Real analysis: continuously differentiable and Lipschitz implies bounded derivatives?

Question

Claim:

If a function $f:\mathbb{R}\rightarrow\mathbb{R}$ is $M$-th continuously differentiable ($f\in C^M$) for some $M\geq 2$ and $f$ is Lipschitz continuos on $\mathbb{R}$, then all the derivatives $f'$, $f''$, ... $f^{(M-1)}$ are bounded on $\mathbb{R}$.

Is this claim true? If not, are there any good counter-examples?

Answer 1

This is not true.

Pick a continuous positive function $g(x)$ which is unbounded, but still $$\int_{- \infty}^{+ \infty}g(t) \mathrm dt < \infty$$ Such an example may be $g(x)=x^2 \exp (-x^8 \sin^2 x)$

Then integrate it twice to get a function $f$ with the property that $f''=g$ and $f(0)=f'(0)=0$. $f$ is twice continuously differentiable.

Now, $f'$ is bounded, so $f$ is Lipschitz. However, $f''$ is not bounded.

Answer 2

Take as a counterexample the function

$$F(x)=\int_0^x\sin{(t^2)}dt$$

This function is Lipschitz and three times continuously differentiable but $F''$ is not bounded.