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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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    Thank you. Please see the updated version.2011-11-10
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    @Patrick Da Silva $\sqrt{t}$ is not $C^1[0,\infty)$ (its derivative cannot be continuously extended to $t=0$). Could the problem still be true in the original form?2011-11-10
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    Usually $C^1$ just means continuous and continuously differentiable in the interior of the domain because it is only there that the derivative is defined. Now you are asking more constraints on $f$ than in the original problem and this is a different question. I'll try thinking about it.2011-11-10
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    @Patrick Da Silva Usually $C^1(\Omega)$ means continuously differentiable on $\Omega$ and if $\Omega$ is closed, as in the problem statement where it is $[0,\infty)$, that would include the boundary as well.2011-11-10
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    many thanks to everyone, please see the revised version.2011-11-10
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    I guess there are many ways to say usually then. Perhaps in more advanced books your notion is better suited and I must say I agree with it, but it's just not every book that likes this for readability for undergrads (especially on first year).2011-11-10
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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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    From the way the problem is order, the constants depend on $f$.2011-11-10
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    Yes, for any fixed f, there exist $C$ and $T$...2011-11-10
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    Thank you, Jeff. here f is fixed. C depends on f.2011-11-10
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    Ok here's a counter example for when $C$ depends on $f$. It is a bit technical so I'll just explain it in words. Start with $f(t)=t+2$ and at each integer $n$, add a small perturbation to $f$ so that $f'(n)=e^{4n}$ and $f(t) \leq 2(t+2)$. This can be done by decreasing the amplitude of the perturbations as $n\to \infty$. Then you are asking that $e^{4n} \leq C e^{2n+4}$ for some constant $C$ and all integers $n$ which is clearly not true.2011-11-10
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    But needs $\frac{df}{dt}>=1$ and $f^'$ is continuous, do you think your example satisfies these?2011-11-10
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    many thanks to Jeff. I add a condition: f is convex. Please prove the problem in stead of give counter examples.2011-11-10
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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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    would you please add some suitable condition so that the inequality is correct? and give the proof. what really counts is in what condition, this inequality is correct in stead of counter example? Thank you so much2011-11-10
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    I'm not sure if convexity is enough or not. I cannot come up with a counter example, but cannot find a proof either.2011-11-10
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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