Expert: Paul Klarreich - 11/18/2008

Question
I am sorry, I'm not trying to annoy you. For some reason it keeps cutting off the end of the problem. And yes, I do proof read after I type.

(a) Let f be differentiable on (a, b) and suppose that there exists
m in R such that abs value f(x) less than or equal to m for all x in (a, b). Prove that f is uniformly
continuous on (a, b).
(b) Find an example of a function that is differentiable and uniformly continuous
on (0, 1), but f prime is unbounded on (0, 1).

Answer
Questioner:   megan
Category:  Advanced Math
Private:  No
 
Subject:  analysis ?
Question:  I am sorry, I'm not trying to annoy you. For some reason it keeps cutting off the end of the problem. And yes, I do proof read after I type.

(a) Let f be differentiable on (a, b) and suppose that there exists
m in R such that abs value f(x) less than or equal to m for all x in (a, b). Prove that f is uniformly continuous on (a, b).
(b) Find an example of a function that is differentiable and uniformly continuous on (0, 1), but f prime is unbounded on (0, 1).
==========================================
Hi, Megan,

YOU DID IT!  You got the whole question through.  I am sure your cat must have jumped on the keyboard the other times and pressed the send button.

Now let's hope I can do this.

(a) Let f be differentiable on (a, b) and suppose that there exists
M in R such that | f(x) | <= M for all x in (a, b). Prove that f is uniformly
continuous on (a, b).
..............
Let's see if I remember --  f(x) is continuous at  x = c  if given e, (epsilon), there exists d (delta) such that:

| f(x) - f(c) | < e, whenever  | x - c | < d.

AND uniformly continuous if given e, we can find d that satisfies that for any c.

HOWEVER, I am not sure about your (a), because I am thinking about:

f(x) = sin(1/x) on  (0,1).

This is certainly bounded: | sin(1/x) | <= 1

This is differentiable: f'(x) = (- cos(1/x)/x^2, and that is defined on (0,1). -- Not at x = 0, so not on [0,1]

Now is this function uniformly continuous?  
............................
(b) Find an example of a function that is differentiable and uniformly continuous on (0, 1), but f prime is unbounded on (0, 1).

Something like this could be, perhaps:

f(x) = x sin (1/x)  on (0,1)

I think this might be U.C. on (0,1), but  

f'(x) = x(- cos (1/x)/x^2 + sin(1/x)

f'(x) = - cos (1/x)/x + sin(1/x)

Now as  x -> 0, the first term is unbounded.

..........................................

I am not sure about this, so I will send it along to my children.  They are much better than I am.  If they come up with something, I'll pass it along.