Is the derivative of a differentiable Lipschitz function also Lipschitz?

Solution 1:

This is false. Consider the case where $f$ is $C^2$. Then, it is lipschitz continuous if, and only if, it's derivative is bounded. Similarly, $f'$ is lipschitz if, and only if, $f''$ is bounded.

However, we can find bounded smooth function with unbounded derivatives : let $\chi:\mathbb R\to \mathbb R$ be smooth step function, ie $\chi$ takes value in $[0,1]$, $\chi(x)$ is $0$ for $x\leq 0$ and $1$ for $x\geq 1$. Then, set : $$g(x) = \sum_{n=1}^{+\infty} \frac{1}{2^n}\chi(4^n(x-n))$$

Since the sum is locally finite, $g$ is smooth and : $$g'(x) = \sum_{n=1}^{+\infty} 2^n\chi'(4^n(x-n))$$ However, we have :$$\sup_\mathbb R g = \lim_{+\infty }g = 1 <+\infty$$ while, if $x\in[0,1]$ has $\chi'(x) >0$, then : $$g'(x+n) = 2^n\chi'(x)$$ so $\sup_{\mathbb R} g' = +\infty$.

Therefore, if $f(x) = \int_0^xg(x) \text dx$, we see that $f$ is Lipschitz, smooth, but its derivative $f'=g$ is not Lipschitz.