- #1

bobbarker

- 8

- 0

## Homework Statement

Prove: If {Fn} converges to F on [a,b] and Fn is nondecreasing for each n[tex]\in[/tex] N, then F is nondecreasing.

## Homework Equations

n/a

## The Attempt at a Solution

First, it doesn't say if Fn converges pointwise or uniformly, so I'm not entirely sure how to deal with that. Just prove uniformly and then it holds for pointwise as well?

My work so far:

Suppose Fn converges to F and Fn is nondecreasing: i.e., for all n [tex]\in[/tex] N, x1 [tex]\leq[/tex] x2 [tex]\Rightarrow[/tex] Fn(x1) [tex]\leq[/tex] Fn(x2).

So [tex]\forall[/tex] [tex]\epsilon[/tex] > 0, [tex]\exists[/tex] N1 [tex]\in[/tex]N, such that [tex]\forall[/tex] n [tex]\geq[/tex] N1, sup{|Fn(x)-F(x)|: x [tex]\in[/tex] [a,b]} < [tex]\epsilon[/tex].

Equivalently, ||Fn(x) - F(x)||[tex]_{}[a,b][/tex] < [tex]\epsilon[/tex]

| ||Fn(x)||[tex]_{}[a,b][/tex] - ||F(x)||[tex]_{}[a,b][/tex] | < [tex]\epsilon[/tex] by the reverse triang. inequal

||F(x)||[tex]_{}[a,b][/tex] - [tex]\epsilon[/tex] < ||Fn(x)||[tex]_{}[a,b][/tex] < ||F(x)||[tex]_{}[a,b][/tex] + [tex]\epsilon[/tex]

I feel like I'm heading down a dead end. Any ideas?