Numerical range and numerical radius

Let $H$ be a nonzero complex Hilbert space and let $T\in \mathcal{B}(H)$ be a bounded linear operator (Hilbert space, The operator norm as the least bound and as the unit-sphere or unit-ball supremum). The numerical range of $T$ is the set of values of its quadratic form on the unit sphere,

$W(T):=\{⟨Tx,x⟩:x\in H,\mathrm{\parallel }x\mathrm{\parallel }=1\}\subseteq \mathbb{C},$

and the numerical radius of $T$ is

$w(T):=\sup \{\mathrm{\mid }z\mathrm{\mid }:z\in W(T)\}\mathrm{.}$

Well-definedness. The unit sphere of a nonzero Hilbert space is nonempty, so $W(T)\mathrm{\ne }\mathrm{\varnothing }$. For $\mathrm{\parallel }x\mathrm{\parallel }=1$ Cauchy–Schwarz gives $\mathrm{\mid }⟨Tx,x⟩\mathrm{\mid }\le \mathrm{\parallel }Tx\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }\le \mathrm{\parallel }T\mathrm{\parallel }$, so $W(T)$ is a nonempty subset of the closed disc of radius $\mathrm{\parallel }T\mathrm{\parallel }$ and the supremum $w(T)$ is a real number satisfying $0\le w(T)\le \mathrm{\parallel }T\mathrm{\parallel }$ (Cauchy–Schwarz: $\mathrm{\mid }⟨x,y⟩\mathrm{\mid }\le \mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$, with equality exactly for dependent pairs). The pairing is linear in its first variable and conjugate-linear in its second, so $W(T)$ is the image of the unit sphere under a continuous map, but no closedness is claimed or used here.

The zero space and the zero operator. On the zero Hilbert space the unit sphere is empty; by convention $W(0):=\{0\}$ and $w(0):=0$ there, so that the numerical radius of the zero operator is $0$ in every dimension. On a nonzero space the zero operator has $W(0)=\{0\}$ and $w(0)=0$ directly from the definition (Real and complex inner-product spaces and their induced length for the pairing convention, which is linear in the first variable throughout this page).

Two elementary facts used later. Since $W(\lambda T)=\lambda W(T)$ for scalars $\lambda$, one has $w(\lambda T)=\mathrm{\mid }\lambda \mathrm{\mid }w(T)$; and $\mathrm{\mid }z\mathrm{\mid }\le \mathrm{\parallel }T\mathrm{\parallel }$ for every $z\in W(T)$, so $w(T)\le \mathrm{\parallel }T\mathrm{\parallel }$ always. Neither definiteness nor the triangle inequality for $w$ is asserted at this point: they are proved on the next page, together with the equality $w(T)=\mathrm{\parallel }T\mathrm{\parallel }$ for normal $T$.