bounded operators on a Hilbert space form a $C^{*}$-algebra

In this entry we show how the algebra $\mathrm{B}(H)$ of bounded linear operators on an Hilbert space $H$ is one of the most natural examples of $C^{*}$-algebras (http://planetmath.org/CAlgebra). In fact, by the Gelfand-Naimark representation theorem, every $C^{*}$-algebra is isomorphic to a *-subalgebra of $\mathrm{B}(H)$ for some Hilbert space $H$.

Lemma If $H$ is a Hilbert space, then $\mathrm{B}(H)$, the algebra of bounded linear operators on $H$, is a $*$-algebra.

Proof: Let $H$ be a Hilbert space. We must prove that the adjugation is an involution. Let $\{P,Q\}\subset\mathrm{B}(H)$ and $l\in\mathbf{C}$. For every $\{x,y\}\subset H$ we have

1. 1.

$\langle P^{**}x|y\rangle=\langle x|P^{*}y\rangle=\langle Px|y\rangle$ so $P^{**}=P$,

2. 2.

$\langle(PQ)^{*}x|y\rangle=\langle x|PQy\rangle=\langle P^{*}x|Qy\rangle=% \langle Q^{*}P^{*}x|y\rangle$ so $(PQ)^{*}=Q^{*}\ P^{*}$ and

3. 3.

$\langle(lP+Q)^{*}x|y\rangle=\langle x|(lP+Q)y\rangle=l\langle x|Py\rangle+% \langle x|Qy\rangle=l\langle P^{*}x|y\rangle+\langle Q^{*}x|y\rangle=\langle(l% ^{*}P^{*}+Q^{*})x|y\rangle$ so $(lP+Q)^{*}=l^{*}P^{*}+Q^{*}$,

so we see that the adjugation is an involution and thus $\mathrm{B}(H)$ is a $*$-algebra. $\Box$

Lemma If $H$ is a Hilbert space, then $\mathrm{B}(H)$ is a Banach algebra.

Proof: Let $H$ be a Hilbert space and let $\{P,Q\}\subset\mathrm{B}(H)$. We have

 $\|PQ\|=\sup_{x\in H\setminus\{0\}}\frac{\|PQx\|_{H}}{\|x\|_{H}}\leq\sup_{x\in H% \setminus\{0\}}\frac{\|P\|\|Qx\|_{H}}{\|x\|_{H}\ }=\|P\|\|Q\|,$

so we see that $\mathrm{B}(H)$ is a Banach algebra. $\Box$

Lemma If $H$ is a Hilbert space, then $\mathrm{B}(H)$ is a $C^{*}$-algebra.

Proof: Let $H$ be a Hilbert space and let $P\in\mathrm{B}(H)$. We have

 $\displaystyle\|P\|^{2}$ $\displaystyle=$ $\displaystyle\sup_{x\in H\setminus\{0\}}\frac{\|Px\|_{H}^{2}}{\|x\|_{H}^{2}}=% \sup_{x\in H\setminus\{0\}}\frac{\langle Px|Px\rangle}{\|x\|_{H}^{2}}=\sup_{x% \in H\setminus\{0\}}\frac{\langle P^{*}Px|x\rangle}{\|x\|_{H}^{2}}$ $\displaystyle\leq$ $\displaystyle\sup_{x\in H\setminus\{0\}}\frac{\|P^{*}Px\|_{H}\|x\|_{H}}{\|x\|_% {H}^{2}}=\sup_{x\in H\setminus\{0\}}\frac{\|P^{*}Px\|_{H}}{\|x\|_{H}}=\|P^{*}P\|$

so $\|P\|^{2}\leq\|P^{*}P\|$ and because of the previous two lemmas say $\mathrm{B}(H)$ is a Banach algebra with involution it is a $C^{*}$-algebra. $\Box$

Lemma If $H$ is a Hilbert space, then every closed $*$-subalgebra of $\mathrm{B}(H)$ is a $C^{*}$-algebra.

Proof: Let $A$ be a closed $*$-subalgebra of $\mathrm{B}(H)$. Because $A$ is a closed subspace of a Banach space it is itself a Banach space and thus a Banach algebra with an involution and also a $C^{*}$-algebra. $\Box$

Title bounded operators on a Hilbert space form a $C^{*}$-algebra BoundedOperatorsOnAHilbertSpaceFormACalgebra 2013-03-22 14:47:12 2013-03-22 14:47:12 HkBst (6197) HkBst (6197) 9 HkBst (6197) Result msc 46L05 RepresentationOfAC_cG_dTopologicalAlgebra