Uniform continuity and derivatives Is there any example of a three times differentiable and uniformly continuous function $f: \Bbb R \rightarrow \Bbb R$ such that $f^{(3)}$ is bounded but $f^{(2)}$ isn't?
 A: Here’s another answer (while I like ajr’s answer a lot): we can write, for real $t,x$, $f(x+t)-f(x)=f’(x)t+\frac{f’’(x)t^2}{2}+\frac{t^3f’’’(c_{x,t})}{6}$ for some $c_{x,t}$ between $x$ and $x+t$. As $f$ is uniformly continuous and $f^{(3)}$ is bounded, it follows that, uniformly in $x$, as $t$ goes to zero, $2f’(x)t+f’’(x)t^2$ goes to zero.
In particular, there is $\epsilon >0$ such that for every $|t| \leq \epsilon$, $2tf’+t^2f’’$ is bounded by $1$. Let $0<u<t<\epsilon$.
Then $f’’=\frac{2tf’+t^2f’’}{t(t-u)}-\frac{2uf’+u^2f’’}{u(t-u)}$ is bounded.
A: There is no such function. Assume that $|f'''|\leq M$ and that $f''$ is unbounded.  Without loss of generality, there exists a sequence $(x_n)$ such that $f''(x_n) > n$.  Since $f'''$ is bounded, by the fundamental theorem of calculus we find that $f''(x_n) \geq n/2$ on $(x_n,x_n + \frac n{2M})$. Hence $f'$ is increasing on $(x_n,x_n + \frac n{2M})$, so it can have at most one zero there. This implies that there exists $(a_n,b_n)\subseteq (x_n,x_n + \frac n{2M})$ such that $b_n - a_n \geq \frac 14\cdot \frac{n}{2M} = \frac{n}{8M} \geq \frac 18$ and  $|f'| \geq  \frac{1}{4}\cdot\frac n2 \cdot \frac {n}{2M} = \frac{n^2}{16M}$ on $(a_n,b_n)$ (in particular $f'$ does not change sign on $(a_n,b_n)$).  Therefore we get a sequence of intervals $(a_n,b_n)$ of length greater than or equal $\frac 18$ on which we have $|f'|\geq \frac{n^2}{16M}$. This means that $|f(x) - f(y)| \geq \frac {n^2}{16M}|x-y|$, whenever $x,y\in (a_n,b_n)$. It follows that $f$ cannot be uniformly continuous.
