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 $l\in 𝐂$. For every $\{x,y\}\subset H$ we have

1. $⟨P^{**}x|y⟩=⟨x|P^{*}y⟩=⟨Px|y⟩$ so $P^{**}=P$,

2. $⟨{(PQ)}^{*}x|y⟩=⟨x|PQy⟩=⟨P^{*}x|Qy⟩=⟨Q^{*}{P}^{*}x|y⟩$ so ${(PQ)}^{*}=Q^{*}{P}^{*}$ and

3. $⟨{(lP+Q)}^{*}x|y⟩=⟨x|(lP+Q)y⟩=l⟨x|Py⟩+⟨x|Qy⟩=l⟨P^{*}x|y⟩+⟨Q^{*}x|y⟩=⟨(l^{*}{P}^{*}+Q^{*})x|y⟩$ so ${(lP+Q)}^{*}=l^{*}{P}^{*}+Q^{*}$,

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

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

$$\parallel PQ\parallel =\underset{x\in H\setminus \{0\}}{sup}\frac{{\parallel PQx\parallel }_H}{{\parallel x\parallel }_H}\le \underset{x\in H\setminus \{0\}}{sup}\frac{\parallel P\parallel {\parallel Qx\parallel }_H}{{\parallel x\parallel }_H}=\parallel P\parallel \parallel Q\parallel ,$$

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

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

${\parallel P\parallel }^2$$=$$\underset{x\in H\setminus \{0\}}{sup}{\displaystyle \frac{{\parallel Px\parallel }_H^2}{{\parallel x\parallel }_H^2}}=\underset{x\in H\setminus \{0\}}{sup}{\displaystyle \frac{⟨Px|Px⟩}{{\parallel x\parallel }_H^2}}=\underset{x\in H\setminus \{0\}}{sup}{\displaystyle \frac{⟨P^{*}Px|x⟩}{{\parallel x\parallel }_H^2}}$

$\le$$\underset{x\in H\setminus \{0\}}{sup}{\displaystyle \frac{{\parallel P^{*}Px\parallel }_H{\parallel x\parallel }_H}{{\parallel x\parallel }_H^2}}=\underset{x\in H\setminus \{0\}}{sup}{\displaystyle \frac{{\parallel P^{*}Px\parallel }_H}{{\parallel x\parallel }_H}}=\parallel P^{*}P\parallel$

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

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.$\mathrm{\square }$