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Limsup h.w. proofs

  • Thread starter aschwartz
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1. Homework Statement

1. A function f(x) is said to be monotonic increasing in A if for all x1, x2 ∈ A, x1≤x2 implies f(x1)≤f(x2).

Prove that if f(x) is monotonic increasing in R [f: R→R] and c is a cluster point of R then the limit of f(x) as x→c[tex]^{-}[/tex] exists (might be +∞).

2. s(δ) = sup{f(x) :0<|x-c|<δ}

s(δ) is a monotonic decreasing function, hence based on previous result 1, lims(δ) as δ→0[tex]^{+}[/tex] = L[tex]^{+}[/tex], which is defined to be the limsupf(x) as x→c.

Prove:

A. If L[tex]^{+}[/tex] = limsupf(x) as x→c, then [tex]\exists[/tex] a sequence x[tex]_{n}[/tex], such that as x[tex]_{n}[/tex]→c, f(x[tex]_{n}[/tex])→L[tex]^{+}[/tex].

B. If x[tex]_{n}[/tex]→c and x[tex]_{n}[/tex]≠c and f(x[tex]_{n}[/tex])→L then L≤L[tex]^{+}[/tex].

C. Similarly define L[tex]^{-}[/tex] = liminff(x) as x→c. [This is a monotonic increasing function.]

Prove that limf(x) as x→c = L if and only if L[tex]^{+}[/tex] = L[tex]^{-}[/tex] = L.


2. The attempt at a solution

Ok, so this is what I have so far - I was able to get #1, but got stuck with the proofs for #2.

1. If f(x) is monotonic increasing (decreasing) then limf(x) as x→c[tex]^{-}[/tex] exists.
f: R→R
c ∈R

Case 1: (Proving a lefthand limit exists for a monotonic increasing function)

Let L = sup{f(x) : x<c}. We want to show that ([tex]\forall[/tex][tex]\epsilon[/tex]>0) ([tex]\exists[/tex][tex]\delta[/tex]) ([tex]\forall[/tex]x ∈ R) (0<c-x<δ ⇒ |f(x) - L|<[tex]\epsilon[/tex].

Consider the interval (L - <[tex]\epsilon[/tex], L). [tex]\exists[/tex]x[tex]_{1}[/tex], such that x[tex]_{1}[/tex]<c and L - [tex]\epsilon[/tex]<f(x[tex]_{1}[/tex])≤L [because otherwise L is not the sup, but L - [tex]\epsilon[/tex] would be the sup! So therefore, f(x[tex]_{1}[/tex]) must exist in between those two numbers.]

Let [tex]\delta[/tex] = c - x[tex]_{1}[/tex].

[tex]\forall[/tex]x ∈ R if 0<c-x<δ → 0<c-x<c - x[tex]_{1}[/tex] ⇒ x>x[tex]_{1}[/tex], so since f is monotonic increasing f(x)>f(x[tex]_{1}[/tex]).

Then, because L - [tex]\epsilon[/tex]<f(x[tex]_{1}[/tex])<f(x)≤L,
L - [tex]\epsilon[/tex]<f(x)≤L and |f(x) - L|<[tex]\epsilon[/tex].


This is all I have so far! Any help or advice in how to solve part 2 of this problem would be greatly appreciated - thanks so much!
 

Answers and Replies

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I would just like to note - I think the latex imaging might have messed this up, but all the epsilons in the problem should not be listed as superscripts, but should be aligned normally and the 1 by the x, should be a subscript. Also, I left out the other 3 cases for fully proving part 1, but the proof is the same - it pretty much follows...Thanks!
 
Last edited:

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