# $\|t\| = \sup_{\|z\| \le 1} \|\langle tz,z\rangle\|$ when $t=t^*$

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?).

• Which proof doesn't generalize? Can't you replace a spectral projection with some functional calculus $f(t)$? Sep 14 at 23:01
• @NarutakaOZAWA I'm not aware of a proof of this that uses spectral projections. Sep 15 at 20:24
• Sketch: By functional calculus you write $t=t_+ - t_-$. Suppose $\| t\| = \|t_+\|$ (the case $\| t\| = \|t_-\|$ is similar). Pick $z\in E$ contractive such that $\| t_+^{1/2} z\|$ is close to $\|t_+^{1/2}\|$. Letting $f:[-1,1] \to [0,1]$ be continuous which is 0 on $[-1,0]$ and 1 on $[\epsilon , 1]$, we have $\| \langle t f(t) z, f(t) z\rangle\|$ is close to $\|t\|$. In the standard Hilbert space proof one would take $f = \chi_{[0,1]}$ but a continuous approximation works as well. Sep 17 at 9:15

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.