Let $A$ be a $C^*$-algebra acting on a Hilbert space that admits a cyclic unit vector $\xi \in H$. Pose $S_\xi = \{\eta \in A \xi: \| \eta \| = 1\}$, and for each $\eta \in S_\xi$, pose $A_\eta = \{a \in A : a \xi = \eta\}$, so $A_\eta \ne \emptyset$ by definition; and let $c_\eta = \inf\{\| a \| : a \in A_\eta\}$. Clearly, $c_\eta \ge 1$ for all $\eta \in S_\xi$. I am interested in the following questions.

Question 1. Is it true that $c_\eta = 1$ for all $\eta \in S_\xi$ ?

Question 1 (weak form). Is it true that $S(A)\xi$ is dense in $S(H)$, where $S(X)$ denotes the closed unit ball of a normed space $X$?

Question 2. If the answer of Question 1 turns out to be negative, can there be any examples for which $\sup\{c_\eta : \eta \in S_\xi\} = +\infty$ ?

Question 3. What if we drop the assumption of $\xi$ being cyclic but merely asks $A$ acts non-degenerately on $H$ ?


2 Answers 2


Set $A = C([0,1])$ and $H = L^2([0,1])$ (with respect to Lebesgue measure) with $\xi=1$ the constant function. Then $A\xi$ is the image of $C([0,1])$ in $L^2([0,1])$, which is a norm-decreasing injective but not bounded below inclusion. Then $S_\xi$ is the intersection of the unit ball of $L^2$ with the continuous functions, and for each $\eta$ we see that $A_\eta$ is the singleton $\{\eta\}$.

Thus, Q1 (both versions) have a negative answer, this example gives a positive answer to Q2, and Q3 can be reduced to the main question by restricting the action of $A$ to the cyclic subspace generated by $\xi$.

However, if $A$ were the compact operators on $H$ then we'd have a positive answer to Q1.


Consider $H=L^2([0,1])$ and $A = C([0,1])$ acting on $H$ by multiplication. Then $\xi = 1$, the constant function $1$, is a cyclic vector. Clearly $A\xi = C([0,1])$, considered as a subset of $L^2([0,1])$ so $S_{\xi}$ is just the set of all continuous functions on $[0,1]$ with unit $L^2$-norm. For each $\eta\in S_{\xi}$, the set $A_{\eta}$ is a singleton $\{\eta\}$, considered as the multiplication operator by $\eta$ on $L^2([0,1])$. Therefore, $c_{\eta} = \|\eta\|_{\infty}$. Recall that on the other hand, $\|\eta\|_{2} = 1$. From here, Question 2 has an affirmative answer because you can make the sup norm as large as you want while keeping $L^2$-norm unchanged.

  • 1
    $\begingroup$ Oops! Just realized that Matthew Daws already answered with the same example while I was typing mine. $\endgroup$
    – T. Le
    Dec 30, 2021 at 12:19

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