[proofplan] We verify the defining properties of a $C^*$-algebra directly from the ambient operator algebra $\mathcal{L}(H)$. The algebraic operations, submultiplicative norm, and involution identities are inherited because $A$ is a complex subalgebra closed under adjoints. Completeness follows from norm-closedness of $A$ inside the [Banach space](/page/Banach%20Space) $\mathcal{L}(H)$. The only special identity to check is the $C^*$-identity, and this is proved from the [Hilbert space](/page/Hilbert%20Space) [inner product](/page/Inner%20Product) formula for the operator norm. [/proofplan]

[step:Inherit the normed algebra and involution structure from $\mathcal{L}(H)$] Let $\|\cdot\|_{\mathcal{L}(H)}$ denote the operator norm on $\mathcal{L}(H)$, and define the norm on $A$ by \begin{align*} \|T\|_A := \|T\|_{\mathcal{L}(H)} \end{align*} for every $T \in A$. Since $A$ is a complex subalgebra of $\mathcal{L}(H)$, it is closed under addition, scalar multiplication, and composition of operators. Hence $A$ is a complex algebra with the inherited operations. For $S,T \in A$, the operator norm on $\mathcal{L}(H)$ gives \begin{align*} \|ST\|_A \leq \|S\|_A \|T\|_A. \end{align*} Thus $A$ is a normed algebra. The adjoint operation on $A$ is the map \begin{align*} *: A &\to A \end{align*} given by $T \mapsto T^*$. This map is well-defined because $A$ is closed under adjoints. For $S,T \in A$ and $\lambda \in \mathbb{C}$, the Hilbert space adjoint identities in $\mathcal{L}(H)$ give \begin{align*} (S+T)^* = S^* + T^*, \end{align*} \begin{align*} (\lambda T)^* = \overline{\lambda}T^*, \end{align*} \begin{align*} (ST)^* = T^*S^*, \end{align*} and \begin{align*} (T^*)^* = T. \end{align*} Therefore $A$ is a normed involutive algebra. [/step]

[step:Use norm-closedness to prove that $A$ is complete]Let $(T_n)_{n \in \mathbb{N}}$ be a [Cauchy sequence](/page/Cauchy%20Sequence) in $A$ with respect to $\|\cdot\|_A$. Since $\|T\|_A=\|T\|_{\mathcal{L}(H)}$ for $T \in A$, the same sequence is Cauchy in $\mathcal{L}(H)$. The space $\mathcal{L}(H)$ is complete in the operator norm, so there exists $T \in \mathcal{L}(H)$ such that \begin{align*} \lim_{n \to \infty} \|T_n - T\|_{\mathcal{L}(H)} = 0. \end{align*} Because $A$ is closed in $\mathcal{L}(H)$ for the operator norm and every $T_n$ belongs to $A$, the limit $T$ belongs to $A$. Hence \begin{align*} \lim_{n \to \infty} \|T_n - T\|_A = 0. \end{align*} Thus every Cauchy sequence in $A$ converges in $A$, so $A$ is a Banach algebra.[/step]

[guided]We must prove completeness for the norm on $A$, not merely for the ambient space. Let $(T_n)_{n \in \mathbb{N}}$ be a Cauchy sequence in $A$ with respect to the inherited norm $\|\cdot\|_A$. By definition of this inherited norm, \begin{align*} \|T_n - T_m\|_A = \|T_n - T_m\|_{\mathcal{L}(H)} \end{align*} for all $m,n \in \mathbb{N}$. Therefore $(T_n)_{n \in \mathbb{N}}$ is also Cauchy in the Banach space $\mathcal{L}(H)$. Since $\mathcal{L}(H)$ is complete in the operator norm, there exists an operator $T \in \mathcal{L}(H)$ such that \begin{align*} \lim_{n \to \infty} \|T_n - T\|_{\mathcal{L}(H)} = 0. \end{align*} At this point the limit is known only to lie in $\mathcal{L}(H)$. The hypothesis that $A$ is norm-closed is exactly what brings the limit back into $A$: because each $T_n$ belongs to $A$ and $T_n \to T$ in the operator norm, closedness gives $T \in A$. Finally, since the norm on $A$ is the restriction of the operator norm, \begin{align*} \lim_{n \to \infty} \|T_n - T\|_A = \lim_{n \to \infty} \|T_n - T\|_{\mathcal{L}(H)} = 0. \end{align*} Thus the original Cauchy sequence converges to an element of $A$. This proves that $A$ is complete in its inherited norm.[/guided]

[step:Prove the $C^*$-identity in the ambient operator algebra] Let $T \in \mathcal{L}(H)$. We first show that \begin{align*} \|T^*T\|_{\mathcal{L}(H)} = \|T\|_{\mathcal{L}(H)}^2. \end{align*} For every $x \in H$, the defining property of the adjoint gives \begin{align*} (T^*Tx,x)_H = (Tx,Tx)_H = \|Tx\|_H^2. \end{align*} If $\|x\|_H \leq 1$, then the operator norm bound applied to $T^*T$ gives \begin{align*} \|Tx\|_H^2 = |(T^*Tx,x)_H| \leq \|T^*T x\|_H \|x\|_H \leq \|T^*T\|_{\mathcal{L}(H)}. \end{align*} Taking the supremum over all $x \in H$ with $\|x\|_H \leq 1$ gives \begin{align*} \|T\|_{\mathcal{L}(H)}^2 \leq \|T^*T\|_{\mathcal{L}(H)}. \end{align*} Conversely, [submultiplicativity of the operator norm](/theorems/1054) and the equality $\|T^*\|_{\mathcal{L}(H)}=\|T\|_{\mathcal{L}(H)}$ give \begin{align*} \|T^*T\|_{\mathcal{L}(H)} \leq \|T^*\|_{\mathcal{L}(H)}\|T\|_{\mathcal{L}(H)} = \|T\|_{\mathcal{L}(H)}^2. \end{align*} The two inequalities prove the $C^*$-identity in $\mathcal{L}(H)$. [/step]

[step:Restrict the $C^*$-identity to $A$] Let $T \in A$. Since $A$ is closed under adjoints, $T^* \in A$. Since $A$ is closed under multiplication, $T^*T \in A$. The norm on $A$ is inherited from $\mathcal{L}(H)$, so the ambient $C^*$-identity gives \begin{align*} \|T^*T\|_A = \|T^*T\|_{\mathcal{L}(H)} = \|T\|_{\mathcal{L}(H)}^2 = \|T\|_A^2. \end{align*} Thus $A$ is a Banach involutive algebra satisfying the $C^*$-identity. Therefore $A$ is a possibly nonunital $C^*$-algebra. [/step]