Let $A$ be a $C^\*$-algebra, $E$ be a (right) Hilbert $A$-module and $t \in \mathcal{L}\_A(E)$ be an adjointable operator satisfying $t=t^\*$. Is it true that
$$\|t\| = \sup\_{z \in E, \|z\| = 1} \|\langle tz,z\rangle\|\_A?$$


Obviously, if we denote the supremum on the right by $M$, we have $M \le \|t\|$ by the Cauchy-Schwarz inequality, so the main interest lies in the other inequality.


When $A= \mathbb{C}$ (and thus $E$ is a Hilbert space), the result is well-known, but neither of the proofs I know seem to generalise to the framework of Hilbert $C^\*$-modules. Perhaps a slick application of the spectral theorem does the job (as seems to be suggested in a comment?).



$\|t\| = \sup_{\|z\| \le 1} \|\langle tz,z\rangle\|$ when $t=t^*$
I'll post an answer based on the comments above.


Since $t$ is self-adjoint, we can write $t= t\_+-t\_-$ where $t\_+$ and $t\_-$ are positive elements with $t\_+ t\_- = 0$. The latter condition ensures that $\|t\| = \max\{\|t\_+\|, \|t\_-\|\}$.


Assume, without loss of generality, that $\|t\| = \|t\_+\|$.


Let $\epsilon > 0$. By definition of the operator norm, there is an element $z$ in the unit ball of $E$ such that $\|t\_+^{1/2}\| \le \epsilon + \|t\_+^{1/2}z\|.$


Consider the isometric unital $\*$-morphism
$$C(\sigma(t)) \to C^\*(1,t): f \mapsto f(t)$$


given by continuous functional calculus. Define $f \in C(\sigma(t))$ to be $f=0$ on $\sigma(t)\cap [-\infty,0]$ and to be $1$ on $\sigma(t)\cap [\delta,\infty]$ and $\operatorname{Im}f\subseteq [0,1]$, where $\delta>0$ is chosen such that $2s^{1/2}< \epsilon$ when $0 \le s \le\delta$, and consider the element $f(t)\in C^\*(1,t)\subseteq \mathcal{L}\_A(E)$. Then we define $x:= f(t)z \in E$, and we note that
$\|x\| \le \|f(t)\|\|z\| \le \|f\|\_\infty =1.$ By definition of $t\_-$, we see that $t\_-f(t) = 0$. Hence,
$$\|\langle tx,x\rangle\| = \|\langle t\_+ x,x\rangle\| = \|t\_+^{1/2}x\|^2.$$


Let $g(s) = \max\{s,0\}$. Then $$\|g^{1/2}f-g^{1/2}\|\_\infty = \sup\_{s \in [0,\delta]} |s^{1/2}f(s)-s^{1/2}| \le \sup\_{s \in [0,\delta]} 2s^{1/2}\le \epsilon.$$


It follows that


$$\|t\_+^{1/2}x-t\_+^{1/2}z\|=\|t\_+^{1/2}f(t)z-t\_+^{1/2}z\| \le \epsilon.$$


Hence, $$\|t\_+^{1/2}\| \le 2\epsilon + \|t\_+^{1/2}x\|=2\epsilon + \|\langle tx,x\rangle\|^{1/2} \le 2 \epsilon + M^{1/2}.$$


Letting $\epsilon \to 0$, we obtain $\|t\_+^{1/2}\| \le M^{1/2}$. Squaring both sides and invoking the $C^\*$-identity, we obtain
$$\|t\| = \|t\_+\| \le M$$
and the other inequality is proven.


