Let $C\_p$, $1<p<\infty$, be the Schatten-$p$-class.
Let $x\in C\_p$ be positive and $\|x\|\_p =1$.
Let $E$ be the conditional expectation onto the diagonal part.
If $\|E(x)\|\_p \ge 1-\delta $ for some small $\delta>0$, then can we get an estimate for
$\|x - E(x)\|\_p$ in terms of $\delta$? In particular, if $\delta\to 0$, then do we have $\|x - E(x)\|\_p\to 0$?


PS. Of course, is it not true for $p=1$.



Norm estimate for the difference between a positive operator and its expectation
Yes. This follows from the uniform convexity of the Schatten $p$ classes.


Indeed, for every unitary diagonal operator $u$, we have $\| (x + u x u^\*)/2\|\_p \geq \| E( x+uxu^\*)/2\|\_p = \|E(x)\|p \geq 1-\delta$ ($E$ is a contraction). Therefore, we have $\|x - uxu^\*\|\_p\leq\varepsilon$ (where $\varepsilon = \varepsilon(p,\delta)$ is given by uniform convexity, and goes to $0$ with $\delta$). This implies that $\|x-y\|\_p \leq\varepsilon$ for every $y$ in the closed convex hull of $\{uxu^\* \mid u \textrm{ unitary diagonal}\}$. But $E(x)$ belongs to this closure, so we have $\|x - E(x)\|\_p \leq \varepsilon$.


The fact that $E(x)$ belongs to the closure of the unitary orbit is a very general fact. It is completely elementary in finite dimension, and it therefore holds by approximation for Schatten classes. It is also true in arbitrary $II\_1$ factors, see [Examples of non-proper conditional expectations onto von Neumann subalgebras of $II\_{1}$ factors](https://mathoverflow.net/questions/273934/examples-of-non-proper-conditional-expectations-onto-von-neumann-subalgebras-of).


**Edit on october 20 2021 :** It was asked in the comments whether the statement holds for general semifinite von Neumann algebras. The answer is yes. That is, if $1<p<\infty$ (and $q$ is the dual exponent), if $\mathcal{M}$ is an arbitrary semifinite von Neumann algebra (with trace $tr$), $\mathcal{N} \subset \mathcal M$ is a sub-von Neumann algebra with trace-preserving conditional expectation $E: \mathcal M \to \mathcal N$, then for every $x \in L\_p(\mathcal M)$ of norm $1$, $\|E(x)\|\_p \geq 1-\delta$ implies $\|x-E(x)\|\_p \leq \varepsilon(p,\delta) = O(\delta^{\min(1/p,1/q)})$.


It might be possible to adapt the preceding proof, at least for factors, but here is a simpler proof (arguably less enjoyable). By duality, there is $y \in L\_q(\mathcal N)$ of norm $1$ such that $tr(y E(x)) = 1-\delta$. $E$ being a trace-preserving conditional expectation, we have $tr(yx) = tr(y E(x))$, so by Hölder
$$ 1-\delta = tr(y(x+E(x))/2) \leq \| (x+ E(x))/2\|\_p.$$
By uniform convexity, $\|x-E(x)\|\_p \leq \varepsilon$.


Observe that, mutatis mutandis (working with Haagerup's $L\_p$ spaces), the same proof applies even when $\mathcal N$ is not semifinite, we only need the existence of a conditional expectation.


