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Dear all,

Suppose we know that $f(x)$ is nonnegative and Hölder continuous at zero with exponents $1/2$. We also know that

$$ f(x) \le g(x) + \int_0^x \frac{f(y)}{y} d y,\quad\forall x>0, $$

where $g(x)$ is some nonnegative nice function, for example, $g(x)=\sqrt{x}$. Is it possible to derive a good upper bound for $f(x)$? Apparently, classical Gronwall's inequality doesn't work since $1/y$ is not integrable around $0$.

EDIT: Just to make it clear, I wish to have a upper bound of the following form: For fixed $c>0$,

$$ \sup_{x\in [0,c] } f(x)\le ? $$

Thank you very much for any hints and help! :-)

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  • $\begingroup$ On your "edit": both answers show that this sup can be as large as you wish. $\endgroup$ Aug 10, 2012 at 21:39
  • $\begingroup$ Dear Professor Eremenko, you are right. Some other conditions are needed.Thanks a lot. $\endgroup$
    – Anand
    Aug 11, 2012 at 10:57

2 Answers 2

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No upper bound can be derived, good or bad. Take $f(x)=cx$ where $c$ is large positive. Your inequality is trivially satisfied.

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  • $\begingroup$ THanks Alexandre Eremenko, What I want is $\sup_{0\le x\le c} f(x)$ for fixed $c>0$. $\endgroup$
    – Anand
    Aug 10, 2012 at 14:05
  • $\begingroup$ To be precise, the function $f(x)=cx$ is not 1/2 Hölder on $[0,+\infty)$ as it was assumed in the question. But $cx^{1/2}$ works, of course. $\endgroup$ Aug 10, 2012 at 18:09
  • $\begingroup$ $f(x)=c x$ is even $1$ Hölder continuous which is stronger than 1/2 Hölder continuous. :-) $\endgroup$
    – Anand
    Aug 10, 2012 at 21:04
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    $\begingroup$ Well, that depends on the definition :-) To me $f:X\to\bf R$ being $\alpha$ Hölder means $|f(x)-f(y)| \le C|x-y|^\alpha$ for all $x$ and $y$ in $X$. I'd call "locally Hölder" a function like $x\mapsto cx$ on $[0,+\infty)$. $\endgroup$ Aug 11, 2012 at 6:54
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Note that the inequality is satisfied by the functions $f(x)=cx^{1/2}$, for any $c\ge0$ and any nonnegative $g$. So, in terms of upper bounds, it doesn't really add anything to the information that $f$ is Hölder continuous of exponent 1/2.

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  • $\begingroup$ Thanks Professor Pietro Majer, do you think that there can be some Gronwall type inequality for this case? Adding Hölder continuity is to make sure the integral is well defined. $\endgroup$
    – Anand
    Aug 10, 2012 at 14:11

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