Let $\mu$ and $\nu$ be two $\sigma$-finite measures, and consider the operator (supposed well-defined) $L^{\infty}(\mu)$ to $L^{\infty}(\nu)$ by $Tg(y) = \int T(x,y)g(x) \mu(dx)$ where the kernel $T(x,y) \geq 0$. Under what conditions on the kernel is the operator uniformly lower bounded : $\|Tg\|_{L^{\infty}(\nu)} \geq \delta \|g\|_{L^{\infty}(\mu)}$ for some $\delta > 0$?
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$\begingroup$ Do you mean $Tg(y) = \int T(x,y) g(x)\mu(dx)$? Or otherwise, how does your definition involve $g$? $\endgroup$– user1149748Commented Jul 8 at 15:28
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$\begingroup$ Yes! Sorry for the mistake $\endgroup$– thibault jeanninCommented Jul 8 at 15:31
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$\begingroup$ Can I also ask, what kind of lower bound are you looking for? A pointwise lower bound is hopeless, since $g\equiv 0$ results in $Tg \equiv 0$. $\endgroup$– user1149748Commented Jul 8 at 15:48
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$\begingroup$ @user1149748 : $\|Tg\|_{L^{\infty}(\nu)} \geq \delta \|g\|_{L^{\infty}(\mu)}$ for $\delta > 0$ $\endgroup$– thibault jeanninCommented Jul 8 at 15:51
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$\begingroup$ This lower bound implies the closedness of the range of $T$, hence $T$ cannot be compact. $\endgroup$– dawCommented Jul 9 at 9:21
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