- #1

Demon117

- 165

- 1

**R**, with the norm(g) = sup|f|. Assume without proof that the norm holds, so that the function d(f,g)=norm(f - g) is a metric.

Prove that the vector subspace F={f in F | f(n) -->0 as n --> infinity} is closed in E.

Here is what I have come up with:

Consider the sequence (f_k) in F, such that f_k --> f. It must be shown that f is in F also.

For such (f_k) in F, it follows that (f_k)(n) -->0 as n --> infinity. Since f_k --> f (by a problem done earlier), it follows that f(n) -->0 as n --> infinity. Since the latter holds, by definition of F it follows that f is contained within. Hence F is closed in E.

This doesn't look right at all. I just wondered what kinds of advice anyone has.