter U that contains
the tail filter. By Exercises 3 and 5, U is Cauchy, so assuming (iii), the ultrafilter U
converges to some x. Prove that xλ → x.)
7. Prove that if F1 and F2 are Cauchy filters, then {F1 + F2 : F1 ∈ F1 , F2 ∈ F2 } is a
Cauchy filter. By an abuse of notation, this filter will be denoted by F1 + F2 .
8. Prove that if F is a Cauchy filter, then for every α ∈ K, {αF : F ∈ F} is a Cauchy
filter. By an abuse of notation, this filter will be denoted by αF.
Notation. We denote by cf(X ) the set of all Cauchy filters in (X , T). Equip cf(X )
with the following relation (use the notations from Exercises 5 and 6):
F ≈ G ⇔ F + (−1)G converges to 0.
Exercise 9. Prove the following
(i) ≈ is an equivalence relation on cf(X ).
(ii) If F is a Cauchy filter, and F0 is a filter that contains F (which is automatically Cauchy
by Exercise 5), then F ≈ F0 .
(iii) If F1 ≈ G1 and F2 ≈ G2 , then F1 + F2 ≈ G1 + G2 .
(iv) If F ≈ G, then αF ≈ αG, ∀ α ∈ K.
Notations. For each F ∈ cf(X ), let [F] denote its equivalence class.
Given a non-empty set S ⊂ X , define
S̃ = {[F] : F ∈ cf(X ), F 3 S}.
Using the preceding Exercise, we see that one has two operations
addition
X̃ × X̃ −−−−→ X̃ ,
multiplication
K × X̃ −−−−−−−→ X̃ ,
correctly defined on equivalence classes by [F1 ] + [F2 ] = [F1 + F2 ] and α[F] = [αF].
Exercise 10. Prove that the above two operations define a vector space structure on
the quotient space X̃ .
Exercise 11. Define, for every x ∈ X , the trivial filter Fx = {x} . Prove that:
(i) For every x ∈ X , the filter Fx is convergent, thus Cauchy.
(ii) The map J : X 3 x 7−→ [Fx ] ∈ X̃ is an injective linear map.
Exercise 12. Consider the collection B = {Ñ : N ⊂ X balanced T-neighborhood of 0}.
Prove that:
3
(i) B is a filter;
(ii) all sets in B are balanced and absorbing;
(iii) for any V ∈ B, there exists W ∈ B, such that W + W ⊂ V;
T
(iv) V∈B V = {0}.
Conclude, using Corollary 1 from TVS I, that there exists a unique Hausdorff linear topology
T̃ on X̃ , so that B is a basic system of T̃-neighborhoods of 0.
Exercise 13. Consider the injective linear map J : X → X̃ defined in Exercise 10.
(i) Prove that the linear subspace R = Range J ⊂ X̃ is dense in X̃ .
(ii) Prove that, if (xλ )λ∈Λ is a Cauchy net in X , and we consider its tail filter F defined in
Exercise 1, then J(xλ ) → [F] in (X̃ , T̃).
(iii) Prove that any point in X̃ can be realized using the construction from (ii), namely, for
every v ∈ X̃ , there exists a Cauchy net (xλ )λ∈Λ in X , such that J(xλ ) → v.
∼
(iv) Prove that, when we equip R with the induced topology T̃R , the map J : X −
→R
becomes a linear homeomorphism.
(v) Prove that, if (X , T) is already complete, then J : (X , T) → (X̃ , T̃) is a linear homeomorphism.
(Hint: For (iv) show that if V is a neighborhood of 0 in X , then V ⊂ J −1 (Ṽ) ⊂ V.)
Exercise 14*. Prove that (X̃ , T̃) is a complete topological vector space. (Hint: Consider the filter V = {Ṽ : V neighborhood of 0 in X }. Show that if G is a Cauchy filter in
X̃ , then G + V is again a Cauchy filter in X̃ , equivalent to G. Using the preceding Exercise,
it follows that the collection F = {J −1 (A) : A ∈ G0 } is a filter in X . Show that F is Cauchy,
and then prove that G0 converges to v = [F] in X̃ . Conclude that, since G is equivalent to
G0 , it also converges to v.)
Definition. The topological vector space (X̃ , T̃) is called the completion of (X , T).
Comment. The construction of the completion is borrowed (word for word) from the
theory of uniform spaces in Topology. Very few of the properties of topological vector spaces
are employed here. In particular, the same construction works for topological groups. (A
topological group is a pair (G, T) consisting of a group G and a topology T on G, such that
the maps G × G 3 (g, h) 7−→ gh ∈ G and G 3 g 7−→ g −1 ∈ G are continuous.)
Exercise 15-16. Suppose (X1 , T1 ) and (X2 , T2 ) are topological vector spaces. Declare a
(not necessarily linear ) map Φ : X1 → X2 uniformly continuous, if it satisfies the following
condition.
(uc) for every T2 -neighborhood V of 0 in X2 , there exists a T1 -neighborhood W of 0 in X1 ,
so that: whenever x, y ∈ X1 are such that x − y ∈ W, it follows that Φ(x) − Φ(y) ∈ V.
It is not hard to show that such a Φ is indeed continuous.
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15. Assume that Φ is uniformly continuous. Let Jk : Xk → X̃k , k = 1, 2 denote the
corresponding injections.
(i) Prove that there exists a unique continuous map Φ̃ : X̃1 → X̃2 , such that Φ̃ ◦ J1 =
J2 ◦ Φ.
(ii) Prove that Φ̃ is in fact uniformly continuous.
(Hint: Either use Exercise 13, by showing that, if (xλ )λ∈Λ is Cauchy in X1 , then
Φ(xλ ) λ∈Λ is Cauchy in X2 , or, for an alternative approach, see the Comment below.)
16. Prove that, for any linear continuous map T : (X1 , T1 ) → (X2 , T2 ), is uniformly continuous, so the above Exercise can be used to obtain a uniformly continuous map
T̃ : (X̃1 , T̃1 ) → (X̃2 , T̃2 ). Prove that T̃ is linear.
Exercise 17. Suppose X is a topological vector space, and S ⊂ X is non-empty. Let
J : X → X̃ denote the map described in Exercise 11. Show that S̃ = J(S) (the closure of
J(S) in X̃ ).
Exercise 18. Suppose X is a topological vector space, and Y ⊂ X is a linear subspace,
equipped with the induced topology, and let X̃ and Ỹ be the completions of X and Y. Let
T : Y ,→ X denote the inclusion map. Prove the following:
(i) The continuous extension T̃ : Ỹ → X̃ is injective.
(ii) If we equip the linear subspace Z = Range T̃ with the induced topology (from X̃ ),
then T̃ : Ỹ → Z is a linear homeomorphism.
(iii) If J : X → X̃ denotes the map described in Exercise 11, then Z = J(Y) (the closure
of J(Y) in X̃ ).
Comments. A. In connection with Exercise 15, perhaps the most optimal way to
construct Φ̃ is as follows. Start with some v ∈ X̃ , represented as v = [F], by a Cauchy filter
F in X1 , and simply notice that the collection Φ(F) = {Φ(F) : F ∈ F} is a Cauchy filter in
X2 , so one can define Φ̃(v) = [Φ(F)]. Of course, one has to check that this construction is
independent on the choice of F, but this is quite obvious.
B. The map Φ̃ can also be constructed in an analogous way to the traditional metric
version, as follows. Start with some v ∈ X̃1 , so by Exercise 13 there exists a Cauchy net
T̃1
(xλ )λ∈Λ ⊂ X1 , such that J1 (xλ ) −→
v. By uniform continuity the net Φ(xλ ) λ∈Λ is Cauchy
in (X2 , T2 ), hence the net J Φ(xλ ) λ∈Λ is convergent in X̃2 , T̃2 ) to some unique vector,
which we declare to be Φ̃(v). Of course, this definition has a slight ambiguity built in (it
appears to depend on the choice of the net), but a posteriori this is easily removed.
C. Using Exercises 14-15, the completion of a topological vector space is unique up to a
linear homeomorphism. The result below provides a direct way of recognizing the completion.
Theorem 1. Suppose (Y, S) is a complete
topological vector space, and X is a linear
subspace. Let T denote the induced topology SX , and let T denote the induced topology SX
on the closure of X . If we consider the completion (X̃ , T̃) and the standard inclusion map
5
∼
J : X ,→ X̃ , there exists a unique linear homeomorphism T : (X̃ , T̃) −
→ (X , T), such that
T (J(x)) = x, ∀ x ∈ X .
Proof. Let I : (X , T) ,→ (Y, S) be the inclusion map, which is obviously continuous, and
let T = I˜ : (X̃ , T̃) → (Y, S) be the extension to the completion (as indicated in Exercises
15-16). By construction, T is linear, continuous, and satisfies T ◦ J = I, Strictly speaking,
T = I˜ takes values in the completion of Y, which is in turn identified with Y, as indicated
in Exercise 13. Using Comment B above, T can be implicitly defined by the implication
T̃
S
J(xλ ) −
→ v (in X̃ ) =⇒ xλ −
→ T (v) (in Y),
(1)
where (xλ )λ∈Λ is a net in X , and v is a vector in X̃ . It is clear that Range T ⊂ X , so we can
think from now on T as a linear continuous map T : (X̃ , T̃) → (X , T).
Claim 1: T is injective.
Start with some v ∈ X̃ , with T (v) = 0, and let us show that v = 0. Choose a net
T̃
S
(xλ )λ∈Λ ⊂ X , with J(xλ ) −
→ v (in X̃ ), so that by (1) it follows that xλ −
→ 0, in Y. By the
T
definition of the induced topology this forces xλ −
→ 0, in X , so by the continuity of J we also
T̃
get J(xλ ) −
→ 0, in X̃ , which yields (X̃ is, of course, Hausdorff) v = 0.
Claim 2: For a vector v ∈ X̃ and a net (xλ )λ∈Λ ⊂ X , the following are equivalent:
T̃
(i) J(xλ ) −
→ v, in X̃ ;
S
(ii) xλ −
→ T (v), in Y;
T
→ T (v), in X .
(ii’) xλ −
The equivalence (ii) ⇔ (ii0 ) is clear, since T is the induced topology. The implication
(i) ⇒ (ii) – which is the same as (1) – was already discussed. To prove the implication
(ii) ⇒ (i), we simply observe that condition (ii), combined with the fact that T is the
induced topology, implies the fact that (xλ )λ∈Λ is Cauchy in (X , T), so the net J(xλ ) λ∈Λ
S
will converge in (X̃, T̃) to some vector w ∈ X̃ . Of course, by (1) this forces xλ −
→ T (w), in
Y. By the uniqueness of limits (Y is, of course, Hausdorff), this yields T (v) = T (w), and
then by Claim 1 we get w = v, thus proving (i).
Based on Claims 1 and 2 it follows that T is a homeomorphism, when regarded as a map
T : X̃ → Range T , where the target space is equipped with the induced topology (by T), so
the proof will be finished, once we prove that Range T = X . Start with some y ∈ X , and let
us indicate how a vector v ∈ X̃ can be found, such that T (v) = y. We know that there exists
S
a net (xλ )λ∈Λ ⊂ X , such that xλ −
→ y. As argued before, it follows that (xλ )λ∈Λ is Cauchy
T̃
in (X , T), and J(xλ ) λ∈Λ is Cauchy in (X̃, T̃), so there exists v ∈ X̃ , such that J(xλ ) −
→ v,
S
in T̃. By (1) it follows that xλ −
→ T (v) in Y, so by the uniqueness of limits, it follows that
T (v) = y.
6
We conclude this sub-section with an important technical result, which will be used in
DT II.
Lemma 1 (Bourbaki’s Embedding Lemma). Suppose T1 ⊃ T2 are two Hausdorff linear
topologies on a vector space X . Let I : (X , T1 ) → (X , T2 ) denote the identity map, and let
(X̃k , T̃k ) denote the completion of (X , Tk ), k = 1, 2.
If there exists a fundamental system V of T1 -neighborhoods of 0, such that all sets V ∈ V
are T2 -closed, then the map I˜ : (X̃1 , T̃1 ) → (X̃2 , T̃2 ) is injective.
Proof. We know already that I˜ is linear (and continuous), so all we need to do is to prove
˜ = 0, and let us prove that v = 0.
that Ker I˜ = {0}. Fix some vector v ∈ X̃1 , such that I(v)
Since the topology T̃1 is Hausdorff, it suffices to prove the following
Claim: For every T̃1 -neighborhood U of 0 (in X̃1 ), the set v + U contains 0.
For the proof of the above statement, we start off by fixing some T1 -Cauchy net (xλ )λ∈Λ
˜
in X that represents v, i.e. J1 (xλ ) → v in (X̃1 , T̃1 ). Since I(v)
= 0 in (X̃2 , T̃2 ), it follows
T2
that xλ −→ 0.
First of all, by the construction of the completion (X̃1 , T̃1 ) we can assume that U =
W̃(⊂ X̃1 ), for some T1 -neighborhood W of 0 in (X , T1 ). Secondly, by the hypothesis, upon
replacing W with a smaller T1 -neighborhood, we can also assume that W is also T2 -closed.
Thirdly, since the net (xλ ) is T1 -Cauchy, there exists λW , such that xλ −xµ ∈ W, ∀ λ, µ λW .
If we fix for the moment µ λW , this means that for every λ λW , there exists zλ ∈ W,
T2
T2
such that xλ − xµ = zλ . Of course, since xλ −→
0, this means that zλ −→
(−xµ ). But the
fact that W is T2 -closed will then force
−xµ ∈ W, ∀ µ λW .
(2)
Since the net (−xµ )µλW represents −v in (X̃1 , T̃1 ), condition (2) forces −v ∈ W̃ = U, and
the Claim is proven.
B. Metrizability
As it turns out, for a large class of topological vector spaces it is possible to carry on the
completion operation by traditional (i.e. metric) means. In preparation for that approach, we
discuss an important topological property: metrizability. As a matter of terminology, we will
say that a topological (vector) space (X , T) is metrizable, if there exists a metric d on X , so
T
that T agrees with the metric topology defined by d. In other words, xλ −
→ x ⇔ d(xλ , x) → 0.
With this terminology, one has the following fundamental result.
Theorem 2 (Metrizability Theorem). For a topological vector space (X , T), the following
conditions are equivalent:
(ii) The topology T is metrizable, i.e. there exists a metric d on X , so that T coincides
with the metric topology defined by d.
(ii) The topology T is first countable4 .
4
This means that every point x ∈ X has a countable fundamental (or basic) T-neighborhood system.
Because of translation invariance, it suffices to check this condition only at x = 0.
7
Moreover, in the case when one of the above conditions is satisfied, the metric d from (i) can
be chosen to be translation invariant, i.e.
d(x + z, y + z) = d(x, y), ∀ x, y, z ∈ X .
Proof. The implication (i) ⇒ (ii) is trivial. To prove the reverse implication, we start off by
choosing basic neighborhood system (Vn )∞
n=0 of 0, with V0 = X , such that:
(a) (−1)Vn = Vn ,
(b) Vn−1 ⊃ Vn + Vn + Vn ,
for all n ∈ N. (This can be easily accomplished using Proposition 2 from TVS I. Condition
(a) is obviously satisfied, if for instance all the Vn ’s are balanced.) Since T is Hausdorff, we
also have:
∞
\
Vn = {0}.
(3)
n=1
Define now ρ : X → [0, 1] by
ρ(x) = inf{2−k : x ∈ Vk }.
Obviously, ρ takes values in {0} ∪ {2−n }∞
n=0 , and using (3) we see that
ρ(x) = 0 ⇔ x = 0.
(4)
Moreover, by construction, for a net (xλ )λ in X , one has the equivalence:
ρ(xλ ) → 0 ⇔ xλ → 0.
(5)
Using condition (a) it also follows that:
ρ(−x) = ρ(x), ∀ x ∈ X .
(6)
Claim 1: ρ(x + y + z) ≤ 2 · max{ρ(x), ρ(y), ρ(z)}, ∀ x, y, z ∈ X .
To prove this we can assume (by symmetry) that the max{ρ(x), ρ(y), ρ(z)} = ρ(z). The
cases when ρ(z) = 0 or ρ(z) = 1, are trivial, so the remaining one is when ρ(z) = 2−k , for
some integer k ≥ 1. But then the inequalities ρ(x), ρ(y) ≤ ρ(z) = 2−k will force x, y, z ∈ Vk ,
so by (b) we get x + y + z ∈ Vk−1 , so ρ(x + y + z) ≤ 2−(k−1) = 2ρ(z).
Claim 2: ρ(x1 + · · · + xn ) ≤ 2 ρ(x1 ) + · · · + ρ(xn ) , ∀ x1 , . . . , xn ∈ X .
We prove the inequality by induction on n. The cases n = 1, 2, 3 are trivial by Claim 1.
Assume the inequality is true for all n with 1 ≤ n ≤ N , for some N ≥ 3, and let us prove it
for n = N + 1. Fix x1 , . . . , xN +1 , and let us prove that
ρ(x1 + · · · + xN +1 ) ≤ 2 ρ(x1 ) + · · · + ρ(xN +1 ) .
(7)
If ρ(xj ) = 0, for some j, then everything is clear, by the inductive hypothesis. Likewise, if
ρ(xj ) = 1, for some j, then the inequality is also trivial, since ρ takes values in [0, 1]. Assume
now 0 < ρ(xj ) < 1, for all j, and also (re-index if necessary), that: ρ(x1 ) ≤ · · · ≤ ρ(xN +1 ).
Equivalently, one has integers k1 ≥ k2 ≥ · · · ≥ kN +1 such that ρ(xj ) = 2−kj , for all j. Let us
consider now two possibilities:
8
(i) There exists j ∈ {2, . . . , N }, such that ρ(xj ) = ρ(xj+1 ). In this case by Claim 1 (note
that ρ(xj+1 ) = max{ρ(xj−1 ), ρ(xj ), ρ(xj+1 )}) we have:
ρ(xj−1 ) + ρ(xj ) + ρ(xj+1 ) ≥ 2ρ(xj+1 ) ≥ ρ(xj−1 + xj + xj+1 ),
so we get
ρ(x1 )+· · ·+ρ(xN +1 ) ≥ ρ(x1 )+· · ·+ρ(xj−2 )+ρ(xj−1 +xj +xj+1 )+ρ(xj+2 )+· · ·+ρ(xN +1 ),
so we can use the inductive hypothesis (for n = N − 1) to get (7)
(ii) ρ(x1 ) ≤ ρ(x2 ) < ρ(x3 ) < · · · < ρ(xN +1 ). In this case we know that xj ∈ Vkj , with
k1 ≥ k2 > k3 > · · · > kN +1 , and then an easy recursive argument shows that
x1 + · · · + xm ∈ Vkm −1 ⊂ Vkm+1 , ∀ m = 1, . . . , N.
In particular x1 + · · · + xN +1 ∈ VkN +1 + VkN +1 ⊂ VkN +1 −1 , which forces
ρ(x1 + · · · + xN +1 ) ≤ 2−(kN +1 −1) = 2ρ(xN +1 ),
and (7) follows immediately.
Define now the map δ : X → [0, 1] by:
δ(x) = inf
n
X
ρ(xj ) : n ∈ N, (xj )nj=1 ∈ X n ,
j=1
n
X
xj = x}.
j=1
By construction and by Claim 2, we clearly have the inequalities:
1
ρ(x)
2
≤ δ(x) ≤ ρ(x), ∀ x ∈ X ,
(8)
thus δ, up to this point, has the following features:
δ(x) = 0 ⇔ x = 0;
δ(xλ ) → 0 ⇔ xλ → 0;
δ(−x) = δ(x), ∀ x ∈ X .
(9)
(10)
(11)
One additional feature, pretty obvious from the definition of δ, is:
δ(x + y) ≤ δ(x) + δ(y), ∀ x, y ∈ X .
(12)
The proof is now finished, by defining d(x, y) = δ(x − y).
Comment. The proof of Theorem 2 is again borrowed (word for word) from the theory
of uniform spaces in Topology, and works equally well for topological groups. (See the
Comment from sub-section A).
Definition. Given a metrizable topological vector space (X , T) a translation invariant
metric d on X , whose metric topology coincides with T, will be declared uniformly compatible
9
with T. In this case, when convenient, we define the d-gauge | · |d : X → [0, ∞) by |x|d =
d(x, 0), so that by translation invariance we have
d(x, y) = |x − y|d , ∀ x, y ∈ X .
Comment. There are three “generic” methods of producing metrizable topological
vector spaces, which are pretty easy to justify (see the three Remarks below).
Remark 1. If X is a metrizable topological vector space, then any linear subspace
Y ⊂ X , equipped with the induced topology is metrizable.
Remark 2.
Q If (Xj )j∈J is a family of metrizable topological vector spaces, then the
product space j∈J Xj , equipped with the product topology, is metrizable, if and only if the
index set J is countable. If we identify, for instance, J as a subset of N, and we pick, for
each j ∈ J, a metric dj on Xj , whose metric topology agrees with the given topology on Xj ,
then the metric
Y
X 1
dj (xj , yj )
·
,
∀
x
=
(x
)
,
y
=
(y
)
∈
Xj
d(x, y) =
j
j∈J
j
j∈J
j
2
1
+
d
j (xj , yj )
j∈J
j∈J
defines the product topology. (See P II.)
In particular, all finite dimensional topological vector spaces are metrizable.
Remark 3. If X is a metrizable topological vector space, and Y ⊂ X is a closed linear
subspace, then the quotient space X /Y, equipped with the quotient topology, is metrizable.
Indeed, if we consider the quotient map π : X → X /Y, then we know (see TVS II) that
π is continuous and open. In particular, if we choose a basic system {Vn }n∈N of open
neighborhoods of 0 in X , then {π(Vn )}n∈N constitutes a basic system of (open) neighborhoods
of 0 in X /Y, thus the quotient topology (which is Hausdorff anyway, since Y is closed) is
first countable.
In connection with the issue of metrizability for quotient spaces, we prove a technical
result below, which will be useful later.
Lemma 2. Let X be a metrizable topological vector space and let d be a uniformly
compatible metric on X Let Y ⊂ X be a closed linear subspace, and let π : X → X /Y denote
the quotient map. The map dˆ : (X /Y) × (X /Y) → [0, ∞), defined by
ˆ w) = inf{d(x, y) : π(x) = v, π(y) = w}, v, w ∈ X /Y,
d(v,
is a metric on X /Y, that is uniformly compatible with the quotient topology.
ˆ w) = d(w,
ˆ v)
Proof. Let us show first that dˆ is indeed a metric. The symmetry property d(v,
ˆ
ˆ
ˆ
is trivial, and so is the triangle inequality d(u, w) ≤ d(u, v) + d(v, w). Suppose v, w ∈ X /Y
ˆ w) = 0, and let us show that v = w (in X /Y). By construction, there exists
are such that d(v,
∞
two sequences (xn )∞
n=1 , (yn )n=1 , such that π(xn ) = v, π(yn ) = w, ∀ n, and d(xn , yn ) → 0. By
translation invariance, the sequence zn = xn − yn satisfies d(zn , 0) = d(xn , yn ), so zn → 0. In
particular, since π is linear and continuous, and we clearly have v−w = π(xn )−π(yn ) = π(zn ),
∀ n, it follows that v − w = 0, i.e. v = w.
ˆ it suffices to show that
To prove the translation invariance of d,
ˆ w) ≥ d(u
ˆ + v, u + w), ∀ u, v, w ∈ X /Y.
d(v,
10
(13)
But this is quite trivial, because if we fix z ∈ X such that π(z) = u, then for any x, y ∈ X
that satisfy π(x) = v and π(y) = w, it follows that π(z + x) = u + v and π(z + y) = w, so
by the translation invariance of d we have
ˆ + v, u + w),
d(x, y) = d(z + x, z + y) ≥ d(u
so if we take the infimum (over all x and y, satisfying π(x) = v and π(y) = w), we get (13)
ˆ coincides
Finally, in order to prove that the metric topology Tdˆ on X /Y, defined by d,
with the quotient topology, all we must show (see TVS II) is that the quotient map π : X →
(X /Y, Tdˆ) is both continuous and open. The continuity is trivial, since by definition we have
the inequality
dˆ π(x), π(y) ≤ d(x, y), ∀ x, y ∈ X
(so not only π is continuous, but in fact Lipschitz.) To prove that π is open, start with some
open set A, and show that π(A) is open in the metric topology Tdˆ. Fix v ∈ π(A), and let
us indicate how to construct r > 0, so that
ˆ w) < r ⇒ w ∈ π(A).
d(v,
(14)
First, choose a ∈ A with π(a) = v, and choose r > 0, such that
d(a, y) < r ⇒ y ∈ A.
(15)
ˆ w) < r, so there exist x, y ∈ X with
To prove (14) start with some w ∈ X /Y with d(v,
π(x) = v, π(y) = w, and d(x, y) < r. Since π(x) = π(a) = v, it follows that a − x ∈ Y, so
the vector y 0 = y + a − x will also satisfy π(y 0 ) = w. But now by translation invariance we
have
d(a, y 0 ) = d(a, y + a − x) = d(0, y − x) = d(y, x) < r,
so using (15) it follows that y 0 ∈ A, thus the vector w = π(y 0 ) indeed belongs to π(A).
In the remainder of this sub-section we revisit the results from sub-section A, under the
metrizability assumption.
Proposition 1. Suppose X1 and X2 are metrizable topological vector spaces. For a (not
necessarily linear) map Φ : X → Y, the following conditions are equivalent.
(i) Φ is uniformly continuous, in the sense that it has property (u) introduced in the
previous sub-section.
(ii) there exist uniformly compatible metrics d1 on X1 and d2 on X2 with the property
(muc) For every ε > 0, there exists δ > 0, so that, whenever x, y ∈ X1 are such that
|x − y|d1 < δ, it follows that |Φ(x) − Φ(y)|d2 < ε.
(ii’) Any uniformly compatible metrics d1 (on X1 ) and d2 (on X2 ) satisfy condition (muc).
Condition (muc) is referred to as the metric uniform continuity condition.
11
Proof. (i) ⇒ (ii0 ). Let d1 and d2 be two uniformly compatible metrics on X1 and X2 ,
respectively. Fix ε > 0, and let us indicate how δ > 0 can be constructed, satisfying the
requirement in (muc). First of all, the set V = {v ∈ X2 , : |v|d2 < ε} is a neighborhood of
0 in X2 , so by (i) there exists a neighborhood W of 0 in X1 , such that: whenever x, y ∈ X1
are such that x − y ∈ W, it follows that Φ(x) − Φ(y) ∈ V, i.e. |Φ(x) − Φ(y)|d2 < ε. Now
we see that δ > 0 can be chosen (use again translation invariance) simply by the condition
W ⊃ {w : |w|d1 < δ}, which is always possible, since the metric topology defined by d1
coincides with T1 .
The implication (ii0 ) ⇒ (ii) is trivial.
To prove the implication (ii) ⇒ (i) we fix d1 and d2 , and argue pretty much as above.
Start with V a neighborhood of 0 in X2 , then there exists ε > 0, such that V ⊃ {v : |v|d2 < ε},
so if we choose δ > 0 as in (muc), the set W = {w : |w|d1 < δ} will do the job.
Corollary 1. If X1 and X2 are as above, and Φ : X1 → X2 is linear, then the conditions
(i), (ii), (ii0 ), from Proposition 1, are also equivalent to the condition that Φ is continuous.
Proof. Immediate from the observation that Φ(x)−Φ(y) ∈ V is equivalent to Φ(x−y) ∈ V, so
condition (i) is obviously equivalent to the fact that Φ is continuous at 0, and this condition
in turn is equivalent to the (global) continuity of Φ.
Proposition 2. Suppose (X , T) is a metrizable topological vector space. For a net
(xλ )λ∈Λ in X , the following are equivalent.
(i) (xλ )λ∈Λ is Cauchy, in the sense introduced in sub-section A.
(ii) There exists a uniformly compatible metric d, such that:
(mc) for every ε > 0, there exists λε ∈ Λ, such that |xλ − xµ |d < ε, ∀ λ, µ λε .
(ii’) Condition (mc) is satisfied. for every uniformly compatible metric d.
Condition (mc) is referred to as the metric Cauchy condition.
Proof. (i) ⇒ (ii0 ). Fix a uniformly compatible metric d, and assume the net (xλ )λ∈Λ is
Cauchy, that is, i.e. it satisfies the condition (c) from sub-section A. The fact that (xλ )λ∈Λ
satisfies condition (mc) is trivial, if we use condition (c) with V = {x : |x|d < ε}, which is
clearly a neighborhood of 0 (for every ε > 0).
The implication (ii0 ) ⇒ (ii) is trivial.
(ii) ⇒ (i). Suppose (xλ )λ∈Λ satisfies condition (mc) for a uniformly compatible metric
d. condition (c) is pretty obvious, since for every neighborhood V of 0, there exists ε > 0,
such that V ⊃ {v : |vd < ε}, so if we choose λε as in (c), then for λ, µ λε , we have
|xλ − xµ |d < ε, which forces xλ − xµ ∈ V.
Proposition 3. For a metrizable topological vector space (X , T), the following are equivalent.
(i) (X , T) is complete.
(ii) Every Cauchy sequence in X is convergent.
12
(iii) Every Cauchy sequence in X has a convergent subsequence.
(iv) There exists a uniformly compatible metric d, such that:
P∞
(st) every sequence (xn )∞
n=1 ⊂ X , with the property
n=1 |xn+1 − xn |d < ∞, is convergent.
(iv’) Every uniformly compatible metric has property (st).
Condition (st) is referred to as the Summability Test.
Proof. (i) ⇒ (iv 0 ).PFix some uniformly compatible metric d, and let (xn )∞
n=1 be a sequence
∞
in
n=1 |xn+1 − xn |d < ∞. In particular, for each k ∈ N the sum tk =
P∞X , such that
n=k |xn+1 − xn |d ≥ 0 is finite,
Pand furthermore limk→∞ tk = 0. Notice now that, for m > n,
we clearly have |xm − xn |d ≤ m−1
j=n |xj+1 − xj |d ≤ tn , which proves that
|xm − xn |d ≤ tk , ∀ m, n ≥ k,
and then the condition tk → 0, combined with Proposition 2, clearly shows that (xn )∞
n=1 is
Cauchy (as a net), thus convergent.
The implication (iv 0 ) ⇒ (iv) is trivial.
(iv) ⇒ (iii). Fix a uniformly compatible metric d with property (st), and a Cauchy
∞
sequence (xn )∞
n=1 ⊂ X . By Proposition 2, we know that (xn )n=1 satisfies the metric Cauchy
condition (mc). In particular, there exists a sequence of (jk )∞
k=1 of positive integers, such
−k
that |xm − xn |d ≤ 2 , m, n ≥ jk . If we define the sequence (kn )∞
n=1 recursively by k1 = j1 ,
and kn+1 = max{k1 , . . . , kn , jn } + 1, then (kn )∞
is
strictly
increasing,
n=1
P∞and the sub-sequence
∞
−n
(xkn )n=1 satisfies the inequalities |xkn+1 −xkn |d ≤ 2 , ∀ n, so clearly n=1 |xkn+1 −xkn |d < ∞.
Now we are done, because (st) forces (xkn )∞
n=1 to be convergent.
∞
(iii) ⇒ (ii). Assume condition (iii) and let (xn )∞
n=1 be a Cauchy sequence. Let (xkn )n=1
be a subsequence which converges to some point x. By the cauchy condition (xn − xkn ) → 0,
so xn → x.
(ii) ⇒ (i). Assume condition (ii) holds, and let us show that every Cauchy net in X
is convergent. Again, as above, this condition is equivalent to the fact that every Cauchy
net in X has a convergent sub-net. Fix then a Cauchy net (xλ )λ∈Λ , and choose a uniformly
compatible metric d, so that, by Proposition 2, we know that the metric Cauchy condition
(mc) is satisfied. In particular, one has a sequence (σn )∞
n=1 ⊂ Λ, such that
|xλ − xµ |d ≤ 2−n , ∀ λ, µ σn .
(16)
Using directedness one can assume that (σn )∞
n=1 is increasing, i.e. σn+1 σn , ∀ n. It is
−k
obvious then that |xσm − xσn |d ≤ 2 , ∀ m, n ≥ k, so the sequence (xσn )∞
n=1 is Cauchy, thus
convergent to some x ∈ X . Taking limit in the preceding inequalities also yields:
|xσn − x|d ≤ 2−n , ∀ n ∈ N.
(17)
Consider now the directed set Λ × N, and the subset Γ = {(µ, k) ∈ Λ × N : λ σk } ⊂ Λ × N.
Remark that Γ is cofinal in Λ × N, which means that, for any (λ, n) ∈ Λ × N, there exists
(µ, k) ∈ Γ, such that µ λ and k ≥ n. (Simply take k = n, and µ ∈ Λ such that
13
µ λ, σk .) If we now consider the map φ : Γ 3 (µ, k) 7−→ µ ∈ Λ, then φ is directed map,
so the net (yγ )γ∈Γ given by yγ = xφ(γ) , is a sub-net of (xλ )λ∈Λ . Notice now that, if we set
γn = (σn , n) ∈ Γ, then for every γ = (µ, k) ∈ Γ with γ γn , using (16) and (17), we have
|xµ − x|d ≤ |xµ − xσn |d + |xσn − x|d ≤ 2 · 2−n . In other words, we have
|yγ − x|d ≤ 2−(n−1) , ∀ γ γn ,
which clearly forces yγ → x.
Comment. Condition (ii) from Proposition 3 is referred to as sequential completeness.
There are many examples of topological vector spaces which are sequentially complete, but
not complete. A general construction of such spaces will be outlined in section B ??.
Remark 4. The completion of a metrizable topological vector space X is constructed in
the “traditional” way (see P II). One fixes a uniformly compatible metric d and constructs
the metric completion (X 0 , d0 ), where X 0 consists of equivalence classes of Cauchy sequences
∞
in X . Here, two Cauchy sequences x = (xn )∞
n=1 and y = (yn )n=1 are declared equivalent, if
limn→∞ d(xn , yn ) = 0. If we denote, for each Cauchy sequence x = (xn )∞
n=1 , its equivalence
class (in X 0 ) by [x], the metric d0 on X 0 is defined by d0 ([x], [y]) = limn→∞ d(xn , yn ). One can
show with little difficulty that X 0 is a vector space5 , and the addition and scalar multiplication
are continuous.
C. Property (F)
As a result of the discussion from the previous sub-section, an important class of topological vector spaces is identified (by the Definition below). As sub-section D will show, these
spaces enjoy some remarkable properties.
Definitions. A topological space X is said to be an (F)-space, it it is complete and
metrizable. (The letter “F” was suggested by Stefan Hanach, in recognition of Maurice
Frechet. As it turns out, a special class of (F)-spaces consists of the so-called Frechet spaces,
which will be introduced in LCVS III. Banach also introduced “property (B)” named after
himself!)
Remarks 1’-3’. In connection with the three “generic” methods discussed in the previous sub-section (Remarks 1-3), we have the following:
1’. If X is an (F)-space, then: a linear subspace Y ⊂ X , equipped with the induced topology is and (F)-space, if and only if Y is closed in X . This follows from the trivial
observation that any Cauchy sequence in Y is convergent in X .
Q
2’. If (Xj )j∈J is a family of metrizable spaces, then: the product space j∈J Xj , equipped
with the product topology, is an (F)-space, if and only if all Xj ’s are (F)-spaces, and
the index set J is countable. (See P II.)
In particular, all finite dimensional topological vector spaces are (F)-spaces.
5
Simply note that the space cs(X ) of all Cauchy sequences is a linear subspace in
X 0 = cs(X )/N , where N is the linear subspace of sequences convergent to 0.
14
Q
N
X , and furthermore,
3’. If X is an (F)-space, and Y ⊂ X is a closed linear subspace, then the quotient space
X /Y, equipped with the quotient topology, is an (F)-space.
To prove this, we use Lemma 1 (and the notations used there) and the Summability
Test. Fix P
d a uniformly compatible metric on X , let (vn )∞
n=1 be a sequence in X /Y,
such that ∞
|v
−v
|
<
∞,
and
let
us
show
that
(v
)
is
convergent in X /Y. Start
n dˆ
n
n=1 n+1
with an arbitrary x1 ∈ X with π(x1 ) = v1 , and construct recursively a whole sequence
(xn )∞
n=1 in X as follows: assuming xn is chosen (such that π(xn ) = vn ), we choose
yn , zn ∈ X such that π(yn ) = vn , π(zn ) = vn+1 , and |zn − yn |d ≤ 2−n + |vn+1 − vn |dˆ,
and we set xn+1 = xn + zn − yn , so that we have:
|xn+1 − xn |d = |zn − yn |d < 2−n + |vn+1 − vn |dˆ,
as well as (by linearity of π):
π(xn+1 ) = π(xn ) + π(zn ) − π(yn ) = vn + vn+1 − vn = vn+1 .
−n
So now our sequence (xn )∞
n=1 satisfies: (a) π(xn ) = vn , and (b)]
P∞|xn+1 − xn |d ≤ 2 +
|vn+1 − vn |dˆ, for all n. On the one hand by (b) it follows that n=1 |xn+1 − xn |d < ∞,
so (xn ) is convergent. On the other hand by (a), combined with the continuity of π,
it follows that (vn ) is convergent too.
Exercise 19*. Suppose X is a metrizable topological vector space, and Y is a linear
subspace. Assume Y is an (F)-space, when equipped with the induced topology.
(i) Prove that Y is closed in X , so in particular the quotient space X /Y is metrizable as
well, when equipped with the quotient topology.
(ii) Prove that, if X /Y is an (F)-space, when equipped with the quotient topology, then
X is also an (F)-space.
(Hint for (ii): Fix a uniformly compatible metric d on X , and let dˆ be the associated metric
on X /Y defined in Lemma 1. As seen in the proof of Proposition 3, it suffices to show that
−n
if a sequence (xn )∞
, ∀ n ∈ N, the (xn ) is convergent. Let
n=1 ⊂ X satisfies |xn+1 − xn |d ≤ 2
−n
vn = π(xn ) ∈ X /Y, and note that |vn+1 − vn |dˆ ≤ 2 , ∀ n, so in particular (vn ) converges in
X /Y to some v, and moreover we have |vn − v|dˆ ≤ 2−n , ∀ n. Fix w ∈ X , such that π(w) = v,
and choose, for every n, some yn ∈ Y, such that |xn − w + yn |d ≤ |vn − v|dˆ + 2−n , so that we
have in fact the inequalities |xn − w + yn | ≤ 2 · 2−n , ∀ n. Prove that |yn+1 − yn |d ≤ 5 · 2−n ,
∀ n, so (yn ) is convergent to some y ∈ X . Conclude that xn → y + w.)
D. Four Principles of Functional Analysis
The four results discussed in this sub-section sit at the foundation of Functional Analysis.
All of them are applications of Baire’s Theorem, concerning complete metric spaces, so these
results hold only for (F)-spaces.
Theorem 3 (Equi-continuity Principle). Suppose X is an (F)-space, Y is an arbitrary
topological vector space, and Θ is a collection of linear continuous maps from X into Y,
which is pointwise bounded, in the sense that:
15
(pb) for every x ∈ X , the set Θx = {T (x) : T ∈ Θ} is bounded (in Y).
Then the collection Θ is equi-continuous6 , in the sense that:
(ec) for every neighborhood W of 0 in Y, there exists a neighborhood V of 0 in X , such that
T (V) ⊂ W, ∀ T ∈ Θ.
Proof. Fix W a neighborhood of 0 in Y. Using Proposition 2 and Exercise 5 from TVS I,
there exists a closed balanced neighborhood U of 0 in Y, such that U + U ⊂ W. Using
property (pb), for every x ∈ X , there exists ρx > 0, such that ρx T (x) ∈ U, ∀ T ∈ Θ. Since U
is balanced, we also have rT (x) ∈ U, ∀ T ∈ Θ, r ∈ [0, ρx ]. Therefore, we have the following
property:
(∗) For every x ∈ X , there exists a positive integer nx , such that
1
T (x)
nx
∈ U, ∀ T ∈ Θ.
Define now, for every n ∈ N, the set
An = {x ∈ X :
1
T (x)
n
∈ U, ∀ T ∈ Θ}.
Since U is closed, andS
all T ∈ Θ are continuous, it follows that An is closed. Furthermore,
by (∗) it follows that n∈N An = X . Using Baire’s Theorem (here it is essential that X is
an (F)-space), there exists n ∈ N, such that Int(An ) 6= ∅. Let then V0 = n1 Int(An ), so that
V0 is open, and T (V0 ) ⊂ U. If we define then the set V = V0 + (−1)V0 , then V is an open
neighborhood of 0, and satisfies:
T (V) ⊂ U + (−1)U ⊂ W.
(The last inclusion follows from the fact that U + U ⊂ W, combined with the fact that U is
balanced, which implies (−1)U = U.)
Comment. A particular case of Theorem 3, known as the Uniform Boundedness Principle, will be treated in B II.
Theorem 4 (Open Mapping Principle). Suppose X and Y are (F)-spaces. Then any
surjective linear continuous map T : X → Y is open, that is:
• whenever A is open in X , it follows that T (A) is open in Y.
Proof. It is pretty clear that the desired conclusion is equivalent to the fact that, for every
x ∈ X , the following condition is satisfied: whenever V is a neighborhood of x in X , it follows
that T (V) is a neighborhood of T (x) in Y. By linearity we only need to check this for x = 0,
that is:
(∗) If V is a neighborhood of 0 in X , then T (V) is a neighborhood of 0 in Y.
6
Condition (ec) implies the fact that: whenever xλ → x (in X ), it follows that T (xλ ) → T (x), uniformly
for all T ∈ Θ.
16
The proof of (∗) will be done in several steps.
Claim 1: If V is a neighborhood of 0 in X , then the closure T (V) has non-empty interior.
Indeed, since for every n ∈ N the dilation y 7−→ ny is a homeomorphism, it follows that
nB = nB, ∀ B ⊂ Y, so by the linearity of T we have:
∞
[
nT (V) =
n=1
∞
[
n=1
nT (V) ⊃
∞
[
T (nV) = T
n=1
∞
[
nV = T (X ).
n=1
S∞
(The last equality follows from the equality n=1 nV = X , which is a consequence of the fact
that limn→∞ ( n1 x) = 0, ∀ x ∈ X .) Since T is surjective (i.e. T (X ) = Y), the above inclusion
S
yields the equality ∞
n=1 nT (V) = Y. Using Baire’s Theorem (for which the hypothesis that
Y is an (F)-space is essential), there exists n such that nT (V) has non-empty interior, so
T (V) itself has non-empty interior.
Claim 2: If V is a neighborhood of 0 in X , then T (V) is a neighborhood of 0 in Y.
To prove Claim 2, we choose U a neighborhood of 0 in X , such that U + (−1)U ⊂ V, and
we use Claim 1, which tells us that T (U) has non-empty interior, which we denote by A. Of
course, A ∩ T (U) 6= ∅, so there exists x ∈ U, such that T x ∈ A. In particular (since A is
open), there exists a neighborhood W of 0 in Y, such that (T x) + W ⊂ A. Observe now
that, since V ⊃ (−1)U + U ⊃ −x + U, and translations are homeomorphisms, we obtain:
T (V) ⊃ T (−x + U) = −T x + T (U) = −T x + T (U) ⊃ −T x + (T x + W) = W,
so T (V) is indeed a neighborhood of 0 in Y.
Having proven Claim 2, we now proceed with the proof of (∗). Fix uniformly compatible
metric d on X . Since we only need to prove (∗) for V in a basic neighborhood system, we can
assume that V = {x ∈ X : |x|d ≤ r}, for some r > 0. Define Vn = {x ∈ X : |x|d < 2−n r},
for all n ∈ N, so that all Vn ’s are again neighborhoods of 0. By Claim 2, we see that, in
order to prove that T (V) is a neighborhood of 0 in Y, it suffices to prove the inclusion
T (V) ⊃ T (V1 ).
(18)
Fix now a uniformly compatible metric d0 on Y, as well as some point w ∈ T (V1 ). We
∞
construct recursively two sequences: (vn )∞
n=1 ⊂ X , (yn )n=1 ⊂ Y, with y1 = w, such that:
(i) yn ∈ T (Vn ),
(ii) vn ∈ Vn ,
(iii) yn+1 = yn − T vn ,
(iv) |yn |d0 ≤ 2−(n−1) |w|d0 ,
for all n ∈ N. This construction is carried on as follows. Assume yn ∈ T (Vn ) and vn−1 ∈ Vn−1
are given7 , and let us indicate how vn and yn+1 are constructed. Since yn ∈ T (Vn ), every
neighborhood of yn intersects T (Vn ). In particular, using Claim 2, the set
B = T (Vn+1 ) ∩ {y ∈ Y : |y|d0 , 0 ≤ 2−n |w|d0 }
7
When n = 1 we do not assume anything: we are simply given y1 = w.
17
is a neighborhood of 0, so yn + (−1)B is a neighborhood of yn , therefore the intersection
(yn + (−1)B) ∩ T (Vn ) is non-empty, hence it contains some point z. On the one hand,
since z ∈ T (Vn ), it follows that z = T vn , for some vn ∈ Vn . On the other hand, since
z ∈ yn + (−1)B, it follows that z = yn − yn+1 , for some yn+1 ∈ B, which means that
yn+1 ∈ T (Vn+1 ), so we have (iii), and also |yn+1 |d0 ≤ 2−n |w|d0 , so we also have (iv).
Having constructed our two sequences, let us remark first that, by (iii), we have
w − yn+1 = T v1 + · · · + T vn = T xn ,
(19)
where xn = v1 + · · · + vn .
Secondly, since xn+1 − xn = vn+1 , by (ii) we know that
|xn+1 − xn |d ≤ 2−(n+1) r, ∀ n ∈ N.
(20)
P∞
On the one hand, since clearly this yields n=1 |xn+1 − xn |d < ∞, by the Summability Test
it follows that (xn ) converges to some point x ∈ X . (The condition that X is an (F)-space
is essential here.) On the other hand, by (ii), the inequalities (20) also give
r
r
r
|xn |d ≤ |x1 |d + |x2 − x1 |d + · · · + |xn − xn−1 |d ≤ + 2 + · · · + n ≤ r, ∀ n ∈ N,
2 2
2
so taking the limit (we know that xn → x) we obtain |x|d ≤ r, which means that x belongs
to V. Going back to (19) it follows that yn = w − T xn−1 is convergent to w − T x. But by
condition (iv) we also know that yn → 0, therefore we must have the equality w = T x, which
proves that w indeed belongs to T (V).
Corollary 1 (Inverse Mapping Principle). If X and Y are (F)-spaces, and T : X → Y
is a continuous linear isomorphism, then its inverse T −1 : Y → X is also continuous.
Proof. The continuity of T −1 is equivalent to the fact that T is open, so everything follows
from Theorem 4.
Corollary 2 (Closed Graph Principle). Suppose X and Y are (F)-spaces. For a linear
map T : X → Y, the following are equivalent:
(i) T is continuous;
(ii) The graph of T , i.e. the set GT = {(x, y) ∈ X × Y : T x = y}, is closed in X × Y, in
the product topology.
Proof. The implication (i) ⇒ (ii) is trivial. (It holds for arbitrary topological Hausdorff
spaces X and Y and arbitrary continuous maps T : X → Y.)
(ii) ⇒ (i). Assume GT is closed in X ×Y. First of all, by linearity, GT is a linear subspace.
Secondly, since X × Y is an (F)-space (by Remark 2’), it follows that GT is also an (F)-space,
when equipped with the induced topology. Thirdly, since both the coordinate maps
PX : X × Y 3 (x, y) 7−→ x ∈ X ,
PY : X × Y 3 (x, y) 7−→ y ∈ Y
are continuous, so will be their restrictions L = PX G : GT → X and M = PY G : GT → Y.
T
T
The point is that L is a linear isomorphism, so by the Inverse Mapping Principle, the inverse
L−1 : X → GT is also continuous. Now we are done, since T can be written as a composition
T = M ◦ L−1 of continuous maps.
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