Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is related to my previous post. Suppose $f:\mathbb{R}\rightarrow\mathbb{R}$ is a $C^1$ function which satisfied the following differential inequality: $$\frac{df}{dt}\leq C(f+f^{\frac{3}{2}}).$$ If $f>0$ and $f(t)\rightarrow 0$ as $t\rightarrow\infty$, then is $\frac{df}{dt}$ bounded as $t\rightarrow\infty$? That is, is $\frac{df}{dt}$ bounded on some interval $(a,\infty)$?

share|cite|improve this question
I think $f'(t) \leq C(f + f^(3/2))$ shows that $f'(t) \rightarrow 0$ as $ t \rightarrow \infty$. Because $f \in C^1$, we know that $f'(t)$ is continuous. This helps see that $f'(t)$ is bounded on some interval $(a,\infty)$. Is there a flaw in my reasoning here? – r.g. Sep 16 '11 at 8:07
@Rohan: It’s bounded above, but $f'(t)$ can be increasingly negative on shorter and shorter intervals. – Brian M. Scott Sep 16 '11 at 8:23
up vote 3 down vote accepted

The derivative need not be bounded below. Imagine first a step function that takes the value $2^{-n}$ on $[n,n+1)$ for $n \ge 0$ and is $1$ when $x \le 0$; clearly this is positive and non-increasing and approaches $0$ as $x\to\infty$. Now smooth out the drops between steps, making the ‘risers’ steeper and steeper. It should be fairly clear intuitively that this can be done, and that the result is a counterexample. Here, if I’ve not loused up the details, is a concrete example:

$$f(x) = \begin{cases} 1,&x \le 2\\ 2^{-n}\cos (2^{2n}\pi(x-n))+3\cdot 2^{-n}, &2 \le n \le x \le n+2^{-2n}\\ 2^{-(n-1)}, &2 < n+2^{-2n} \le x \le n+1, \end{cases}$$

and $$f'(x) = \begin{cases} 0,&x\le 2\\ -2^n\pi \sin(2^{2n}\pi(x-n)),&2 \le n \le x \le n+2^{-2n}\\ 0,&2 < n+2^{-2n}\le x \le n+1. \end{cases}$$

For integer $n \ge 2$, $f'(x)$ runs from $0$ to $-2^n$ and back to $0$ on the interval $[n,n+2^{-2n}]$, so it’s continuous and non-positive everywhere, and for $n\ge 2$, $f'(n+2^{-(2n+1)})=-2^n\pi\to -\infty$ as $n\to\infty$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.