Norm of a self adjoint operator from its quadratic form

The quadratic form is dominated by $M$. For $z\in H$, if $z=0$ then $\mathrm{\mid }q(z)\mathrm{\mid }=0=M\mathrm{\parallel }z{\mathrm{\parallel }}^2$, while if $z\mathrm{\ne }0$ then [A2] gives $q(z)=\mathrm{\parallel }z{\mathrm{\parallel }}^{2}q(u)$ for the unit vector $u=z\mathrm{/}\mathrm{\parallel }z\mathrm{\parallel }$, hence $\mathrm{\mid }q(z)\mathrm{\mid }=\mathrm{\parallel }z{\mathrm{\parallel }}^2\mathrm{\mid }q(u)\mathrm{\mid }\le M\mathrm{\parallel }z{\mathrm{\parallel }}^2$.

$M\le \mathrm{\parallel }T\mathrm{\parallel }$. For every unit vector $x$, Cauchy–Schwarz and the norm bound give $\mathrm{\mid }q(x)\mathrm{\mid }=\mathrm{\mid }⟨Tx,x⟩\mathrm{\mid }\le \mathrm{\parallel }Tx\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }\le \mathrm{\parallel }T\mathrm{\parallel }$, so $\mathrm{\parallel }T\mathrm{\parallel }$ is an upper bound of the set whose supremum is $M$; hence $M\le \mathrm{\parallel }T\mathrm{\parallel }$, and when $H=\{0\}$ both numbers are $0$.

A sesquilinear bound. For all $x,y\in H$ one has $\mathrm{\mid }⟨Tx,y⟩\mathrm{\mid }\le \frac{M}{2}(\mathrm{\parallel }x{\mathrm{\parallel }}^2+\mathrm{\parallel }y{\mathrm{\parallel }}^2)$: if $x=0$ or $y=0$ then $⟨Tx,y⟩=0$ and the right side is $\ge 0$; otherwise put $c:=⟨Tx,y⟩$, and if $c\mathrm{\ne }0$ choose the unit scalar $\omega$ with $\bar{\omega }c=\mathrm{\mid }c\mathrm{\mid }$ (namely $\omega =c\mathrm{/}\mathrm{\mid }c\mathrm{\mid }$ over $\mathbb{C}$, $\omega =\mathrm{sign}(c)$ over $\mathbb{R}$) and set $u:=\omega y$, so that $\mathrm{\parallel }u\mathrm{\parallel }=\mathrm{\parallel }y\mathrm{\parallel }$ and $⟨Tx,u⟩=\bar{\omega }c=\mathrm{\mid }c\mathrm{\mid }\ge 0$ is real; then [A2] gives $4\mathrm{\mid }⟨Tx,y⟩\mathrm{\mid }=4⟨Tx,u⟩=q(x+u)-q(x-u)\le \mathrm{\mid }q(x+u)\mathrm{\mid }+\mathrm{\mid }q(x-u)\mathrm{\mid }\le M(\mathrm{\parallel }x+u{\mathrm{\parallel }}^2+\mathrm{\parallel }x-u{\mathrm{\parallel }}^2)$ by [step 1.1], and the parallelogram law [A3] turns the last factor into $2\mathrm{\parallel }x{\mathrm{\parallel }}^2+2\mathrm{\parallel }u{\mathrm{\parallel }}^2=2\mathrm{\parallel }x{\mathrm{\parallel }}^2+2\mathrm{\parallel }y{\mathrm{\parallel }}^2$.

Removing the norms. For $x,y\mathrm{\ne }0$ and every real $t>0$, [step 2.1] applied to the pair $(tx,t^{-1}y)$ together with the scaling identity of [A2] gives $\mathrm{\mid }⟨Tx,y⟩\mathrm{\mid }\le \frac{M}{2}(t^2\mathrm{\parallel }x{\mathrm{\parallel }}^2+t^{-2}\mathrm{\parallel }y{\mathrm{\parallel }}^2)$; the right side is minimised at $t^2=\mathrm{\parallel }y\mathrm{\parallel }\mathrm{/}\mathrm{\parallel }x\mathrm{\parallel }>0$, where it equals $M\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$, so $\mathrm{\mid }⟨Tx,y⟩\mathrm{\mid }\le M\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$ for all $x,y\in H$ (the zero cases being trivial).

$\mathrm{\parallel }T\mathrm{\parallel }\le M$. For $x\mathrm{\ne }0$ the dual norm formula [A4] gives $\mathrm{\parallel }Tx\mathrm{\parallel }={\sup }_{\mathrm{\parallel }y\mathrm{\parallel }=1}\mathrm{\mid }⟨Tx,y⟩\mathrm{\mid }\le M\mathrm{\parallel }x\mathrm{\parallel }$ by [step 3.1] with $\mathrm{\parallel }y\mathrm{\parallel }=1$, and the inequality also holds at $x=0$; thus $M$ is a uniform bound for $T$ on the unit ball, so $\mathrm{\parallel }T\mathrm{\parallel }\le M$ by the unit-ball characterisation of the operator norm in [A4].

Conclusion. Steps 1.2 and 4.1 give $\mathrm{\parallel }T\mathrm{\parallel }=M$; if $H=\{0\}$ then $T=0$, $M=0$ by the empty-supremum convention and $\mathrm{\parallel }T\mathrm{\parallel }=0$, so the identity holds there as well.