Download MAT 578 Functional Analysis

MAT 578 Functional Analysis
John Quigg
Fall 2008
revised September 16, 2008
Topological vector spaces
All vector spaces will have scalar field F = R or C.
Definition 1. A topological vector space is a vector space X which is also a topological space
such that:
(1) addition is a continuous map X × X → X;
(2) scalar multiplication is a continuous map F × X → X;
(3) X is Hausdorff.
If X only satisfies (1)–(2) we call X a pre-topological vector space.
We are primarily interested in topological vector spaces, but for the beginning development
of the theory we sometimes ignore the Hausdorff property and focus on pre-topological vector
spaces. The term “pre-topological vector space” would seem to imply that such a space could
be somehow “extended” to a topological vector space; although this actually is true (more
precisely, the quotient by the closure of {0} would be a topological vector space), we do not
have any need for this — I just couldn’t think of a better term than pre-topological vector
space. And this term will only be temporary, to get some foundational results out of the
way efficiently.
Definition 2.
(1) Ba := {t ∈ F : |t| < a}
(2) B̄a := {t ∈ F : |t| ≤ a}
Definition 3. A subset A of a vector space X is:
(1) balanced if B̄1 A ⊂SA;
(2) absorbing if X = t>0 tA.
Some authors use a stronger condition in the definition of absorbing, namely that for all
x ∈ X there exists s > 0 such that x ∈ tA for all t ≥ s. However, when we combine
absorbing with balanced (which we will most of the time), our weaker condition will in fact
be equivalent to the stronger one.
Observation 4. Every balanced set contains 0.
Lemma 5. Every neighborhood of 0 in a pre-topological vector space is absorbing.
Proof. Let U be a neighborhood of 0 in a pre-topological vector space X, and let x ∈ X.
By continuity of scalar multiplication, there exists a > 0 such that ax ∈ U , and then
x ∈ a−1 U .
Definition 6. A local base for a topological vector space is a neighborhood base at 0.
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Lemma 7. Every pretopological vector space has a balanced local base.
Proof. Let U be a neighborhood of 0 in a pre-topological vector space X. It suffices to show
that U contains a balanced neighborhood of 0. By continuity of scalar multiplication, there
exist a > 0 and a neighborhood V of 0 such that Ba V ⊂ U . Then Ba V is a neighborhood of
0, because
[
Ba V =
tV
t∈Ba
and each tV for t 6= 0 is a neighborhood of 0 since multiplication by a nonzero scalar is a
homeomorphism on X. Also Ba V is balanced, because B̄1 Ba ⊂ Ba , so that
B̄1 Ba V ⊂ Ba V.
Definition 8. A local base of a topological vector space is balanced if it consists of balanced
sets.
Lemma 9. If X is a pre-topological vector space, and B is a balanced local base. Then:
(1)
(2)
(3)
(4)
for all U, V ∈ B there exists W ∈ B such that W ⊂ U ∩ V ;
for all U ∈ B there exists V ∈ B such that V + V ⊂ U ;
U is balanced for all U in B;
U is absorbing for all U in B.
Conversely, if X is a vector space and B is a family of subsets of X satisfying (1)–(4), then
B is a local base for a unique topology making X into a pre-topological vector space.
Proof. First assume that X is a pre-topological vector space and B is a balanced local base.
Property (3) holds by hypothesis (and is only included for the purposes of the converse),
and (1) is a property of neighborhoods in any topological space. (2) follows from continuity
of addition at 0. For (4), let U ∈ B and x ∈ X. By continuity of scalar multiplication, there
exists a > 0 such that ax ∈ U , and then x ∈ a−1 U .
Conversely, suppose X is a vector space and B is a family of subsets of X satisfying (1)–(4).
For each x ∈ X put
Bx = {x + U : U ∈ B}.
Then for each x ∈ X we have:
• x ∈ U for all U ∈ Bx (because U contains 0 since it is balanced);
• for all U, V ∈ Bx there exists W ∈ Bx such that W ⊂ U ∩ V .
From general topology we know that there is a unique topology T on X such that Bx is a
neighborhood base at x for each x ∈ X. Once we show that (X, T ) is a pre-topological vector
space, the uniqueness assertion of T will be clear, because if T 0 is any topology making X
into a pre-topological vector space and having B as a local base then the families Bx will
be neighborhood bases at each x ∈ X. Thus it remains to show that addition and scalar
multiplication are continuous. Addition is easy: let x, y ∈ X and U ∈ B. It suffices to show
that there exists V ∈ B such that
(x + V ) + (y + V ) ⊂ x + y + U.
But this follows immediately from property (2).
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Scalar multiplication is a little harder. Basically, we an need abstract version of the standard
triangle inequality argument showing that scalar multiplication is continuous in any normed
space. Let t ∈ F, x ∈ X, and U ∈ B. It suffices to show that there exist a > 0 and V ∈ B
such that if |s − t| < a and y − x ∈ V then sy − tx ∈ U . First choose W ∈ B such that
W + W ⊂ U.
Claim 1. There exists a > 0 such that Ba x ⊂ W . To see this, first note that since W is
absorbing there exists t > 0 such that x ∈ tW , so if we let a = t−1 then a > 0 and ax ∈ W .
Then since W is balanced we have
Ba x = B1 ax ⊂ B1 W ⊂ W.
Claim 2. Put b = |t| + a. Then there exists V ∈ B such that Bb V ⊂ W . This one is a little
tricky: first choose n ∈ N such that b ≤ 2n . Using property (2) and induction we can find
V ∈ B such that
2n
2n
X
}|
{
z
V := V + · · · + V ⊂ W,
1
and then since V is balanced we have
n
n
Bb V = bB1 V ⊂ bV ⊂ 2 V ⊂
2
X
V ⊂ W.
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Now suppose |s − t| < a and y − x ∈ V . Then
sy − tx = sy − sx + sx − tx = s(y − x) + (s − t)x.
Since |s| < |t| + a we have s ∈ Bb , so
s(y − x) ∈ Bb V ⊂ W,
and since s − t ∈ Ba we have (s − t)x ∈ W . Therefore
sy − tx ⊂ W + W ⊂ U,
as desired.
Corollary 10. Let X be a vector space, and let S be a family of balanced absorbing sets
in X such that for all U ∈ S there exists V ∈ S such that V + V ⊂ U . Then the family
of all finite intersections of sets in S is a local base for a unique topology making X into a
pre-topological vector space.
Proof. Let B be the family of all finite intersections of sets in S. Then it is easy to see that
B satisfies properties (1)–(4) of Lemma 9 (property (1) holds by construction, and the other
properties continue to hold for B because they hold for S).
Now we put the Hausdorff condition back:
Lemma 11. Let B be a balanced local base for a pre-topological vector space X, and assume
that for all x ∈ X \ {0} there exists U ∈ B such that x ∈
/ U . Then X is a topological vector
space.
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Proof. Recall that, by definition, addition and scalar multiplication are continuous on X.
We only need to show that X is Hausdorff. Let x, y ∈ X, with x 6= y. It suffices to find
U ∈ B such that
(1)
(x + U ) ∩ (y + U ) = ∅.
First put z = x − y. Then z 6= 0, so we can choose V ∈ B such that z ∈
/ V . Next choose
U ∈ B such that U + U ⊂ V . We will show that this U satisfies (1). Suppose not, so that
there exist u, v ∈ U such that
x + u = y + v.
Then since U is balanced we have
z = x − y = v − u ∈ U − U = U + U ⊂ V,
which is a contradiction.
Remark 12. It is a corollary of the above proof that if a pre-topological vector space is T1 ,
that is, points are closed, then it is Hausdorff.
Proposition 13. A linear map between topological vector spaces is continuous if it is continuous at 0.
Proof. Let X and Y be topological vector spaces, and let T : X → Y be linear. Assume
that T is continuous at 0, and let {xi } be a net converging to x in X. Then xi − x → 0, so
T (xi ) − T (x) = T (xi − x) → 0.
hence T (xi ) → T (x).
Corollary 14. Let f be a continuous linear functional on a topological vector space X. Then
f is continuous if Ref < 1 on some neighborhood U of 0.
Proof. Let ε > 0. Then Ref < ε on the neighborhood εU of 0. Choose a balanced neighborhood V of 0 such that V ⊂ U . Then A := f (V ) is a balanced subset of F such that
Re(t) < ε for all t ∈ A. Thus we must have |t| < ε for all t ∈ A, so for all x ∈ V we have
|f (x)| < ε. Therefore f is continuous.
Definition 15. The dual space of a topological vector space X, denoted by X ∗ , is the vector
space of all continuous linear functionals on X.
Lemma 16. Every nonzero continuous linear functional on a topological vector space is an
open map.
Proof. Let X be a topological vector space, and let f ∈ X ∗ with f 6= 0. Then f is surjective.
Let U ⊂ X be open, and let x ∈ U . Choose a balanced neighborhood V of 0 such that
x + V ⊂ U . Then
f (x) + f (V ) = f (x + V ) ⊂ f (U ).
Since V is balanced, so is f (V ). Since V is a neighborhood of 0, it is absorbing, hence so is
f (V ) since f is surjective. Since f (V ) is balanced and absorbing, it is a neighborhood of 0
in F. Thus f (U ) is open.
Definition 17. Suppose X is a vector space and {Yi }i∈I is a family of vector spaces, and
for each i ∈ I we have a linear map Ti : X → Yi . Then the set {Ti : i ∈ I} is separating if
for all x ∈ X \ {0} there exists i ∈ I such that Ti (x) 6= 0.
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Lemma 18. Suppose X is a vector space, {Yi }i∈I is a family of topological vector spaces,
and for each i ∈ I we have a linear map Ti : X → Yi . Assume that the set {Ti : i ∈ I}
is separating, and give X the weakest topology making every Ti continuous. Then X is a
topological vector space.
Proof. Since the Ti ’s are separating, X is Hausdorff. We only need to check continuity of
addition and scalar multiplication. Let {xj } and {yj } be two nets in X indexed by the same
directed set J, and assume that xj → x and yj → y. Then for every i ∈ I we have
lim Ti (xj ) = Ti (x) and
j
lim Ti (yj ) = Ti (y),
j
so
lim Ti (xj + yj ) = lim Ti (xj ) + Ti (yj )
j
j
= T (x) + T (y)
because Yi is a topological vector space
= T (x + y)
Thus xj + yj → x + y by definition of the topology. Similarly, if also cj → c in F, then for
every i we have
lim Ti (cj xj ) = lim cj Ti (xj )
j
j
= cTi (x)
because Yi is a topological vector space
= Ti (cx),
and consequently cj xj → cx.
Corollary 19. Let {Xi }i∈I be a family of topological vector spaces, and give X :=
the product topology. Then X is a topological vector space.
Q
i∈I
Xi
Proof. This follows from Lemma 18 because the product topology is the weakest one making
the coordinate projections πi : X → Xi continuous.