ABSTRACT
This proves that if E is bounded in 2ll(!l) then condition (i) must hold. Condition (ii) follows from the fact that the topology of rz!JK is the topology it inherits as a subspace of 2ll(!l). 0
Conversely, if <Pk-- 0 in rz!J(!l) then (<f>k) is a bounded sequence in Qn(!l), as we have seen in Section 1.2. From Theorem 2.2 it lies in rz!JK for some K C !1 and (i) follows. But then <Pk-- 0 in the subspace topology of rz!JK and (ii) follows. 0
The disadvantage of the topology that we have defined on C;'(ll) is that it is not metrizable. This may be seen from the following argument: Assuming that d is a metric which defines the topology of 2ll(0), let !1 = U Kn with Kn compact and Kn C kn+l for all n. Choose <f>n E £ll(!l) such that supp <Pn $Z' Kw Since multiplication by a constant is a continuous mapping from 2ll(ll) into 2ll(0), we can find An > 0 small enough so that d(O,A,.<j>,.) < lin for every n. This means that the sequence A,.<f>,.- 0 in 2li(!l). But this is not possible since supp(An<f>n) cannot be contained in a single compact subset of n.
