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!



Is a unital $*$-morphism from a unital $C^*$-algebra $A$ to $\operatorname{End}_{\mathbb{C}}(K)$ automatically contractive?
$\newcommand{\End}{\operatorname{End}}$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](https://en.wikipedia.org/wiki/Russo%E2%80%93Dye_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 $u\in A$.


