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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 $b,c$.

Gary
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Aastha Choudhary
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2 Answers2

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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.

Kavi Rama Murthy
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There are a number of issues with your reasoning for $(a)$ and $(b)$.

  1. So, a counterexample here is to consider something like: $$f(x)= \sqrt{1-x}$$ This is bounded on $(0,1)$ but its derivative is unbounded. Now, the post that you linked attempts to show that your proposed function is unbounded over $\mathbb{R}$, not over $(0,1)$.

  2. Similarly, let: $$f(x) = (1-x)^{\frac{3}{2}}$$ This is bounded on $(0,1)$ and its derivative is bounded on $(0,1)$. But its second derivative is unbounded.

  3. That doesn't work as a counterexample. You could use something like $f(x) = 2x$.

  4. You can just prove that its true. So, let $f''$ be a polynomial. Then, find two of its antiderivatives and you'll get a family of solutions that are all polynomials.

Mousedorff
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