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MAT 578 Functional Analysis
John Quigg
Fall 2008
revised September 16, 2008
Locally convex spaces
This section establishes the fundamental properties of locally convex spaces. Acknowledgment: although I wrote these notes to reflect how I think of these things, some of the
development is essentially the same as in Jack Spielberg’s notes, and in fact his notes were
a valuable source for me.
Reminder: all vector spaces will have scalar field F = R or C.
Definition 1. Let X be a vector space and x, y ∈ X. The line segment joining x and y is
the set
{(1 − t)x + ty : 0 ≤ t ≤ 1}.
Definition 2. A subset A of a vector space X is convex if it contains the line segment
joining any two of its elements
Observation 3. The intersection of any family of convex sets is convex.
Definition 4. A local base of a topological vector space is convex if it consists of convex
sets.
By a reasonable use of the English language, a balanced convex local base will mean a local
base which is both balanced and convex.
Definition 5. A locally convex space is a topological vector space which has a convex local
base.
Definition 6. Let X be a vector space.
(1) A convex combination of x1 , . . . , xn ∈ X is any vector of the form
n
n
X
X
ti = 1;
ti xi
with ti ≥ 0,
1
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(2) The convex hull of a subset A of X, denoted by co A, is the intersection of all convex
sets containing A.
Lemma 7. If A ⊂ X, then:
(1) co A is the smallest convex set containing A;
(2) co A coincides with the set of all convex combinations of elements of A.
Proof. For (1), just note that co A is convex since it is an intersection of convex sets.
For (2), let B denote the set of all convex combinations of elements of A. Then co A ⊃ B,
since (by an easy induction argument) every convex set is closed under convex combinations.
On the other hand, a straightforward computation shows that a convex combination of
convex combinations of elements of A is again a convex combination of elements of A, so B
is convex. Since A ⊂ B, we have co A ⊂ B. Thus we must have co A = B, as desired.
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Lemma 8. Every locally convex space has a balanced convex local base.
Proof. Let B be the family of all convex neighborhoods of 0 in a locally convex space X,
so that B is a convex local base. It suffices to show that for all U ∈ B there exists V ∈ B
such that V is balanced and V ⊂ U . Since the balanced neighborhoods of 0 form a local
base, we can choose a balanced neighborhood W of 0 such that W ⊂ U . Put V = co W .
Then W ⊂ V ⊂ U (because U is convex), so V is a convex neighborhood of 0. We finish
by showing that V is balanced.
Let |s| ≤ 1 and
P
P x ∈ V . Then there exist t1 , . . . , tn ≥ 0 and
y1 , . . . , yn ∈ W such that n1 ti = 1 and x = n1 ti yi . Since W is balanced we have syi ∈ W
for all i. Thus
n
X
sx =
ti syi
1
is a convex combination of elements of W , hence is an element of V .
We now investigate how locally convex spaces can be characterized by seminorms. This is
in fact the reason why local convexity is so important.
Definition 9. Let p be a seminorm on a vector space X, and let a > 0. Put
[p < a] = p−1 (Ba )
[p ≤ a] = p−1 (B̄a )
Thus [p < a] = {x ∈ X : p(x) < a} and [p ≤ a] = {x ∈ X : p(x) ≤ a}.
Observation 10. If p is a seminorm on a vector space X, then:
(1) for every a > 0 the set [p < a] is balanced, convex, and absorbing;
(2) [p < a] = a[p < 1] = [ap < 1].
The reason why seminorms are so closely tied to locally convex spaces is that, conversely,
balanced convex absorbing sets give rise to seminorms:
Definition 11. Let U be an absorbing set in X. The Minkowski functional associated to U
is the function pU : X → R defined by
pU (x) = inf{t > 0 : x ∈ tU }.
Note that pU is well-defined since U is absorbing.
Theorem 12. Let U be a balanced, convex, and absorbing subset of a vector space X. Then
the Minkowski functional pU is a seminorm, and
(1)
[pU < 1] ⊂ U ⊂ [pU ≤ 1].
Proof. For this proof let p = pU . It follows straight from the definition that p is nonnegative.
Let x ∈ X and c ∈ F. We must show that p(cx) = |c|p(x). First suppose c = 0. Since U is
absorbing, we have 0 ∈ tU for some t > 0, and then in fact 0 ∈ tU for every t > 0. Thus
p(0x) = p(0) = 0 = 0p(x).
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Suppose next that c > 0. Then
{t > 0 : cx ∈ tU } = c{t > 0 : x ∈ tU },
and taking the inf on both sides gives
p(cx) = cp(x).
Finally, if c 6= 0 then |c| > 0, and for all t > 0 we have cx ∈ tU if and only if |c|x ∈ tU , since
U is balanced. Thus
p(cx) = p(|c|x) = |c|p(x).
For the triangle inequality, let x, y ∈ X, and let
p(x) < a and p(y) < b.
Then there exist s ∈ (0, a) and t ∈ (0, b) such that
x ∈ sU
and y ∈ tU.
Thus
x + y ∈ sU + tU = (s + t)
s
t
U+
U
s+t
s+t
⊂ (s + t)U
because U is convex. Therefore p(x + y) ≤ s + t. Taking the inf over s and t gives
p(x + y) ≤ p(x) ≤ p(y).
For (1), let x ∈ [p < 1], so that p(x) < 1. Then there exists t ∈ (0, 1) such that x ∈ tU .
Then t−1 x ∈ U , hence x ∈ U since U is convex and x is in the line segment joining two of
its elements 0 and t−1 x. Finally, if x ∈ U then p(x) ≤ 1 by definition.
Later, for the proof of the Hahn-Banach Separation Theorem, it will be necessary to have
the following technical variation of Theorem 12:
Lemma 13. In the above theorem, if we only assume U is convex and absorbing, then pU is
still a sublinear functional, and (1) still holds.
Proof. This follows from a mildly careful examination of the above proof.
Lemma 14. A seminorm on a topological vector space X is continuous if it is continuous
at 0.
Proof. Let p be a seminorm on X, and let {xi } be a net converging to x in X. Since p
satisfies the alternate form of the triangle inequality, we have
|p(xi ) − p(x)| ≤ p(xi − x) → 0
because xi − x → 0.
Corollary 15. A seminorm p on a topological vector space X is continuous if [p < 1] is a
neighborhood of 0.
Proof. For all a > 0 the set [p < a] is a neighborhood of 0 (because multiplication by a is a
homeomorphism). Thus p is continuous at 0, hence is continuous.
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Lemma 16. If U is a convex open neighborhood of 0 in a topological vector space X, then
U = [pU < 1].
Proof. By Lemma 13, it suffices to show that if x ∈ U then pU (x) < 1. Since U is open and
scalar multiplication is continuous there exists t ∈ (0, 1) such that t−1 x ∈ U , hence x ∈ tU .
Thus
pU (x) ≤ t < 1.
Observation 17. If p and q are seminorms on a vector space X, then p ≤ q if and only if
[q < a] ⊂ [p < a]
for all a > 0.
Corollary 18. Let p and q be seminorms on a topological vector space X. If p ≤ q and q is
continuous, then p is continuous.
Proof. This is immediate from Observation 17 and Corollary 15.
Definition 19. A family P of seminorms on a vector space X is separating if for all x ∈
X \ {0} there exists p ∈ P such that p(x) 6= 0.
Lemma 20. The family of all continuous seminorms on a locally convex space is separating.
Proof. Let X be a locally convex space and x ∈ X \{0}. Then there exists a convex balanced
neighborhood U of 0 such that x ∈
/ U . Then pU is a continuous seminorm, and pU (x) ≥ 1
since x ∈
/ U.
We now show that, conversely, a separating family of seminorms gives a locally convex
topology:
Theorem 21. Let P be a separating family of seminorms on a vector space, and put
(n
)
\
B=
[pi < ai ] : p1 , . . . , pn ∈ P, a1 , . . . , an > 0 .
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Then B is a local base for a unique topology making X into a locally convex space.
Proof. Note that B coincides with the family of all finite intersections of sets in the family
S := {[p < a] : p ∈ P, a > 0}.
S is a family of convex balanced absorbing sets in X such that for all U ∈ S there exists
V ∈ S such that V + V ⊂ U (if U = [p < a] take V = [p < a/2]). Thus the family B is
a local base for a unique topology making X into a pre-topological vector space. For each
x ∈ X \ {0} there exists U ∈ S such that x ∈
/ U , since the family P is separating. Thus X
is Hausdorff. Since every set in S is convex, so is every set in B. Therefore X is a locally
convex space.
Definition 22. In the notation of Theorem 21, P is a generating family of seminorms for
the locally convex space X. Also, we say the unique locally convex topology on X with local
base B is generated by P.
Thus, a family P of seminorms of a locally convex space X is generating if and only if the
family B associated to P as in Theorem 21 is a local base.
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Corollary 23. Let X be a locally convex space with a generating family P of seminorms.
Then a seminorm q on X is continuous if and only if there exist p1 , . . . , pn ∈ P and a > 0
such that
q ≤ a max pi .
i
Proof. If q ≤ a maxi pi with p1 , . . . , pn ∈ P, then by Lemma 18 q is continuous because
the seminorm a maxi pi is. For the converse, assume that q is continuous. Then the set
U = [q < 1] is a neighborhood of 0. Since the family B of Theorem 21 is a local base, there
exist p1 , . . . , pn ∈ P and a1 , . . . , an > 0 such that
\
[pi < ai ] ⊂ U.
i
Put a = mini ai . Then a > 0 and
[a max pi < 1] = [max pi < a] ⊂
i
i
\
[pi < ai ] ⊂ U,
i
so
q ≤ a max pi .
i
The following lemma shows how a generating family of seminorms is used to characterize
convergence:
Lemma 24. Let X be a locally convex space with a generating family P of seminorms. Then
a net {xi }i∈I in X converges to x if and only if p(xi − x) → 0 for all p ∈ P.
Proof. Put yi = xi − x. Then xi → x if and only if yi → 0. Since each p ∈ P is continuous,
if yi → 0 then p(yi ) → 0. For the converse, assume that p(yi ) → 0 for all p ∈ P. Let U be a
neighborhood of 0. We must show that there exists k ∈ I such that
yi ∈ U
for all i ≥ k.
We can choose p1 , . . . , pn ∈ P and a1 , . . . , an > 0 such that
n
\
[pj < aj ] ⊂ U.
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Since limi∈I pj (yi ) = 0 for all j, we can choose k ∈ I such that for all i ≥ k we have pj (yi ) < aj
for every j = 1, . . . , n, hence yi ∈ U .
Next we characterize locally convex topologies in terms of maps into normed spaces. Suppose
p is a seminorm on a vector space X. Put
N = p−1 (0).
Then N is a subspace of X by the properties of seminorms. Let Q : X → X/N be the
quotient map. It is easy to see that there is a norm on X/N defined by
kQ(x)k = p(x).
Lemma 25. With the above notation, if X is a topological vector space then p is continuous
if and only if Q is continuous into the normed space X/N .
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Proof. Each of p and Q is continuous if and only if it is continuous at 0. Let xi → 0 in the
topological vector space X. Then by construction p(xi ) → 0 if and only if kQ(xi )k → 0.
The result follows.
Now we play the above game with a separating family P of seminorms: for each p ∈ P
we have a quotient map Qp : X → X/Np into a normed space X/Np with norm given by
kQp (x)k = p(x).
Lemma 26. With the above notation, the topology on X generated by the family P coincides
with the weakest topology making every Qp continuous.
Proof. Let T be the topology on X generated by the family P, and let T 0 be the weakest
topology making every Qp continuous. Both topologies make X into a topological vector
space, so it suffices to show that xi → 0 in T if and only if xi → 0 in T 0 . For each p ∈ P we
have p(xi ) → 0 if and only if kQp (xi )k → 0, equivalently Qp (xi ) → 0 in the normed space
X/Np . By Lemma 24, xi → 0 in T if and only if p(xi ) → 0 for all p ∈ P, and by definition
xi → 0 in T 0 if and only if Qp (xi ) → 0 for all p ∈ P. The result follows.
Finally, we characterize locally convex topologies in terms of continuity of seminorms:
Corollary 27. Let X be a vector space with a separating family P of seminorms. Then
the topology on X generated by P in the sense of Definition 22 coincides with the weakest
topology on X making each seminorm in P continuous.
Proof. By Lemma 26 the topology on X generated by P coincides with the weakest topology
making each quotient map Qp : X → X/Np continuous (in the notation of the above discussion), which by Lemma 25 coincides with the weakest topology on X making each seminorm
in P continuous.