Proof of Score Bounds at Envelope Test Inequalities from the Absorption Slack

Each result cited is universally quantified over the data in its own statement. For $L^{2}$ $d$-tuples of one tracial W*-probability space, sums and real multiples are the vector operations of $H^{d}$ and $\langle\cdot,\cdot\rangle_{2}$ is the real inner product of $H^{d}$ with norm $\lVert\cdot\rVert_{2}$, by Sums, Real Multiples and the Pairing of Square-Integrable Tuples in a Tracial W*-Probability Space §operations and Sums, Real Multiples and the Pairing of Square-Integrable Tuples in a Tracial W*-Probability Space §pairing; the difference $P-\delta Q$ of Tracial W*-Probability Spaces and Square-Integrable Tuples: Standing Notation §tuples is entrywise, hence equals $P+\delta(-Q)$ with $-Q=(-1)Q$, and $\lVert-Q\rVert_{2}=\lVert Q\rVert_{2}$. By Cauchy-Schwarz Inequality in a Complex Inner Product Space in $H^{d}$ (with the vectors $Q$ and $P$), $|\langle Q,P\rangle_{2}|\le\lVert Q\rVert_{2}\lVert P\rVert_{2}$.

Order of choice of constants. Let $r>0$. By Absorption of Shifts by a Hamiltonian on Phase-Space Noncommutative Laws §absorption, applied with this $r$, there are a real $\delta_{0}>0$ and a nondecreasing function $\omega:[0,\infty)\to[0,\infty)$ (Monotone Real Function §nondecreasing) with the absorption inequality for all tracial W*-probability spaces, all $L^{2}$ $d$-tuples $X,P,Q$ with $\mathrm{law}(X)\in\kappa_{d}(\Sigma_{d,r})$ and all $0<\delta\le\delta_{0}$. Put $\delta_{1}=\delta_{0}$. By Hamiltonians on Phase-Space Noncommutative Laws that are Bounded at Zero Momentum at Bounded Positions §bounded, applied with the same $r$, there is a real $K$ with $|\mathcal{H}_{M}(X,0)|\le K$ for every tracial W*-probability space $(H,M,\Omega)$ and every $L^{2}$ $d$-tuple $X$ of it with $\mathrm{law}(X)\in\kappa_{d}(\Sigma_{d,r})$; put $K_{+}=\max\{K,0\}$. Now let $m\ge0$ and $b\ge0$, and put

$$c=K_{+}+\delta_{1}\,\omega(0)+\frac{\sigma^{2}m^{2}}{4\delta_{1}},\qquad A=\rho b+c+\delta_{1}\,\omega(m),\qquad C=4m+\frac{\sqrt{8A\delta_{1}}}{\sigma}.$$

Since $\omega\ge0$, $K_{+}\ge0$, $\rho,\sigma,\delta_{1}>0$ and $b,m\ge0$, we have $c\ge0$, $A\ge0$ and $C\ge0$. Fix $(H,M,\Omega)$, $X,P,Q$, $a$ and $\delta$ as in the statement, and write $q=\lVert Q\rVert_{2}$. Then $|\langle Q,P\rangle_{2}|\le mq$ and $\rho|a|\le\rho b$.

Step 1 (the Hamiltonian at momentum $P$). Apply the absorption inequality with its momentum the zero tuple $0=0X$, its shift direction $\delta_{1}^{-1}P$ and its step $\delta_{1}$ (admissible, as $0<\delta_{1}\le\delta_{0}$). Since $0+\delta_{1}(\delta_{1}^{-1}P)=P$, $\lVert0\rVert_{2}=0$ and $\lVert\delta_{1}^{-1}P\rVert_{2}^{2}=\delta_{1}^{-2}\lVert P\rVert_{2}^{2}\le\delta_{1}^{-2}m^{2}$, it gives

$$\bigl|\mathcal{H}_{M}(X,P)-\mathcal{H}_{M}(X,0)\bigr|\le\delta_{1}\,\omega(0)+\frac{\sigma^{2}m^{2}}{4\delta_{1}},$$

and with $|\mathcal{H}_{M}(X,0)|\le K\le K_{+}$ we get $|\mathcal{H}_{M}(X,P)|\le c$.

Step 2 (a quadratic inequality solved). We show: if

$$\frac{\sigma^{2}\delta}{4}q^{2}\le A+\frac{\sigma^{2}m}{2}q,\qquad(\ast)$$

then $q\le C/\delta$. If $\frac{\sigma^{2}m}{2}q\ge A$, then $(\ast)$ gives $\frac{\sigma^{2}\delta}{4}q^{2}\le\sigma^{2}mq$, so $q=0$ or $q\le4m/\delta$; in both cases $q\le4m/\delta\le C/\delta$. If $\frac{\sigma^{2}m}{2}q<A$, then $(\ast)$ gives $\frac{\sigma^{2}\delta}{4}q^{2}<2A$, so, using $\delta\le\delta_{1}$,

$$q^{2}<\frac{8A}{\sigma^{2}\delta}=\frac{8A\delta}{\sigma^{2}\delta^{2}}\le\frac{8A\delta_{1}}{\sigma^{2}\delta^{2}},$$

hence $q\le\sqrt{8A\delta_{1}}/(\sigma\delta)\le C/\delta$, by monotonicity of squares of nonnegative reals (Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field).

Step 3 (Claim 1, subsolution side). The absorption inequality with $X,P,Q$ and the step $\delta$ (admissible, as $0<\delta\le\delta_{1}=\delta_{0}$) gives $\mathcal{H}_{M}(X,P+\delta Q)\ge\mathcal{H}_{M}(X,P)-\delta\,\omega(\lVert P\rVert_{2})-\frac{\sigma^{2}\delta}{4}q^{2}$. Since $\omega$ is nondecreasing and nonnegative, $\lVert P\rVert_{2}\le m$ and $\delta\le\delta_{1}$, we have $\delta\,\omega(\lVert P\rVert_{2})\le\delta_{1}\,\omega(m)$; with Step 1, $\mathcal{H}_{M}(X,P+\delta Q)\ge-c-\delta_{1}\omega(m)-\frac{\sigma^{2}\delta}{4}q^{2}$. Inserting this, $\rho a\ge-\rho b$ and $\langle Q,P\rangle_{2}\ge-mq$ into the hypothesis of Claim 1 gives

$$0\ge-\rho b-c-\delta_{1}\omega(m)-\frac{\sigma^{2}\delta}{4}q^{2}-\frac{\sigma^{2}m}{2}q+\frac{\sigma^{2}\delta}{2}q^{2},$$

which is $(\ast)$. By Step 2, $\lVert Q\rVert_{2}\le C/\delta$.

Step 4 (Claim 2, supersolution side). The absorption inequality with $X$, $P$, the shift direction $-Q$ and the step $\delta$ gives, since $P+\delta(-Q)=P-\delta Q$ and $\lVert-Q\rVert_{2}=q$, $\mathcal{H}_{M}(X,P-\delta Q)\le\mathcal{H}_{M}(X,P)+\delta\,\omega(\lVert P\rVert_{2})+\frac{\sigma^{2}\delta}{4}q^{2}\le c+\delta_{1}\omega(m)+\frac{\sigma^{2}\delta}{4}q^{2}$, as in Step 3. Inserting this, $\rho a\le\rho b$ and $\langle Q,P\rangle_{2}\le mq$ into the hypothesis of Claim 2 gives

$$0\le\rho b+c+\delta_{1}\omega(m)+\frac{\sigma^{2}\delta}{4}q^{2}+\frac{\sigma^{2}m}{2}q-\frac{\sigma^{2}\delta}{2}q^{2},$$

which is again $(\ast)$. By Step 2, $\lVert Q\rVert_{2}\le C/\delta$.