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Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuously differentiable real function such that

  1. $f(0) = 0$
  2. $f'(x) \le -\frac{1}{2}$ for every $x \in \mathbb{R}$

Then it is always true that $\lim_{x \rightarrow \infty} f^{2}(x) = +\infty$?

I could not find a counterexample neither prove this statement. I only could prove that if a counterexample $g$ exists, then $\lim_{x \rightarrow \infty} g(x)$ is not finite or $\lim_{x \rightarrow \infty} g'(x)$ is not finite. For otherwise, it can be proved that $\lim_{x \rightarrow \infty} g'(x) = 0$, contradicting (2).

Similarly, if a counterexample $g$ exists, then $\lim_{x \rightarrow \infty} g(x)$ is not finite or $g'$ is not uniformly continuous. For otherwise, by Barbalat's lemma, then $\lim_{x \rightarrow \infty} g'(x) = 0$, again contradicting (2).

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  • $\begingroup$ Source of the problem, please? $\endgroup$ Commented Aug 1, 2015 at 5:45
  • $\begingroup$ @GerryMyerson This is a special case of Exercise 4.10(b) of Khalil's book Nonlinear Systems. I was thinking that $V(x)$ was not going to be radially unbounded and tried to produce a counterexample using a scalar valued function $f$. $\endgroup$
    – shamisen
    Commented Aug 1, 2015 at 19:54

2 Answers 2

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Before editing the second condition was $f'\le 1/2$. In that case the statement is false. As a counterexample consider the function $x\mapsto \sin(0.5x)$ or $x\mapsto 0.5\sin x$


Now, with the new statement. For every $x\ge 0$ we have: $$ f(x)=\int_0^xf'(t)\,dt\le -\frac{x}{2} \le 0. $$ It follows that $$ f^2(x)\ge \frac{x^2}{4} \quad \forall x\ge 0, $$ and therefore $$ \lim_{x\to\infty}f^2(x)\ge \lim_{x\to\infty}\frac{x^2}{4}=\infty. $$

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Its almost obvious. Clearly the derivative is negative so as $x$ increases $f(x)$ decreases and hence $f(x) \to -\infty$ or $f(x) \to L$ as $x \to \infty$. If $f(x) \to L$ then by MVT $f'(c) = f(x + 1) - f(x) \to 0$ and this contradicts $f'(x) \leq -1/2$. Hence it follows that $f(x) \to -\infty$ and hence $f^{2}(x) \to \infty$. The result holds even if $(-1/2)$ of the question is replaced by any negative number.

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  • $\begingroup$ Is the use of MVT to justify that $f'(x) \to 0$ correct? It is very well possible for a continuously differentiable function to tend to a limit without the derivative tending to 0 (see bullet point 2: en.wikipedia.org/wiki/…) $\endgroup$
    – user217285
    Commented Aug 2, 2015 at 6:57
  • $\begingroup$ @Nitin: I agree that it is very well possible for a continuously differentiable function to tend to a limit without the derivative tending to $0$. What I am saying is that $f(x + 1) - f(x) \to 0$ and hence for every $\epsilon > 0$ we have an $N > 0$ such that $|f(x + 1) - f(x)| < \epsilon$ for all $x > N$. This means that there will infinitely many values of $c$ such that $N < x < c < x + 1$ such that $|f'(c)| < \epsilon$ and this contradicts $f'(x) < -1/2$ as soon as $0 < \epsilon < 1/2$. I am sorry if my wording suggested that $f'(x) \to 0$. $\endgroup$
    – Paramanand Singh
    Commented Aug 2, 2015 at 7:03

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