# 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! :-)

• On your "edit": both answers show that this sup can be as large as you wish. Aug 10, 2012 at 21:39
• Dear Professor Eremenko, you are right. Some other conditions are needed.Thanks a lot. Aug 11, 2012 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.
• THanks Alexandre Eremenko, What I want is $\sup_{0\le x\le c} f(x)$ for fixed $c>0$. Aug 10, 2012 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. Aug 10, 2012 at 18:09
• $f(x)=c x$ is even $1$ Hölder continuous which is stronger than 1/2 Hölder continuous. :-) Aug 10, 2012 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 $[0,+\infty)$. Aug 11, 2012 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.