Return to Revisions

With $C=\lVert a\rVert_{p,\omega}'$ there holds $a_j^*\le j^{-1/p}C$ for every $j$. Hence, $$\lVert a\rVert_{p,\omega}\le \sup_nn^{-1+1/p}\sum_{j=1}^nj^{-1/p}C\le \sup_nn^{-1+1/p}\Bigl(1+\int_1^nx^{-1/p}dx\Bigr)C=\sup_nn^{-1+1/p}\frac1{1-1/p}n^{1-1/p}C=\frac p{p-1}\lVert a\rVert_{p,\omega}'.$$