Let $f\in L^p(\Omega, \nu).$ Let $L$ be a self-adjoint operator on $L^p.$ Suppose we want, for every $p>1,$ to prove an inequality $$||Lf||_p\leq C(p)||f||_p,$$ where $C(p)$ is some function depending on $p.$ Why is it by duality enought to prove the inequality for the case $p>2?$
Thank you for your answers.
The assertion "$L$ is self-adjoint on $\mathbb L^p$" means that for each $g\in \mathbb L^{p'}$ (where $1/p+1/p'=1$) and $f\in\mathbb L^p$, the following equality holds: $$\langle Lf,g\rangle_{p,p'}=\langle L^*g,f\rangle_{p',p}=\langle Lg,f\rangle_{p',p}.$$ Assume that we proved that $\lVert Lf\rVert_p\leqslant C(p)\lVert f\rVert_p$ for $p>2$. Then if $1\lt p\leqslant 2$, $p'>2$ and we conclude noticing that $$\lVert Lf\rVert_p=\sup_{g\in\mathbb L^{p'}, \lVert g\rVert_{p'}=1}\langle Lf,g\rangle= \sup_{g\in\mathbb L^{p'}, \lVert g\rVert_{p'}=1}\langle Lg,f\rangle\leqslant C(p)\lVert f\rVert_{p}.$$