Download Topological Vector Spaces I: Basic Theory

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Topological Vector Spaces I: Basic Theory
Notes from the Functional Analysis Course (Fall 07 - Spring 08)
Convention. Throughout this note K will be one of the fields R or C. All vector spaces
mentioned here are over K.
Definitions. Let X be a vector space. A linear topology on X is a topology T such that
the maps
X × X 3 (x, y) 7−→ x + y ∈ X ,
K × X 3 (α, x) 7−→ αx ∈ X ,
(1)
(2)
are continuous. (For the map (1) we use the product topology T × T. For the map (2) we
use the product topology TK × T, where TK is the standard topology on K.
A topological vector space is a pair (X , T) consisting of a vector space X and a Hausdorff
linear topology1 T on X .
Example 1. The field K, viewed as a vector space over itself, becomes a topological
vector space, when equipped with the standard topology TK .
Exercise 1. Let X be a vector space. Prove that the trivial topology T = {∅, X} is
linear, but the discrete topology T = P(X ) is not.
Remark 1. In terms of net convergence, the continuity requirements for a linear topology
on X read:
(i) Whenever (xλ ) and (yλ ) are nets in X , such that xλ → x and yλ → y, it follows that
(xλ + yλ ) → (x + y).
(ii) Whenever (αλ ) and (xλ ) are nets in K and X , respectively, such that αλ → α (in K))
and xλ → x (in X )), it follows that (αλ xλ ) → (αx).
Exercise 2. Show that a linear topology on X is Hausdorff, if and only if the singleton
set {0} is closed.
Example 2. Let I be an arbitrary non-empty set. The product space KI (defined as
the space of all functions I → K) is obviously a vector space (with pointwise addition and
scalar multiplication). The product topology turns KI into a topological vector space. This
can be easily verified using Remark 1.
Remark 2. If X is a vector space, then the following maps are continuous with respect
to any linear topology on X :
(i) The translations Ty : X → X , y ∈ Y, defined by Ty x = x + y.
(ii) The dilations Dα : X → X , α ∈ K, defined by Dα x = αx.
1
Many textbooks assume linear topologies are already Hausdorff.
1
The translations are in fact homeomorphisms. So are the non-zero dilations Dα , α 6= 0.
Notations. Given a vector space X , a subset A ⊂ X , and a vector x ∈ X , we denote
the translation Tx (A) simply by A + x (or x + A), that is,
A + x = x + A = {a + x : a ∈ A}.
Likewise, for an α ∈ K we denote the dilation Dα (A) simply by αA, that is,
αA = {αa : a ∈ A}.
Given another subset B ⊂ X , we define
A + B = {a + b : a ∈ A, b ∈ B} =
[
(a + B) =
a∈A
[
(A + b).
b∈B
Warning! In general we only have the inclusion 2A ⊂ A + A.
Proposition 1. Let T be a linear topology on the vector space X .
(i) For every T-neighborhood V of 0, there exists a T-neighborhood W of 0, such that
W + W ⊂ V.
(ii) For every T-neighborhood V of 0, and any compact set C ⊂ K, there exists a Tneighborhood W of 0, such that γW ⊂ V, ∀ γ ∈ C.
Proof. (i) Let A : X × X → X denote the addition map (1). Since A is continuous at
(0, 0) ∈ X × X , the pre-image A−1 (V) is a neighborhood of (0, 0) in the product topology.
In particular, there exists T-neighborhoods W1 , W2 of 0, such that W1 × W2 ⊂ A−1 (V), so if
we take W = W1 ∩ W2 , then W is still a T-neighborhood of 0, satisfying W × W ⊂ A−1 (V),
which is precisely the desired inclusion W + W ⊂ V.
(ii) Let M : K × X → X denote the multiplication map (2). Since M is continuous at
(0, 0) ∈ K × X , the pre-image M −1 (V) is a neighborhood of (0, 0) in the product topology.
In particular, there exists a neighborhood M of 0 in K and a T-neighborhood W0 of 0 in X ,
such that M × ×W0 ⊂ M −1 (V). Let then ρ > 0 be such that M contains the closed “disk”
B ρ (0) = {α ∈ K : |α ≤ ρ}, so that we still have the inclusion B ρ (0) × W0 ⊂ A−1 (V) i.e.
α ∈ K & |α| ≤ ρ ⇒ αW0 ⊂ V.
(3)
Since C ⊂ K is compact, there is some R > 0, such that
|γ| ≤ R, ∀ γ ∈ C.
(4)
Let us then define W = (ρ/R)W0 . First of all, since W is a non-zero dilation of W0 , it is a
T-neighborhood of 0. Secondly, if we start with some γ ∈ C and some w ∈ W, written as
w = (ρ/R)w0 with w0 ∈ W0 , then
γw = (ρα/R)w0 .
By (4) we know that |ρα/R| ≤ ρ, so by (3) we get γw ∈ V.
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Definitions. Let X be a vector space.
(i) A subset A ⊂ X is said to be absorbing, if for every x ∈ X , there exists some scalar
ρ > 0, such that ρx ∈ A.
(i) A subset A ⊂ X is said to be balanced, if for every α ∈ K with |α| ≤ 1, one has the
inclusion αA ⊂ A.
Remark 3. Given T a linear topology of a vector space X , all T-neighborhoods of 0 are
absorbing. Indeed, if we start with some x ∈ X , the sequence xn = n1 x clearly converges to
0, so every T-neighborhood of 0 will contain (many) terms xn .
Notation. Given a vector space X , and some non-empty subset A ⊂ X , the set
[
bal A =
αA
α∈K
|α|≤1
is clearly balanced. Furthermore, if B ⊂ X is a balanced subset that contains A, then
B ⊃ bal A. In other words, bal A is the smallest balanced subset in X , that contains A, so
we will call it the balanced hull of A.
Remark 4. If X is equipped with a linear topology, and if A is an open set that contains
0, then bal A is also open. Indeed, since q ∈ A, we also have the equality
[
bal A =
αA,
α∈K
0<|α|≤1
which is now a union of (non-zero dilations of) open sets.
Proposition 2. Let X be a vector space and let T be a linear topology on X .
A. If V is a basic system of T-neighborhoods of 0, then:
(i) For every V ∈ V, there exists W ∈ V, such that W + W ⊂ V.
(ii) For every V ∈ V and every compact set C ⊂ K, there exists W ∈ V, such that
γW ⊂ V, ∀ γ ∈ C.
(iii) For every x ∈ X , the collection Vx = {V + x : V ∈ V} is a basic system of Tneighborhoods for x.
T
(iv) The topology T is Hausdorff, if and only if V∈V V = {0}.
B. There exists a basic system of T-neighborhoods of 0, consisting of T-open balanced sets.
Proof. A. Statements (i) and (ii) follow immediately from Proposition 1. Statement (iii) is
clear, since translations are homeomorphisms. T
(iv). Denote for simplicity the intersection V∈V V by J , so clearly 0 ∈ J . Assume first
T is Hausdorff. In particular, for each x ∈ X r {0}, the set X r {x} is an open neighborhood
of 0, so there exists some V x ∈ V with V x ⊂ X r {x}. We then clearly have the inclusion
\
\
J ⊂
Vx ⊂
(X r {x}) = {0},
x6=0
x6=0
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so J = {0}. Conversely, assume J = {0}, and let us show that T is Hausdorff. Start with
two points x, y ∈ X with x 6= y, so that x − y 6= 0, and let us indicate how to construct two
disjoint neighborhoods,
one for x and one for y. Using translations, we can assume y = 0.
T
Since 0 6= x 6∈ V∈V V, there exists some V ∈ V, such that x 6∈ V. Using (i), there is some
W ∈ V, such that W + W ⊂ V, so we still have x 6∈ W + W. This clearly forces2
x + ((−1)V) ∩ V = ∅.
(5)
Since V is a neighborhood of 0, so is (−1)V (non-zero dilation), thus by (iii) the left-hand
side of (5) is indeed a neighborhood of x.
B. Let us take the V to be the collection of all open balanced sets that contain 0. All we
have to prove is the following statement: for every neightborhood V of 0, there exists some
B ∈ V, such that B ⊂ V. Using (ii) there exists some open set W 3 0, such that
γW ⊂ V, ∀ γ ∈ K s.t. |γ| ≤ 1.
(6)
In particular, the balanced hull B = bal W is contained in V. By Remark 4, B is also open,
so B ∈ V.
The following definition is a generalization of a well known notion in Calculus.
Definition. Assume T is a linear topology on a vector space X . A subset B ⊂ X is said
to be T-bounded, if it satisfies the following condition:
(b) for every T-neighborhood V of 0, there exists ρ > 0, such that B ⊂ ρV.
Proposition 3 (“Zero · Bounded” Rule). Suppose T is a linear topology on a vector
space X . If the net (αλ )λ∈Λ ⊂ K converges to 0, and the net (xλ )λ∈Λ ⊂ X is T-bounded, then
(αλ xλ )λ∈Λ is T-convergent to 0.
Proof. Start with some T-neighborhood V of 0. We wish to construct an index λV ∈ Λ, such
that
αλ xλ ∈ V, ∀ λ λV .
(7)
Using Proposition 2.B, we can assume that V is balanced (otherwise we replace it with a
balanced open set V 0 ⊂ V). Using the boundedness condition (b) we find ρ > 0, such that
xλ ∈ ρV, ∀ λ ∈ Λ.
(8)
Using the condition αλ → 0, we then choose λV ∈ Λ, so that
1
|αλ | ≤ , ∀ λ λV .
ρ
(9)
To check (7), start with some λ λV and apply (8) to write xλ = ρv, for some v ∈ V. Now
we have
αλ xλ = (αλ ρ)v ∈ (αλ ρ)V,
with |αλ ρ| ≤ 1, so using the fact that V is balanced, it follows that αλ xλ ∈ V.
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We deliberately avoid the shotrcut notation −V in place of (−1)V.
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Before we continue, the reader is urged to solve the following:
Exercises 3-22. Assume (X , T) is a topological vector space. (In some of the exercises
below, T need not be Hausdorff. Identify them!)
3. Prove that if A ⊂ X is open, then A + S is open, for any subset S ⊂ X .
4. Prove that if V ⊂ X is a neighborhood of 0, then for any subset S ⊂ X , the closure S
is contained in S + V.
5. Prove that the collection of all balanced closed neighborhoods of 0 constitutes a basic
neighborhood system for 0. In other words, show that for any neighborhood V of 0,
there exists a balanced closed neighborhood of W of 0, such that W ⊂ V.
6. Consider the set C = {0} – the closure of the singleton {0}. Prove that
(i) C equals the intersection of all neighborhoods of 0;
(ii) C is a closed linear subspace;
(iii) C is compact.
7. Prove that if A ⊂ X is closed and C ⊂ X is compact, then A + C is closed.
S
8. Prove that if A ⊂ X and C ⊂ K are compact, then so is γ∈C γA. In particular, the
balanced hull bal A is compact.
S
9. Prove that, A ⊂ X is closed and C ⊂ K r {0} is compact, then γ∈C γA is closed.
Give a counterexample that shows that the condition C 63 0 is essential
10. Prove that if A, B ⊂ X are compact, then so is A + B.
11. Prove that if A ⊂ X is closed, C ⊂ X is compact, and A ∩ C = ∅, then there exists a
neighborhood V of 0, such that: (A + V) ∩ (C + V) = ∅.
12. Prove that, if in Proposition 3, the net (xλ ) is only assumed to be eventually bounded
(i.e. there exists λ0 , such that (xλ )λλ0 is bounded), then again (αλ xλ ) converges to 0.
13. Prove that if a net (αλ )λ ⊂ K is eventually bounded (i.e. there exists λ0 , such that
supλλ0 |αλ | < ∞) and the net (xλ )λ ⊂ X converges to 0, then (αλ xλ )λ converges to 0.
14. Prove that, if B ⊂ X is not bounded, there exists a net (xλ ) in B, and a net (αλ ) ∈ K,
such that αλ → 0, but (αλ xλ ) does not converge to 0.
15. Prove that if B ⊂ X is bounded, then it satisfies the following stronger condition:
(b’) for every T-neighborhood V of 0, there exists ρ > 0, such that B ⊂ rV, ∀ r ≥ ρ.
16. Prove that all compact subsets in X are bounded.
17. Prove that every convergent sequence in X is bounded.
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18. Give an example of a topological vector space in which there exists nets convergent to
0, that are not eventually bounded, as defined in Exercise 3. (Hint: Try KI , with I
infinite. For every F ∈ Pfin (I) let xF = (xiF )i∈I ∈ KI be defined by xiF = 0, if i ∈ I,
and xiF = card F , otherwise. Analyze the net (xF )F ∈Pfin (I) .)
19*. Prove that, if B ⊂ X is bounded, then its closure B is also bounded. (Hint: Use
Exercise 4.)
20. Prove that if A, B ⊂ X are bounded, then so is A + B.
21. Prove that if A ⊂ X and C ⊂ K are bounded, then so is
balanced hull bal A is bounded.
S
γ∈C
γA. In particular, the
22. Assume Y is a topological vector space and T : X → Y is linear. Prove that the
following are equivalent:
(i) T is continuous;
(ii) T is continuous at 0.
23. Suppose Y is a topological vector space and T : X → Y is linear and continuous. Show
that, if A ⊂ X is bounded, then T (A) is bounded in Y.
The next two results (Theorem 1 and its Corollary) provide the main method for constructing linear topologies.
Theorem 1. Suppose X is a K-vector space and V is a filter3 in X , with the following
properties:
(i) V 3 0, ∀ V ∈ V;
(ii) every V ∈ V is absorbing;
(iii) for every V ∈ V, there exists W ∈ V, such that W + W ⊂ V;
(iv) for every V ∈ V, there exist W ∈ V and r > 0, such that ρW ⊂ V, for all ρ ∈ K with
|ρ| ≤ r.
Then there exists a unique linear topology T on X , so that
T V is a basic system of Tneighborhoods of 0. Moreover, T is Hausdorff, if and only if V∈V V = {0}.
Proof. Declare a set A ⊂ X open, if it has the following property:
(∗) For every a ∈ A, there exists V ∈ V, such that V + a ⊂ A,
and let T be the collection of all open sets. In order to show that T is a topology, we need
to check the following axioms:
(t0 ) T contains the empty set ∅ and the total set X ;
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This means that all sets in V are non-empty, and for any V1 , V2 ∈ V, there exists V ∈ V, such that
V ⊂ V1 ∩ V2 .
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(t1 ) for any collection (Ai )i∈I ⊂ T, the union
S
i∈I
Ai also belongs to T;
(t2 ) if A1 and A2 belong to T, then the intersection A1 ∩ A2 also belongs to T.
The axioms (t0 ) and (t1 ) are trivially verified. To check (t2 ) one uses the filter property.
One useful feature of the topology T is translation invariance, i.e.
(**) for any given x ∈ X , a subset A ⊂ X is open, if and only if A + x is open.
(This is pretty obvious from the definition of T.)
The next important step in the proof is contained in the following statement:
Claim 1. For every M ∈ V, there exists an open set A, such that M ⊃ A 3 0.
To prove Claim 1, we start off by defining
A = x ∈ X : there exists V ∈ V, such that V + x ⊂ M .
Since by condition (i) we know that V + x 3 x, ∀ x ∈ X , V ∈ V, it is clear that A ⊂ M.
Since M + 0 = M ∈ V, it follows that A contains 0. To prove that A is open, start with
some point a ∈ A and check condition (∗). On the one hand, by construction, there exists
V0 ∈ V, such that
V0 + a ⊂ M.
(10)
On the other hand, by (iii) there exists V ∈ V such that V + V ⊂ V0 , so by (10) we also get
V + V + a ⊂ M.
(11)
In particular, for every x ∈ V + a we have the inclusion V + x ⊂ M, which means that
x ∈ A. Of course, this means that V + a ⊂ A, and we are done.
Claim 2. For every x ∈ X , the collection V(x) = (V + x)V∈V constitutes a basic system of
T-neighborhoods of x.
Indeed, by translation invariance (∗∗), it suffices to consider only the case x = 0, when
V(0) = V. On the one hand, by Claim 1, it follows that every V ∈ V is indeed a Tneighborhood of 0. On the other hand, by the definition (∗), for every open set A ⊂ X the
condition A 3 0, implies the existence of some V ∈ V with 0 3 V ⊂ A, so V is indeed a
basic system of T-neighborhoods of 0.
We now proceed with the proof of the continuity of the addition map (1) and the multiplication map (2). To prove the continuity of addition, we start with two nets (xλ )λ∈Λ and
(yλ )λ∈Λ in X , xλ → x, and yλ → y, and we prove that (xλ + yλ ) → (x + y). By Claim 2, all
we need to prove is that:
• for every V ∈ V, there exists λV ∈ Λ, such that:
xλ + yλ ∈ V + (x + y), ∀ λ λV .
(12)
To get this we first choose, using (iii), some W ∈ V, such that W + W ⊂ V, and then using
the hypotheses xλ → x and yλ → y, we choose µ, ν ∈ Λ, such that
xλ ∈ W + x, ∀ λ µ,
yλ ∈ W + y, ∀ λ ν.
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Now we are done by choosing λV ∈ Λ to be any element such that λV µ, ν, because the
condition λ λV will force
xλ + yλ ∈ (W + x) + (W + y) = (W + W) + (x + y) ⊂ V + (x + y).
To prove the continuity of multiplication, we start with two nets (xλ )λ∈Λ ⊂ X and (αλ )λ∈Λ ⊂
K, xλ → x, and αλ → α, and we prove that (αλ xλ ) → (αx). Using the continuity of addition,
it suffices to prove that
(a) αλ (xλ − x) → 0 and
(b) (αλ − α)x → 0.
Start with some V ∈ V, and let us indicate how to produce indices λV , νV ∈ Λ, such that
αλ (xλ − x) ∈ V, ∀ λ λV ,
(αλ − α)x ∈ V ∀ λ νV .
(13)
(14)
For both conditions we start by fixing r > 0 and W ∈ V as in condition (iv).
For (a) we notice that, since αλ → α, there is some index λ0 , such that the quantity
M = supλλ0 |αλ | is finite. Let n ∈ N be such that nr ≥ M , and let Z ∈ V be such that
nZ ⊂ |Z + Z +
{z· · · + Z} ⊂ W.
(15)
n times
(The first inclusion is always true, for any subset Z.) By the choice of λ0 (which implies
|αλ /n| ≤ r, ∀ λ λ0 ), by the construction of r and n, and by (15), we now have
αλ Z = (αλ /n)(nZ) ⊂ (αλ /n)W ⊂ V, ∀ λ λ0 .
(16)
Since xλ → x, there exists λ1 ∈ Λ, such that
xλ − x ∈ Z, ∀ λ λ1 ,
so using (16) we see that any λV λ1 , λ2 will satisfy (a)
For (b) we use condition (ii) (for the first time!) to choose some t > 0, such that tx ∈ W
and we choose νV such that
|αλ − α| ≤ rt, ∀ λ νV .
(17)
To check (b) we simply notice that, if λ νV , then |t−1 (αλ − α)| ≤ r, so by condition (iv)
we have t−1 (αλ − α)W ⊂ V, ∀ λ νV . Of course, since tx ∈ W, this will imply
(αλ − α)x = t−1 (αλ − α)(tx) ∈ t−1 (αλ − α)W ⊂ V, ∀ λ νV .
The uniqueness part, as well as the characterization of the Hausdorff property, are quite
obvious.
Corollary. Suppose X is a K-vector space and V is a filter in X , with the following
properties:
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(i) V 3 0, ∀ V ∈ V;
(ii) every V ∈ V is absorbing;
(iii) for every V ∈ V, there exists W ∈ V, such that W + W ⊂ V;
(iv) every V ∈ V is balanced.
Then there exists a unique linear topology T on X , so that
T V is a basic system of Tneighborhoods of 0. Moreover, T is Hausdorff, if and only if V∈V V = {0}.
Proof. Simply notice that condition (iv) in the statement implies condition (iv) from Theorem 1, by taking W = V and r = 1.
Example 3. Fix some positive real number p, and an infinite set S. Consider the space
X
`pK (S) = x = (xs )s∈S :
|xs |p < ∞ .
s∈S
(When K = C, the subscript is omitted from the notation. When S = N the set is also
omitted from the notation.) It is not too hard to show, using for instance the inequalities
(a + b)p ≤ 2p (ap + bp ), ∀ a, b ≥ 0,
(18)
that `pK (S) is a K-vector space. Define now, for every r > 0, the set
X
Vr = x = (xs )s∈S ∈ `pK (S) :
|xs |p < r .
s∈S
It is not hard to check that the collection V = {Vr }r>0 satisfies conditions (i)-(iv) from the
above Corollary. The only non-trivial condition is (iii). Using the inequality (18), however,
we immediately get the inclusions
Vr/2p+1 + Vr/2p+1 ⊂ Vr ,
and (iii) is now clear. The unique Hausdorff linear topology T on `pK (S), given by the
Corollary, is referred to as the standard metric topology. The reason for this terminology is
the fact that this topology can be in fact defined by a metric Dp , given by
 P
p
, if 0 < p < 1

s∈S |xs − ys |
Dp (x, y) =
 P
p 1/p
|x
−
y
|
, if 1 ≤ p < ∞
s
s
s∈S
(The case 0 < p < 1 relies on the inequality: (a + b)p ≤ ap + bp , ∀ a, b ≥ 0. The case p ≥ 1
will be discussed in anothe section. It relies on the Minkowski Inequality.)
Example 4. Fix some positive real number p, and a measure space (X, A, µ). (This
means that A is a σ-algebra on X and µ is a measure on A.) Consider the space
Z
p
LK (X, A, µ) = f : X → K : f A-measurable, with
|f |p dµ < ∞ .
X
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(When K = C, the subscript is omitted from the notation.) Exactly as above, one can show
that that LpK (X, A, µ) is a K-vector space. Define now, for every r > 0, the set
Z
p
Vr = f ∈ LK (X, A, µ) :
|f |p dµ < r .
X
Exactly as above, the collection V = {Vr }r>0 satisfies conditions (i)-(iv) from the above
Corollary. The unique (non-Hausdorff!) linear topology T on LpK (X, A, µ), given by the
Corollary, is referred to as the standard semi-metric topology. As before, the Dp , given by
 R
|f − g|p dµ
, if 0 < p < 1

X
Dp (f, g) =
1/p
 R
|f − g|p dµ
, if 1 ≤ p < ∞
X
is a semi-metric, in the sense that
• Dp (f, g) = Dp (g, f ) ≥ 0,
• Dp (f, h) ≤ Dp (f, g) + Dp (g, h),
for all f, g, h ∈ LpK (X, A, µ). (The implication Dp (f, g) = 0 ⇒ f = g fails! The correct
implication is Dp (f, g) = 0 ⇒ f = g, µ-a.e.)
Comment. The `p - and Lp -spaces for 0 < p < 1 are quite pathological, as will be shown
in another section. The case p ≥ 1 is much nicer, as will be explained later.
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