Basic Analysis – Gently Done Topological Vector Spaces

74 Basic Analysis
For the last part, we note that, by definition, in a locally convex topological
vector space, 0 has a neighbourhood base of convex sets. By the above, each of
these contains a balanced, convex neighbourhood of 0. The proof is completed by
observing that any neighbourhood of 0 is absorbing.
Example 7.14 Wehavediscussedcontinuouslinearmapssoitisnaturaltoconsider
the possibility of discontinuous ones. In this connection, the existence of a Hamel
basis proves useful in the construction of such various ‘pathological’ examples.
We first consider the existence of discontinuous or, equivalently, by Theorem 7.10,
unbounded linear functionals on a normed space. In some cases such functionals
are not difficult to find. For example, let X be the linear space of those complex
sequences which are eventually zero—thus (a n ) ∈ X if and only if a n = 0 for
all sufficiently large n (depending on the particular sequence). Equip X with the
norm �(a n )� = sup|a n |, and define φ : X → C by
(an ) ↦→ φ((an )) = �
an .
Evidently φ isan unbounded linear functional on X. Another exampleis furnished
by the functional f ↦→ f(0) on the normed space C([0,1]) equipped with the norm
�f� = � 1
0 |f(s)|ds.
It is not quite so easy, however, to find examples of unbounded linear functionals
or everywhere defined unbounded linear operators on Banach spaces. To do
this, we make use of a Hamel basis. We shall consider a somewhat more general
setting. Suppose, then, that (X,T) is an infinite dimensional topological vector
space whose topology T is determined by a countable family P of seminorms and
let Y be any topological vector space possessing at least one continuous seminorm
q, say, such that q(y) �= 0 for some y ∈ Y. Then there is a linear map T : X → Y
such that T is not continuous at any point of X.
To construct such a map, let M be a Hamel basis of X. Then there is a
sequence {un : n ∈ N} of distinct elements in M, since X is infinite dimensional.
Let {yn : n ∈ N} be any sequence in Y, with q(yn ) �= 0. Define T on the un by
T(u n ) = n � p 1 (u n )+···+p n (u n ) � (1/q(y n ))y n
and set Tv = 0 for v ∈ M with v �= u k for any k ∈ N. The map T is extended
to the whole of X by linearity, thus giving a linear map from X into Y. Now,
evidently, q(T(u n )) = n(p 1 (u n )+···+p n (u n )) and so, by Theorem 7.10, T fails to
be continuous, even at 0. T is then discontinuous at every point since continuity at
some z ∈ X is easily seen to imply continuity at 0 (and hence at every point, again
by Theorem 7.10). If we set Y = K, then T is an unbounded linear functional.
One might then ask whether the existence of such discontinuous linear functionals
also holds for any infinite dimensional topological vector space. It turns
Department of Mathematics King’s College, London
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