Let $f:(0,1)\to \mathbb R$ be a twice differentiable function. Which of the following is/are FALSE?
$a).$ If $f$ is bounded then $f'$ is bounded.
$b).$ If $f$ and $f'$ are bounded then $f''$ is bounded.
$c).$ If $f(x)>0$ and $f'(x)>0$, $\forall x \in (0,1)$ then $f''(x)>0$, $\forall x \in (0,1)$.
$d).$ If $f''$ is a polynomial then $f$ is a polynomial.
For option $a)$. if $f$ is a constant function then it is trivially true. Visited here Can the graph of a bounded function ever have an unbounded derivative? for the counters. I chose the one who satified the given criteria i.e. $f(x)=\sin(x\sin x)$ and $f(x)=\sin x^2$. For $f(x)=\sin x^2$ we have $f'(x)=2x\cos x^2$. But for $x\in (0,1)$, I plotted it on desmos and it was bounded! Couldn't even search for the sequences that will claim $f'$ is unbounded. I believe $a).$ is true as given that $f(x)$ is twice differentiable, then it itself implies that $f'$ exist finitely and is continuous as well on $(0,1)$. One doubt here is that the property satisfied by this $f(x)$ implies that $f(x)$ is uniformly continuous, since it has a bounded derivative,.
For option $b).$ Now given that $f$ and $f'$ both are bounded, but I don't think $f''$ needs to be bounded.The counter $f(x)=(x-\frac{1}{2})^{3/2}$
For option $c).$ Take counter $f(x)=\log x$.
For option $d).$ I guess that's also true.
So the false statements are $b,c$.
Best Answer
For a) $f(x)=\sqrt x$ is a counter-example.
For b) $f(x)=x^{3/2}$ is a counter-example.
For c) $f(x)=3x-x^{2}$ is a counter-example.
d) is true: Just integrate twice.