The Resolvent Transform of Square-Integrable Tuples: Consistency, Concatenation, the $L^2$ Lipschitz Bound and Universal Polynomial Recovery

In the setting of Tracial W*-Probability Spaces and Square-Integrable Tuples: Standing Notation, let $d,m\in\mathbb{N}$. Resolvent transforms $\mathbf{R}(X)$ of $L^{2}$ tuples and $\mathbf{R}(\xi)$ of self-adjoint vectors are those of The Resolvent of a Self-Adjoint Vector and the Resolvent Transform of a Square-Integrable Tuple, and resolvent transforms $\mathbf{R}(s)$ of self-adjoint tuples $s$ in $M$ are those of Resolvents of Bounded Self-Adjoint Operators and the Resolvent Transform of a Self-Adjoint Tuple §transform; vacuum tuples $t\Omega$ of self-adjoint tuples $t$ in $M$ are those of Tracial W*-Probability Spaces and Square-Integrable Tuples: Standing Notation §tuples; and $p(t)$ is the value of $p\in\mathcal{P}_{n}$ at an $n$-tuple $t$ in $M$ as in Noncommutative Laws, Couplings and the Wasserstein Distance: Standing Notation §polynomials.

1. (Consistency and concatenation)¶ Let $(H,M,\Omega)$ be a tracial W*-probability space and let $X$ be an $L^{2}$ $d$-tuple and $Y$ an $L^{2}$ $m$-tuple of it. Then $\mathbf{R}(X)$ is a self-adjoint $2d$-tuple in $M$ whose entries have operator norm at most $1$; $\mathbf{R}(s\Omega)=\mathbf{R}(s)$ for every self-adjoint $d$-tuple $s$ in $M$; and $\mathbf{R}((X,Y))$ is the $2(d+m)$-tuple $(\mathbf{R}(X)_{1},\dots,\mathbf{R}(X)_{2d},\mathbf{R}(Y)_{1},\dots,\mathbf{R}(Y)_{2m})$.

2. (Lipschitz bound)¶ For every tracial W*-probability space $(H,M,\Omega)$ and all $L^{2}$ $d$-tuples $X,Y$ of it, $\mathbf{R}(X)\Omega$ and $\mathbf{R}(Y)\Omega$ are $L^{2}$ $2d$-tuples and $\lVert\mathbf{R}(X)\Omega-\mathbf{R}(Y)\Omega\rVert_{2}\le\lVert X-Y\rVert_{2}$.

3. (Universal polynomial recovery)¶ There is a sequence $(P_{n})_{n\in\mathbb{N}}$ in $\mathcal{P}_{2,\mathrm{sa}}$ such that, for every tracial W*-probability space $(H,M,\Omega)$ and every $\xi\in H_{\mathrm{sa}}$, the sequence $\bigl(P_{n}(\mathbf{R}(\xi))\Omega\bigr)_{n\in\mathbb{N}}$ converges to $\xi$ in $H$.