Let $A \cong L^\infty[0,1]$ be a non-atomic 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 non-zero 
projections $p,q \in A$ such that $\| qTp \| \le \epsilon \| T \|$. 



A small corner w.r.t. a masa in a von Neumann algebra
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 $\left\| 1\_F T 1\_E \right\| = 1$.


