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 $0$.

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

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! :-)

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 $f(x)$?

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

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 $0$.

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