 Solution

Because $f^{\prime \prime }$ exists everywhere, $f^{\prime }$ is differentiable everywhere and therefore continuous everywhere; continuity of $f^{\prime \prime }$ is not required. For $x<y$, apply the mean value theorem to $f^{\prime }$ on $[x,y]$. It gives $\xi \in (x,y)$ withFor $x=y$ equality is immediate. Hence $f^{\prime }$ is nondecreasing: $x\le y$ implies $f^{\prime }(x)\le f^{\prime }(y)$.