most recent 30 from http://mathoverflow.net 2013-05-19T00:25:59Z http://mathoverflow.net/feeds/question/104402 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104402/how-to-solve-a-singular-integral-equation-involving-the-kernel-1-x Anand 2012-08-10T11:07:03Z 2012-08-10T14:07:19Z

<p>Dear all,</p> <p>Suppose we know that $f(x)$ is nonnegative and Hölder continuous at zero with exponents $1/2$. We also know that </p> <p>$$ f(x) \le g(x) + \int_0^x \frac{f(y)}{y} d y,\quad\forall x>0, $$</p> <p>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 <a href="http://en.wikipedia.org/wiki/Gronwall%27s_inequality" rel="nofollow">Gronwall's inequality</a> doesn't work since $1/y$ is not integrable around $0$.</p> <p>EDIT: Just to make it clear, I wish to have a upper bound of the following form: For fixed $c>0$,</p> <p>$$ \sup_{x\in [0,c] } f(x)\le ? $$</p> <p>Thank you very much for any hints and help! :-)</p>

http://mathoverflow.net/questions/104402/how-to-solve-a-singular-integral-equation-involving-the-kernel-1-x/104408#104408 Pietro Majer 2012-08-10T12:53:11Z 2012-08-10T13:03:43Z

<p>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. </p>

http://mathoverflow.net/questions/104402/how-to-solve-a-singular-integral-equation-involving-the-kernel-1-x/104409#104409 Alexandre Eremenko 2012-08-10T12:59:09Z 2012-08-10T12:59:09Z

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