Proving Continuity of Monotone Functions on Interval Domains

[math8]

Suppose f:A-->R is monotone (ACR: reals)
and suppose the range of f is an interval, show f is continuous on A.

By drawing a picture, I can see the conclusion. Since f is monotone, the only type of discontinuity it may have is a jump discontinuity. But since the range of f is an interval, this cannot happen.

I would like to have a more consistent proof (analytical proof), I mean, I know that since f is monotone, for each point a of A, f(a+) and f(a-) exist. Now f(a+)> or eq. to f(a) and f(a-)< or eq. to f(a). Now how would I show that f(a+)< or eq. to f(a) and f(a-)> or eq. to f(a) using the fact that range(f) is an interval? Hence I would be able to conclude that f(a-)= f(a+)=f(a), therefore, f continuous on A.

 

[Dick]

Just take the direct approach. Pick an x and an epsilon e>0. You want to show that there is a delta d>0 such that |f(x)-f(y)|<e if |x-y|<d, right? Think about x=f^(-1)(f(x)), f^(-1)(f(x)+e) and f^(-1)(f(x)-e). Can you figure out how to define a d using those?

 

[math8]

I would use d such that |f^(-1)(f(x)+e) - f^(-1)(f(x)-e)|> d. Is that correct? How do we use the fact that range of f is an interval?
I see that you're using f^(-1), in fact the problem has a subproblem that says to use the result to prove the continuity of inverse functions. How do we conclude that?

 

[Dick]

If there is a jump discontinuity, then the range isn't an interval. I use that the range is an interval to show that a point like f^(-1)(f(x)+e) even exists. You've probably drawn a picture of a function with a jump discontinuity, so you should be able to see how the proof breaks down. I would say d should be smaller than the maximum of |f^(-1)(f(x)+e)-x| and and |f^(-1)(f(x)-e)-x|. Do you see why? Try doing the subproblem about f^(-1) and see how far you get.

 

[math8]

shouldn't d be smaller than the minimum of |f^(-1)(f(x)+e)-x| and |f^(-1)(f(x)-e)-x| instead?

 

[Dick]

Sure. Absolutely. I was just testing you. :)

 

[math8]

oh, good, then, at least I understand what's going on...

 

[Dick]

That's what I was checking. Actually 'maximum' was just a typo...

 

[math8]

I am trying for the f^-1 part, but I cannot get anywhere, can you give me another hint?

 

[Dick]

What exactly is the f^(-1) part?

 

[math8]

the subproblem that says to use the result to prove the continuity of inverse functions.

 

[Dick]

The inverse of a continuous function is not necessarily continuous even if it's monotone and the range is an interval. Can you state the full problem? Do you know something about the domain?

 

[math8]

that's it, that's how it is stated:

"Suppose f:A-->R is monotone (ACR: reals)
and suppose the range of f is an interval, show f is continuous on A. Use this result to prove the continuity of inverse functions".

 

[Dick]

Let f(x)=x for x in [0,1) and f(x)=x-1 for x in [2,3]. f is continuous. The domain is [0,1)U[2,3] which is a subset of the reals. The range is [0,2]. It's monotone. The inverse has a jump discontinuity at 1. Do you see my problem?

 

[math8]

yeah, I guess, there is a problem or maybe something missing in the question, anyways, thanks for your help.