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. 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. $\langle P^{**} x | y \rangle = \langle x | P^*y \rangle = \langle P x | y \rangle$ so $P^{**} = P$ ,
2. $\langle (PQ)^* x | y \rangle = \langle x | PQ y \rangle = \langle P^* x | Q y \rangle = \langle Q^* P^* x | y \rangle$ so $(PQ)^* = Q^*\ P^*$ and
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 $(lP+Q)^* = l^*P^* + Q^*$ ,

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 $$\|PQ\| = \sup_{x \in H \setminus \{0\} } \frac{\| PQx \|_H}{\| x \|_H} \le \sup_{x \in H \setminus \{0\} } \frac{\|P\| \| Qx \|_H}{\| x \|_H\ } = \|P\| \|Q\|,$$ so we see that $\mathrm{B}(H)$ is a Banach algebra.

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

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.