# Limsup h.w. proofs

1. Mar 24, 2008

### aschwartz

1. The problem statement, all variables and given/known data

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$$^{-}$$ 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$$^{+}$$ = L$$^{+}$$, which is defined to be the limsupf(x) as x→c.

Prove:

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

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

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

Prove that limf(x) as x→c = L if and only if L$$^{+}$$ = L$$^{-}$$ = 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$$^{-}$$ 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 ($$\forall$$$$\epsilon$$>0) ($$\exists$$$$\delta$$) ($$\forall$$x ∈ R) (0<c-x<δ ⇒ |f(x) - L|<$$\epsilon$$.

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

Let $$\delta$$ = c - x$$_{1}$$.

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

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

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!

2. Mar 24, 2008

### aschwartz

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: Mar 24, 2008