Let $A \cong L^\infty[0,1]$ be a nonatomic maximal abelian *subalgebra in $M \cong B(L^2[0,1])$ (or any von Neumann algebra $M$). Is the following true? For every $T \in M$ and $\epsilon>0$, there are nonzero projections $p,q \in A$ such that $\ qTp \ \le \epsilon \ T \$.
This is not true. Here's an example for which it fails:
Let $\{ r_n \}_{n \in \mathbb N}$ be an enumeration of the rationals. Construct an orthogonal set $\left\{ f_{j, k}^{l, m} \right\}_{j, k, l, m \in \mathbb N} \subset L^2 (\mathbb R)$ such that each $f_{j, k}^{l, m}$ is valued in $\{ 1, 0, 1 \}$, has only finitely many discontinuity points, and satisfies $\left f_{j, k}^{l, m} \right = 1_{(r_j  \frac{1}{k}, r_j + \frac{1}{k})}$. (This is easy enough to do inductively on an enumeration of the quadruples $(j, k, l, m)$). Define a partial isometry $T \in \mathcal B(L^2(\mathbb R) )$ such that $T\left( \sqrt{ \frac{k}{2} } f_{j, k}^{l, m} \right) = \sqrt{ \frac{m}{2} } f_{l, m}^{j, k}$.
If we're given $\varepsilon > 0$ and $E, F \subset \mathbb R$ having positive measure then by Lebesgue's density theorem there exists a quadruple $(j, k, l, m)$ such that \begin{equation} \left \left(r_j  \frac{1}{k}, r_j + \frac{1}{k} \right) \setminus E \right <\varepsilon^2 \frac{2}{k} \ \ \ {\rm and} \ \ \ \left \left(r_l  \frac{1}{m}, r_l + \frac{1}{m} \right) \setminus F \right < \varepsilon^2 \frac{2}{m}. \end{equation} Therefore, $\sqrt{ \frac{k}{2} } \left\ 1_E f_{j, k}^{l, m}  f_{j, k}^{l, m} \right\_2 < \varepsilon$, and $\sqrt{ \frac{m}{2} } \left\ 1_F f_{l, m}^{j, k}  f_{l, m}^{j, k} \right\_2 < \varepsilon$. Hence,
\begin{equation} \left\ \sqrt{ \frac{m}{2} } f_{l, m}^{j, k}  1_F T 1_E \left( \sqrt{ \frac{k}{2} } f_{j, k}^{l, m} \right) \right\_2 < 2\varepsilon, \end{equation} so that $\left\ 1_F T 1_E \right\ = 1$.

$\begingroup$ Thanks. I hoped the other way, but this is a nice example. $\endgroup$ Sep 15 '17 at 20:45