uniformly continuous and bounded derivative When I learn uniformly continuous function, I used to view uniformly continuous function as a function with bounded derivative. 
Most of time up to now it seems right. But can we prove or disprove that the uniformly continuous functions are functions with bounded derivative?
 A: No, this is quite false. The absolute value function $|x|$ is uniformly continuous, but isn't differentiable (the derivative, however, does exist almost everywhere and is bounded). This can be easily extended to give a uniformly Lipschitz function whose derivative fails to exist at a given finite set.
For a better counterexample, note that the function $x^{1/3}$ is uniformly continuous, but its derivative doesn't exist at $0$, and is unbounded in every neighborhood of zero.
For an even more spectacular example, the Weierstrass function is uniformly continuous on $[-1,1]$, but is differentiable nowhere.
A: First of all, a continuous function need not have a derivative at all.
But assuming $f$ is differentiable and has bounded derivative everywhere, we can say $$|f'(x)|<M$$ so by the Mean Value Theorem $$\left|\frac{f(b)-f(a)}{b-a}\right|=|f'(c)|<M\implies |f(b)-f(a)|<M|b-a|$$ Therefore given $\epsilon >0$, we take $\delta=\epsilon/M$ to get $$|f(b)-f(a)|<M{\epsilon\over M}=\epsilon$$ and the function is uniformly continuous.
