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 .

*Proof*.

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

implies

Hence,

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 ,

Therefore, .

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 .

So

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 .

Then,

by the Dominated Convergence Theorem. If , we see that .

Thus , and is a complete metric.