properties of bounded function's second derivative

Question

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?

Answer 1

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. $\qquad\square$

Answer 2

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.