MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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! :-)

share|cite|improve this question
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.

share|cite|improve this answer
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.