[proofplan]
The easy inequality follows from the automatic contractivity of $*$-homomorphisms between $C^*$-algebras. For the reverse inequality, we avoid spectral permanence and instead use the uniqueness of the $C^*$-norm. Faithfulness lets us transport the operator norm from $\pi(A)$ back to the same involutive algebra $A$; the additional closed-range hypothesis makes this transported norm complete. Once the transported norm is verified to be a complete $C^*$-norm, uniqueness forces it to agree with the original norm.
[/proofplan]
[step:Use contractivity to obtain the upper bound]
By [citetheorem:8547], every $*$-homomorphism between $C^*$-algebras is contractive. Since $\pi:A\to \mathcal{L}(H)$ is a $*$-homomorphism, for every $a\in A$ we have
\begin{align*}
\|\pi(a)\|_{\mathcal{L}(H)}\le \|a\|_A.
\end{align*}
[guided]
The [first inequality](/theorems/2897) uses only that $\pi$ is a $*$-homomorphism. The theorem [citetheorem:8547] applies because both $A$ and $\mathcal{L}(H)$ are $C^*$-algebras, and $\pi:A\to \mathcal{L}(H)$ preserves multiplication, addition, scalar multiplication, and the involution. Its conclusion is precisely the contractive estimate
\begin{align*}
\|\pi(a)\|_{\mathcal{L}(H)}\le \|a\|_A
\end{align*}
for every $a\in A$.
This proves one direction of the desired equality. The remaining work is to show that no norm is lost by applying the faithful representation.
[/guided]
[/step]
[step:Transport the operator norm back to $A$]
Define a function
\begin{align*}
\|\cdot\|_\pi:A\to [0,\infty)
\end{align*}
by
\begin{align*}
\|a\|_\pi:=\|\pi(a)\|_{\mathcal{L}(H)}
\end{align*}
for $a\in A$. Since $\pi$ is injective, $\|a\|_\pi=0$ implies $\pi(a)=0$ and hence $a=0$. Linearity of $\pi$ and the operator norm triangle inequality and homogeneity show that $\|\cdot\|_\pi$ is a norm on the complex [vector space](/page/Vector%20Space) $A$.
The norm $\|\cdot\|_\pi$ is submultiplicative because, for all $a,b\in A$,
\begin{align*}
\|ab\|_\pi=\|\pi(ab)\|_{\mathcal{L}(H)}=\|\pi(a)\pi(b)\|_{\mathcal{L}(H)}\le \|\pi(a)\|_{\mathcal{L}(H)}\|\pi(b)\|_{\mathcal{L}(H)}=\|a\|_\pi\|b\|_\pi.
\end{align*}
It is compatible with the involution because
\begin{align*}
\|a^*\|_\pi=\|\pi(a^*)\|_{\mathcal{L}(H)}=\|\pi(a)^*\|_{\mathcal{L}(H)}=\|\pi(a)\|_{\mathcal{L}(H)}=\|a\|_\pi.
\end{align*}
It satisfies the $C^*$-identity because
\begin{align*}
\|a^*a\|_\pi=\|\pi(a^*a)\|_{\mathcal{L}(H)}=\|\pi(a)^*\pi(a)\|_{\mathcal{L}(H)}=\|\pi(a)\|_{\mathcal{L}(H)}^2=\|a\|_\pi^2.
\end{align*}
[guided]
The aim is to turn the represented norm into a second $C^*$-norm on the same algebraic object $A$. Define
\begin{align*}
\|\cdot\|_\pi:A\to [0,\infty)
\end{align*}
by
\begin{align*}
\|a\|_\pi:=\|\pi(a)\|_{\mathcal{L}(H)}.
\end{align*}
Faithfulness is used exactly here: if $\|a\|_\pi=0$, then $\pi(a)=0$, and injectivity of $\pi$ gives $a=0$. Thus the transported seminorm is a genuine norm. The remaining norm axioms follow from linearity of $\pi$ and the corresponding properties of the operator norm on $\mathcal{L}(H)$.
We now check that this norm respects the algebra structure. Since $\pi$ is multiplicative, for $a,b\in A$ we have
\begin{align*}
\|ab\|_\pi=\|\pi(ab)\|_{\mathcal{L}(H)}=\|\pi(a)\pi(b)\|_{\mathcal{L}(H)}\le \|\pi(a)\|_{\mathcal{L}(H)}\|\pi(b)\|_{\mathcal{L}(H)}=\|a\|_\pi\|b\|_\pi.
\end{align*}
Since $\pi$ preserves the involution and the operator norm is invariant under adjoint, we also have
\begin{align*}
\|a^*\|_\pi=\|\pi(a^*)\|_{\mathcal{L}(H)}=\|\pi(a)^*\|_{\mathcal{L}(H)}=\|\pi(a)\|_{\mathcal{L}(H)}=\|a\|_\pi.
\end{align*}
Finally, the $C^*$-identity in the operator algebra $\mathcal{L}(H)$ gives
\begin{align*}
\|a^*a\|_\pi=\|\pi(a^*a)\|_{\mathcal{L}(H)}=\|\pi(a)^*\pi(a)\|_{\mathcal{L}(H)}=\|\pi(a)\|_{\mathcal{L}(H)}^2=\|a\|_\pi^2.
\end{align*}
Thus $\|\cdot\|_\pi$ has all algebraic and $C^*$-norm properties required of a $C^*$-norm.
[/guided]
[/step]
[step:Use the closed range hypothesis to verify completeness of the transported norm]
Let $(a_n)_{n\in\mathbb N}$ be a [Cauchy sequence](/page/Cauchy%20Sequence) in $(A,\|\cdot\|_\pi)$. By definition of $\|\cdot\|_\pi$, the sequence $(\pi(a_n))_{n\in\mathbb N}$ is Cauchy in the [Banach space](/page/Banach%20Space) $\mathcal{L}(H)$. Hence there exists $T\in\mathcal{L}(H)$ such that
\begin{align*}
\|\pi(a_n)-T\|_{\mathcal{L}(H)}\to 0.
\end{align*}
The statement assumes that $\pi(A)$ is norm-closed in $\mathcal{L}(H)$, and every $\pi(a_n)$ lies in $\pi(A)$. Therefore $T\in\pi(A)$. Choose $a\in A$ such that $T=\pi(a)$. Then
\begin{align*}
\|a_n-a\|_\pi=\|\pi(a_n)-\pi(a)\|_{\mathcal{L}(H)}=\|\pi(a_n)-T\|_{\mathcal{L}(H)}\to 0.
\end{align*}
Thus every $\|\cdot\|_\pi$-Cauchy sequence converges in $A$, so $(A,\|\cdot\|_\pi)$ is complete.
[guided]
We must prove completeness for the transported norm, because the uniqueness theorem for $C^*$-norms applies only to complete $C^*$-algebra norms. Let $(a_n)_{n\in\mathbb N}$ be Cauchy in $(A,\|\cdot\|_\pi)$. The definition of the transported norm says
\begin{align*}
\|a_m-a_n\|_\pi=\|\pi(a_m)-\pi(a_n)\|_{\mathcal{L}(H)}.
\end{align*}
Therefore $(\pi(a_n))_{n\in\mathbb N}$ is Cauchy in $\mathcal{L}(H)$. Since $\mathcal{L}(H)$ is a Banach space, there is an operator $T\in\mathcal{L}(H)$ with
\begin{align*}
\|\pi(a_n)-T\|_{\mathcal{L}(H)}\to 0.
\end{align*}
The point of the closed-range hypothesis is exactly to identify this operator limit as another represented element. Since each $\pi(a_n)$ belongs to $\pi(A)$ and $\pi(A)$ is norm-closed in $\mathcal{L}(H)$, the limit $T$ also belongs to $\pi(A)$. Hence there exists $a\in A$ such that $T=\pi(a)$. Substituting this into the transported norm gives
\begin{align*}
\|a_n-a\|_\pi=\|\pi(a_n)-\pi(a)\|_{\mathcal{L}(H)}=\|\pi(a_n)-T\|_{\mathcal{L}(H)}\to 0.
\end{align*}
So the original Cauchy sequence converges in $(A,\|\cdot\|_\pi)$. This proves that the transported norm is complete.
[/guided]
[/step]
[step:Apply uniqueness of the $C^*$-norm]
By the previous two steps, the original involutive algebra underlying $A$ admits two complete norms, namely $\|\cdot\|_A$ and $\|\cdot\|_\pi$, that make it into complex $C^*$-algebras with the same multiplication and involution. The hypotheses of [citetheorem:8546] are therefore satisfied. Hence, for every $a\in A$,
\begin{align*}
\|a\|_A=\|a\|_\pi.
\end{align*}
By the definition of $\|\cdot\|_\pi$, this is exactly
\begin{align*}
\|a\|_A=\|\pi(a)\|_{\mathcal{L}(H)}.
\end{align*}
Since $a\in A$ was arbitrary, $\pi$ is isometric on all of $A$.
[guided]
We now invoke [citetheorem:8546], the uniqueness of the $C^*$-norm. Its hypotheses require two norms on the same complex involutive algebra, with the same multiplication and involution, and each norm must make that algebra into a $C^*$-algebra. The original norm $\|\cdot\|_A$ has this property by the assumption that $A$ is a complex $C^*$-algebra. The transported norm $\|\cdot\|_\pi$ has the norm, submultiplicative, involution-invariant, and $C^*$-identity properties by the transported-norm step, and it is complete by the closed-range completeness step.
Therefore the theorem applies to the pair $\|\cdot\|_A$ and $\|\cdot\|_\pi$. For every $a\in A$ it gives
\begin{align*}
\|a\|_A=\|a\|_\pi.
\end{align*}
Finally, the definition of the transported norm is
\begin{align*}
\|a\|_\pi=\|\pi(a)\|_{\mathcal{L}(H)}.
\end{align*}
Combining these two equalities yields
\begin{align*}
\|a\|_A=\|\pi(a)\|_{\mathcal{L}(H)}.
\end{align*}
Because $a\in A$ was arbitrary, $\pi$ is isometric on all of $A$.
[/guided]
[/step]
