Proving Nondecreasingness of Fn When Converging to F

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SUMMARY

The discussion focuses on proving that if a sequence of functions {Fn} converges to a function F on the interval [a,b] and each Fn is nondecreasing, then F is also nondecreasing. The participants clarify that proving uniform convergence is not necessary; instead, pointwise convergence suffices for a more general theorem. A suggested approach involves selecting two points c and d within the interval such that c < d, and using the definition of pointwise convergence to establish the nondecreasing property of F.

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  • Understanding of pointwise and uniform convergence of functions
  • Familiarity with nondecreasing functions and their properties
  • Basic knowledge of limits and epsilon-delta definitions in analysis
  • Experience with sequences of functions and their convergence behavior
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  • Study the definitions and differences between pointwise and uniform convergence
  • Learn about the properties of nondecreasing functions in real analysis
  • Explore epsilon-delta proofs in the context of function limits
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Homework Statement


Prove: If {Fn} converges to F on [a,b] and Fn is nondecreasing for each n\in 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 \in N, x1 \leq x2 \Rightarrow Fn(x1) \leq Fn(x2).

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

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

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

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

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


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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