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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?).

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    $\begingroup$ Which proof doesn't generalize? Can't you replace a spectral projection with some functional calculus $f(t)$? $\endgroup$ Sep 14 at 23:01
  • $\begingroup$ @NarutakaOZAWA I'm not aware of a proof of this that uses spectral projections. $\endgroup$
    – user839372
    Sep 15 at 20:24
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    $\begingroup$ 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. $\endgroup$
    – Jamie Gabe
    Sep 17 at 9:15
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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.

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