2. Topological Vector Spaces
We begin with a summary of some essential notions and facts about topological vector spaces. We refer to [
11] for proofs.
By a topological vector space we mean a real or complex vector space X, equipped with a Hausdorff topology for which the operations of addition and multiplication are continuous.
A set C in a vector space (topological vector space) is said to be convex if given any , the linear combination for all . A locally convex space is a topological vector space in which every neighborhood of 0 contains a convex neighborhood of 0. A convex neighborhood V of 0 is balanced if it closed under multiplication by scalars of magnitude . Every convex neighborhood of 0 contains a balanced convex neighborhood of 0, and it is the latter type of neighborhood that is most useful.
2.1. Seminorms from Neighborhoods
Let V is a balanced convex neighborhood of 0 and define by way of
Since
V is a neighborhood of 0 there is a small enough positive scaling of
x that makes it fall within
V, and so there is a large enough scaling of
V that includes
x; thus
. Since
V is balanced and convex it follows that
is a
semi-norm:
for all
and scalar
λ.
A set is said to be bounded if D lies inside a suitably scaled up version of any given neighborhood of 0; thus, D is bounded if for any neighborhood V of 0 there is a such that . If X is locally convex then is bounded if and only if is a bounded subset of for every convex balanced neighborhood U of 0.
As usual, a set K in the topological vector space X is said to compact if every open cover of K has a finite subcover. A sequence in X is said to be Cauchy if the differences eventually lie in any given neighborhood of 0, and X is said to be complete if every Cauchy sequence converges. Also, X is said to be separable if there exists a countable set such that every nonempty open subset of X contains at least one element of D.
If the topology on X is metrizable then there is a metric that induces the topology, is translation-invariant, and for which open balls are convex. If, moreover, X is also complete then such a metric can be chosen for which every metric-Cauchy sequence is convergent.
2.2. Dual Spaces
The dual space of a topological vector space X is the vector space of all continuous linear functionals on X (these functionals take values in the field of scalars, or ).
If X is locally convex then the Hahn-Banach theorem guarantees that if X itself is not zero.
There are several topologies of interest on
. For now let us note the two extreme ones:
Using
to denote the field of scalars, the
weak topology on
is the smallest topology on
for which the evaluation map
is continuous for all
. This topology consists of all unions of translates of sets of the form
with
D running over finite subsets of
X, and
ϵ over
.
The
strong topology on
consists of all unions of translates of sets of the form
with
D running over all bounded subsets of
X and
ϵ over
.
The remainder of this section provides context for the rest of the paper but is not actually used later.
2.3. An Associated Chain of Banach Spaces
It is clear that the set
of vectors of semi-norm zero form a vector subspace of
X. The quotient space
is a normed linear space, with norm
induced from
:
where
denotes any element in
X that projects down to
a; the value
is independent of the specific choice of
. We denote by
the Banach space obtained by completion of this space:
and let
denote the quotient projection, viewed as a map of
X into
:
In some cases of interest itself is a norm, in which case is actually an injection into the completion of X relative to this norm.
Thus for any locally convex topological vector space X there is associated a system of Banach spaces , these arising from convex balanced neighborhoods U of 0 in X.
If
U and
V are convex, balanced neighborhoods of 0 in
X, and
, then
and so there is a well-defined contractive linear mapping
specified uniquely by requiring that it maps
to
for all
.
A complete, metrizable, locally convex space X is obtainable as a ‘projective limit’ of the Banach spaces and the system of maps . We will not need a general understanding of a projective limit; for our purposes let us note that if for each convex balanced neighborhood U of 0 in X an element is given, such that whenever , then there is an element such that for all .
With X as above, let us choose a sequence of neighborhoods of 0, with each balanced and convex, such that every neighborhood of 0 contains some and
By (
6) we have the chain of spaces
where the mappings are of the form
. Thus we can view
X as obtained from this sequence of spaces.
Complete metrizable nuclear spaces arise in the case where the spaces are Hilbert spaces and the mappings are all Hilbert-Schmidt, mapping the closed unit ball in into an ellipsoid in whose semiaxes lengths form a square-summable sequence.
3. The Nuclear Space Structure
We work with a real or complex infinite dimensional topological vector space equipped with additional structure as we now describe. We assume that there is a sequence of inner-products for , on such that
The completion of in the norm is denoted , and inside this Hilbert space the completion of with respect to is a dense subspace of denoted by . We assume that is separable, and that is the intersection of all the spaces . Thus,
(This is a special form of the chain of spaces seen in (
7).) Furthermore, we assume that each inclusion
is a Hilbert-Schmidt operator (such operators are discussed below in
Section 3.1). Then there is an orthonormal basis
in
for which
We denote by
is the topology on
given by
. By (
8), the identity map
is continuous, for
, and so
The inclusions here are strict because of the Hilbert-Schmidt assumption made above (bearing in mind that is infinite-dimensional).
Because of the relations (
11), the union
τ of all the topologies
is also a topology. Thus a subset of
is open if and only if it is the union of open
-balls with
p running over some subset of
. This topology makes
a topological vector space. Moreover,
, with this topological vector space structure, is the
projective limit from the inclusions
. We shall not use any general facts or theories of projective limits.
We take as the nuclear space we work with; thus in our discussions the nuclear space comes equipped with the additional structure of the Hilbert norms .
An infinite-dimensional topological vector space is never locally compact. That is, points do have necessarily have compact neighborhoods. However, a nuclear space is an excellent substitute in the infinite-dimensional case because it satisfies the Heine-Borel property as we shall see in Fact 14.
3.1. Hilbert-Schmidt Operators
We will use Hilbert-Schmidt operators, and recall here some standard facts about them. For a linear operator
between Hilbert spaces, define the Hilbert-Schmidt norm to be
where
is some orthonormal basis of
H; if
the Hilbert-Schmidt norm of
T is, by definition, 0. If
then
T is said to be a
Hilbert-Schmidt operator. The following facts are well known (see, for instance, [
12,
13,
14,
15]):
Fact 1. For a linear operator between Hilbert spaces, the Hilbert-Schmidt norm given in (12) is independent of the choice of the orthonormal basis . If T is Hilbert-Schmidt, then: - (i)
T is bounded and ;
- (ii)
the adjoint is also Hilbert-Schmidt and ;
- (iii)
the image of any closed ball in H under T is a compact subset of K;
- (iv)
if T is injective then H is separable;
- (v)
if then there is an orthonormal basis of H consisting of eigenvectors of (see [15] (p. 28)).
In the context of (v) consider an orthonormal sequences
in
H consisting of eigenvectors of
TT:
for all
, where
are eigenvalues of
TT; then
and, similarly,
where
is the inner-product on
H.
Some unexpected facts about nuclear spaces are connected with the following simple observation:
Fact 2. Let be Hilbert-Schmidt map between Hilbert spaces. Then there is a sequence of points in H which does not converge in H but whose image by j in K is convergent.
Proof. An orthonormal sequence in H is not Cauchy and hence not convergent, but the sequence converges to 0 in K because is convergent. ☐
3.2. Dual of a Nuclear Space
A linear functional on
is continuous with respect to
if and only if it extends to a (unique) continuous linear functional on
. Thus we may and will make the identification
where the left side is the vector space of all linear functionals on
that are continuous with respect to the norm
. To stress that we treat the dual space
as a Hilbert space, i.e., having the
strong topology, we denote it by
; the norm on this space can also be obtained as
The topological dual of
is
where we are viewing each
as
.
It is readily checked that the inclusion is the adjoint of the inclusion map and is thus also Hilbert Schmidt.
In addition to the weak and strong topologies, there is also the inductive limit topology on , which is the largest locally convex topology for which the inclusions are all continuous.
Fact 3. The strong topology and the inductive limit topology on are the same.
4. Balls and Cubes
In this section,
p denotes an element of
. We work, as before, with the space
, equipped with inner-products
, with the Hilbert-Schmidt condition explained after (
9).
4.1. Open and Closed Balls
Letting
R denote any positive real number, we examine properties of the ‘open’ balls
and the “closed” balls
for every
. Any open or closed ball in a normed linear space is clearly bounded.
Fact 4. For all , is open in .
Proof. The set is in fact just the open R-ball in the norm on and hence is in , and therefore also in the topology τ on . ☐
Fact 5. The set is not bounded in -norm. It is not bounded in .
Proof. Let be an orthonormal basis of lying inside (a maximal -orthonormal set in , which is dense inside , is necessarily also maximal in and is hence an orthonormal basis of ). Then, since the inclusion is Hilbert-Schmidt, the sum is finite, and so the lengths tend to 0. Thus the vectors
which are all in
, have
-norm going to
∞. In particular,
is not bounded in
. Consequently
is also not bounded in
. ☐
The next observation provides some compact sets in .
Fact 6. The set is compact in , for any .
Proof. A closed ball is closed in the topology , and so it is a closed subset of . If W is a neighborhood of 0 in , then there is a and an such that ; hence, if . Thus, is both closed and bounded and hence, by the Heine-Borel property proved below in Fact 14, it is compact in ☐
We turn to balls in the dual spaces and . Unless stated otherwise, we equip each Hilbert space dual with the strong topology, which is the same as the Hilbert-space topology.
Fact 7. The set , with any topology making it a topological vector space, contains no non-empty open subset lying entirely inside , for . More generally, in a topological vector space a proper subspace has empty interior.
Proof. If a proper subspace Y of a topological vector space X contains an open set U with a point y lying in U, then is a neighborhood of 0 in X lying entirely inside Y. Then the union of all multiples of would be all of X, and hence Y would be all of X. ☐
Fact 8. The set is open in . More generally, is open in for , with .
Proof. The inclusion induces the adjoint which is also continuous. So is open in . Identifying the dual spaces and , the map is just the inclusion map. Thus is the part of the ball that lies inside ; that is, , which is thus open in . ☐
Fact 9. The set is not open in in the strong (Hilbert) topology. More generally, is not strongly open in for with .
Proof. Note that
is a proper subspace of
for any
by (
17). By Fact 7,
must have an empty interior in
. Since
,
must also have an empty interior in
. Thus
is not open in
. ☐
Fact 10. The set is weakly, and hence also strongly, closed in . If , with , and , then is compact in the Hilbert-space . Moreover, is strongly, and hence weakly, compact as well as sequentially compact in .
Proof. First, since we have
Thus is the the intersection of weakly closed sets and is thus weakly closed. It follows that is also strongly closed in .
The inclusion map being Hilbert-Schmidt, so is the adjoint inclusion , and thus is compact inside with the strong (Hilbert-space) topology. By continuity of the inclusion map it follows that is compact in with respect to the strong topology.
Continuity of the inclusion map also implies that is compact in with respect to the strong topology.
Since is compact in the metric space it is sequentially compact in , i.e., any sequence on has a subsequence which is convergent in . From continuity of the inclusion it follows that such a subsequence also converges in . Thus, is strongly, and hence also weakly, sequentially compact in . ☐
4.2. Cubes
We use the the term “cube” for what might more properly be called a “box” in a linear space, bounded by “walls”. For the following observation about small closed cubes with nonempty interior we focus on a nuclear space
for which there is an orthonormal basis
of
, the vectors of which all lie in
and are orthogonal within each
, with
for all
and
. This is the structure in many applications, including the Schwartz space (see
Section 8).
Fact 11. Suppose is an orthonormal basis of which lies in , and assume that the vectors form an orthonormal basis of , where . Assume also that
Let C be the closed cube in given by
Then C contains a strong neighborhood of 0 and hence is not compact. However, is compact for every finite-dimensional subspace of .
Proof. Let , for some , and . Then, for every n,
To make this
, we should take
and this should hold for all
. Thus, we could take
, and, for
:
wherein we have dropped all the later terms
with
, as these are all
. Thus, with this choice of
, the cube
contains each open ball
and hence also the convex hull of their union, which is a neighborhood of 0 in the inductive limit topology (see [
6], Chapter V,
Section 2).
Now consider a finite-dimensional subspace
F of
, and
a point in
. Then
for all
. Let us
assume, for the moment, that there is an orthonormal basis of and a fixed such that each on F is a linear combination of the functionals , say
This shows that is a bounded subset of F. Since C is weakly closed and so is any finite-dimensional subspace, is weakly closed in and hence weakly closed in the induced topology on F; but on the finite-dimensional subspace F there is only one topological vector space structure, and so is closed. Being closed and bounded in F, it is compact.
It remains to prove the algebraic statement assumed earlier. Suppose that the algebraic linear span of the functionals on F is not the entire dual of F; then they span a proper subspace of and hence there is a non-zero vector on which they all vanish, but this would then be a vector in whose inner-product with every is 0 and hence f would have to be 0. Thus the algebraic linear span of the functionals on F is the entire dual , and so, in particular, any element of is a finite linear combination of the functionals . This establishes the assumption made earlier. ☐
5. Facts about the Nuclear Space Topology
We work with an infinite-dimensional nuclear space
with structure as detailed in
Section 3. While there are some treacherous features, such as the one in Fact 12, a nuclear space has some very convenient properties that make them almost as good as finite-dimensional spaces.
Fact 12. There is no non-empty bounded open set in .
Proof. Suppose U is a bounded open set in containing some point y. Then is a bounded open neighborhood of 0 in and so contains some ball , with . By Fact 5 this is impossible. ☐
Fact 13. The topology on is metrizable but not normable.
Proof. If there were a norm then the unit ball in the norm would be a bounded neighborhood of 0, contradicting Fact 12. The translation-invariant metric on given by
induces the topology on
. ☐
One immediate consequence of the preceding observations is that an infinite dimensional Banach space is not a nuclear space.
If is a continuous linear map between topological vector spaces and B is a bounded subset of X then is a bounded subset of Y, for if V is a neighborhood of 0 in Y then for some scalar t and hence . This observation is used to prove the following useful fact about nuclear spaces.
Fact 14. The nuclear space has the Heine-Borel property: every closed and bounded set is compact.
Proof. Let B be a closed and bounded subset of . Take any . Since the inclusion is continuous, B is bounded in . Since the inclusion is Hilbert-Schmidt it follows that B, as a subset of , is contained in a compact set. Let be a sequence of points in B. Then there is a subsequence that converges in . Applying the Cantor process of extracting repeated subsequences, there is a subsequence that is convergent in for every . By continuity of the inclusions it follows that the limit is the same for every p, i.e., there is a point , such that the subsequence converges to y in all the . If V is a neighborhood of 0 in then there is a and an such that . Since in we have for large n, and so in . Thus, a closed and bounded subset of is sequentially compact; since is metrizable, sequential compactness is equivalent to compactness for any subset of . ☐
6. Facts about the Dual Topologies
We turn now to the dual of an infinite-dimensional nuclear space
, and continue with the notation explained earlier in the context of
Section 3.
Let us recall [
11] the
Banach-Steinhaus theorem (uniform boundedness principle): If
S is a non-empty set of continuous linear functionals on a complete, metrizable, topological vector space
X, and if, for each
the set
is bounded, then for every neighborhood
W of 0 in the scalars, there is a neighborhood
U of 0 in
X such that
for all
. As consequence,
if are such that exists for all then f is in .The following facts in this section are standard and can be found in any of [
5,
6,
7,
8,
10], among other places.
Fact 15. If X is a complete, metrizable topological vector space, then the dual space is complete with respect to both strong and weak topologies. In particular, the dual of a nuclear space is complete with respect to both the strong and weak topologies.
Proof. Let be a non-empty collection of subsets of X, closed under finite unions and whose union is all of X. Let be the topology on X whose open sets are unions of translates of sets of the form
with
A running over
and
ϵ over
. If
is the set of all finite subsets of
X then
is the weak topology; if
is the set of all bounded subsets of
X then
is the strong topology.
Suppose is Cauchy sequence in for the topology . This means that, for any , the sequence of functions is uniformly Cauchy, and hence uniformly convergent, for every . Let . By Banach-Steinhaus, f is continuous. A -neighborhood U of f contains a set of the form , and the uniform convergence on A implies that for large n, and so for large n. Thus, in . ☐
Fact 16. Any weakly open neighborhood of 0 in the dual of an infinite dimensional locally convex topological vector space contains an infinite-dimensional subspace.
Proof. If Y is a vector space over a field F, and are linear functionals, then the mapping
has kernel infinite-dimensional if
Y is infinite-dimensional. We apply this to
. A weakly open neighborhood of 0 in
contains contains a set of the form
for some
, and open neighborhoods
of 0 in the scalars, and so
B contains the
. ☐
The weak dual of an infinite-dimensional Banach space is not metrizable. In a similar vein there is the following negative result:
Fact 17. The weak topology and the strong topology on are not metrizable.
Proof. Consider the nuclear space
with topology generated by norms
for
as in
Section 3. Let
be the subset of
given by the closed ball of radius
R, center 0, for the
-norm, and let
be the set of all linear functionals
f in
which map
into scalars of magnitude
. Each set
, being the intersection of weakly closed sets, is a weakly closed set, and hence also strongly closed, in
. The union of the sets
with
and
is all of
. Now
lies in the proper subspace
of
, and so has empty interior in any topology on
which makes
a topological vector space by Fact 7. Thus,
, which is a complete topological vector space with respect to both weak and strong topologies, is the countable union of nowhere dense sets, and hence the weak and strong topologies on
are not metrizable. ☐
In a sense, sequences do not detect the difference between the weak and strong topologies on the dual space:
Fact 18. A sequence in is weakly convergent if and only if it is strongly convergent.
Proof. Since the weak topology is contained in the strong topology, strong convergence implies weak convergence. For the converse, let
be a sequence in the dual
of the nuclear space
, converging weakly to
. By Banach-Steinhaus,
is a uniformly continuous set of functions on
and hence, as may be checked, the convergence
is uniform on compact subsets of
. The closure
of the bounded set
D is bounded (see Rudin [
11] (Theorem 1.13(f))) and hence also compact by the Heine-Borel property for nuclear spaces Fact 14. So the sequence
is uniformly convergent on
, and hence also on
D, and therefore it is strongly convergent as a sequence in
. ☐
6.1. Bounded and Compact Sets in the Dual of Nuclear Space
We now examine bounded and compact sets in . In particular, we will see that weakly and strongly bounded sets are one in the same. The same goes for weakly and strongly compact sets.
Fact 19. A set is strongly bounded if and only if B is bounded on each bounded set in .
Proof. Let be strongly bounded and let D be a bounded set of . Consider the local base set of (for the strong topology) given by
Since B is bounded there exists an such that or, equivalently, . Then for any we have that . Thus for any . Therefore B is bounded on the set D.
Suppose B is bounded on each bounded set . Consider the open set containing 0 in . By hypothesis, . So, for any we have that when . Therefore or, equivalently, . Hence B is bounded. ☐
Fact 20. A set B in is strongly bounded if and only if there exists p such that B is bounded on .
Proof. Consider the local base for given by
By contradiction suppose that B is not bounded on for any integer . Then for every p there exists and a such that . By construction, the sequence goes to 0 in and thus must be bounded. So there must exists a positive number M such that for all and all . This contradicts the way by which and were chosen.
Conversely, let be bounded on some . Take a bounded set D in . Then for some . Thus B is bounded on D and is therefore bounded on . ☐
The following result provides us with a way of studying bounded sets in in terms of the norms on .
Fact 21. A set is strongly bounded if and only if there exists an integer such that and B is bounded in the norm on .
Proof. The converse direction is the easier of the two. If and for all , then B is bounded on and consequently bounded on . Thus Fact 20 applies to tell us that B is strongly bounded.
Now for the forward direction: suppose B is a strongly bounded set in . Then by Fact 20 there exists an integer such that B is bounded on the set . That is, there is an such that for all and all .
Consider the set in given by Since is dense in we have that
From the above we see that for any and any nonzero vector we have
Therefore for any . Thus for any we have that and . ☐
Fact 22. A set is weakly bounded if and only if B is strongly bounded.
Proof. The converse direction is straightforward: Assuming a set B is strongly bounded, then it must also be weakly bounded since the strong topology on is finer than the weak topology on .
For the forward direction, we let
be a weakly bounded set. Define the set
as follows:
Note that C is the intersection of convex, closed, and balanced sets. Therefore C is also a convex, closed, and balanced set.
Also note that C is absorbent: Take . Since B is weakly bounded there must exist such that where . For any α satisfying we have that for all . Hence or equivalently . This gives us that C is absorbing.
Since
C is absorbing we have that
. Knowing that
is a complete metric space (Facts 13 and 15) we can apply the Baire category theorem to see that
C is not nowhere dense. Thus the interior of
C,
, is not empty. Take
and a set
such that
. Because
C is balanced we have
and since
we have
Now C is convex with and . So we must have that the convex hull of is contained in C. But this convex hull contains : Observe, for any we have that
Therefore .
Since B is bounded on C and we have that B is bounded on and is thus strongly bounded. ☐
Fact 23. A set is weakly compact if and only if K is strongly compact.
Proof. Since the strong topology is finer than the weak topology, this immediately gives us that if K is strongly compact, then K is weakly compact.
For the other direction suppose K is weakly compact. Then K is weakly bounded. Thus we can apply Fact 22 to obtain that K is strongly bounded. By Fact 21 there exists an integer and a real number R such that and for all x in K.
Since K is weakly compact in , K is also weakly closed in Therefore, because the inclusion map is continuous we have is closed in . Also, using that we have is closed in , where again we use the continuity of the inclusion map.
Because K is bounded in , we get has compact closure in . (Here we have used that the inclusion map from to is Hilbert-Schmidt and is thus a compact operator). However, we have seen K is closed in . Hence K is compact in . The continuity of the inclusion map from to gives us the continuous image is compact in . ☐
6.2. The Dual Space is Sequential
Now we examine the viability of using sequences to study the dual space . It turns out that the topology on the dual space does allow us to study topological properties such as closure and continuity in terms of sequences. We have the following result:
Fact 24. The dual space , equipped with the strong/inductive limit topology, is a sequential space: if a subset C of is such that every convergent sequence in C has limit lying in C then C is closed.
To prove this result we make use of the following Lemma about Hilbert spaces that is fairly straightforward to prove.
Lemma 1. Let H be a Hilbert space. Suppose K is a compact set and C is a closed set in H. There exists and corresponding closed (or open) ball such that
Proof. (of Fact 24) Suppose C is sequentially closed in . We show C is closed by selecting a point outside of C and showing that this point is also not in the closure of C. Let . Because , for some integer . Since C is sequentially closed, then using the inclusion map we have is sequentially closed in . Thus is closed in the Hilbert space . Since there exists such that
where
.
Now consider the inclusion map . Then is closed and compact in (see Fact 8).
By Lemma 1 there exists such that
Repeating this process we obtain such that
Consider the set N given by
We now show
N is convex:
Take
. Recalling that
when
we have the union of sets making up
N is an increasing union. Thus there exists a
k such that
Note that the above is a sum of convex sets and is consequently convex. Hence, for any
, we have
is also in
and therefore also in
N. So
N is convex.
N contains where for :
For we have and . When we have for some positive integer k and thus . For we have using that and .
Therefore N contains the convex hull of (where for ), which is a neighborhood of 0 in the inductive limit topology.
By construction we must have . Thus f is not a member of the closure of C. Hence C is closed. ☐
An argument very similar to the above can be used to demonstrate that is a k-space. That is, is closed if and only if is compact for any compact set .
6.3. Separability
We now examine the separability of the dual space .
Fact 25. The space is separable when is endowed with the strong, inductive limit, or weak topology.
Proof. We first need a candidate for a countable dense set. Since and each is a separable Hilbert space we let be a countable dense subset of and form the countable set
We will show Q is dense in .
Take
. Then
for some
p. Let
U be an open set containing
f in
(
can be endowed with the weak, strong, or inductive limit topology). Then
is an open set about 0. We know the inclusion maps from
are continuous when
has the weak, strong, or inductive limit topology. Therefore
is an open set about 0 in
and as such there exists an
such that
or equivalently
Since is dense in there exists such that
That is, . Thus showing that Q is dense in and is separable. ☐
6.4. First Countability (Lack Thereof)
A topological space is said to be first countable if each point has countable local base. It is said to be second countable if the topology has a countable base. In this section we seek to determine if the dual space is first countable. From previous results we know is separable (Fact 25). Also, we saw that can be constructed as the inductive limit of separable Hilbert spaces, which are first (and second) countable. Unfortunately though, does not retain the property of first countability.
Fact 26. The dual space is not first countable.
Proof. Assume, by contradiction, that
is first countable. By Fact 25
is separable. The following Lemma tells us that under these assumptions
must be second countable. This lemma, along with its proof, is a standard result that can be found in many treatments of topological vector spaces, including [
5,
6,
7,
8,
10].
Lemma 2. If X is a separable first countable topological vector space, then X is second countable.
Proof. Let be a countable dense subset of X and let be a balanced countable local base. We assert that the collection is a countable base for the topology on X. To see this let U be an nonempty open set in X with . Since is a base, there exists an integer such that . Using continuity of addition there exists an integer such that
Consider the open set about x given by . Because is a dense set there exists an integer such that .
We first show that . Since we have . Since is balanced we must also have . Thus .
Now we demonstrate that . Take where . Then write z as
Noting that
and
gives us that
using (
22). ☐
Under the assumption that
is first countable, we have established that
is second countable. Also, we recall that
is a Hausdorff topological vector space and thus regular in the strong or weak topology. Therefore Urysohn’s Metrization theorem [
17] tells us
is metrizable. This contradicts Fact 17. ☐
Recall that in Fact 24 we saw the dual space is a sequential space. Thus, is a (somewhat rare) example of a topological space that is sequential but not first countable.
7. Continuous Functions
Linear functionals automatically stand a better chance of having continuity properties in locally convex spaces because they map convex sets to convex sets. In this section we look at some examples of nonlinear functions.
Fact 27. If X is an infinite-dimensional topological vector space (such as ) then the only continuous function on X having compact support is 0.
Proof. This is because any compact set in X has empty interior, since X, being infinite-dimensional, is not locally compact. ☐
Fact 28. There is a weakly continuous function S on which satisfies , and S equals 1 exactly on .
which is clearly weakly continuous. Then
is also weakly continuous, lies in
, and is equal to 1 if and only if
. Then each finite sum
is weakly continuous, has values in
, and has maximum possible value
exactly when
is
. Hence the uniform limit
is a weakly continuous function, with values in
, and is equal to 1 if and only if all
are
, i.e., on
. ☐
For the following recall Fact 11. We assume that there are vectors which form an orthonormal basis of and, moreover, form an orthonormal basis of , for all , where
We require monotonicity and a Hilbert-Schmidt condition
Fact 29. For the cube C in Fact 11, with assumptions as stated in Fact 11, the function f given on by
is continuous but not weakly continuous. Proof. Suppose and where x is on the ‘boundary’ of the cube, i.e., for some . Since in we have in for some . Hence .
Now
and the last term goes to 0 as
.
Now consider in . Then for some N we have for all , is outside the cube. Thus for all and hence as .
Finally, consider in . Then in for some p. In particular,
Lemma 3. There exists an such that for any with we have
Proof. Since we have,
Since
p and
M are fixed, there must be an integer
N such that for all
because
Hence, the infimum is realized in the first
N terms:
☐
Returning to the proof for Fact 28, by Lemma 3 to we have
Thus, f is sequentially continuous and hence continuous. ☐
8. Schwartz Space
The paper [
18] examines the Schwartz space of test functions and the dual space of distributions. There are, of course, many other references for the Schwartz space, including, but not limited to [
15,
19,
20,
21]. The Schwartz space is perhaps the best example of a nuclear space to which all the results of this article apply. In particular, the Schwartz space satisfies the structure introduced in Fact 11. In this section we briefly provide an overview of the Schwartz space from [
18].
We denote the Schwartz space by and define it as the space of real-valued infinitely differentiable rapidly decreasing functions on . However, here we outline how to reconstruct the Schwartz space as a nuclear space arising from and the number operator .
In [
18] we see that orthonormal basis for
is formed by the functions
which are eigenfunctions of
N. In particular,
Using this orthonormal basis and the operator
N, an inner-product can be defined for any
for
. (Note:
is a Hilbert-Schmidt operator.) Using
to denote the norm corresponding to the above inner product we complete
with respect to these norms to form the spaces
. These observations give us
and
just as in (
8) and (
9).
Also, based on the equation in (
27) and (
26),
forms an orthogonal basis for each
. In particular the structure introduced in Fact 11 is satisfied with
. Hence all of the results of this article can be applied to the Schwartz space and its dual space.