Monotone Limit at Endpoints a and b?

[kathrynag]

Homework Statement

let f: [a,b] ---> R be monotone. Prove that f has a limit at a and b.

Homework Equations

The Attempt at a Solution

here is my proof:
We need to show that f has a limit at a and b.
Consider f:[a,b]--> R where f is increasing. Now for all x elements of [a,b] f(a)<f(x)<f(b), hence f is bounded and by the monotonocity, f cannot be oscillatory. Suppose x1<x2<...xk. Then f(a)<f(x1)<f(x2)...<f(xk)<f(b). So limx-->xk f(x)=f(xk)<f(b)and f has a limit at b. Then Limx-->x1 f(x)=f(x1)>f(a) and f has a limit at a.


I'm not so sure about this one...

 

[Pere Callahan]

I'm not so sure about this one...

Me neither.

What you want to show is that if you have a sequence $(x_k)_{k\geq 0}$ with $x_k\to b$ then the sequence $(f(x_k))_{k\geq 0}$ converges. This sequence does not need to converge to f(b) which would mean that f is continuous at b - a fact which might not generally be true and which you are not asked to prove.

As for what you have done so far: I can see only few things there which are correct.

You said $\lim_{x\to x_k}f(x)=f(x_k)$. This in general not true unless f is continuous at $x_k$. Moreover you conclude from this (erroneous) statement that f should have a limit at b. How is the behaviour of f at $x_k$ related to the behaviour of f at b? - Not at all.

I suggest you give it another try

 

[boombaby]

you can define (why?) A = sup{f(x)|a<=x<b}, which is the key to this question IMO, the rest of the proof becomes easy: prove such A is exactly f(b-) by a simple "epsilon-delta" argument

 

[kathrynag]

Me neither.

What you want to show is that if you have a sequence $(x_k)_{k\geq 0}$ with $x_k\to b$ then the sequence $(f(x_k))_{k\geq 0}$ converges. This sequence does not need to converge to f(b) which would mean that f is continuous at b - a fact which might not generally be true and which you are not asked to prove.

As for what you have done so far: I can see only few things there which are correct.

You said $\lim_{x\to x_k}f(x)=f(x_k)$. This in general not true unless f is continuous at $x_k$. Moreover you conclude from this (erroneous) statement that f should have a limit at b. How is the behaviour of f at $x_k$ related to the behaviour of f at b? - Not at all.

I suggest you give it another try

$\lim_{x\to x_k}f(x)=f(x_k)$. I'm still not quite sure what to do here instead.

 

[Office_Shredder]

If you have a sequence (x_k) that approaches b on the left, then assuming x_k increases monotonically, you have that f(x_k is increasing and bounded above. Hence?

Then see what happens if different sequences have different limits

 

[kathrynag]

If you have a sequence (x_k) that approaches b on the left, then assuming x_k increases monotonically, you have that f(x_k is increasing and bounded above. Hence?

Then see what happens if different sequences have different limits

Where did you get the approaching b from the left? Because of the fact that we have a<x₁< x₂...<x_k<b?

 

[Office_Shredder]

The question is prove f has a limit at b. It makes sense to pick an arbitrary sequence that approaches b to investigate what the limit would be (often, instead of trying to prove that it as a limit, you start by finding what the limit should be, and then proving that is the limit).

 

[kathrynag]

The question is prove f has a limit at b. It makes sense to pick an arbitrary sequence that approaches b to investigate what the limit would be (often, instead of trying to prove that it as a limit, you start by finding what the limit should be, and then proving that is the limit).

Ok, so because the sequence approaches b, then the limit will be b? Then, do the same with x_k approaching a from the right?