Noncommutative maximal weak $L_1$ norms with respect to sub algebra

Question

Let $(\mathcal M,\tau)$ be a von Neumann algebra with normal finite faithful trace $\tau.$ For any sequence $(x_n)_{n\geq 1}\in \mathcal M$ define $\|(x_n)\|_{\Lambda_{1,\infty}(\mathcal M;\ell_\infty)}:=\sup_{\lambda>0}\lambda\inf\{\tau(1-e):e\in\mathcal P(\mathcal M)\ \text{such that}\sup_{n\geq 1}\|ex_ne\|_\infty<\lambda\}$ where $\mathcal{P}(\mathcal M)$ is the lattice of projections. Let $\mathcal N$ be a von Neumann sub algebra of $\mathcal M.$ Is it true that that for all $(x_n)\in\mathcal N$ we have $\|(x_n)\|_{\Lambda_{1,\infty}(\mathcal N;\ell_\infty)}\leq C \|(x_n)\|_{\Lambda_{1,\infty}(\mathcal M;\ell_\infty)}$ for some positive constant $C$ which does not depend on the sequence? This is true when the sequence $(x_n)$ is positive. The proof can be seen through the following fact: if $e\in\mathcal P(\mathcal M)$ be such that $\sup_{n\geq 1}\|ex_ne\|_\infty<\lambda$ and $\tau(1-e)\leq \frac{C}{\lambda}$ for some $C$ and $\lambda$ positive, there exists $\widetilde{e}\in \mathcal P(\mathcal N)$ be such that $\sup_{n\geq 1}\|\widetilde{e}x_n\widetilde{e}\|_\infty<4\lambda$ and $\tau(1-\widetilde{e})\leq \frac{2C}{\lambda}$. Take $\mathcal E$ to be the canonical trace preserving conditional expectation. Then $\|\mathcal E(e)x_n^{1/2}\|^2\leq \lambda.$ Consider $a=\mathcal E(e).$ Define $\widetilde{e}=\chi_{[1/2,1]}(a)$.If we define $g(r)=1/r\chi_{[1/2,1]}(r)$ then $\|\widetilde{e}x_n\widetilde{e}\|_{\infty}=\|\widetilde{e}g(a)ax_na\widetilde{e}g(a)\|_{\infty}\leq\frac{2C}{\lambda}.$