Continuous Functional Calculus for Self Adjoint and Normal Operators

Assume Countable Choice. For every nonzero complex Hilbert space $H$, $\mathcal{B}(H)$, with the operator norm, composition, identity, and Hilbert adjoint, is a unital C*-algebra and $\mathrm{\parallel }{T}^{\ast }T\mathrm{\parallel }=\mathrm{\parallel }T{\mathrm{\parallel }}^2$.

Assume Countable Choice. If $T$ is a bounded self-adjoint operator on a nonzero complex Hilbert space, then $\sigma (T)\subseteq \mathbb{R}$, and $\mathrm{\parallel }(T-zI)x\mathrm{\parallel }\ge \mathrm{\mid }\mathrm{Im}z\mathrm{\mid }\mathrm{\parallel }x\mathrm{\parallel }$ for every $z\mathrm{\notin }\mathbb{R}$ and every $x\in H$.

Assume Countable Choice. If $T$ is a bounded positive operator on a nonzero complex Hilbert space, then $\sigma (T)\subseteq [0,+\mathrm{\infty })$.

Assume Countable Choice and let $H$ be a nonzero complex Hilbert space. Write

$\mathcal{B}(H{)}_{\mathrm{s}\mathrm{a}}:=\{T\in \mathcal{B}(H):T^{\ast }=T\}$

for the set of bounded self-adjoint operators on $H$ (Self-adjoint, positive, unitary and normal operators). For $S,T\in \mathcal{B}(H{)}_{\mathrm{s}\mathrm{a}}$ define

$S\le T⟺⟨(T-S)x,x⟩\ge 0\text{for every }x\in H\mathrm{.}$

Equivalently $T\ge S$; the notation $T\ge 0$ is exactly the positivity of Self-adjoint, positive, unitary and normal operators. Operators are compared only when both are self-adjoint: the relation is not defined for a general pair in $\mathcal{B}(H)$, and no conjugate-linear or non-real quadratic form is admitted by the definition.

$\mathcal{B}(H{)}_{\mathrm{s}\mathrm{a}}$ is a real vector space and $\le$ is a partial order on it. Sums and real scalar multiples of self-adjoint operators are self-adjoint, because the adjoint is conjugate-linear and $T^{\ast \ast }=T$ (Hilbert-adjoint identities); the zero operator is self-adjoint. So the comparisons below are between elements of a real vector space.

• Reflexivity. $S-S=0$ and $⟨0x,x⟩=0$, so $S\le S$.
• Transitivity. If $S\le T$ and $T\le U$, then for every $x$ $⟨(U-S)x,x⟩=⟨(U-T)x,x⟩+⟨(T-S)x,x⟩\ge 0,$ by additivity of the pairing in its first argument, so $S\le U$.
• Antisymmetry. If $S\le T$ and $T\le S$, then $⟨(T-S)x,x⟩=0$ for every $x$. The sesquilinear form $B(x,y):=⟨(T-S)x,y⟩$ is linear in $x$ and conjugate-linear in $y$ and satisfies $B(z,z)=0$ for every $z$, so the four-term expansion $4B(x,y)=B(x+y,x+y)-B(x-y,x-y)+iB(x+iy,x+iy)-iB(x-iy,x-iy)$ vanishes for all $x,y$. Hence $⟨(T-S)x,y⟩=0$ for all $x,y$, and fixing $x$ and taking $y=(T-S)x$ gives $\mathrm{\parallel }(T-S)x{\mathrm{\parallel }}^2=0$, so $T-S=0$ and $S=T$. The expansion is the displayed consequence of additivity and conjugate-linearity alone, so antisymmetry consumes no completeness of $H$ (Hilbert space).

Two conventions. First, the order is a partial order on the real vector space of self-adjoint operators; it is not a total order, and the extrema of the spectrum in the later results are taken in $\mathbb{R}$, not by comparing operators. Second, $T\ge 0$ refers to the quadratic form of a self-adjoint operator; the counterexample on the companion page shows that nonnegativity of the spectrum alone does not define positivity for operators that are not self-adjoint (The operator norm as the least bound and as the unit-sphere or unit-ball supremum for the operator data used throughout).

Assume Countable Choice and let $H$ be a nonzero complex Hilbert space with $T\in \mathcal{B}(H)$ (The spaces (\mathcal B(X,Y)) and (\mathcal B(X)) of bounded linear operators). The unital $\ast$-algebra generated by $T$ is the smallest subset $P(T)\subseteq \mathcal{B}(H)$ containing the identity $I$ and $T$ that is closed under addition, scalar multiplication, multiplication and the adjoint. The C*-algebra generated by $T$ is its norm closure

$C^{\ast }(I,T):={\overline{P(T)}}^{\mathrm{\parallel }\cdot \mathrm{\parallel }}\subseteq \mathcal{B}(H)\mathrm{.}$

For an arbitrary $T$, the concrete elements of $P(T)$ are finite linear combinations of finite words in the two letters $T$ and $T^{\ast }$. When $T$ is normal, the two letters commute and every such word can be reordered, so in that case

$P(T)=\{{\sum }_{j,k}{c}_{jk}{T}^j(T^{\ast }{)}^k:\text{only finitely many }{c}_{jk}\mathrm{\ne }0\}\mathrm{.}$

Well-definedness and the algebra structure. The set $P(T)$ is a complex $\ast$-subalgebra of $\mathcal{B}(H)$ containing $I$, and the closure of a $\ast$-subalgebra of a C*-algebra is again a unital $\ast$-subalgebra: sums, products and adjoints of limits are the limits of the corresponding sums, products and adjoints, because the algebra operations and the adjoint are continuous (Hilbert-adjoint identities, Bounded Hilbert operators form a C star algebra, C star algebra). A closed subset of the complete space $\mathcal{B}(H)$ is complete, so $C^{\ast }(I,T)$ with the inherited norm, multiplication, unit and adjoint is itself a unital complex C*-algebra, a unital C*-subalgebra of $\mathcal{B}(H)$ with the same identity $I$.

Commutativity is exactly normality. If $T$ is normal (Self-adjoint, positive, unitary and normal operators), then $T$ commutes with $T^{\ast }$, hence any two words in $T$ and $T^{\ast }$ commute, hence any two $\ast$-polynomials commute, so $P(T)$ is commutative; commutativity passes to the closure because if $A_n{B}_n=B_n{A}_n$ and $A_n\to A$, $B_n\to B$, then $AB=BA$ by continuity of multiplication. Conversely, if $C^{\ast }(I,T)$ is commutative, then the elements $T$ and $T^{\ast }$ of it commute, that is $T{T}^{\ast }=T^{\ast }T$ and $T$ is normal. In particular $C^{\ast }(I,T)$ is a nonzero commutative unital C*-algebra exactly when $T$ is normal, and only that case is fed to the Gelfand theory later on this page.

Minimality convention. $C^{\ast }(I,T)$ is the smallest closed unital $\ast$-subalgebra of $\mathcal{B}(H)$ containing $T$: every closed unital $\ast$-subalgebra contains $P(T)$ and hence its closure. No generator other than $I$ and $T$ is adjoined, and the definition does not presuppose any particular representation of the generated algebra (The operator norm as the least bound and as the unit-sphere or unit-ball supremum for the norm used in the closure).

Assume AC. For a bounded normal operator $T$ on a nonzero complex Hilbert space, $\mathrm{\parallel }T\mathrm{\parallel }=r(T)=\max \{\mathrm{\mid }\lambda \mathrm{\mid }:\lambda \in \sigma (T)\}$.

Assume AC. A bounded normal operator whose spectrum is $\{0\}$ is the zero operator.

Assume Countable Choice and let $H,K$ be nonzero complex Hilbert spaces with $U\in \mathcal{B}(H,K)$ a bounded linear operator (The operator norm as the least bound and as the unit-sphere or unit-ball supremum).

• $U$ is an isometry when $\mathrm{\parallel }Ux\mathrm{\parallel }=\mathrm{\parallel }x\mathrm{\parallel }$ for every $x\in H$; equivalently $U^{\ast }U=I_H$.
• $U$ is a coisometry when $U^{\ast }$ is an isometry; equivalently $U{U}^{\ast }=I_K$.
• $U$ is a partial isometry when $U$ vanishes on its kernel and is isometric on the orthogonal complement of its kernel: $x\in \ker U⟹Ux=0,x\in (\ker U{)}^{\perp }⟹\mathrm{\parallel }Ux\mathrm{\parallel }=\mathrm{\parallel }x\mathrm{\parallel }\mathrm{.}$ The closed subspace $(\ker U{)}^{\perp }$ is the initial space of $U$, and the closed subspace $\mathrm{ran}U$ is its final space.

The equivalence in the isometry clause. If $U^{\ast }U=I$ then $\mathrm{\parallel }Ux{\mathrm{\parallel }}^2=⟨U^{\ast }Ux,x⟩=⟨x,x⟩$. Conversely, if $\mathrm{\parallel }Ux\mathrm{\parallel }=\mathrm{\parallel }x\mathrm{\parallel }$ for every $x$, then $S:=U^{\ast }U-I$ is self-adjoint and $⟨Sx,x⟩=\mathrm{\parallel }Ux{\mathrm{\parallel }}^2-\mathrm{\parallel }x{\mathrm{\parallel }}^2=0$ for every $x$; the four-term expansion of the sesquilinear form $(x,y)\mapsto ⟨Sx,y⟩$ applied to the vanishing diagonal values gives $⟨Sx,y⟩=0$ for all $x,y$, hence $S=0$, that is $U^{\ast }U=I$ (Hilbert-adjoint identities for the adjoint identities).

Well-definedness of the subspaces. The kernel $\ker U$ is a closed linear subspace because $U$ is bounded and linear, so its orthogonal complement is a closed subspace and the orthogonal-decomposition theorem gives $H=\ker U\oplus (\ker U{)}^{\perp }$ (Orthogonal decomposition by a closed subspace, Orthogonality and the orthogonal complement). The range of a partial isometry is closed: $U$ is isometric on the closed subspace $(\ker U{)}^{\perp }$ and vanishes on its orthogonal complement, so $U$ carries the unit sphere of $(\ker U{)}^{\perp }$ to a closed set and $\mathrm{ran}U=U[(\ker U{)}^{\perp }]$ is closed. In the terminology of The Hilbert orthogonal projection onto a closed subspace, the orthogonal projection onto the initial space is an orthogonal projection in the sense of that item, and the partial isometry restricted to it is an isometry onto $\mathrm{ran}U$.

Immediate cases and conventions. Every isometry and every coisometry is a partial isometry: an isometry has $\ker U=\{0\}$ and is isometric on $H=(\ker U{)}^{\perp }$, and a coisometry has $(\ker U{)}^{\perp }=\mathrm{ran}{U}^{\ast }$ on which it is isometric (Hilbert-adjoint identities). The zero operator is a partial isometry, with $\ker U=H$ and $\mathrm{ran}U=\{0\}$ (Hilbert space). A partial isometry need not be an isometry and need not be unitary; the unilateral shift on the companion page is an isometry that is not a coisometry.

Assume Countable Choice. For a bounded operator $U$ on a nonzero complex Hilbert space, the partial-isometry condition is equivalent to $U^{\ast }U$ being the orthogonal projection onto $(\ker U{)}^{\perp }$, and then $U{U}^{\ast }$ is the orthogonal projection onto $\mathrm{ran}U$; equivalently $U^{\ast }$ is a partial isometry.

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 $T$.

Assume AC. On a complex Hilbert space, $w$ is a norm with $w(T)\le \mathrm{\parallel }T\mathrm{\parallel }\le 2w(T)$ for every bounded $T$; if $T$ is normal, then $w(T)=\mathrm{\parallel }T\mathrm{\parallel }$.