Is the domain of an operator valued weight closed under Hahn-Jordan decomposition?

Here is a counter-example. Which is sadly rather long. I use Haagerup's original paper for the definition of $T$ below.

Let $M$ be the $\ell^\infty$ direct sum of the matrix algebras $\mathbb M_n$ with $\tau_M$ being the sum of the un-normalised traces on $\mathbb M_n$. Let $N$ be the subalgebra of all diagonal matrices, and give $N$ the ``weighted'' trace $$ \tau_N(x) = \sum_n \frac{1}{n} \sum_i x^{(n)}_i. $$ where $x = (x_n)\in N$ and each $x_n$ has diagonal entries $(x^{(n)}_i)$. Below I'll regard $N$ as the direct sum of $\ell^\infty_n$.

Let $T:M_+\rightarrow \widehat{N_+}$ be the operator valued weight, so $$ \tau_M(y^{1/2}xy^{1/2}) = \tau_N(y^{1/2} T(x) y^{1/2}) \qquad (x\in M, y\in N). $$ In general, $T(x)$ is not bounded, so a little care is needed to define the right-hand side. However, it follows from this that $T$ must respect the direct-sum decomposition, and so we obtain maps $T_n:\mathbb M_n \rightarrow \ell^\infty_n$ which satisfy $$ \sum_i y_i x_{i,i} = \frac{1}{n} \sum_i y_i T_n(x)_i \qquad (x=(x_{i,j})\in \mathbb M_n^+, y=(y_i)\in(\ell^\infty_n)^+) $$ Thus $T_n(x)_i = n x_{i,i}$, that is, project onto the diagonal and multiply by $n$.

Suppose we can find positive matrices $a_n, b_n$ in $\mathbb M_n$ so that $\|a_n\|\leq 1, \|b_n\|\leq 1$, with the diagonal entries bounded by $1/n$, but with $n |a_n - b_n|$ having diagonal entries which are unbounded, as $n\rightarrow\infty$. Set $a=(a_n) \in M$ and $b=(b_n)$. By the assumptions on $(a_n)$ and $(b_n)$ we see that $a,b \in \mathfrak{m}_T^+$ but that $|a-b|\not\in\mathfrak{m}_T^+$. This is equivalent to $x=a-b \in \mathfrak{m}_T$ but $x_+\not\in\mathfrak{m}_T^+$.

How do we find $a_n,b_n$? Here's one construction. Let $(\delta_i)_{i=1}^n$ be the usual orthonormal basis of $\mathbb C^n$ and let $$ \xi = \frac{1}{\sqrt n} \sum_{i=1}^n \delta_i, \qquad \eta = \frac{1}{\sqrt n} \sum_{i=1}^{n-1} \delta_i. $$ Let $a_n = a = \theta_{\xi,\xi}$ the rank-one positive operator, and let $b_n = b = \theta_{\eta,\eta}$. As matrices, $a$ has all entries $1/n$ and $b$ has all entries $1/n$ except for the last row and column which are 0. We claim that $|a-b|$ has diagonal entries of size order $\sqrt n$, to be precise, $|a-b|_{n,n}$ is about $\sqrt n$.

I don't see an easy way to show this. I spent ages working out a formula for $|\theta_{\xi,\xi} - \theta_{\eta,\eta}|$ for arbitrary vectors $\xi,\eta$ in a Hilbert space, and then performed the particular calculation here. If anyone has a nice argument, I'd like to see it.