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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 $(a,\infty)$?

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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

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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 $n\to\infty$.

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