6

Let $f\in C^2(\mathbb{R},\mathbb{R})$.

Assume there exist $(a,b) \in \mathbb{R}^2$ such that $\forall x \in \mathbb{R}, (f(x))^{2} \leqslant a$ and $(f'(x))^{2} + (f''(x))^{2} \leqslant b$.

Prove that: $$(f(x))^{2} + (f'(x))^{2} \leqslant \max(a,b)$$ for all $x \in \mathbb{R}$.

  • I tried to use taylor lagrange with the integral form of the remainder but I did not succeed. Ì really don't know how to tackle the problem.

Any suggestion, Hint (or Answer) will be very helpful,

Thank you in advance,

  • This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. – Nick Peterson Feb 15 '14 at 15:21

1 Answers1

4

For typographical reasons, I'm going to switch to Newton-style $\dot f$ notation for derivatives.

It is natural to consider the function $F=f^2+\dot f^2$ and its derivative $\dot F = 2\dot f (f+\ddot f)$. At any point of extremum of $F$ we have either

  • $\dot f=0$ — hence $F=f^2\le a$, or
  • $f+\ddot f=0$ — hence $F= \ddot f^2+\dot f^2 \le b$

Unfortunately this isn't enough, because it is conceivable that $F$ could grow at infinity without having any points of extremum. We have to work harder.

Suppose the conclusion is false. Then there is $\epsilon>0$ such that the set $$U=\{f^2+\dot f^2> \max(a,b)+\epsilon\}$$ is nonempty. Being open, $U$ can be written as the disjoint union of open intervals. Let $I$ be such an interval. At every point of $I$ we have $\dot f^2>\epsilon$. By the mean value theorem, $$|f(u)-f(v)|\ge \sqrt{\epsilon}\,|u-v|,\qquad \forall u,v\in I$$ Since $f$ is bounded, it follows that $I$ is a finite interval. By its construction, $F=\max(a,b)+\epsilon$ at the endpoints of $I$. Therefore, $F$ attains a maximum inside of $I$, to which the reasoning from the beginning of the post applies.

user127096
  • 9,683