Coefficient variation controls a weighted-shift operator

If the source basis is empty, both shifts are the zero map on the zero domain, so both norms are zero by [F2] and the claimed bound holds. Otherwise the source basis is nonempty. Put $S=\{x:a_x\mathrm{\ne }0\}$. All terms outside $S$ vanish, and injectivity on $S$ makes the destination vectors indexed by $S$ pairwise orthogonal. Thus $\mathrm{\parallel }Tv{\mathrm{\parallel }}^2={\sum }_{x\in S}\mathrm{\mid }{a}_x{\mathrm{\mid }}^2\mathrm{\mid }{v}_x{\mathrm{\mid }}^2={\sum }_x\mathrm{\mid }{a}_x{\mathrm{\mid }}^2\mathrm{\mid }{v}_x{\mathrm{\mid }}^2$ for every vector in the source span, including when $S$ is empty. This is bounded by $({\max }_x\mathrm{\mid }{a}_x\mathrm{\mid }{)}^2\mathrm{\parallel }v{\mathrm{\parallel }}^2$, and equality is attained on a source basis vector with maximal coefficient, giving $\mathrm{\parallel }T\mathrm{\parallel }={\max }_x\mathrm{\mid }{a}_x\mathrm{\mid }$ by the operator-norm definition. The coefficients $a_x{e}_x$ of $T_s-T$ have support contained in $S$, so their nonzero destinations are also distinct; the same argument gives $\mathrm{\parallel }{T}_s-T\mathrm{\parallel }={\max }_x\mathrm{\mid }{a}_x{e}_x\mathrm{\mid }$.

For a nonempty source basis, this maximum is at most $ϵ{\max }_x\mathrm{\mid }{a}_x\mathrm{\mid }$. For a fixed $\delta$, equality of two destinations $(J-1,m+\delta )$ forces both $J$ and $m$ equal, so the claimed angular-momentum application is injective. If all $a_x$ vanish, both shifts vanish.