{"paper":{"title":"Boundedness of Hardy-type operators with a kernel: integral weighted conditions for the case $0<q<1\\le p<\\infty$","license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","headline":"","cross_cats":[],"primary_cat":"math.FA","authors_text":"Martin K\\v{r}epela","submitted_at":"2016-02-02T08:02:11Z","abstract_excerpt":"Let $U:[0,\\infty)^2 \\to [0,\\infty)$ be a~measurable kernel satisfying:\n  (i) $U(x,y)$ is nonincreasing in $x$ and nondecreasing in $y$;\n  (ii) there exists a~constant $\\theta>0$ such that $U(x,z) \\le \\theta\\left( U(x,y)+U(y,z) \\right)$ for all $0\\le x<y<z<\\infty$;\n  (iii) $U(0,y)>0$ for all $y>0$.\n  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 \\[ \\le"},"claims":{"count":0,"items":[],"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"source":{"id":"1602.00820","kind":"arxiv","version":1},"verdict":{"id":null,"model_set":{},"created_at":null,"strongest_claim":"","one_line_summary":"","pipeline_version":null,"weakest_assumption":"","pith_extraction_headline":""},"references":{"count":0,"sample":[],"resolved_work":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57","internal_anchors":0},"formal_canon":{"evidence_count":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"author_claims":{"count":0,"strong_count":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"builder_version":"pith-number-builder-2026-05-17-v1"}