I'm trying to prove that if $f$ is twice differentiable in $(a,b)$ and there exists $M\ge 0$ such that $|f''(x)| \le M$ for each $x \in (a,b)$, then $f$ is uniformly continuous. Let $x,y \in (a,b)$ and assume $x<y$. By the mean value theorem, there exists $c_{x,y} \in (x,y)$ such that $|f'(x)-f'(y)|=|x-y|\cdot |f'(c_{x,y})| \le M|x-y|$; this proves that $f'$ is Lipschitz continuous in $(a,b)$.
Hence, $f'$ is bounded by some $K\ge 0$. Using again the mean value theorem $|f(x)-f(y)|=|x-y| \cdot |f'(d_{x,y})| \le K|x-y|$ for some $d_{x,y} \in (x,y)$; so, $f$ is Lipschitz continuous as well an…