Weak stability of $\|a\|-M\|a\|_\phi$ on a $C^*$-algebra

Question

Let $A$ be a $C^*$-algebra, $\phi$ a state on $A$, and $M\geq 1$. Define $U=\{b\in A:\|b\|\leq M\phi(b^*b)^\frac{1}{2}\}$.

Based on some preliminary exploration, it seems that there always exists a bound $R>0$ s.t. for all $a\in A$ with $\|a\|\geq M\phi(a^*a)^\frac{1}{2}$,

$d(a,U)\leq R(\|a\|-M\phi(a^*a)^\frac{1}{2})$. However, I can't find the right approach for a proof. Any ideas? If not, any ideas for a counterexample?

Answer 1

This is not true. For example, $A = L^\infty([0, 1])$, $\phi$ is the state on $A$ induced by the Lebesgue measure, and $M = 1$. Then $U$ consists of those $f \in L^\infty([0, 1])$ s.t. $|f|$ is a.s. constant. Moreover, all $f \in A$ satisfy $\|f\| \geq M\phi(f^\ast f)^{\frac{1}{2}}$. Now, for any $0 < \epsilon < 1$, one can pick $X \subset [0, 1]$ of measure $\epsilon$ and let $f = 1_X$. Then $d(f, U) = \frac{1}{2}$ but $\|f\| - M\phi(f^\ast f)^{\frac{1}{2}} = 1 - \sqrt{\epsilon}$. Hence, for any $R > 0$ the bound will be violated as $\epsilon \to 1$.

Remark: I won't write down the details here since it's really just an annoyingly long computation, but the same $A$ and $\phi$ work as a counterexample with any $M \geq 1$. The contradiction again comes from choosing $f = 1_X$ with $X \subset [0, 1]$ of measure $\epsilon$, but with $\epsilon \to \frac{1}{M}$ from below.

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This is not true even when $A$ is finite-dimensional and $\phi$ is pure. Indeed, let $A = M_2(\mathbb{C})$ and $\phi((a_{ij})) = a_{11}$. Let $M = 1$, in which case all $a \in A$ satisfy $\|a\| \geq M\phi(a^\ast a)^{\frac{1}{2}}$. We first consider what the elements of $U$ are. Clearly, $x \in U$ iff $|x| \in U$. Now, let the eigenvalues of $|x|$ be $\lambda_1, \lambda_2$, with $\lambda_1 \geq \lambda_2 \geq 0$; and let the corresponding normalized eigenvectors be $v_1, v_2$. Then $\|x\| = \lambda_1$ and $\phi(x^\ast x)^{\frac{1}{2}} = \sqrt{\lambda_1^2|\langle v_1, e_1 \rangle|^2 + \lambda_2^2|\langle v_2, e_1 \rangle|^2}$. Multiplying $v_1, v_2$ by complex phase factors if necessary, we may assume $\langle v_1, e_1 \rangle, \langle v_2, e_1 \rangle \geq 0$. Thus, $x \in U$ iff,

$$\lambda_1^2 \leq \lambda_1^2\langle v_1, e_1 \rangle^2 + \lambda_2^2\langle v_2, e_1 \rangle^2$$

Our goal now is to lower bound $d(p, U)$ for a rank-one projection $p$. Write $p = |v\rangle\langle v|$ for some unit vector $v$. Per this paper, we have, for any $y \in U$, $\||p| - |y|\|_{HS} \leq \sqrt{2}\|p - y\|_{HS}$, so $\|p - |y|\| \leq 2\|p - y\|$. As $y \in U$ implies $|y| \in U$, we thus have,

$$d(p, U) \geq \frac{1}{2}d(p, U_+)$$

Fix $x \in U_+$. We shall retain the previous notations on $\lambda_i$ and $v_i$. Fix $\frac{1}{2} > \epsilon > 0$. Now, we have three cases,

1. $\|x\| < 1 - \epsilon$ or $\|x\| > 1 + \epsilon$. In this case, we have $\|p - x\| > \epsilon$ as $p$ has norm $1$.
2. $\lambda_2 > \epsilon$. Per Eckart-Young-Mirsky theorem, $\|p - x\| > \epsilon$ as well, as $p$ has rank one.
3. Assuming both $\lambda_1 = \|x\| \in [1 - \epsilon, 1 + \epsilon]$ and $\lambda_2 \leq \epsilon$. In this case, we have,

$$\lambda_1^2 \leq \lambda_1^2\langle v_1, e_1 \rangle^2 + \lambda_2^2\langle v_2, e_1 \rangle^2 \leq \lambda_1^2\langle v_1, e_1 \rangle^2 + \epsilon^2$$

Dividing by $\lambda_1^2$ on both sides,

$$1 \leq \langle v_1, e_1 \rangle^2 + \frac{\epsilon^2}{\lambda_1^2} \leq \langle v_1, e_1 \rangle^2 + \frac{\epsilon^2}{(1 - \epsilon)^2} \leq \langle v_1, e_1 \rangle^2 + 4\epsilon^2$$

So $\langle v_1, e_1 \rangle \geq \sqrt{1 - 4\epsilon^2}$. Hence,

$$\|v_1 - e_1\|^2 = 2 - 2\langle v_1, e_1 \rangle \leq 2 - 2\sqrt{1 - 4\epsilon^2}$$

Therefore,

$$\|x - |e_1\rangle\langle e_1|\| \leq \lambda_2 + |1 - \lambda_1| + 2\|v_1 - e_1\| \leq 2\epsilon + 2\sqrt{2 - 2\sqrt{1 - 4\epsilon^2}}$$

Thus,

$$d(p, U_+) \geq \min\left\{\epsilon, \|p - |e_1\rangle\langle e_1|\| - 2\epsilon - 2\sqrt{2 - 2\sqrt{1 - 4\epsilon^2}}\right\}$$

We have,

$$\|p - |e_1\rangle\langle e_1|\|^2 = \|(p - |e_1\rangle\langle e_1|)^2\| \geq \frac{1}{2}\operatorname{Tr}((p - |e_1\rangle\langle e_1|)^2) = 1 - |\langle v, e_1 \rangle|^2$$

So,

$$d(p, U) \geq \frac{1}{2}d(p, U_+) \geq \frac{1}{2}\min\left\{\epsilon, \sqrt{1 - |\langle v, e_1 \rangle|^2} - 2\epsilon - 2\sqrt{2 - 2\sqrt{1 - 4\epsilon^2}}\right\}$$

On the other hand,

$$\|p\| - \phi(p^\ast p)^{\frac{1}{2}} = 1 - |\langle v, e_1 \rangle|$$

Set $1 - |\langle v, e_1 \rangle| = \delta$ and $\epsilon = \frac{\sqrt{\delta}}{6}$, then,

$$d(p, U) \geq \frac{1}{2}\min\left\{\frac{\sqrt{\delta}}{6}, \sqrt{1 - (1 - \delta)^2} - \frac{\sqrt{\delta}}{3} - 2\sqrt{2 - 2\sqrt{1 - \frac{\delta}{9}}}\right\}$$

Performing a series expansion, it is not hard to see that the RHS behaves as $\frac{\sqrt{\delta}}{12}$ as $\delta \to 0$. But $d(p, U) \leq R(\|p\| - \phi(p^\ast p)^{\frac{1}{2}}) = R\delta$. This is a contradiction as $\delta \to 0$.