Download B Basic facts concerning locally convex spaces

B Basic facts concerning locally convex spaces
In this appendix, we recall the definition of locally convex topological vector
spaces, describe various basic constructions, and prove all results required for
our purposes. Throughout the following, K ∈ {R, C}. All vector spaces will
be K-vector spaces and all linear maps are K-linear, unless the contrary is
stated. We assume that the reader is familiar with some basic facts concerning
normed spaces (like the uniform boundedness principle).
B.1 Basic definitions and first properties
Definition B.1.1. A topological vector space is a vector space E, equipped
with a Hausdorff topology turning both the addition map
E×E →E,
(x, y) 7→ x + y
and scalar multiplication
K×E →E,
(z, x) 7→ zx
into continuous mappings.1 A topological vector space is called locally convex
if every 0-neighbourhood contains a convex 0-neighbourhood.2
Given subsets U and V of a vector space E, and t ∈ K, we write U + V :=
{x + y : x ∈ U, y ∈ V } and tU := {tx : x ∈ U }. We also abbreviate
Dr := {z ∈ K : |z| ≤ r}
for r > 0 and D := D1 .
Lemma B.1.2. For each topological vector space E, the following holds:
(a) For each x ∈ E, the translation τx : E → E, τx (y) := x + y is a homeomorphism.
1
2
Here R is equipped with its usual topology, and the products are equipped with
the product topology.
A subset U of a vector space is called convex if it contains with any two points
x, y ∈ U also the line segment joining them, viz. tx+(1−t)y ∈ U for all t ∈ [0, 1].
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B Locally convex spaces
c Helge Glöckner
(b) For each t ∈ K× , the map ht : E → E, ht (x) := tx is a homeomorphism.
(c) Every 0-neighbourhood U ⊆ E contains a 0-neighbourhood V which is
“balanced,” i.e., tV ⊆ V for each t ∈ K such that |t| ≤ 1. Then V = −V
in particular.
(d) For each 0-neighbourhood U ⊆ E, there exists a 0-neighbourhood V ⊆ E
such that V + V ⊆ U .
S
(e) Each 0-neighbourhood U ⊆ E is “absorbing,” i.e., E = r>0 rU .
(f) If B is a basis of 0-neighbourhoods in E, then for each x ∈ E the set
{x + U : U ∈ B} is a basis of neighbourhoods of x in E.
(g) Each 0-neighbourhood contains a closed 0-neighbourhood.
Proof. (a) The addition map α : E × E → E, α(u, v) := u + v and the
map ix : E → E × E, ix (y) := (x, y) being continuous, so is τx = α ◦ ix .
Clearly τx is invertible, with inverse τ−x which is continuous. Hence τx is a
homeomorphism.
(b) Likewise ht = µ(t, •) is continuous being a partial map of the continuous scalar multiplication µ : K × E → E, and so is (ht )−1 = ht−1 .
(c) By continuity of the scalar multiplication µ, the preimage µ−1 (U ) is
open in K × E, entailing that there exists ε > 0 and a 0-neighbourhood
W ⊆ E such that Dr × W ⊆ µ−1 (U ) and thus V := Dr W = µ(Dr × W ) ⊆ U .
Then V ⊆ U and DV = V because DDr = Dr .
(d) Since addition α : E × E → E is continuous and α(0, 0) = 0, the
preimage α−1 (U ) is a (0, 0)-neighbourhood. Now E × E being equipped with
the product topology, there exist 0-neighbourhoods V1 , V2 ⊆ E such that
V1 × V2 ⊆ α−1 (U ). Then V := V1 ∩ V2 has the desired property.
(e) Given x ∈ E, the map f : K → E, f (t) := tx is continuous, and
f (0) = 0. Hence f −1 (U ) is a 0-neighbourhood in K, and so there exists
n ∈ N such that n1 ∈ f −1 (U ). Then n1 x ∈ U and thus x ∈ nU .
(f) follows from the fact that τx is a homeomorphism (by (a)).
(g) Given a 0-neighbourhood U ⊆ E, let V ⊆ E be a 0-neighbourhood
such that V − V ⊆ U (see (d) and (c)). Given w ∈ V , by (f) the set w + V is
a neighbourhod of w in E. Hence, there exists v1 ∈ V such that v1 ∈ w + V ,
whence in turn there is v2 ∈ V such that v1 = w + v2 . Then w = v1 − v2 ∈ U .
Thus V ⊆ U .
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Lemma B.1.3. Let E be a topological vector space and U ⊆ E be convex.
Then the following holds:
(a) The sets tU and U + x are convex, for each t ∈ K and x ∈ E.
(b) The interior U 0 and the closure U of U are convex.
(c) If U 0 6= ∅, then U 0 is dense in U .
Proof. (a) Is trivial.
(b) Given t ∈ [0, 1], consider φ : E × E → E, φ(x, y) := tx + (1 − t)y. Since
U is convex, we have φ(U × U ) ⊆ U and hence φ(U × U ) ⊆ U , by continuity
of φ. Hence U is convex. Given x ∈ U and t ∈ ]0, 1[, the set tx+(1−t)U 0 ⊆ U
B.1 Basic definitions and first properties
213
open in E by Lemma B.1.2 (a) and (b), and thus tx + (1 + t)U 0 ⊆ U 0 . Hence
tU 0 + (1 − t)U 0 ⊆ U 0 for each t ∈ ]0, 1[, and thus U 0 is convex.
(c) Given x ∈ U and y ∈ U 0 , we have ut := tx+(1−t)y ∈ tx+(1−t)U 0 ⊆
0
U for each t ∈ ]0, 1[ as just shown, where ut → x as t → 1. Thus x ∈ U 0 . u
t
Remark B.1.4. Likewise, tU , x + U and U are balanced if so is U ⊆ E. If
U ⊆ E is a balanced 0-neighbourhood, then so is U 0 .
Recall that a subset U of a vector space E is called absolutely convex if U is
both balanced and convex.
Lemma B.1.5. Let U be a subset of a vector space E. Then
(a) There exists a smallest convex subset conv(U ) ⊆ E containing U .
(b) An element x ∈ E belongs to conv(U ) if and only if there exists k ∈ N,
Pk
elements x1 , . . . , xk ∈ U and t1 , . . . , tk ∈ ]0, ∞[ such that j=1 tj = 1
Pk
and x = j=1 tj xj .
(c) absconv(U ) := conv(DU ) is the smallest absolutely convex subset of E
which contains U .
Pk
Proof. Let C be the set of all elements x = j=1 tj xj as described in (b). We
claim that if D ⊆ E is convex and U ⊆ D, then C ⊆ D. ToPsee this, we show
k
by induction on k that each element of C of the form x = j=1 tj xj is in D.
Pk+1
The case k = 1 is trivial. If the assertion holds for k and x = j=1 tj xj ∈ C,
Pk
t
then x = (1 − tk+1 )y + tk+1 xk+1 where y := j=1 1−tjk+1 xj ∈ D by induction
and xk+1 ∈ U ⊆ D and thus x ∈ D, the set D being convex. Thus C ⊆ D.
To see that C is convex, note that we may assume that the elements xj are
Pk
pairwise distinct in the formula x = j=1 tj xj . Furthermore, clearly we get
the same set C if we allow tj = 0. Hence, given x, y ∈ C, we may assume that
Pk
Pk
x = j=1 tj xj and y = j=1 sj xj , using the same elements x1 , . . . , xk ∈ C
Pk
Pk
and certain sj , tj ∈ [0, ∞[ such that j=1 tj = j=1 sj = 1. Given t ∈ [0, 1],
Pk
we then have tx + (1 − t)y =
j=1 rj xj where rj := ttj + (1 − t)sj ≥ 0
Pk
Pk
Pk
and
j=1 rj = t
j=1 tj + (1 − t)
j=1 sj = t + (1 − t) = 1 and thus
tx + (1 − t)y ∈ C. Thus C is convex and hence C = conv(U ).
Since every absolutely convex set D containing U also contains DU , we
must have conv(DU ) ⊆ D. On the other hand, it is clear from the formula from (b) that conv(DU ) is balanced and thus absolutely convex. Hence
conv(DU ) is the smallest absolutely convex subset of E containing U .
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Lemma B.1.6. If E is a locally convex space, then every 0-neighbourhood
U ⊆ E contains an open, absolutely convex 0-neighbourhood and a closed,
absolutely convex 0-neighbourhood. Furthermore, each neighbourhood of a
point x ∈ E contains an open x-neighbourhood and a closed, convex xneighbourhood.
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B Locally convex spaces
c Helge Glöckner
Proof. By Lemma B.1.2 (g), there exists a closed 0-neighbourhood V ⊆ E
such that V ⊆ U . By definition of a locally convex space, there exists a
convex 0-neighbourhood C ⊆ E such that C ⊆ V . By Lemma B.1.2 (c),
there exists a balanced 0-neighbourhood B ⊆ C. Then Q := absconv(B) =
conv(DB) = conv(B) ⊆ C is absolutely convex and a 0-neighbourhood. By
Lemma B.1.3 (b) and Remark B.1.4, the sets Q0 and Q ⊆ V ⊆ U are absolutely convex 0-neighbourhoods of E contained in U which are open and
closed, respectively. Hence the set B of all open (resp., closed) absolutely
convex 0-neighbourhoods is a basis of 0-neighbourhoods. Given x ∈ E, the
set {x + W : W ∈ B} is a basis of x-neighbourhoods, by Lemma B.1.2 (f).
Here x + W is open (resp., closed) since translation by x is a homeomorphism
(Lemma B.1.2 (a)). Also, W being convex, so is x+W (Lemma B.1.3 (a)). u
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B.2 Continuity and openness of linear maps
Proposition B.2.1. Let f : E → F be a linear map between topological vector spaces. Then the following holds:
(a) f is continuous if and only if f is continuous at 0.
(b) f is open if and only if f (U ) is a 0-neighbourhood in F , for each 0neighbourhood U ⊆ E.
Proof. (a) Assume that f is continuous at 0; given x ∈ E, we show that f is
continuous at x. Using the translations τx and τf (x) on E and F , respectively,
we have f ◦ τx = τf (x) ◦ f as f (x + y) = f (x) + f (y). Hence
f = τf (x) ◦ f ◦ τx−1 .
(B.1)
Now τx−1 being continuous with τx−1 (x) = 0, the right hand side of (B.1) is
continuous at x and hence so is the left hand side, f .
(b) Let U ⊆ E be an open set. For each x ∈ U , the set U − x is a 0neighbourhood in E (see Lemma B.1.2 (a)) and hence f (U −x) = f (U )−f (x)
is a 0-neighbourhood in F , by the hypothesis. As a consequence, f (U ) =
f (U − x) + f (x) is a neighbourhood of f (x) in F , and thus f (x) ∈ f (U )0 .
Thus f (U ) = f (U )0 and thus f (U ) is open.
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Of course, it suffices to consider U in a basis of 0-neighbourhoods of E in
Proposition B.2.1 (b).
B.3 Vector subspaces, quotients and direct products
Proposition B.3.1. Let E be a topological vector space and (Ei )i∈I be a
family of topological vector spaces.
B.3 Vector subspaces, quotients and direct products
215
(a) If F ⊆ E is a vector subspace, then the induced topology makes F a
topological vector space. Furthermore, the closure F of F is a vector
subspace of E. If E is locallyQconvex, then so is F .
(b) The cartesian product P := i∈I Ei , equipped with the product topology
and componentwise addition, is a topological vector space. If each Ei is
locally convex, then so is P .
(c) If F ⊆ E is a closed vector subspace, then the quotient topology with
respect to q : E → E/F , q(x) := x + F makes the quotient vector space
E/F a topological vector space, which is locally convex if so is E. The
quotient map q : E → E/F is open.
Proof. (a) Since addition α : E × E → E is continuous, so is its restriction
α|F ×F : F × F → E and the co-restriction α|F
F ×F : F × F → F , which is the
addition on F . Similarly, continuity of scalar multiplication on F is inherited
from that on E. Thus F is a topological vector space. Since α is continuous
and F +F = α(F ×F ) ⊆ F , we deduce that F +F = α(F ×F ) ⊆ F . Likewise,
K F ⊆ F , and thus F is a vector subspace of E. If E is locally convex and
V ⊆ F is a 0-neighbourhood, by definition of the induced topology there
exists a 0-neighbourhood U ⊆ E such that U ∩ F ⊆ V . There exists a convex
0-neighbourhood W ⊆ E such that W ⊆ U . Then F ∩ W is convex (as an
intersection of convex sets), and it is a 0-neighbourhood of F contained in V .
Hence F is locally convex.
(b) Let pri : P → Ei be the coordinate projection for i ∈ I. The addition
α : P × P → P is given by α((xi )i∈I , (yi∈I ) = (αi (xi , yi ))i∈I in terms of the
addition αi of Ei . Hence pri ◦ α = αi ◦ (pri × pri ) is continuous for each i ∈ I.
By ?? in Appendix A, this entails that α is continuous. Similarly, we see
that scalar multiplication on P is continuous. Hence P is a topological vector
space. If each Ei is locally convex and U ⊆ P is a 0-neighbourhood, there
exists
a finite subset J ⊆ I and 0-neighbourhoods Ui ⊆ Ei such that V :=
T
−1
pr
i (Ui ) ⊆ U , by definition of the product topology. After shrinking Ui ,
i∈J
we may assume that each Ui is convex. Then also the preimage pr−1
i (Ui )
under the linear map pri is convex (exercise), and hence the intersection V
of convex sets is convex, whence V is a convex 0-neighbourhood contained
in U .
(c) We equip Q := E/F with the quotient topology. If U ⊆ E is open,
then also U + x is openSin E for each x ∈ F (cf. Lemma B.1.2 (a)) and hence
q −1 (q(U )) = U + F = x∈F (U + x) is open as a union of open sets, whence
q(U ) is open in Q by definition of the quotient topology. Now q being open
and surjective, so is q × q : E × E → Q × Q, (x, y) 7→ (q(x), q(y)), whence also
q × q is a quotient map. For the addition maps α on E and β on Q, we have
β ◦ (q × q) = q ◦ α .
Here the right hand side is continuous and hence so is β ◦ (q × q). Since
q × q is a quotient map, this entails that β is continuous (see ?? in Appendix A). Similarly, we find that scalar multiplication is continuous on Q,
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c Helge Glöckner
B Locally convex spaces
and hence Q is a topological vector space. If E is locally convex and U ⊆ Q
is a 0-neighbourhood, then q −1 (U ) is a 0-neighbourhood in E and hence contains a convex, open 0-neighbourhood W of E. Then q(W ) ⊆ U is open in Q
and is convex, being the image of a convex set under a linear map (exercise).
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B.4 Neighbourhoods of 0 and continuous seminorms
Although a locally convex space need not be normable (i.e., there need not
be a norm on it which defines its topology), locally convex spaces are not too
far away from normed spaces: At least, their topology can always be defined
by a family of continuous semi norms, as we recall now.
Definition B.4.1. Let E be a vector space. A map p : E → [0, ∞[ is called
a seminorm on E if it satisfies the following conditions:
(a) p(x + y) ≤ p(x) + p(y) for all x, y ∈ E (triangle inequality);
(b) p(zx) = |z|p(x) for all z ∈ K and x ∈ E.
Given x ∈ E and ε > 0, we write Bεp (x) := {y ∈ E : p(y − x) < ε}.
B.4.2. In contrast to the case of a norm, there may be non-zero vectors x ∈ E
such that p(x) = 0. It readily follows from (a) and (b) that Np := {x ∈ E :
p(x) = 0} is a vector subspace of E, for each seminorm p on E (for example,
if x, y ∈ Np , we have 0 ≤ p(x + y) ≤ p(x) + p(y) = 0, whence p(x + y) = 0 and
thus x + y ∈ Np ). We let Ep := E/Np be the quotient vector space, formed
by the cosets x + Np with x ∈ E. Then
k.kp : Ep → [0, ∞[ ,
kx + Np kp := p(x)
is well defined (if x + Np = y + Np , then x = y + z for some z ∈ Np
and thus p(x) ≤ p(y) + p(z) = p(y); likewise p(y) ≤ p(x)). Furthermore,
since p is a seminorm, so is k.kp , and it is in fact a norm on Ep because
0 = kx + Np kp = p(x) entails x ∈ Np . We let αp : E → Ep , αp (x) := x + Np
be the canonical quotient map.3
Convex 0-neighbourhoods and continuous seminorms are closely related.
Proposition B.4.3. Let E be a topological vector space and U ⊆ E be a
absolutely convex 0-neighbourhood. Then the Minkowski functional
µU : E → [0, ∞[ ,
µU (x) := inf {t > 0 : x ∈ tU }
is a continuous seminorm on E. If U is closed, then U coincides with the
closed unit ball {x ∈ E : µU (x) ≤ 1} of µU . Conversely, for every continuous
seminorm p : E → [0, ∞[, its closed unit ball B is an absolutely convex 0neighbourhood in E and µB = p.
3
Occasionally, we shall also write k.kp instead of p for a seminorm on E, to increase
the readability. No confusion with k.kp on Ep just defined is likely.
B.4 Neighbourhoods of 0 and continuous seminorms
217
Proof. Since U is absorbing, µU (x) < ∞ for each x ∈ E. Clearly µU (0) = 0
and thus µU (zx) = 0 if z = 0. Now suppose z ∈ K× . If t > 0 such that
zx ∈ tU , then x ∈ tz −1 U = t|z|−1 U and thus t|z|−1 ≥ µU (x). Passing to
the infimum we obtain µU (zx)|z|−1 ≥ µU (x) and thus µU (zx) ≥ |z|µU (x).
Similarly, µU (x) = µU (z −1 zx) ≥ |z|−1 µU (zx) and thus µU (zx) ≤ |z|µU (x),
whence equality holds.
If x, y ∈ E and
t, s > 0 such that x ∈ tU and y ∈ sU , then x + y ∈
s
t
U + t+s
U ) = (t + s)U , entailing that µU (x) + µU (y) ≥
tU + sU = (t + s) t+s
µU (x + y). Hence µU is a seminorm.
We now assume that U is closed. If x ∈ U , then t−1 x ∈ U for each
t > 1 and thus x ∈ tU , whence µU (x) ≤ 1. Conversely, let x ∈ E such that
µU (x) ≤ 1. If µU (x) < 1, then there is t ∈ ]0, 1[ such that x ∈ tU ⊆ U and
thus x ∈ U . If µU (x) = 1, then µU (tx) = t < 1 for each t ∈ ]0, 1[ and hence
tx ∈ U , by what has just been shown. Letting t → 1, we find that x ∈ U = U .
Hence indeed U is the closed unit ball of µU .
As µ−1
U ([0, ε]) = εU for each ε > 0, the map µU is continuous at 0. Given
x ∈ E, we have |µU (x)−µU (y)| ≤ |µU (x−y)| for all x, y ∈ E as a consequence
of the triangle inequality. Hence µU (x + εU ) ⊆ [µU (x) − ε, µU (x) + ε] for each
ε > 0, entailing that µU is continuous at x.
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Definition B.4.4. Let E be a vector space.
(a) A set P of seminorms on E is called separating if p(x) = 0 for all p ∈ P
implies x = 0.
(b) If P is a separating set of seminorms on E, then the map
Y
α := (αp )p∈P : E →
Ep , α(x) := (αp (x))p∈P
p∈P
is injective and linear (where Ep and αp are as in B.4.2). We equip the
product of normed spaces with the product topology. There is a uniquely
determined topology O on E which makes α a homeomorphism onto
α(E), equipped with the induced topology; it is called the locally convex
topology on E defined by the set of seminorms P.
Thus O is the initial topology on E with respect to the family (αp )p∈P .
Remark B.4.5. Since α(E) is a locally convex space (being a vector subspace of a locally convex space) and the co-restriction α|α(E) : E → α(E) is
an isomorphism of vector spaces, it readily follows that O makes E a locally
convex space. Furthermore, each p ∈ P is a continuous seminorm on (E, O),
as it can be written as the composition p = k.kp ◦ prp ◦ α of continuous maps
(where prp is the projection from the cartesian product onto the factor Ep ).
Remark B.4.6. Note that, in the situation of Definition B.4.4 (b), a basis of
0-neighbourhoods for the locally convex topology on E defined by P is given
by the sets of the form
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c Helge Glöckner
B Locally convex spaces
Bεp1 (0) ∩ · · · ∩ Bεpn (0) ,
(B.2)
where ε > 0, n ∈ N and p1 , . . . , pn ∈ P. In fact, clearly each of the sets in (B.2)
is an open 0-neighbourhood in E, as the locally convex topology defined by P
makes each p ∈ P a continuous seminorm. Given a 0-neighbourhood U ⊆ E,
there exists a finite subsetTF ⊆ P and 0-neighbourhoods Up ⊆ Ep for p ∈ F
such that α(U ) ⊇ α(E) ∩ ( p∈F pr−1
p (Up ) (by definition of the product topology and definition of the induced topology on α(E)). Now, Ep being normed,
we find εp such
T that {x ∈ Ep : kxkp < εp } ⊆ Up . SetT ε := min{εp : p ∈ F }.
Then V := p∈F Bεp (0) ⊆ U because α(V ) ⊆ α(E) ∩ p∈F pr−1
p (Up ) ⊆ α(U ),
which in T
turn holds because kprp (α(x))kp = kαp (x)kp = p(x) < ε for each
x ∈ V = p∈F Bεp (0) and each p ∈ F .
Definition B.4.7. A locally convex space E is called normable if its locally
convex vector topology can be defined using a single norm.
The following fact follows from Proposition B.4.3:
Proposition B.4.8. If E is a locally convex topological vector space, then
the set P of all continuous seminorms on E is separating. For each p ∈ P,
the map αp : E → Ep is continuous, and
Y
α := (αp )p∈P : E →
Ep =: P
p∈P
is linear and a topological embedding (i.e., a homeomorphism onto its image).
In other words, P defines the given locally convex topology on E.
k.k
Proof. For each p ∈ P, the linear map αp is continuous as α−1 (Bε p (0)) =
Bεp (0) is a 0-neighbourhood for each ε > 0. Let prp : P → Ep be the canonical
projection. Since prp ◦ α = αp is continuous for each p ∈ P, the map α is
continuous (see ?? in App A). Furthermore, clearly α is linear since so is
each αp . Given 0 6= x ∈ E, there exists a 0-neighbourhood U ⊆ E such
that x 6∈ U . By Lemma B.1.6, there exists a closed, aboslutely convex 0neighbourhood B ⊆ E such that B ⊆ U . By Proposition B.4.3, there is
a continuous seminorm p ∈ P with closed unit ball B. Since x 6∈ B, we
deduce that 1 < p(x) = kαp (x)kp , whence αp (x) 6= 0 and hence α(x) 6= 0.
Therefore α is injective. It remains to show that α is open onto its image. By
Proposition B.2.1 (b), we only need to show that α(V ) is a 0-neighbourhood
in α(E), for each 0-neighbourhood V ⊆ E. As before, we see that V contains
the closed unit ball Q of some continuous seminorm q ∈ P. Given y ∈ α(E),
we have y = α(x) for some x ∈ E; due to the injectivity of α,
y = α(x) ∈ α(Q)
holds if and only if y ∈ Q. The latter holds if and only if 1 ≥ q(x) =
kαq (x)kq = kprq (y)kq , i.e., if and only if y ∈ pr−1
q (R), where R ⊆ Eq is the
closed unit ball of k.kq . Hence α(V ) ⊇ α(Q) = α(E) ∩ pr−1
q (R), which is a
0-neighbourhood.
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B.4 Neighbourhoods of 0 and continuous seminorms
219
Remark B.4.9. Let E be a vector space and p, q : E → [0, ∞[ be seminorms
such that p ≤ q (pointwise). Then q(x) = 0 entails p(x) = 0, whence Nq ⊆ Np .
By the homomorphism theorem, there is a unique linear map
αpq : Eq → Ep
such that αpq ◦ αq = αp . Given y ∈ Eq , we have y = αq (x) for some x ∈ E.
Then kαpq (y)kp = kαpq (αq (x))kp = kαp (x)kp = p(x) ≤ q(x) = kαq (x)kq =
kykq shows that αpq is a continuous linear map, of operator norm kαpq k ≤ 1.
We deduce:
If O is a topology on E which makes αq continuous, then also αp = αpq ◦ αq
is continuous on (E, O).
Example B.4.10. Let r ∈ N0 ∪ {∞}. For k ∈ N0 such that k ≤ r, the map
pk : C r ([0, 1], R) → [0, ∞[ ,
pk (γ) := sup{|γ (k) (t)| : t ∈ [0, 1]}
is a seminorm on the space C r ([0, 1], R) of real-valued C r -maps on [0, 1]. The
set P of all pk ’s as before is a separating set of seminorms on C r ([0, 1], R).
We give C r ([0, 1], R) the locally convex topology defined by P.
It can be shown that C ∞ ([0, 1], R) is not normable.
Corollary B.4.11. On Rn , there is only one locally convex vector topology.
Any norm on Rn defines this vector topology.
Proof. Let O be a locally convex vector topology on Rn . Choose a continuous
seminorm q1 6= 0 on Rn ; then its 0-space Nq1 is a proper vector subspace
of Rn . If E2 := Nq1 6= {0}, we choose 0 6= v2 ∈ Nq2 and a continuous
seminorm on (Rn , O) such that q2 (v2 ) 6= 0; then E3 := Nq1 ∩ Nq2 is a proper
vector subspace of E2 . Proceeding in thisTway, after r ≤ n steps we obtain
r
continuous
seminorms q1 , . . . , qr such that i=1 Nqi = {0}, entailing that q :=
Pr
n
n
i=1 qi is a continuous norm on (R , O). Each norm on R being equivalent,
we deduce that q defines the usual locally convex vector topology T on Rn .
Hence T ⊆ O. Now, if p is any continuous seminorm on (Rn , O), then p + q is
a continuous norm on (Rn , O) and hence defines the usual topology, entailing
that αp+q : Rn → (Rn )p−q (and hence also αp , by Remark B.4.9) is continuous
on (Rn , T ). The topology O being the coarsest topology making all of the
maps αp continuous, we deduce that O ⊆ T . Hence O = T .
t
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Remark B.4.12. It can be shown with slightly more effort that there is only
one vector topology on Rn .
Recall that a topological space X is called metrizable if there exists a metric d
on X whose associated topology coincides with the given one.
Corollary B.4.13. For a locally convex topological vector space E, the following conditions are equivalent:
220
c Helge Glöckner
B Locally convex spaces
(a) E is metrizable;
(b) E is first countable, i.e., for each x ∈ E there exists a countable basis of
x-neighbourhoods;
(c) There exists a countable basis of 0-neighbourhoods in E;
(d) There exists a countable set P of continuous seminorms on E which defines its topology.
Proof. (a)⇒(b): Let d be a metric on E defining its topology. Then, for each
x ∈ E, the balls B1/n (x) := {y ∈ E : d(x, y) < 1/n} (indexed by n ∈ N) form
a countable basis of x-neighbourhoods.
(b)⇒(c) is trivial.
(c)⇒(d): Let U be a countable basis of 0-neighbourhoods in E. For each
U ∈ U, there exists a closed absolutely convex 0-neighbourhood BU ⊆ U
(see Lemma B.1.6) and a continuous seminorm pU on E with closed unit
ball BU (see Proposition B.4.3). Then P := {pU : U T∈ U } is a countable
set of continuous seminorms which is separating as U ∈U BU = {0}. Set
B := {BU : U ∈ U}. For each finite sequence U1 , . . . , Un in U and r > 0, the
set rBU1 ∩ · · · ∩ rBUn is a 0-neighbourhood in E and hence contains some
U ∈ U and hence also BU . Therefore Remark B.4.6 implies that B is a basis
of 0-neighbourhoods for the locally convex topology on E defined by P. But
U also is a basis of 0-neighbourhoods for the original topology, entailing that
the two vector topologies coincide.
(d) Suppose that (pn )n∈N be a sequence of continuous seminorms on E
which defines the locally convex topology O on E. Then
d : E × E → [0, ∞[ ,
d(x, y) :=
∞
X
2−n pn (x − y)
1 + pn (x − y)
n=1
is a continuous function, because the series converges uniformly on E × E.
Furthermore, using that
h : [0, ∞[ → [0, 1[ ,
h(t) :=
1
t
= 1−
1+t
t
is monotonically increasing, it is easy to see that d is a metric (exercise).
Since d is continuous on (E, O), the topology T determined by d is coarser
then the given one, T ⊆ O. Given x ∈ E and a neighbourhood U of x
in (E, O), using Lemma
TnB.1.2 (f) and Remark B.4.6, we find ε > 0 and
n ∈ N such that B := k=1 Bεpk (x) ⊆ U . Set δ := 2−n h(ε). If y ∈ E such
that d(x, y) < δ, then 2−k h(pk (x − y)) < δ for each k ∈ {1, . . . , n}, whence
h(pk (x − y)) < 2k δ ≤ 2n δ = h(ε) and hence pk (x − y) < ε as h is strictly
monotonically increasing. Therefore {y ∈ E : d(x, y) < δ} ⊆ B ⊆ U . Since,
for each x ∈ E, every x-neighbourhood in (E, O) also is an x-neighbourhood
in (E, T ), we deduce that O ⊆ T . Hence both topologies coincide.
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B.4.14 (Further examples). Here are further examples of (non-normable)
locally convex spaces.
B.5 Linear functionals and the Hahn-Banach theorem
221
Q
(a) The product topology makes RN = n∈N R a metrizable locally convex
space.
(b) We equip C(R, R), the space of continuous real-valued functions on R,
with the locally convex topology defined by the set of all seminorms of
the form
pK : C(R, R) → [0, ∞[ ,
pK (γ) := sup{|γ(x)| : x ∈ K} ,
where K ranges through the set of all compact subsets K of R. The
(same) locally convex topology on C(R, R) is defined by the countable
set {p[−n,n] : n ∈ N} of seminorms (exercise), showing that C(R, R) is
metrizable. Following the same pattern, C(X, K) can be made a locally
convex K-vector space, for each topological space X.
(c) Given an open subset Ω ⊆ C, the set Hol(Ω) of all holomorphic functions γ : Ω → C is a vector subspace of the locally convex space C(Ω, C)
(defined as in (b)), which is metrizable since we can find an ascending
seS
quence K1 ⊆ K2 ⊆ · · · of compact subsets of Ω such that Ω = n∈N Kn
and each compact subset of K ⊆ Ω is contained in some Kn (exercise).
The induced topology makes Hol(Ω) a metrizable locally convex space.
B.5 Linear functionals and the Hahn-Banach theorem
Given a topological K-vector space E, we let E 0 be the set of all continuous
linear maps (“functionals”) λ : E → K. Since sums and scalar multiples of
continuous linear functionals are such, E 0 is a vector subspace of the space
KE of all K-valued functions on E. We call E 0 the dual space of E.
The Hahn-Banach theorems provide continuous linear functionals with specific properties. They are powerful tools of functional analysis. We shall follow
a geometric, calculation-free approach to the Hahn-Banach theorems.
Two preliminary lemmas will be used:
Lemma B.5.1. Let E be a real topological vector space. If E \ {0} is disconnected, then E is 1-dimensional.
Proof. If dim(E) > 1, let x, y ∈ E \ {0}. If x and y are linearly independent,
then [0, 1] → E, t 7→ x + t(y − x) is a path from x to y in E \ {0}. If x and y
are linearly dependent, we choose ∈ E \ span({x, y}). Then x and z as well
as z and y are linearly independent, whence x and z (resp., z and y) can be
joined by a path γ (resp., η) in E \ {0}, as just shown. Concatenating the
two paths we obtain a path from x to y. We have shown that E \ {0} is path
connected and hence connected.
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Lemma B.5.2. Let E be a real topological vector space. If there exists an
open, non-empty convex subset U ⊆ E \ {0} such that U ∩ Rx 6= ∅ for each
x ∈ E \ {0}, then E is 1-dimensional.
222
c Helge Glöckner
B Locally convex spaces
Proof. Abbreviate P := ]0, ∞[. The set P U is stable under multiplication
with positive scalars (as P P = P). Furthermore, P U + P U ⊆ P U because
a
b
av + bw = (a + b) a+b
v + a+b
w ∈ (a + b)U ⊆ P U for all a, b ∈ P and
v, w ∈ U , exploiting the convexity of U . Hence P U is what one calls a “cone”
(a set S
stable under addition and multiplication with positive scalars). As
P U = r>0 rU , the set P U is open. Furthermore, 0 6∈ P U , since 0 6∈ U . We
also have P U ∩ (−P U ) = ∅, because au = −bv with a, b > 0, u, v ∈ U would
entail that 0 = au + bv ∈ P U + P U ⊆ P U , which we just excluded. Since U
meets Rx for each 0 6= x ∈ E, we have
E = P U ∪ (−P U ) ∪ {0} .
(B.3)
By the preceding, the union in (B.3) is disjoint. Therefore E \ {0} =
P U ∪ (−P U ) as a disjoint union, where both P U and −P U are open and
non-empty. Hence E \ {0} is not connected, whence E is 1-dimensional by
Lemma B.5.1.
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Theorem B.5.3 (Hahn-Banach theorem, real case). If E is a real topological vector space and U ⊆ E an open, convex subset such that 0 6∈ U , then
there exists λ ∈ E 0 such that
λ(U ) ⊆ ]0, ∞[ .
Proof. Without loss of generality U 6= ∅ (otherwise, take λ := 0). We consider
the set A of all vector subspaces F ⊆ E such that F ∩ U = ∅. Inclusion of
sets provides a partial order on A which makes A an inductive set. Indeed,
let Γ ⊆ A be a chain (i.e., a totally ordered subset). If Γ = ∅, then the
S trivial
vector space {0} ∈ A is an upper bound for Γ . Otherwise, clearly Γ ∈ A,
and this union is an upper bound for Γ . Hence, by Zorn’s lemma, there exists
a maximal element H ∈ A. We claim that H is a closed hyperplane, i.e., a
closed vector subspace such that dim(E/H) = 1.
H is closed: Since the closure H is a vector subspace and H ⊆ E \ U
(as the latter set is closed and contains H), we deduce H = H from the
maximality of H.
E/H is 1-dimensional. To see this, let q : E → E/H be the quotient map
and set V := q(U ). Since q is linear and open, the set V is an open, convex
subset of E/H. Note that V does not contain the 0-element H of the quotient
space, because H ∈ V would entail ∅ =
6 q −1 ({H}∩V ) = q −1 ({H})∩q −1 (V ) =
H ∩ (U + H) whence there exist h1 , h2 ∈ H and u ∈ U such that h1 = u + h2
and therefore u = h1 − h2 ∈ U ∩ H, which is a contradiction. We now show
that every vector subspace K ⊆ E/H such that K ∩ V = ∅ is trivial. To
see this, note that q −1 (K) is a vector subspace of E such that H ⊆ q −1 (K).
Furthermore, U ∩ q −1 (K) ⊆ q −1 (q(U )) ∩ q −1 (K) = q −1 (V ) ∩ q −1 (K) =
q −1 (V ∩ K) = ∅. Hence H = q −1 (K), by maximality of H, and thus K =
q(q −1 (K)) = {H} is the trivial vector subspace. Using Lemma B.5.2, we see
that the locally convex space E/H is 1-dimensional and hence isomorphic
B.5 Linear functionals and the Hahn-Banach theorem
223
to R as a consequence of Corollary B.4.11. We choose an isomorphism of
topological vector spaces φ : E/H → R and set λ := φ ◦ q : E → R. Then
λ is a continuous linear functional such that ker λ = H does not meet U ,
whence 0 6∈ λ(U ). Being the image of a convex set under a linear map,
λ(U ) is a convex subset of R. As 0 6∈ λ(U ), we must have λ(U ) ⊆ ]0, ∞[ or
λ(U ) ⊆ ]−∞, 0[. In the first case, we are home; in the second case, we replace
λ with −λ.
t
u
To transfer the Hahn-Banach theorem from the real case to the complex
case, we first establish a one-to-one correspondence between real linear and
complex linear functionals on a complex topological vector space.
Lemma B.5.4. Let E be a complex topological vector space. If λ : E → C
is a continuous complex linear functional on E, then Re ◦ λ : E → R is a
continuous real linear functional. Conversely, for each continuous real linear
functional u : E → R, there is a unique continuous complex linear functional
λ : E → C such that u = Re ◦ λ; it is given by the formula
λ(x) = u(x) − iu(ix)
for all x ∈ E.
(B.4)
Proof. It is obvious that u := Re ◦ λ is continuous real linear if λ ∈ E 0 .
Conversely, let u : E → R be a continuous real linear functional. If there
exists λ ∈ E 0 such that u = Re ◦ λ, then λ = u + iv with v := Im ◦ λ. For
each x ∈ E, we have
u(ix) + iv(ix) = λ(ix) = iλ(x) = i(u(x) + iv(x)) = iu(x) − v(x) ;
comparing the real and imaginary parts, we find that v(x) = −u(ix). Hence
λ(x) = u(x) − iu(ix) indeed, showing in particular that λ is uniquely determined by its real part u. To see that a complex linear functional with
real part u exists, we simply define a map λ : E → C via (B.4). Then λ is
continuous, real linear, and u = Re ◦ λ. Being real linear, clearly λ will be
complex linear if we can show that λ(ix) = iλ(x). This is the case: We have
λ(ix) = u(ix) − iu(i(ix)) = i(u(x) − iu(ix)) = iλ(x).
t
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Theorem B.5.5 (Hahn-Banach Theorem, complex case). If E is a
complex topological vector space, and U ⊆ E an open, convex subset such
that 0 6∈ U , then there exists λ ∈ E 0 such that
Re(λ(U )) ⊆ ]0, ∞[ .
Proof. By Theorem B.5.3, there exists a continuous real linear functional
u : E → R such that u(U ) ⊆ ]0, ∞[. Lemma B.5.4 provides λ ∈ E 0 such that
Re ◦ λ = u. Then Re(λ(U )) = u(U ) ⊆ ]0, ∞[.
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Theorem B.5.6. If E is a locally convex topological K-vector space, then E 0
separates points on E, i.e., for all x, y ∈ E such that x 6= y, there exists
λ ∈ E 0 such that λ(x) 6= λ(y).
224
c Helge Glöckner
B Locally convex spaces
Proof. Since x − y 6= 0, there exists an open, convex 0-neighbourhood U ⊆ E
such that x − y 6∈ U . Then U − (x − y) is an open, convex subset of E such
that 0 6∈ U − (x − y). The Hahn-Banach theorem provides λ ∈ E 0 such that
Re(λ(U − (x − y))) ⊆ ]0, ∞[. Then Re(λ(y)) − Re(λ(x)) = Re(λ(−(x − y))) ∈
]0, ∞[ and thus Re(λ(x)) 6= Re(λ(y)), whence also λ(x) 6= λ(y).
t
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The following observation is useful:
Lemma B.5.7. If E is a topological K-vector space and λ : E → K a linear
functional such that λ 6= 0, then λ is an open map.
Proof. By Proposition B.2.1 (b), we only need to show that λ(U ) is a 0neighbourhood in K for each 0-neighbourhood U in E. After shrinking U ,
we may assume that U is balanced. Then also λ(U ) ⊆ K is balanced. Since
λ 6= 0, there exists x0 ∈ E such that λ(x0 ) 6= 0; after replacing x0 with
tx0 for sufficiently small 0 6= t ∈ K, we may assume that x0 ∈ U . Setting
r := |λ(x0 )| > 0, we then have Dr = Dλ(x0 ) ⊆ λ(U ), showing that indeed
λ(U ) is a 0-neighbourhood.
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Theorem B.5.8 (Hahn-Banach separation theorem). If E is a real or
complex locally convex space, A ⊆ E a closed convex subset and x0 ∈ E \ A,
then there exists λ ∈ E 0 and r ∈ R such that
Re(λ(A)) ⊆ ]−∞, r[
and
Re(λ(x0 )) ∈ ]r, ∞[ .
(B.5)
Proof. Since x0 ∈ E \ A where E \ A is open (as A is closed), there exists an
absolutely convex, open 0-neighbourhood W ⊆ E such that x0 + W ⊆ E \ A.
Then x0 6∈ A+W (otherwise, x0 = a+w and thus a = x0 −w ∈ A∩(x0 +W ),
contradiction). Thus, U := A + W − x0 is an open, convex subset of E such
that 0 6∈ U . Theorem B.5.3 (resp., Theorem B.5.5) provide µ ∈ E 0 such that
u(A) + u(W ) − u(x0 ) = u(U ) ⊆ ]0, ∞[ ,
(B.6)
abbreviating u := Re ◦ µ : E → R. Since u(W ) is an open 0-neighbourhood
in R (see Lemma B.5.7), there exists w ∈ W such that u(w) < 0. Now (B.6)
shows that u(A) ⊆ ]u(x0 ) − u(w), ∞[, whence u(A) ⊆ ]r, ∞[ and u(x0 ) ∈
t
u
]−∞, r[ with r := u(x0 ) − 21 u(w). Now set λ := −µ.
Thus, in the real case, A and x0 are contained in the disjoint open half-spaces
λ−1 (]−∞, r[) and λ−1 (]r, ∞[), respectively. In other words, A and x0 can be
separated by the closed hyperplane λ−1 ({r}).
Theorem B.5.9 (Bipolar Theorem). Let E be a locally convex space, A ⊆
E be an absolutely convex, closed, non-empty subset, and x0 ∈ E such that
x0 6∈ A. Then there is λ ∈ E 0 such that |λ(x)| ≤ 1 for all x ∈ A, but
|λ(x0 )| > 1.
B.6 Completeness and sequential completeness
225
Proof. By the Hahn-Banach separation theorem, there exists µ ∈ E 0 and
r ∈ R such that Re(µ(A)) ⊆ ]−∞, r[ and Re(µ(x0 )) > r. Let x ∈ A. There
exists z ∈ K such that |z| = 1 and µ(zx) = zµ(x) ∈ [0, ∞[. Now zx ∈ A as
A is balanced. Hence 0 ≤ |µ(x)| = zµ(x) = µ(zx) = Re(µ(zx)) ∈ ]−∞, r[,
entailing that r > 0 and |µ(x)| < r. For λ := 1r µ, we then have |λ(x)| < 1 for
each x ∈ A, and furthermore Re(λ(x0 )) > 1 and thus |λ(x0 )| > 1.
t
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The name of the Bipolar Theorem has the following origin.
Remark B.5.10. Given a subset A of a locally convex space E, one calls
A◦ := {λ ∈ E 0 : |λ(x)| ≤ 1 for all x ∈ A}
the polar of A in E 0 . Similarly, given B ⊆ E 0 one defines the polar of B in E
as the set B◦ := {x ∈ E : |λ(x)| ≤ 1 for all λ ∈ B}. It is obvious from these
definitions that A ⊆ (A◦ )◦ for each subset A ⊆ E. The Bipolar Theorem
entails that
A = (A◦ )◦ ,
(B.7)
for each absolutely convex, closed, non-empty subset A ⊆ E (exercise).
We shall not use polars in this course; Theorem B.5.9 is all we need.
B.6 Completeness and sequential completeness
We now study completeness properties of locally convex spaces, which are
useful for analysis as they can frequently be used to show that certain limits of
interest exist. Although the results concerning sequences are usually sufficient
for our purposes, it is more natural to discuss convergence of nets as well. The
definitions of nets and convergence of nets can be looked up in Appendix A.
Definition B.6.1. Let E be a locally convex topological K-vector space.
(a) A sequence (xn )n∈N in E is called a Cauchy sequence if, for every zeroneighbourhood U in E, there exists N ∈ N such that
xn − xm ∈ U
for all n, m ≥ N .
(b) The space E is called sequentially complete if every Cauchy sequence
(xn )n∈N in E converges.
(c) A net (xα )α∈A in E is called a Cauchy net if, for every 0-neighbourhood U
in E, there exists α0 ∈ A such that xα − xβ ∈ U for all α, β ≥ α0 .
(d) E is called complete if every Cauchy net in E converges.
Note that a sequence (xn )n∈N in E is a Cauchy sequence if and only if it is
a Cauchy net. Consequently, every complete locally convex space E also is
sequentially complete.
226
c Helge Glöckner
B Locally convex spaces
Proposition B.6.2. A metrizable locally convex space E is complete if and
only if it is sequentially complete.
Proof. In view of the observation preceding the proposition, we only need
to show that every metrizable, sequentially complete locally convex space is
complete. Thus, assume that E is metrizable and sequentially complete.
Since E is metrizable, there exists a sequence (Un )n∈N of 0-neighbourhoods
U
Tn ⊆ E which form a basis of 0-neighbourhoods. After replacing Un with
k≤n Uk , we may assume that U1 ⊇ U2 ⊇ · · · .
Let (xα )α∈A be a Cauchy net in E. Since (xα ) is a Cauchy net, there exists
an element α1 ∈ A with xα − xβ ∈ U1 for all α, β ≥ α1 . Next, we find α2 ∈ A
such that xα − xβ ∈ U2 for all α, β ≥ α2 . Note that, since A is directed, there
is α20 ∈ A such that α20 ≥ α1 and α20 ≥ α2 . Replacing α2 with α20 , we may
assume that α2 ≥ α1 . Proceeding in this way, we find an increasing sequence
α1 ≤ α2 ≤ α3 ≤ · · · in A such that, for each n ∈ N,
xα − xβ ∈ Un
for all α, β ≥ αn .
For all k, m ≥ n, we have αk , αm ≥ αn and hence xαk − xαm ∈ Un ; thus
(xαn )n∈N is a Cauchy sequence and therefore convergent to some x ∈ E
by hypothesis. Let U be any neighbourhood of x in E. Then there exists
some n ∈ N such that Un + x ⊆ U . As a consequence of Lemma B.1.2 (d),
there exists some m ∈ N with Um + Um ⊆ Un . The sequence (xαk )k∈N being
convergent to x, there exists a natural number ` ≥ m with xαk ∈ Um + x for
all k ≥ `. For every α ≥ α` , we obtain
xα = (xα − xα` ) + xα` ∈ U` + Um + x ⊆ Um + Um + x ⊆ Un + x ⊆ U .
Thus the net (xα ) converges to x.
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For example, every Banach space is a complete locally convex space by the
preceding proposition. Complete, metrizable locally convex spaces are encountered frequently and deserve their own name.
Definition B.6.3. A complete, metrizable locally convex space is called a
Fréchet space.
Lemma B.6.4. If λ : E → F is a continuous linear map between locally
convex spaces and (xα )α∈A a Cauchy net in E, then (λ(xα ))α∈A is a Cauchy
net in F .
Proof. In fact, given a 0-neighbourhood U ⊆ F , due to the continuity of λ the
pre-image λ−1 (U ) is a 0-neighbourhood in E. Now (xα )α∈A being a Cauchy
net, there exists α0 ∈ A such that xα − xβ ∈ λ−1 (U ) for all α, β ≥ α0 . Then,
for all α, β ≥ α0 , we have
U ⊇ λ(λ−1 (U )) 3 λ(xα − xβ ) = λ(xα ) − λ(xβ ) ,
whence indeed (λ(xα ))α∈A is a Cauchy net.
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B.7 The completion of a locally convex space
227
Definition B.6.5. Let X be a Hausdorff topological space. A subset A ⊆ X
is called sequentially closed if limn→∞ an ∈ A, for each sequence (an )n∈N in A
which converges in X.
Note that if A ⊆ X is closed, then A is sequentially closed.
New complete locally convex spaces can be constructed from given ones.
Proposition B.6.6. (a) If (Ei )i∈I is a family of complete (resp., sequentially
complete) locally convex spaces, then also the direct product P :=
Q
E
i∈I i is complete (resp., sequentially complete).
(b) Let E be a complete (resp., sequentially complete) locally convex space
and F ⊆ E be a closed (resp., sequentially closed) vector subspace. Then
also F is complete (resp., sequentially complete).
Proof. We only discuss completeness; to prove the analogues concerning sequential completeness, simply replace nets by sequences.
(a) Let (xα )α∈A be a Cauchy net in P . Given i ∈ I, the canonical projection pri : P → Ei is continuous linear map, and thus xi,α := pri (xα ) is
a Cauchy net in Ei (Lemma B.6.4) and thus convergent to some yi ∈ Ei as
Ei is assumed complete. Then (xα )α∈A converges to y := (xi )i∈I ∈ P , since
xi,α → yi for each i (Appendix A, Lemma ??).
(b) If (xα )α∈A is a Cauchy net in F , then (xα )α∈A also is a Cauchy net
in E (apply Lemma B.6.4 to the continuous linear inclusion map F → E)
and hence converges in E, to x ∈ E say. Now F being closed in E, we deduce
from {xα : α ∈ A} ⊆ F that x ∈ F = F (see Appendix A, Proposition ??).
Then xα → x in F , equipped with the topology induced by E (exercise). u
t
B.7 The completion of a locally convex space
It is useful that every locally convex space can be completed.
Definition B.7.1. A completion of a locally convex space E is a complete
e together with a linear topological embedding κE :
locally convex space E,
e with dense image.
E→E
If a completion exists, then standard set theoretic arguments can be used to
e κE ) such that E is a vector subspace of E
e and
manufacture a completion (E,
κE the inclusion map, x 7→ x.
Proposition B.7.2. Every locally convex space E has a completion.
Proof. Let P be the set of all continuous seminorms on E. For each p ∈ P,
fp be the Banach space obtained by completing the normed space
we let E
(Ep , k.kp ), where Ep := E/Np with Np := {x ∈ E : p(x) = 0} and the norm
fp .
is defined via kx + Np kp := p(x) for x ∈ E. We may assume that Ep ⊆ E
228
c Helge Glöckner
B Locally convex spaces
fp , αp (x) := x + Np be the natural map and α := (αp )p∈P :
Let αp : E → E
Q
fp =: P . Being a direct product of complete locally convex
E → p∈P E
e := α(E)
spaces, P is complete (Proposition B.6.6 (a)). Then the closure E
of the image of α is a complete locally convex space, being a closed vector
subspace of the complete locally convex space P (Proposition B.6.6 (b)). We
e
e be the co-restriction of α to E.
e As a consequence
let κE := α|E : E → E
of Proposition B.4.8, the linear map α, and hence also κE , is a topological
e by construction, (E,
e κE ) is a
embedding. The image of κE being dense in E
completion of E.
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B.8 Continuous extension of linear maps
Proposition B.8.1. Let E be a locally convex space, E0 ⊆ E be a dense
vector subspace, F a complete locally convex space, and λ : E0 → F be a
continuous linear map. Then there exists a unique continuous linear map
Λ: E → F
which extends λ, i.e., Λ|E0 = λ.
Proof. Step 1: Definition of Λ. Given x ∈ E, Proposition ?? from Appendix A provides a net (xα )α∈A in E0 converging to x in E, since E0 = E.
By Lemma B.6.4, (λ(xα ))α∈A is a Cauchy net in F and thus convergent, as
F is assumed complete. We want to define
Λ(x) := lim λ(xα )
(B.8)
as the limit of (λ(xα ))α∈A . To see that this definition is independent of the
choice of net (xα )α∈A , let also (yβ )β∈B be a net in E0 converging to x
in E. Let W ⊆ F be an open 0-neighbourhood. Then λ−1 (W ) is an open
0-neighbourhood in E0 . As E0 is equipped with the induced topology, we
find an open 0-neighbourhood V ⊆ E such that V ∩ E0 = λ−1 (W ). Let
U ⊆ E be a 0-neighbourhood such that U − U ⊆ V . We find α0 ∈ A and
β0 ∈ B such that x − xα ∈ U and x − yβ ∈ U , for all A 3 α ≥ α0 and
B 3 β ≥ β0 . Thus, for any such α and β, noting that E0 is a vector subspace
we get
xα − yβ = x − yβ − (x − xα ) ∈ (U − U ) ∩ E0 ⊆ V ∩ E0 = λ−1 (W )
and thus λ(xα ) − λ(yβ ) = λ(xα − yβ ) ∈ W . Passing to the limit in α first and
then to the limit in β, we deduce that
lim λ(xα ) − lim λ(yβ ) ∈ W .
Since sets of the form W constitute a basis of 0-neighbourhoods, and F is
Hausdorff, we deduce that lim λ(xα ) − lim λ(yβ ) = 0, whence a mapping
B.8 Continuous extension of linear maps
229
Λ : E → F is well-defined by (B.8).
Step 2: Λ is linear. In fact, given x, y ∈ E and z ∈ K, let (xα )α∈A and
(yβ )β∈B be nets in E0 converging to x and y in E, respectively. Then C :=
A × B becomes a directed set by declaring (α1 , β1 ) ≤ (α2 , β2 ) for α1 , α2 ∈ A,
β1 , β2 ∈ B if and only if α1 ≤ α2 and β1 ≤ β2 (exercise). Furthermore, the net
(xα )(α,β)∈C converges to x and (yβ )(α,β)∈C converges to y (exercise). Using
these nets instead of (xα )α∈A and (yβ )β∈B , we may assume without loss of
generality that (A, ≤) = (B, ≤). Then (xα + zyα )α∈A is easily seen to be a
Cauchy net in E0 converging to x + zy in E, and thus
Λ(x + zy) = lim λ(xα + zyα ) = lim (λ(xα ) + zλ(yα ))
= lim λ(xα ) + z lim λ(yα ) = Λ(x) + zΛ(y) ,
using Proposition ?? and Lemma ?? in Appendix A (exploiting that addition
and scalar multiplication by z are continuous maps). Thus Λ is indeed linear.
Step 3: Λ is continuous. Given any 0-neighbourhood Q ⊆ F , there exists
an open 0-neighbourhood W ⊆ F whose closure W is contained in Q. Then
V := λ−1 (W ) is an open 0-neighbourhood in E0 . Let U ⊆ E be an open
0-neighbourhood in E such that U ∩ E0 = V . Since E0 is dense is E, the
set U ∩ E0 = V is dense in U , and thus U = V , the closure of V in E.
Given x ∈ U , we have x ∈ V , whence there exists a net (xα )α∈A in V such
that xα → x in E. Since λ(xα ) ∈ W ⊆ W for each α, we deduce from
Proposition ?? in Appendix A that
Λ(x) = lim λ(xα ) ∈ W ⊆ Q .
Thus Λ(U ) ⊆ Q, showing that the linear map Λ is continuous at 0 and thus
continuous. Since E0 is dense in E, the continuous map Λ extending λ is
uniquely determined.
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e be a completion
Example B.8.2. Let E be a locally convex space and E
e
of E, together with the canonical embedding κ : E → E. We may assume
e and κ(x) = x. As K is a complete locally convex space,
that E ⊆ E
Proposition B.8.1 shows that every continuous linear functional λ : E → K
e → E. Conversely,
extends uniquely to a continuous linear functional E
0
e 0.
κ (λ) := λ ◦ κ = λ|E is a continuous linear functional on E, for each λ ∈ (E)
Hence, as it is also linear, the bijection
e 0 → E0 ,
κ0 : (E)
is an isomorphism of vector spaces.
λ 7→ κ0 (λ) = λ ◦ κ
230
B Locally convex spaces
c Helge Glöckner
B.9 Bounded sets and the weak topology
Definition B.9.1. A subset B ⊆ E of a topological vector space E is
called bounded if it is absorbed by each 0-neighbourhood, i.e., for each 0neighbourhood U ⊆ E there exists r > 0 such that B ⊆ rU .
Lemma B.9.2. If α : E → F is a continuous linear map between topological
vector spaces and B ⊆ E is bounded, then α(B) ⊆ F is bounded.
Proof. If U is a 0-neighbourhood in F , then α−1 (U ) is a 0-neighbourhood
in E. Since B is bounded, we find r > 0 such that B ⊆ rα−1 (U ) = α−1 (rU ).
Consequently, B ⊆ rU .
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Lemma B.9.3. If E is a topological vector space, F ⊆ E a vector subspace
and B ⊆ F , then B is bounded in F if and only if B is bounded in E.
Proof. If B is bounded in F , then B = λ(B) is bounded in E, the inclusion
map λ : F → E, λ(x) := x being continuous linear (Lemma B.9.2). Conversely, assume that B is bounded in E, and let U ⊆ F be a 0-neighbourhood.
Then there exists a 0-neighbourhood V ⊆ E such that F ∩ V ⊆ U . The
set B being bounded in E, we find r > 0 such that B ∈ rV . Thus
B ∈ F ∩ rV = r(F ∩ V ) ⊆ rU , showing that B is bounded in F .
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Lemma
B.9.4. Let (Ei )j∈J be a family of topological vector spaces and E :=
Q
E
j∈J j be their direct product, with canonical projections prj : E → Ej .
Then a subset B ⊆ E is bounded if and only if prj (B) is bounded in Ej for
each j ∈ J.
Proof. If B is bounded, then so is prj (B) for each j ∈ J, by Lemma B.9.2 (the
projection prj being continuous linear). Conversely, suppose that prj (B) is
bounded for each j ∈ J. If U ⊆ E is a 0-neighbourhood, thenTthere exists a finite subset F ⊆ J and 0-neighbourhoods Uj ⊆ Ej such that j∈F pr−1
j (Uj ) ⊆
U . After shrinking Uj , we may assume that each Uj is balanced. Since prj (B)
is bounded, we find rj > 0 such that prj (B) ⊆ rj Uj . Set r := max{rj : j ∈ F }.
−1
Then prj (B) ⊆ rj Uj ⊆ rUj and hence B ⊆ pr−1
j (rUj ) = rprj (Uj ) for each
T
j ∈ F , using that Uj is balanced. Thus B ⊆ r j∈F pr−1
j (Uj ) ⊆ rU , showing
that B is bounded.
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Lemma B.9.5. If E is a locally convex space, then a subset B ⊆ E is bounded
if and only if p(B) is a bounded subset of K, for each continuous seminorm
p on E.
Proof. If B is bounded and p : E → [0, ∞[ is a continuous seminorm, then the
ball U := {x ∈ E : p(x) ≤ 1} is a 0-neighbourhood, whence B ⊆ rU for some
r > 0 and thus p(B) ⊆ p(rU ) = rp(U ) ⊆ [0, r]. Conversely, suppose that p(B)
is bounded for each continuous seminorm p. If U ⊆ E is a 0-neighbourhood,
Lemma B.1.6 and Proposition B.4.3 provide a continuous seminorm q such
that V := {x ∈ E : q(x) ≤ 1} ⊆ U . By hypotheses, q(B) ⊆ [0, r] for some
r > 0, whence q(r−1 B) ⊆ [0, 1] and thus r−1 B ⊆ V and so B ⊆ rV ⊆ rU . u
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B.9 Bounded sets and the weak topology
231
Definition B.9.6. The weak topology on a locally convex space E is the
initial topology with respect to the set E 0 of all continuous linear functionals,
i.e., the topology Ow which makes
Y
0
η : (E, Ow ) →
K = KE ,
x 7→ (λ(x))λ∈E 0
(B.9)
λ∈E 0
a topological embedding, where the right hand side is equipped with the
product topology. We write Ew for E, equipped with the weak topology Ow .
The product on the right hand side of (B.9) being a locally convex space,
so is Ew . A subset of E is called weakly open (weakly closed , resp., weakly
bounded ) if it is open (closed, resp., bounded) in Ew .
Note that each λ ∈ E 0 is continuous on Ew , by definition of the weak topology; hence E 0 ⊆ (Ew )0 . The weak topology being coarser than the original
topology, we also have E 0 ⊇ (Ew )0 and thus
E 0 = (Ew )0 .
(B.10)
Lemma B.9.7. Let E be a locally convex space. Then a subset B ⊆ E is
weakly bounded if and only if λ(B) is a bounded subset of K, for each λ ∈ E 0 .
Proof. If B ⊆ Ew is bounded, then λ(B) ⊆ K is bounded for each λ ∈ E 0 =
(Ew )0 , by Lemma B.9.2. If, conversely, λ(B) is bounded in K for each λ ∈ E 0 ,
0
then evλ (η(B)) = λ(B) is bounded for each λ ∈ E 0 (where η : E → KE is
0
as in (B.9), and evλ = prλ the evaluation map KE 3 ξ 7→ ξ(λ)). Therefore
0
λ(B) is bounded in KE by Lemma B.9.4. Since λ(B) ⊆ λ(E), this entails
that λ(B) is bounded in η(E). Therefore B = (η|η(E) )−1 (η(B)) is bounded
in Ew (by Lemma B.9.2).
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Theorem B.9.8 (Mackey’s theorem). A subset B of a locally convex
space E is bounded if and only if it is weakly bounded.
Proof. The map Λ : E → Ew , x 7→ x being continuous, boundedness of B
in E entails boundedness of B = Λ(B) in Ew . For the converse, assume that
B ⊆ E is weakly bounded. Let P be the set of all continuous seminorms on E.
By Lemma B.9.5, B will be bounded if we can show that p(B) is bounded
in R for each p ∈ P. We define Ep := E/Np and let k.kp be the norm on
Ep determined by kαp (x)kp = p(x), as in B.4.2. We choose a Banach space
fp , k.ke ) which is a completion of the normed space (Ep , k.kp ), i.e., we are
(E
p
fp with dense image. Then βp :=
also given a linear isometry γp : Ep → E
f
γp ◦ αp : E → Ep is a continuous linear map such that kβp (x)kep = p(x) for
each x ∈ E. Hence B will be bounded if we can show that kβp (B)kep = p(B)
is bounded, for each p ∈ P. Note that, for each continuous linear functional
ep → K, the composition λ ◦ βp is a continuous linear functional on E,
λ: E
whence λ(βp (B)) = (λ ◦ βp )(B) ⊆ K is bounded. Hence βp (B) is weakly
232
c Helge Glöckner
B Locally convex spaces
fp , and it only remains to show that βp (B) is bounded in E
fp .
bounded in E
We may therefore assume now without loss of generality that E is a Banach
space, and that B ⊆ E is weakly bounded. As we know from functional
analysis on normed spaces, E 0 becomes a Banach space when equipped with
the operator norm. Repeating this procedure, also the bi-dual E 00 becomes a
Banach space. It is well known that the evaluation homomorphism
η : E → E 00
taking x ∈ E to η(x) : E 0 → K, λ 7→ λ(x) is a linear isometry.4 As a consequence of Lemma B.9.3, B will be bounded if we can show that η(B) is
bounded in E 00 . Given λ ∈ E 0 , let evλ : E 00 → K, evλ (ξ) = ξ(λ) be the point
evaluation. Since evλ (η(x)) = η(x)(λ) = λ(x) for each x ∈ E, we obtain
evλ (η(B)) = λ(B), which is bounded in K. Hence, applying the Uniform
Boundedness Principle (as in Rudin, “Real and Complex Analysis,” Theorem 5.8) to the set η(B) of continuous linear maps E 0 → K on the Banach
space E 0 , we find that η(B) is bounded in E 00 , as required.
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B.10 Projective limits
We now define projective limits of topological vector spaces, which are useful
tools in later stages of this course. Our present approach is quite pedestrian;
we postpone a more illuminating definition.
Definition B.10.1. Let (I, ≤) be a directed set, i.e., a partially ordered set
such that, for each finite subset F ⊆ I, there exists j ∈ I such that i ≤ j
for all i ∈ F . A projective system of topological vector spaces (over I) is
a family (Ei )i∈I of topological vector spaces indexed by I, together with a
family (qij )i≤j of continuous linear maps qij : Ej → Ei , indexed by pairs
(i, j) ∈ I × I such that i ≤ j, such that
(a) qii = idEi for each i ∈ I and
(b) qij ◦ qjk = qik for all i, j, k ∈ I such that i ≤ j ≤ k.
Given a projective
system S = ((Ei )i∈I , (qij )i≤j ) of topological vector spaces,
Q
let P := i∈I Ei be their cartesian product (equipped with the product
topology) and define
n
o
E := (xi )i∈I ∈ P : qij (xj ) = xi for all i, j ∈ I such that i ≤ j .
Then E is a vector subspace of P (exercise); equipped with the induced
topology, it becomes a topological vector space called the projective limit
4
This also follows from the Bipolar theorem, noting that the polar U ◦ of the closed
unit ball U ⊆ E is the closed unit ball in E 0 , whence U ◦◦ is the closed unit ball
in E 00 .
B.10 Projective limits
233
of the projective system S. The mappings πi := pri |E : E → Ei (where
pri : P → Ei is the respective coordinate projection) are called the limit
maps. More generally, if F is a topological vector space isomorphic to E and
φ : F → E an isomorphism of topological vector spaces, we call F , together
with the maps qi := πi ◦ φ, a projective limit of S and also write lim Ei := F .
←−
Clearly, projective limits of locally convex spaces are locally convex.
Example B.10.2. N0 , equipped with its usual order, is a directed set. The
sequence (C i ([0, 1], R))i∈N0 , together with the inclusion maps
qij : C j ([0, 1], R) → C i ([0, 1], R) ,
qij (γ) := γ
(B.11)
for i ≤ j, apparently is a projective system. We claim that
C ∞ ([0, 1], R) = lim C k ([0, 1], R) .
←−
(B.12)
Q
i
To see this, let P :=
i∈I C ([0, 1], R). As qij (γ) = η holds for γ ∈
j
i
C ([0, 1], R) and η ∈ C ([0, 1], R) if and only if γ = η (see (B.11)), we get
E := (γi )i∈N0 ∈ P : qij (γj ) = γi if i ≤ j = {(γ)i∈N0 : γ ∈ C ∞ ([0, 1], R)} .
Thus φ : C ∞ ([0, 1], R) → E, φ(γ) := (γ)i∈N0 = (γ, γ, . . .) is an isomorphism of
vector spaces. Then qi := pri ◦φ : C ∞ ([0, 1], R) → C i ([0, 1], R) is the inclusion
map for each i ∈ I, which is continuous by definition of the topologies. Each
component qi of φ being continuous, φ is continuous. Closer inspection shows
that φ is also open (exercise). Hence φ is an isomorphism of topological vector
spaces, and thus (B.12) holds.