1. ## Uniformly continuous function

Let
f : R--->R be differentiable with bounded derivative. Show that f is uniformly continuous.

(f(x) - f(y)) / (x - y) = f'(p)
for some p in (x,y) (by the mean value theorem) and if M > 0 is the bound
on all values of the derivative, so that
|f(x) - f(y) | / |x - y| <= M
or |f(x) - f(y)| <= M*|x-y| for all x,y.

2. Originally Posted by charikaar
Let
f : R--->R be differentiable with bounded derivative. Show that f is uniformly continuous.

(f(x) - f(y)) / (x - y) = f'(p)
for some p in (x,y) (by the mean value theorem) and if M > 0 is the bound
on all values of the derivative, so that
|f(x) - f(y) | / |x - y| <= M
or |f(x) - f(y)| <= M*|x-y| for all x,y.

You're on the right track...

Fix e>0, we want to show there exists a d such that:

lx-yl < d implies lf(x) - f(y)l < e

lf(x) - f(y)l <= lx-yl *M

if lx-yl < d, then

lf(x) - f(y)l < dM

Finding the right d is not hard to get lf(x) - f(y)l < e.

3. Originally Posted by charikaar
(f(x) - f(y)) / (x - y) = f'(p)
for some p in (x,y) (by the mean value theorem) and if M > 0 is the bound
on all values of the derivative, so that
|f(x) - f(y) | / |x - y| <= M
or |f(x) - f(y)| <= M*|x-y| for all x,y.

the last line of your work implies that $\displaystyle f$ is Lipschitz, and every Lipschitz function is uniformly continuous.