Let be separable Banach spaces over .
Show that the following is a complete, compatible metric for the strong topology on :
where is a dense sequence in the unit ball of .
It is not difficult to check that distance function given above is a metric. Denote the topology induced by the metric by .
Lemma 1. Let . Given , there is a ball around in , such that
, implies .
Proof of Lemma 1.
Let . If is a positive real number, then
which is the same as
Hence, implies .
So if we let , we have found a ball with the above property.
Lemma 2. .
Proof of Lemma 2.
Let be a basis element for .
Let . Let . Let . Since is dense in the unit ball of ,
there exists such that .
Now, for any ,
By Lemma 1, there is a ball around such that implies .
Then, for , , which yields .
Therefore, if we choose , we have such that implies .
Let . Then .
Lemma 3. .
Let be an arbitrary ball of . Choose such that
Let . Then .
That is, , so .
By Lemmas 2 and 3, .
Finally, we check that is a complete metric. Let be a Cauchy sequence in . That is, for all there exists
such that implies . That is,
So, in particular, if . So is a Cauchy sequence in .
So it converges to some , which we will name . This defines . Now, consider the set . It is closed under -linear combinations, and we easily extend to , by defining , for ; it is easily verified satisfies , for . Furthermore, it is clear that is dense in . Then, by this result, extends to a linear map in , which we will also call .
by the Dominated Convergence Theorem. If , we see that .
Thus , and is a complete metric.