Does bounded derivative always imply uniform continuity?

[lonewolf5999]

I'm working on a problem for my analysis class. Here it is:

Let f be differentiable on an open subset S of R. Suppose there exists M > 0 such that for all x in S, |f'(x)| ≤ M, i.e. the derivative is bounded. Show that f is uniformly continuous on S.

I'm not too sure that this question is correct though, as I think I have a counterexample. Let S be the union of (-1,0) and (0, 1), then clearly S is open. Define f(x) = |x| / x for x in S.
Then if x < 0, f(x) = -1, and if x > 0, f(x) = 1.
f'(x) = 0 at every x in S, since 0 is not in S, so f is differentiable on S and the derivative is bounded.

And now f is not uniformly continuous on S, since if we set ε= 1, let δ be arbitrary, and pick x,y close to 0 such that x<0, y>0, and |x - y| < δ, it does not follow that |f(x) - f(y)| < ε. So no δ will work for this ε.

I'd really appreciate any feedback on my reasoning. Thanks for your time!

 

[micromass]

This indeed looks like a valid counterexample!

 

[Deveno]

indeed, the problem should read:

Let f be differentiable on an open interval S of R. Suppose there exists M > 0 such that for all x in S, |f'(x)| ≤ M, i.e. the derivative is bounded. Show that f is uniformly continuous on S.

so there's something special about intervals and continuity. intervals are connected.

 

[lonewolf5999]

Thanks for the replies!