Boundedness of Hardy-type operators with a kernel: integral weighted conditions for the case $0

Research paper by Martin Křepela

Indexed on: 02 Feb '16Published on: 02 Feb '16Published in: Mathematics - Functional Analysis


Let $U:[0,\infty)^2 \to [0,\infty)$ be a~measurable kernel satisfying: (i) $U(x,y)$ is nonincreasing in $x$ and nondecreasing in $y$; (ii) there exists a~constant $\theta>0$ such that $U(x,z) \le \theta\left( U(x,y)+U(y,z) \right)$ for all $0\le x<y<z<\infty$; (iii) $U(0,y)>0$ for all $y>0$. Let $0<q<1< p <\infty$. We prove that the weighted inequality \[ \left( \int_0^\infty \left( \int_0^t f(x)U(x,t) dx \right)^q w(t) dt \right)^\frac 1q \le C \left( \int_0^\infty f^p(t)v(t)dt \right)^\frac 1p \] holds for all nonnegative measurable functions $f$ on $(0,\infty)$ if and only if \[ \left( \int_0^\infty \left( \int_t^\infty w(x)dx \right)^\frac{r}{p} w(t) \left( \int_0^t U^{p'}(z,t)v^{1-p'}(z) dy \right)^\frac{r}{p'} dt \right)^\frac 1r <\infty \] and \[ \left( \int_0^\infty \left( \int_t^\infty w(x) U^q(t,x) dx \right)^\frac{r}{p} w(t) \sup_{z\in(0,t)} U^q(z,t)\left( \int_0^z v^{1-p'}(s) ds \right)^\frac{r}{p'} dt \right)^\frac 1r <\infty, \] where $p':=\frac{p}{p-1}$ and $r:=\frac{pq}{p-q}$. Analogous conditions for the case $p=1$ and for the dual version of the inequality are also presented.