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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 $f$ bounded?

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up vote 2 down vote accepted

No, any monotonically decreasing positive $C^1$ function that goes to infinity for $t\to-\infty$ is a counterexample, for instance

$$f(t)=\begin{cases}1-t&t\lt0\\\mathrm e^{-t}&t\ge0\;.\end{cases}$$

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Thank you so much. Then I should modify my question then. – Paul Sep 16 '11 at 6:38
@Paul, You have already given that $f(t) \to 0$ as $t \to \infty$. So it is clearly bounded as $t \to \infty$. More precisely, the function is bounded in the interval $(a, \infty)$ for any $a \in \mathbb R$. – Srivatsan Sep 16 '11 at 6:42
@Srivatsan Narayanan: Yes, you are right! Actually I want to see if $\frac{df}{dt}$ is bounded or not as $t\rightarrow\infty$. – Paul Sep 16 '11 at 6:45

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