Suppose $V$ is a von-Neumann algebra with a trace $\tau$ so that $\tau(I) = 1$. Suppose $W$ is a sub-von Neumann algebra of $V$. Then we can define a conditional expection to be a projection $\mathbb E:V \to W$ so that $\tau(\mathbb E(x) y) = \tau(x \mathbb E(y))$.

For commutative von Neumann algebras, this corresponds to the standard definition of conditional expection from $L^\infty(\Omega,\mathcal F,\mu) \to L^\infty(\Omega,\mathcal G,\mu)$, where $(\Omega,\mathcal F,\mu)$ is a probability space, and $\mathcal G$ is a sub sigma-algebra of $\mathcal F$.

Question: to what extent does the following inequality hold? For $x,y \in V$, and $p,q \in (1,\infty)$ with $\frac1p + \frac1q = 1$, we have $$ \mathbb E(xy) \le (\mathbb E|x|^p)^{1/p} (\mathbb E|y|^q)^{1/q} ?$$ It would be enough for me if I had to assume that $x$ and $y$ commute, $\mathbb E|x|^p$ and $\mathbb E|x|^q$ commute, $x$ and $y$ are self-adjoint non-negative elements of $V$, and even that $V$ is finite dimensional.

References are preferred to proofs.