uniform continuity and derivatives

[daniel_i_l]

1. The problem statement, all variables and given/known data
1) f is some function who has a bounded derivative in (a,b). In other words, there's some M>0 so that |f'(x)|<M for all x in (a,b).
Prove that f is bounded in (a,b).
2) f has a bounded second derivative in (a,b), prove that f in uniformly continues in (a,b).

2. Relevant equations



3. The attempt at a solution

1) For all x in (a,b), (f(x)-f(a))/(x-a) = f'(c) =>
|f(x)-f(a)| = |f'(c)||x-a| =< M|b-a| (c is in (a,x))
and so
f(a) - M|b-a| =< f(x) >= f(a) + M|b-a| and so f(x) is bounded in (a,b).

2)Using (1) we see that f'(x) is bounded in (a,b). So there's some M>0 so that |f'(x)|<=M. And so for all x,y in (a,b)
(f(x)-f(y))/(x-y) = f'(c) =< |M| and so
|f(x)-f(y)| =< |x-y|M and it's easy to see that f is UC in (a,b).

Did I do that right? It some what bothers me that in (1) I also could have proved that f is UC which is a stronger result than the fact that it's bounded the same way that I proved it in (2). Is that right? If so then why did they only ask to prove that it was bounded?

Thanks.

[Kummer]

Theorem: Let f be continous on a (possibly infinite) interval I. Let J be the open interval without the endpoints of I. If f is differentiable on J and |f'|\leq M then f is uniformly continous on I.

Proof:Chose M>0 so that |f'(x)|\leq M for all x\in J. For any \epsilon > 0 choose \delta = \frac{\epsilon }{M}.
Choose a<b with b-a<\delta. By Mean Value Theorem: \exists x \in (a,b) so that f'(x) = \frac{f(b)-f(a)}{b-a}.
Thus,
|f(b)-f(a)| = |f'(x)||b-a|\leq M|b-a| < M\delta = M\cdot \frac{\epsilon}{M} = \epsilon.

-Wolfgang.