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Let $f\in C[0,\infty)\cap C^1(0,\infty)$ be an increasing convex function with $f(0)=0$ $\lim_{t->\infty}\frac{f(t)}{t}=+\infty$ and $\frac{df}{dt} \ge 1$. Then there exists constants $C$ and $T$ such that for any $t\in [T,\infty)$, $\frac{df}{dt}\le Ce^{f(t)}.$

Is it correct? If the conditions are not enough, please add some condition and prove it. Thank you

3 Answers 3

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It is false if there is no constraints on the limit of the derivative when $t$ goes to $0$. Let $f(t) = \sqrt t$. We see that $f(t)$ is an increasing function. Then f'(t) = \frac 1{2\sqrt t} and since $f(t)$ is bounded in the interval say $[0,1]$, for any constant $C$, $Ce^{f(t)}$ is bounded in $[0,1]$. But f'(t) is not, so this is not possible.

Maybe the actual question puts some more constraints on $f$ : here I used the fact that it is possible that the derivative of $f$ goes to infinity as $t \to 0$. Perhaps you might ask for the derivative to have a limit when $t \to 0^+$?

Hope that helps,

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    what really count is when the inequlity is correct? could you please add some condition and then prove it?2011-11-10
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It is false as can be seen by the following proof by contradiction. Suppose it is true, and such a $C$ and $T$ exist. Then consider the following sequence of functions $f_n(t)=n(t-T) + 1+2T$ for $t\geq T$ and then extend $f_n$ smoothly for $t < T$ while keeping $f_n > 1$ and f_n'\geq 1. Then we have that f_n'(T) \leq Ce^{f(T)}, and f_n'(T)=nT and $e^{f(T)}=e^{1+2T}$, hence the inequality would say that $nT \leq Ce^{1+2T}$ or $C \geq nTe^{-(1+2T)}$ which is a contradiction as $n\to\infty$.

NOTE: I assumed that you do not want the constants to depend on $f$ (as this would change the problem).

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    I guess you could use a bump function to do this trick at each integer $n$. Make this into a complete answer and this could become an accepted answer very easily =) +1!2011-11-10
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Here are some more details on my answer in the comment above. Let

$ g_n(t) = 2^{-n}\left(\frac{1}{\pi}\arctan\left(2^ne^{4n}\pi(t-n)\right) + \frac{1}{2}\right).$

Note that $|g_n(t)| \leq 2^{-n}$ and g_n'(t) \geq 0 for all $t$ and in particular g_n'(n) = e^{4n}.

Now define

$f(t) = t + 2 + \sum_{n=1}^\infty g_n(t).$

You should verify that this series actually converges and $f\in C^1$ (this is not hard). Then $|f(t)| \leq t + 3$ and f'(n) = 1 + e^{4n}. If the statement were true, then there would exist $C$ such that

$e^{4n} \leq Ce^{n+3},$

for all integers $n$ which is a contradiction.

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    If you want to get better answers, you should try to avoid constantly modifying your question. If someone gives a correct answer, then mark it as correct and ask a new question with different conditions.2011-11-10