how to solve a singular integral equation involving the kernel $1/x$

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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On your "edit": both answers show that this sup can be as large as you wish. –  Alexandre Eremenko Aug 10 '12 at 21:39
Dear Professor Eremenko, you are right. Some other conditions are needed.Thanks a lot. –  Anand Aug 11 '12 at 10:57

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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THanks Alexandre Eremenko, What I want is $\sup_{0\le x\le c} f(x)$ for fixed $c>0$. –  Anand Aug 10 '12 at 14:05
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. –  Pietro Majer Aug 10 '12 at 18:09
$f(x)=c x$ is even $1$ Hölder continuous which is stronger than 1/2 Hölder continuous. :-) –  Anand Aug 10 '12 at 21:04
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 $[0,+\infty)$. –  Pietro Majer Aug 11 '12 at 6:54

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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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. –  Anand Aug 10 '12 at 14:11