Is a unital $*$-morphism from a unital $C^*$-algebra $A$ to $\operatorname{End}_{\mathbb{C}}(K)$ automatically contractive? Let $A$ be a unital $C^*$-algebra and let $K$ be an inner product space (not necessarily complete!). Let $\pi: A \to \operatorname{End}_{\mathbb{C}}(K)$ be a unital algebra homomorphism such that
$$\langle \pi(a)\xi, \eta\rangle = \langle \xi, \pi(a^*)\eta\rangle$$
for all $a \in A$ (i.e. the adjoint of $\pi(a)$ exists and equals $\pi(a^*)$). Is it true that $\|\pi(a)\|\le \|a\|$ for all $a \in A$? If $K$ is a Hilbert space, this result is well-known. However, since $K$ is no longer complete, $B(K)$ is not Banach and in particular not a $C^*$-algebra. Does the result remain true?
I'm mainly interested in knowing the answer for the $C^*$-algebra $A= \ell^\infty\prod_{i \in I} M_{n_i}(\mathbb{C})$.
Thanks for your help!
 A: I think the answer is yes for your space, if I understand correctly that you mean what we call the $l^\infty$ direct sum of a sequence of matrix algebras.
Given $\pi$, let $\pi_k$ be its restriction to the $k$th summand. This is a homomorphism from $M_{n_k}$ into ${\rm End}_\mathbb{C}(K)$. Now $M_{n_k}$ is the linear span of its matrix units $(e_{ij})$, and for each $i$ we have $\pi_k(e_{ii})^2 = \pi_k(e_{ii}^2) = \pi_k(e_{ii})$, so $\pi_k(e_{ii})$ is a projection, and also $\langle \pi_k(e_{ii})\xi,\eta\rangle = \langle \xi, \pi_k(e_{ii})\eta\rangle$ which forces $\pi_k(e_{ii})$ to be the orthogonal projection onto its range. Now for $i \neq j$ we get that $\pi_k(e_{ij})$ is a partial isometry from the range of $e_{jj}$ to the range of $e_{ii}$, and this shows that we can continuously extend each $\pi_k(e_{ij})$ to the completion of $K$, which yields that $\pi_k$ is either an isometry or zero.
Letting $P_k = \pi_k(I_k)$, where $I_k$ is the unit of $M_{n_k}$, we get that the $P_k$ are mutually orthogonal projections, which lets us extend the previous result to any finite number of summands. Finally, for any $A \in \bigoplus M_{n_k}$ observe that $I_k\pi(A) = \pi(I_kA)$. This means that $\pi(A)$ restricts on the range of each $P_k$ to something of norm at most $\|A\|$, which means that $\pi$ is nonexpansive.
A: $\newcommand{\End}{\operatorname{End}}$For $T\in\End_{\mathbb C}(H)$ I write $T^*$ if the adjoint exists.  Given the hypotheses in the question, if $u\in A$ is an isometry, then $1 = \pi(1) = \pi(u^*u) = \pi(u)^*\pi(u)$.  Thus, for $\xi\in H$,
$$ \|\xi\|^2 = (\xi|\xi) = (\pi(u)^*\pi(u)\xi|\xi) = (\pi(u)\xi|\pi(u)\xi)
= \|\pi(u)\xi\|^2. $$
Hence $\pi(u)$ is an isometry, and so extends to the completion of $H$.
In particular, this applies to any unitary $u\in A$.  By the Russo-Due Theorem it follows that $\pi(a)$ is bounded for any $a\in A$, and so extends to the completion of $H$.  Alternatively, the link to wikipedia shows that $\| \pi \| \leq 1$ as $\pi(u)$ is a contraction for each unitary $u\in A$.
