Numerical radius is an equivalent operator norm

For arbitrary vectors $x,y$ the expansion of the complex sesquilinear form $B(x,y):=⟨Tx,y⟩$ gives $4B(x,y)=B(x+y,x+y)-B(x-y,x-y)+iB(x+iy,x+iy)-iB(x-iy,x-iy)$.

$\sup \{\mathrm{\mid }⟨Tx,y⟩\mathrm{\mid }:\mathrm{\parallel }x\mathrm{\parallel }\le 1,\mathrm{\parallel }y\mathrm{\parallel }\le 1\}=\mathrm{\parallel }T\mathrm{\parallel }$: the upper bound is Cauchy–Schwarz and the operator-norm bound, and taking $y=Tx\mathrm{/}\mathrm{\parallel }Tx\mathrm{\parallel }$ when $Tx\mathrm{\ne }0$ recovers $\mathrm{\parallel }Tx\mathrm{\parallel }$. The same supremum over unit $x,y$ equals $\mathrm{\parallel }T\mathrm{\parallel }$: [A3] gives the unit-sphere formula for the operator norm on nonzero $H$, and the preceding choice of $y$ works whenever $Tx\mathrm{\ne }0$; when $T=0$ all values are zero.

If $w(T)=0$ then $⟨Tx,x⟩=0$ for every $x$, and applying the expansion of the form $(x,y)\mapsto ⟨Tx,y⟩$ to the zero diagonal values gives $⟨Tx,y⟩=0$ for all $x,y$, hence $T=0$.

If $S$ is normal, then $\mathrm{\parallel }{S}^{\ast }x\mathrm{\parallel }=\mathrm{\parallel }Sx\mathrm{\parallel }$ for every $x$, because $\mathrm{\parallel }{S}^{\ast }x{\mathrm{\parallel }}^2=⟨S{S}^{\ast }x,x⟩=⟨S^{\ast }Sx,x⟩=\mathrm{\parallel }Sx{\mathrm{\parallel }}^2$.

$w$ is a norm on $\mathcal{B}(H)$: homogeneity is $w(\lambda T)=\mathrm{\mid }\lambda \mathrm{\mid }w(T)$ from $W(\lambda T)=\lambda W(T)$, the triangle inequality follows from $\mathrm{\mid }⟨(S+T)x,x⟩\mathrm{\mid }\le \mathrm{\mid }⟨Sx,x⟩\mathrm{\mid }+\mathrm{\mid }⟨Tx,x⟩\mathrm{\mid }\le w(S)+w(T)$ on unit vectors, and definiteness is step 1.3.

For unit vectors $x,y$ one has $\mathrm{\mid }⟨Tx,y⟩\mathrm{\mid }\le 2w(T)$: the expansion of step 1.1 writes $4⟨Tx,y⟩$ as a signed sum of the four values $⟨Tz,z⟩$ at $z=x+y,x-y,x+iy,x-iy$, so $4\mathrm{\mid }⟨Tx,y⟩\mathrm{\mid }\le {\sum }_z\mathrm{\mid }⟨Tz,z⟩\mathrm{\mid }\le w(T){\sum }_z\mathrm{\parallel }z{\mathrm{\parallel }}^2$, and the four squared norms sum to $\mathrm{\parallel }x+y{\mathrm{\parallel }}^2+\mathrm{\parallel }x-y{\mathrm{\parallel }}^2+\mathrm{\parallel }x+iy{\mathrm{\parallel }}^2+\mathrm{\parallel }x-iy{\mathrm{\parallel }}^2=4(\mathrm{\parallel }x{\mathrm{\parallel }}^2+\mathrm{\parallel }y{\mathrm{\parallel }}^2)=8$, whence $4\mathrm{\mid }⟨Tx,y⟩\mathrm{\mid }\le 8w(T)$.

If $T$ is normal and $\lambda \in \sigma (T)$, then $T-\lambda I$ is not bounded below: if $\mathrm{\parallel }(T-\lambda I)x\mathrm{\parallel }\ge c\mathrm{\parallel }x\mathrm{\parallel }$ for some $c>0$, its kernel would be zero and its range would be closed. To see closedness, for any point $v$ in its range closure, AC chooses $u_n$ with $\mathrm{\parallel }(T-\lambda I)u_n-v\mathrm{\parallel }<1\mathrm{/}(n+1)$. The lower bound makes $(u_n)$ Cauchy; completeness gives a limit $u$, and boundedness gives $(T-\lambda I)u=v$. Furthermore, normality gives $\ker (T-\lambda I{)}^{\ast }=\ker (T-\lambda I)=\{0\}$ by equality of the two kernel norms, so the range would be dense, hence all of $H$, making $T-\lambda I$ invertible with inverse bound $1\mathrm{/}c$, contrary to $\lambda \in \sigma (T)$.

Hence $w(T)\le \mathrm{\parallel }T\mathrm{\parallel }$ and $\mathrm{\parallel }T\mathrm{\parallel }\le 2w(T)$: the first is the definition, and the second follows by taking the supremum of $\mathrm{\mid }⟨Tx,y⟩\mathrm{\mid }\le 2w(T)$ over unit $x,y$ and using step 1.2, which identifies that supremum with $\mathrm{\parallel }T\mathrm{\parallel }$.

If $T$ is normal then $\sigma (T)\subseteq \overline{W(T)}$: for each $n$ the failure of a lower bound in step 2.3 gives a unit vector at tolerance $1\mathrm{/}(n+1)$, and AC selects unit vectors $x_n$ with $\mathrm{\parallel }(T-\lambda I)x_n\mathrm{\parallel }<1\mathrm{/}(n+1)$, and then $\mathrm{\mid }⟨T{x}_n,x_n⟩-\lambda \mathrm{\mid }=\mathrm{\mid }⟨(T-\lambda I)x_n,x_n⟩\mathrm{\mid }\le \mathrm{\parallel }(T-\lambda I)x_n\mathrm{\parallel }\to 0$, so $\mathrm{\mid }\lambda \mathrm{\mid }\le w(T)$ for every $\lambda \in \sigma (T)$ and $r(T)\le w(T)$.

If $H=\{0\}$, its operator space consists only of $0$; [A1] and [A3] give $w(0)=\mathrm{\parallel }0\mathrm{\parallel }=0$, which defines a norm on this zero vector space and proves both estimates and the normal equality there, without any spectral maximum. For $H\mathrm{\ne }\{0\}$, therefore $w$ is a norm with $w(T)\le \mathrm{\parallel }T\mathrm{\parallel }\le 2w(T)$, and for normal $T$ the chain $\mathrm{\parallel }T\mathrm{\parallel }=r(T)\le w(T)\le \mathrm{\parallel }T\mathrm{\parallel }$ gives $w(T)=\mathrm{\parallel }T\mathrm{\parallel }$.