How do you prove there's a subinterval where f is Lipschitz?

Question

I think this problem is kinda interesting:

Let I be an interval of R and suppose f:I -> R is differentiable on I. Prove that there exists a subinterval of I where f is Lipschitz.

Answer 1

If f'(x) = 0 on I, then the result is trivially true.

Otherwise, assume there is a point x0 in I for which f'(x0) > 0. If there is no such point, then work with -f and show it's Lipschitz on some subinterval, which would immediately imply f is too.

Let L=f'(x0). Then, for any e > 0, there exists a d > 0 such that |(f(x)-f(x0))/(x-x0) - L| < e whenever |x-x0| < d.

Also, there exists a d' > 0 such that f'(x) > 0 whenever |x-x0| < d'.

Let h=min(d,d'), and let I0 = (x0-h,x0+h). Then, for x in I0,

|f(x)-f(x0) - L(x-x0)| < e|x-x0|, so
-e|x-x0| + L(x-x0) < f(x)-f(x0) < L(x-x0) + e(x-x0).

Handle the cases xx0 separately. In each case, because you know the sign of f(x)-f(x0) by virtue of f being increasing on I0, you can bound on |f(x)-f(x0)| by C|x-x0| for a constant C dependent on L and e.

Or something like that :)