Infinite-dimensional submajorization conjecture for spectral suprema

Complete proof for arbitrary semifinite algebras, unbounded measurable operators, and finite families.

Bourin–Lee, arXiv:2606.15624, Conjecture 3.10, asks whether the finite-dimensional submajorization

$$A\vee B\prec\!\prec A+B$$

extends to all positive measurable operators affiliated with an arbitrary semifinite von Neumann algebra. The finite-dimensional case is already proved in the source’s Corollary 3.8, and the source credits the classical finite-matrix convex trace inequality to Rotfel’d. We prove the full requested unbounded extension directly, together with its arbitrary finite-family strengthening.

Let $(\mathcal M,\tau)$ be any semifinite von Neumann algebra with faithful normal semifinite trace. For positive $\tau$-measurable affiliated operators $A_1,\ldots,A_m$, put

$$X=\bigvee_{i=1}^{m}A_i, \qquad C=\sum_{i=1}^{m}A_i,$$

where the join is in Olson’s spectral order. We prove

$$\boxed{ \tau((X-s)_+) \le\sum_{i=1}^{m}\tau((A_i-s)_+) \le\tau((C-s)_+) \qquad(s\ge0).}$$

Write $e_Z(u)=\mathbf1_{(u,\infty)}(Z)$. Spectral joins satisfy

$$e_X(u)=\bigvee_{i=1}^{m}e_{A_i}(u).$$

For projections,

$$\tau\left(\bigvee_{i=1}^{m}p_i\right) \le\sum_{i=1}^{m}\tau(p_i).$$

This also shows that $X$ remains $\tau$-measurable. The layer-cake formula immediately gives

$$\tau((X-s)_+) =\int_s^\infty\tau(e_X(u))\,du \le \sum_{i=1}^{m}\int_s^\infty\tau(e_{A_i}(u))\,du = \sum_{i=1}^{m}\tau((A_i-s)_+).$$

For the reverse comparison, semifiniteness and normality give the finite-projection variational formula

$$\tau((Z-s)_+) = \sup_{\substack{p\in\mathcal M\ \mathrm{projection}\\tau(p)<\infty}} \bigl\{\tau(pZp)-s\tau(p)\bigr\} \qquad(Z\ge0).$$

All expressions are understood in $[0,\infty]$; since $\tau(p)<\infty$, there is no indeterminate subtraction.

Choose arbitrary finite-trace projections $p_1,\ldots,p_m$, and set

$$r=\bigvee_{i=1}^{m}p_i.$$

Then $\tau(r)\le\sum_i\tau(p_i)<\infty$, and positivity plus traciality imply

$$\tau(p_iA_ip_i)\le\tau(rA_ir).$$

Consequently,

$$\begin{aligned} \sum_{i=1}^{m} \bigl(\tau(p_iA_ip_i)-s\tau(p_i)\bigr) &\le \sum_{i=1}^{m}\tau(rA_ir)-s\tau(r)\\ &= \tau(rCr)-s\tau(r)\\ &\le\tau((C-s)_+). \end{aligned}$$

Taking the independent suprema over all $p_i$ yields

$$\sum_{i=1}^{m}\tau((A_i-s)_+) \le\tau((C-s)_+),$$

completing both stop-loss inequalities without any boundedness, integrability, finite-factor, or separability assumption.

Finally, generalized singular numbers satisfy

$$\int_0^t\mu_u(Z)\,du = \inf_{s\ge0} \left\{ts+\tau((Z-s)_+)\right\} \qquad(t>0).$$

Therefore, for every $t>0$,

$$\boxed{ \int_0^t\mu_u\left(\bigvee_{i=1}^{m}A_i\right)\,du \le \int_0^t\mu_u\left(\sum_{i=1}^{m}A_i\right)\,du.}$$

Equivalently,

$$\boxed{\bigvee_{i=1}^{m}A_i\prec\!\prec\sum_{i=1}^{m}A_i.}$$

For $m=2$, this proves the complete original conjecture in precisely its semifinite and potentially unbounded generality. Every norm respecting submajorization therefore also satisfies

$$\boxed{\|A\vee B\|\le\|A+B\|.}$$