Prove: {Fn} converges to F on [a,b] and Fn is nondecreasing for all n; show F is too

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  • #1
bobbarker
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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?
 

Answers and Replies

  • #2
LeonhardEuler
Gold Member
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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?

Since a sequence of functions that converges uniformly also converges pointwise, you actually want to prove it for the case of pointwise convergence (A weaker hypothesis gives a more general theorem). It will also be much more direct to prove this way since you will be comparing two points and taking limits at those points.
 
  • #3
Dick
Science Advisor
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Here's a hint. Suppose f(c)>f(d) with c<d. Pick epsilon equal to |f(c)-f(d)|/2. Now remember what pointwise convergence means.
 

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