[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.
[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]
[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]
