properties of bounded function's second derivative Is there a bounded function $f:\mathbb R\to\mathbb R$, such that $\forall_{t\in\mathbb R}$ $f''(t)\ne0$? If not, how to prove that?
 A: The answer there is no bounded function $f:\Bbb{R}\to\Bbb{R}$ with second derivative that does not vanish. Indeed, suppose that $f$ is a bounded function with $f''(t)\ne 0$ for every $t$.
By Darboux's theorem $f''$ has the intermediate value property, so if it does not vanish, then it must keep a constant sign, and replacing $f$ by $-f$ if necessary we may,(and will), suppose that $f''(t)>0$ for every $t$. This implies that $f$ is strictly convex.
Now consider $a\in\Bbb{R}$. Using the convexity of $f$ we conclude that for $x<a<y$ we have
$$
\frac{f(a)-f(x)}{a-x}\leq f'(a)\leq \frac{f(y)-f(a)}{y-a}
$$
Using the boundedness of $f$ and letting $x$ tend to  $-\infty$ and $y$ tend to $+\infty$ we conclude that $f'(a)=0$. But $a$ is arbitrary, so $f'=0$ which is absurd. $\qquad\square$
A: If $f''(t) \ne 0$  for all $t$ and $f''$ is continuous, then without loss of
generality  $f''(t) > 0$  for all $t$, which means that $f$ is convex.
Now it follows from


*

*Show bounded and convex function on $\mathbb R$ is constant
that $f$ is constant, which is a contradiction to $f'' \ne 0$.
I do not know if the assumption on the continuity of $f''$ can be dropped.
