Proof of The Amalgamated Free Product Space: Multilinearity of the Tuple Vectors and the Bounded Left Actions of the Two Tracial Algebras

Preliminaries. Let $\varepsilon\in\{1,2\}$. By Two Noncommutative Laws with a Common Marginal: Standing Notation for Their Amalgamated Free Product §embeddings, Marginals of a Noncommutative Law: the Isometry of GNS Spaces, the Trace-Preserving Embedding of Tracial Algebras and the Conditional Expectation §homomorphism and Marginals of a Noncommutative Law: the Isometry of GNS Spaces, the Trace-Preserving Embedding of Tracial Algebras and the Conditional Expectation §expectation (for $\gamma=\gamma_{\varepsilon}$), $\pi_{\varepsilon}$ is linear with $\pi_{\varepsilon}(I)=I$, $\pi_{\varepsilon}(ST)=\pi_{\varepsilon}(S)\pi_{\varepsilon}(T)$ and $\pi_{\varepsilon}(T^{*})=\pi_{\varepsilon}(T)^{*}$, and $E_{\varepsilon}$ is linear with $E_{\varepsilon}(\pi_{\varepsilon}(S)\,c\,\pi_{\varepsilon}(T))=S\,E_{\varepsilon}(c)\,T$ and $E_{\varepsilon}(\pi_{\varepsilon}(T))=T$ for $S,T\in N$ and $c\in A_{\varepsilon}$. All sums, scalar multiples, products and adjoints formed below lie in $A_{\varepsilon}$ or $N$ by The Tracial Algebra of a Noncommutative Law: a Norm-Closed Unital *-Algebra with a Faithful Positive Trace, Determined by Vacuum Vectors, Closed under Square Roots §star-algebra, and products of operators distribute over sums and commute with scalar multiples by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §operations. For $c\in A_{\varepsilon}$ and $y\in N$ we record four consequences. (F1) Taking $S=I$ or $T=I$: $E_{\varepsilon}(c\,\pi_{\varepsilon}(y))=E_{\varepsilon}(c)\,y$ and $E_{\varepsilon}(\pi_{\varepsilon}(y)\,c)=y\,E_{\varepsilon}(c)$. (F2) Since $\pi_{\varepsilon}$ and $E_{\varepsilon}$ are linear, $(c+c')^{\circ}=c^{\circ}+c'^{\circ}$ and $(\lambda c)^{\circ}=\lambda c^{\circ}$ for $c'\in A_{\varepsilon}$, $\lambda\in\mathbb{C}$; and $c=c^{\circ}+\pi_{\varepsilon}(E_{\varepsilon}(c))$ by definition of the centred part (The Formal Amalgamated Free Product over a Common Marginal: Alternating Tuples, the Free Vector Space, the Nested-Expectation Form and the Formal Actions §centred). (F3) $\pi_{\varepsilon}(y)^{\circ}=\pi_{\varepsilon}(y)-\pi_{\varepsilon}(y)=0$, in particular $I^{\circ}=0$ as $I=\pi_{\varepsilon}(I)$; and $a^{\circ}=a-\pi_{\varepsilon}(0)=a$ for $a\in A_{\varepsilon}^{\circ}$. (F4) $(c\,\pi_{\varepsilon}(y))^{\circ}=c\,\pi_{\varepsilon}(y)-\pi_{\varepsilon}(E_{\varepsilon}(c)\,y)=c\,\pi_{\varepsilon}(y)-\pi_{\varepsilon}(E_{\varepsilon}(c))\pi_{\varepsilon}(y)=c^{\circ}\pi_{\varepsilon}(y)$, by (F1). By The Formal Amalgamated Free Product over a Common Marginal: Alternating Tuples, the Free Vector Space, the Nested-Expectation Form and the Formal Actions §centred, $b^{\circ}$, $\pi_{\varepsilon}(x)\,a$ and $a\,\pi_{\varepsilon}(x)$ are centred for $b\in A_{\varepsilon}$, $a\in A_{\varepsilon}^{\circ}$, $x\in N$, and sums and scalar multiples of centred elements are centred since $E_{\varepsilon}$ is linear; this is why every tuple written below is an alternating tuple: each new entry below is a centred part $c^{\circ}$, a product $\pi(y)\,a$ or $a\,\pi(y)$ with $a$ centred, or a sum or scalar multiple of centred elements, and it stands in the position of an entry of the same algebra. We also use three general facts.

(P0) Vanishing on basis vectors. Let $W$ be a complex vector space and $D:\mathcal{F}\to W$ additive with $D(c\,\delta_{u})=0$ for all $c\in\mathbb{C}$ and $u\in\mathcal{T}$. Then $D=0$. Indeed, $D(0)=D(0+0)=D(0)+D(0)$ gives $D(0)=0$, which settles $\xi$ with empty support, as then $\xi=0$. If $\operatorname{supp}\xi$ has $n\in\mathbb{N}$ elements and the claim holds for supports with $n-1$ elements, pick $u\in\operatorname{supp}\xi$ and let $\xi'=\xi+(-\xi(u))\delta_{u}$ (operations of The Formal Amalgamated Free Product over a Common Marginal: Alternating Tuples, the Free Vector Space, the Nested-Expectation Form and the Formal Actions §space); then $\xi'(u)=0$ and $\xi'(s)=\xi(s)$ for $s\neq u$, so $\operatorname{supp}\xi'=\operatorname{supp}\xi\setminus\{u\}$ has $n-1$ elements, $\xi=\xi'+\xi(u)\delta_{u}$, and $D\xi=D\xi'+D(\xi(u)\delta_{u})=0$. In particular two linear maps $\mathcal{F}\to W$ that agree on every $\delta_{u}$ are equal (apply (P0) to their difference).

(P1) The canonical map. By The Amalgamated Free Product Space of Two Noncommutative Laws over a Common Marginal, Its Vacuum Vector and Vacuum State §space, $\mathcal{H}$ is a complex Hilbert space whose inner product is the pairing of the completion and $J$ is its canonical map, so by The Complex Hilbert Completion is a Complex Hilbert Space Containing a Dense Isometric Image, and Bounded Complex-Linear Maps Extend to It §isometry $J$ is complex-linear and $\langle J\xi,J\eta\rangle_{\mathcal{H}}=h(\xi,\eta)$ for $\xi,\eta\in\mathcal{F}$. Hence $\lVert J\xi\rVert_{\mathcal{H}}^{2}=h(\xi,\xi)$ by definition of the induced norm, and $J\zeta=0$ if $h(\zeta,\zeta)=0$, by definiteness (claim 4 of Elementary Properties of a Complex Inner Product).

(P2) Comparison. If $\xi,\eta\in\mathcal{F}$ satisfy $h(\delta_{\sigma},\xi)=h(\delta_{\sigma},\eta)$ for every $\sigma\in\mathcal{T}$, then $J\xi=J\eta$. Indeed, let $\zeta=\xi+(-1)\eta$; by the linearity of $h$ in its second argument (The Formal Amalgamated Free Product: the Free Vector Space, Positivity of the Nested-Expectation Form, and Linearity and Formal Adjoints of the Actions §form), $h(\delta_{\sigma},\zeta)=0$ for every $\sigma$. The map $D(\chi)=h(\chi,\zeta)$ is additive and satisfies $D(c\,\chi)=\overline{c}\,D(\chi)$, because $h(\chi,\zeta)=\overline{h(\zeta,\chi)}$ and $h(\zeta,\cdot)$ is linear (The Formal Amalgamated Free Product: the Free Vector Space, Positivity of the Nested-Expectation Form, and Linearity and Formal Adjoints of the Actions §form); so $D(c\,\delta_{\sigma})=0$ for all $c,\sigma$, and $D=0$ by (P0). Thus $h(\zeta,\zeta)=0$, $J\zeta=0$ by (P1), and $J\xi=J\eta$ since $J$ is linear. We shall use (P2) with $\xi,\eta$ sums of vectors $\delta_{\tau}$, computing $h(\delta_{\sigma},\delta_{\tau})=h_{0}(\sigma,\tau)$ by The Formal Amalgamated Free Product: the Free Vector Space, Positivity of the Nested-Expectation Form, and Linearity and Formal Adjoints of the Actions §form and expanding by linearity in the second argument. Recall from The Formal Amalgamated Free Product over a Common Marginal: Alternating Tuples, the Free Vector Space, the Nested-Expectation Form and the Formal Actions §form that $h_{0}(\sigma,(k,t))=0$ unless $\sigma=(k,s)$ with $s$ of the same length and type as $t$, and then $h_{0}(\sigma,(k,t))=\tau_{\mu}(X_{k}(s,t))$; $\tau_{\mu}$ is linear by The Tracial Algebra of a Noncommutative Law: a Norm-Closed Unital *-Algebra with a Faithful Positive Trace, Determined by Vacuum Vectors, Closed under Square Roots §trace.

Claim 1 (Multilinearity). For $x,y\in N$, $c\in\mathbb{C}$ and $\sigma\in\mathcal{T}$: if $\sigma=(0,z)$ then $h_{0}(\sigma,(0,x+y))=\tau_{\mu}(z^{*}x)+\tau_{\mu}(z^{*}y)$ and $h_{0}(\sigma,(0,cx))=c\,\tau_{\mu}(z^{*}x)$ by linearity of $\tau_{\mu}$, and for every other $\sigma$ all these pairings vanish. So $h(\delta_{\sigma},\delta^{N}_{x+y})=h(\delta_{\sigma},\delta^{N}_{x}+\delta^{N}_{y})$ and $h(\delta_{\sigma},\delta^{N}_{cx})=h(\delta_{\sigma},c\,\delta^{N}_{x})$ for every $\sigma$, and (P2) with the linearity of $J$ gives $\Xi_{N}(x+y)=\Xi_{N}(x)+\Xi_{N}(y)$ and $\Xi_{N}(cx)=c\,\Xi_{N}(x)$.

For the tuple vectors, let $t=(a_{1},\dots,a_{k})$, let $t'$ be $t$ with $a_{j}$ replaced by $a_{j}'$, $t''$ the same with $a_{j}+a_{j}'$, and $t_{c}$ the same with $c\,a_{j}$. Let $s=(d_{1},\dots,d_{k})$ have the same type. We show by induction on $i\in[k]$ that $X_{i}(s,t'')=X_{i}(s,t)=X_{i}(s,t')$ and $X_{i}(s,t_{c})=X_{i}(s,t)$ for $i<j$, and $X_{i}(s,t'')=X_{i}(s,t)+X_{i}(s,t')$ and $X_{i}(s,t_{c})=c\,X_{i}(s,t)$ for $i\ge j$ (The Formal Amalgamated Free Product over a Common Marginal: Alternating Tuples, the Free Vector Space, the Nested-Expectation Form and the Formal Actions §nested). For $i<j$ the recursion for $X_{i}$ involves only entries with index at most $i$, which coincide in the four tuples, and previous terms which coincide by induction. For $i=j$ the entry $a_{j}+a_{j}'$ (resp. $c\,a_{j}$) enters linearly: $X_{j}(s,t'')=E_{e_{j}}(d_{j}^{*}P(a_{j}+a_{j}'))=E_{e_{j}}(d_{j}^{*}Pa_{j})+E_{e_{j}}(d_{j}^{*}Pa_{j}')$, where $P=I$ if $j=1$ and $P=\pi_{e_{j}}(X_{j-1}(s,t))$ otherwise, which is common to the tuples; likewise for $t_{c}$. For $i>j$, $X_{i}(s,t'')=E_{e_{i}}(d_{i}^{*}\pi_{e_{i}}(X_{i-1}(s,t)+X_{i-1}(s,t'))a_{i})=X_{i}(s,t)+X_{i}(s,t')$ by linearity of $\pi_{e_{i}}$ and $E_{e_{i}}$, and likewise $X_{i}(s,t_{c})=c\,X_{i}(s,t)$. With $i=k$ and linearity of $\tau_{\mu}$, $h_{0}(\sigma,(k,t''))=h_{0}(\sigma,(k,t))+h_{0}(\sigma,(k,t'))$ and $h_{0}(\sigma,(k,t_{c}))=c\,h_{0}(\sigma,(k,t))$ for $\sigma=(k,s)$ with $s$ of the type of $t$, and all these pairings vanish for every other $\sigma$, since $t,t',t'',t_{c}$ share length and type. Hence $h(\delta_{\sigma},\delta_{t''})=h(\delta_{\sigma},\delta_{t}+\delta_{t'})$ and $h(\delta_{\sigma},\delta_{t_{c}})=h(\delta_{\sigma},c\,\delta_{t})$ for all $\sigma$, and (P2) and linearity of $J$ give the two identities. Finally, if $a_{j}=0$ then $a_{j}=0\cdot a_{j}$, so $\Xi(a_{1},\dots,a_{k})=0\cdot\Xi(a_{1},\dots,a_{k})=0$; in the same way $\Xi_{N}(0)=0$.

Claim 2 (Balance over $N$). Let $t=(a_{1},\dots,a_{k})$, $j<k$, $x\in N$, let $t'$ be $t$ with $a_{j}$ replaced by $a_{j}\pi_{e_{j}}(x)$ and $t''$ be $t$ with $a_{j+1}$ replaced by $\pi_{e_{j+1}}(x)a_{j+1}$; both are alternating of the type of $t$. Let $s=(d_{1},\dots,d_{k})$ have this type. For $i<j$, $X_{i}(s,t')=X_{i}(s,t'')=X_{i}(s,t)$ as in Claim 1. With $P$ as in Claim 1, (F1) gives $X_{j}(s,t')=E_{e_{j}}(d_{j}^{*}Pa_{j}\pi_{e_{j}}(x))=X_{j}(s,t)\,x$, while $X_{j}(s,t'')=X_{j}(s,t)$. Then, $\pi_{e_{j+1}}$ being multiplicative,

and for $i>j+1$ the recursions for $t'$ and $t''$ use the same entries and, by induction, equal previous terms, so $X_{k}(s,t')=X_{k}(s,t'')$. Hence $h_{0}(\sigma,(k,t'))=h_{0}(\sigma,(k,t''))$ for all $\sigma\in\mathcal{T}$ (both vanish unless $\sigma=(k,s)$ as above), and (P2) gives $\Xi(t')=\Xi(t'')$.

Claim 3 (Relations). For each identity, both sides are linear maps $\mathcal{F}\to\mathcal{H}$ of $\xi$, as composites, sums and scalar multiples of the linear maps $J$ (P1) and $\ell_{\varepsilon}(\cdot)$ (The Formal Amalgamated Free Product: the Free Vector Space, Positivity of the Nested-Expectation Form, and Linearity and Formal Adjoints of the Actions §actions); so by (P0) it suffices to take $\xi=\delta_{u}$, $u\in\mathcal{T}$. By The Formal Amalgamated Free Product: the Free Vector Space, Positivity of the Nested-Expectation Form, and Linearity and Formal Adjoints of the Actions §actions, $\ell_{\varepsilon}(b)\delta_{u}=\rho_{\varepsilon,b}(u)$, and $J\rho_{\varepsilon,b}(u)$ is the sum of the two vectors $\Xi(\cdot)$ or $\Xi_{N}(\cdot)$ (The Amalgamated Free Product Space of Two Noncommutative Laws over a Common Marginal, Its Vacuum Vector and Vacuum State §vectors) of the two labels in The Formal Amalgamated Free Product over a Common Marginal: Alternating Tuples, the Free Vector Space, the Nested-Expectation Form and the Formal Actions §actions, as $J$ is linear. We write $\pi=\pi_{\varepsilon}$, $E=E_{\varepsilon}$, $\rho_{b}=\rho_{\varepsilon,b}$, and let $t=(a_{1},\dots,a_{k})$ denote a tuple of type $(e_{1},\dots,e_{k})$.

Linearity in $b$. In each case (a), (b), (c) of The Formal Amalgamated Free Product over a Common Marginal: Alternating Tuples, the Free Vector Space, the Nested-Expectation Form and the Formal Actions §actions the two labels of $\rho_{b}(u)$ depend on $b$ through one entry each, namely $(b\pi(x))^{\circ}$ and $E(b\pi(x))$; $b^{\circ}$ and $\pi_{e_{1}}(E(b))a_{1}$; $(ba_{1})^{\circ}$ and $\pi_{e_{2}}(E(ba_{1}))a_{2}$ (or $E(ba_{1})$ if $k=1$). By (F2) and the linearity of $E$ and $\pi_{e_{1}}$, $\pi_{e_{2}}$, each of these entries for $b+b'$ (resp. $cb$) is the sum of the entries for $b$ and $b'$ (resp. $c$ times the entry for $b$). Claim 1 applied in that entry, and rearranging the four resulting vectors, gives $J\rho_{b+b'}(u)=J\rho_{b}(u)+J\rho_{b'}(u)$ and $J\rho_{cb}(u)=c\,J\rho_{b}(u)$.

Unit. For $u=(0,x)$: $I\pi(x)=\pi(x)$, $\pi(x)^{\circ}=0$ (F3) and $E(\pi(x))=x$, so $J\rho_{I}(u)=\Xi((\varepsilon,0))+\Xi_{N}(x)=\Xi_{N}(x)=J\delta_{u}$ by Claim 1, where $\Xi((\varepsilon,0))$ denotes the vector of the one-entry tuple $((\varepsilon,0))$. For $u=(k,t)$ with $e_{1}=\bar{\varepsilon}$: $I^{\circ}=0$ (F3) and $\pi_{e_{1}}(E(I))a_{1}=\pi_{e_{1}}(I)a_{1}=a_{1}$, so $J\rho_{I}(u)=\Xi(0,a_{1},\dots,a_{k})+\Xi(t)=\Xi(t)$ by Claim 1. For $u=(k,t)$ with $e_{1}=\varepsilon$: $(Ia_{1})^{\circ}=a_{1}$ (F3) and $E(a_{1})=0$, so $J\rho_{I}(u)$ is $\Xi(t)+\Xi(\pi_{e_{2}}(0)a_{2},a_{3},\dots,a_{k})=\Xi(t)+\Xi(0,a_{3},\dots,a_{k})=\Xi(t)$ if $k\ge2$ and $\Xi(t)+\Xi_{N}(0)=\Xi(t)$ if $k=1$, by Claim 1.

Amalgamation. Let $x\in N$ and $b_{\varepsilon}=\pi_{\varepsilon}(x)$, so $b_{\varepsilon}^{\circ}=0$ by (F3) and $E_{\varepsilon}(b_{\varepsilon})=x$ by Marginals of a Noncommutative Law: the Isometry of GNS Spaces, the Trace-Preserving Embedding of Tracial Algebras and the Conditional Expectation §expectation. For $u=(0,y)$ and either $\varepsilon$: $b_{\varepsilon}\pi_{\varepsilon}(y)=\pi_{\varepsilon}(xy)$, so $J\rho_{\varepsilon,b_{\varepsilon}}(u)=\Xi((\varepsilon,0))+\Xi_{N}(xy)=\Xi_{N}(xy)$ by (F3) and Claim 1; the two sides agree. For $u=(k,t)$ with $e_{1}=\varepsilon$ (so $e_{2}=\bar{\varepsilon}$ if $k\ge2$): the action of $b_{\varepsilon}$ is case (c); $\pi_{\varepsilon}(x)a_{1}$ is centred, so $(\pi_{\varepsilon}(x)a_{1})^{\circ}=\pi_{\varepsilon}(x)a_{1}$ (F3) and $E_{\varepsilon}(\pi_{\varepsilon}(x)a_{1})=x\,E_{\varepsilon}(a_{1})=0$ (F1); by Claim 1, $J\rho_{\varepsilon,b_{\varepsilon}}(u)=\Xi(\pi_{\varepsilon}(x)a_{1},a_{2},\dots,a_{k})+\Xi(0,a_{3},\dots,a_{k})$ if $k\ge2$, and $\Xi(\pi_{\varepsilon}(x)a_{1})+\Xi_{N}(0)$ if $k=1$, which in both cases equals $\Xi(\pi_{\varepsilon}(x)a_{1},a_{2},\dots,a_{k})$. The action of $b_{\bar{\varepsilon}}$ is case (b), with $\bar{\varepsilon}$ in the role of $\varepsilon$: $J\rho_{\bar{\varepsilon},b_{\bar{\varepsilon}}}(u)=\Xi(0,a_{1},\dots,a_{k})+\Xi(\pi_{\varepsilon}(x)a_{1},a_{2},\dots,a_{k})$, the same vector by Claim 1. Thus $J\ell_{1}(\pi_{1}(x))\delta_{u}=J\ell_{2}(\pi_{2}(x))\delta_{u}$ for every $u$.

Multiplicativity. Let $b,b'\in A_{\varepsilon}$ and write $\rho_{b'}(u)=\delta_{v}+\delta_{w}$ as in The Formal Amalgamated Free Product over a Common Marginal: Alternating Tuples, the Free Vector Space, the Nested-Expectation Form and the Formal Actions §actions; by linearity of $\ell_{\varepsilon}(b)$ and The Formal Amalgamated Free Product: the Free Vector Space, Positivity of the Nested-Expectation Form, and Linearity and Formal Adjoints of the Actions §actions, $J\ell_{\varepsilon}(b)\ell_{\varepsilon}(b')\delta_{u}=J\rho_{b}(v)+J\rho_{b}(w)$, which we compare with $J\rho_{bb'}(u)$.

Case $u=(0,x)$, or $u=(1,t)$ with $e_{1}=\varepsilon$. Let $g=\pi(x)$, resp. $g=a_{1}$, and $f=(b'g)^{\circ}$, $y=E(b'g)$; then $v$ is the one-entry tuple $((\varepsilon,f))$ and $w=(0,y)$, and $b'g=f+\pi(y)$ (F2). By cases (c) with $k=1$ and (a), $J\rho_{b}(v)+J\rho_{b}(w)=\Xi((bf)^{\circ})+\Xi_{N}(E(bf))+\Xi((b\pi(y))^{\circ})+\Xi_{N}(E(b\pi(y)))$, while $J\rho_{bb'}(u)=\Xi((bb'g)^{\circ})+\Xi_{N}(E(bb'g))$ in both cases. Since $bb'g=bf+b\pi(y)$, (F2) and linearity of $E$ give $(bb'g)^{\circ}=(bf)^{\circ}+(b\pi(y))^{\circ}$ and $E(bb'g)=E(bf)+E(b\pi(y))$, and Claim 1 gives equality.

Case $u=(k,t)$ with $e_{1}=\bar{\varepsilon}$. Here $v=(b'^{\circ},a_{1},\dots,a_{k})$, of type $(\varepsilon,e_{1},\dots,e_{k})$ and length $k+1\ge2$, and $w=(\pi_{e_{1}}(E(b'))a_{1},a_{2},\dots,a_{k})$. Case (c) for $v$ and case (b) for $w$ give

and $J\rho_{bb'}(u)=\Xi((bb')^{\circ},a_{1},\dots,a_{k})+\Xi(\pi_{e_{1}}(E(bb'))a_{1},a_{2},\dots,a_{k})$. By (F2), $bb'=bb'^{\circ}+b\,\pi(E(b'))$, so by (F2), (F4) and (F1), $(bb')^{\circ}=(bb'^{\circ})^{\circ}+b^{\circ}\pi(E(b'))$ and $E(bb')=E(bb'^{\circ})+E(b)E(b')$, whence $\pi_{e_{1}}(E(bb'))a_{1}=\pi_{e_{1}}(E(bb'^{\circ}))a_{1}+\pi_{e_{1}}(E(b))\pi_{e_{1}}(E(b'))a_{1}$. Claim 1 in the first entry splits the first vector of $J\rho_{bb'}(u)$ into $\Xi((bb'^{\circ})^{\circ},a_{1},\dots,a_{k})+\Xi(b^{\circ}\pi(E(b')),a_{1},\dots,a_{k})$, and Claim 2 (with $j=1<k+1$) turns the latter into $\Xi(b^{\circ},\pi_{e_{1}}(E(b'))a_{1},a_{2},\dots,a_{k})$; Claim 1 in the first entry splits the second vector into the second and fourth vectors above. So the two sides agree.

Case $u=(k,t)$ with $e_{1}=\varepsilon$ and $k\ge2$. Let $f=(b'a_{1})^{\circ}$ and $y=E(b'a_{1})$, so $b'a_{1}=f+\pi(y)$ (F2). Here $v=(f,a_{2},\dots,a_{k})$, of length $k\ge2$ starting with $\varepsilon$, and $w=(\pi_{e_{2}}(y)a_{2},a_{3},\dots,a_{k})$, of length $k-1$ starting with $e_{2}=\bar{\varepsilon}$. Case (c) for $v$ and case (b) for $w$ give

and $J\rho_{bb'}(u)=\Xi((bb'a_{1})^{\circ},a_{2},\dots,a_{k})+\Xi(\pi_{e_{2}}(E(bb'a_{1}))a_{2},a_{3},\dots,a_{k})$. As $bb'a_{1}=bf+b\,\pi(y)$, (F2), (F4) and (F1) give $(bb'a_{1})^{\circ}=(bf)^{\circ}+b^{\circ}\pi(y)$ and $E(bb'a_{1})=E(bf)+E(b)\,y$. Exactly as in the previous case, Claim 1 in the first entry and Claim 2 (with $j=1<k$) identify the two sides.

Claim 4 (Bound). Let $b\in A_{\varepsilon}$ and $\beta=\lVert b\rVert_{\mathrm{op}}$. Since $A_{\varepsilon}$ is the tracial algebra of $\gamma_{\varepsilon}\in\Sigma_{m_{\varepsilon}}$ (Two Noncommutative Laws with a Common Marginal: Standing Notation for Their Amalgamated Free Product §data, Two Noncommutative Laws with a Common Marginal: Standing Notation for Their Amalgamated Free Product §algebras), The Tracial Algebra of a Noncommutative Law: a Norm-Closed Unital *-Algebra with a Faithful Positive Trace, Determined by Vacuum Vectors, Closed under Square Roots §positivity gives $T=\beta^{2}I-b^{*}b\ge0$, where $T\in A_{\varepsilon}$ by The Tracial Algebra of a Noncommutative Law: a Norm-Closed Unital *-Algebra with a Faithful Positive Trace, Determined by Vacuum Vectors, Closed under Square Roots §star-algebra, and then $q\in A_{\varepsilon}$ with $q\ge0$ and $qq=T$. As $q\ge0$, $q$ is self-adjoint (Complex Hilbert Spaces and Bounded Linear Maps: Standing Notation §maps), hence an adjoint of itself (Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-calculus), so $q^{*}=q$ by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-unique and $q^{*}q=T$. For $d\in A_{\varepsilon}$ and $\xi\in\mathcal{F}$, (P1), the adjoint relation of The Formal Amalgamated Free Product: the Free Vector Space, Positivity of the Nested-Expectation Form, and Linearity and Formal Adjoints of the Actions §actions and multiplicativity in Claim 3 give

Taking $d=q$ and using linearity in $b$ and the unit relation of Claim 3, $J\ell_{\varepsilon}(T)\xi=\beta^{2}J\xi-J\ell_{\varepsilon}(b^{*}b)\xi$, so, taking also $d=b$,

Thus $\lVert J\ell_{\varepsilon}(b)\xi\rVert_{\mathcal{H}}^{2}\le(\beta\lVert J\xi\rVert_{\mathcal{H}})^{2}$; both $\lVert J\ell_{\varepsilon}(b)\xi\rVert_{\mathcal{H}}$ and $\beta\lVert J\xi\rVert_{\mathcal{H}}$ are nonnegative reals ($\beta\ge0$ as $\beta$ is a bound, Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §least-bound and Bounded Linear Maps between Complex Normed Spaces and the Operator Norm §bounded), so claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field gives $\lVert J\ell_{\varepsilon}(b)\xi\rVert_{\mathcal{H}}\le\beta\lVert J\xi\rVert_{\mathcal{H}}$.

Claim 5 (Actions). Let $b\in A_{\varepsilon}$, $\beta=\lVert b\rVert_{\mathrm{op}}$, and $T_{b}:\mathcal{F}\to\mathcal{H}$, $T_{b}\xi=J\ell_{\varepsilon}(b)\xi$. Then $T_{b}$ is complex-linear, being a composite of linear maps, and $\lVert T_{b}\xi\rVert_{\mathcal{H}}^{2}\le\beta^{2}\lVert J\xi\rVert_{\mathcal{H}}^{2}=\beta^{2}h(\xi,\xi)$ by Claim 4 and (P1). By The Complex Hilbert Completion is a Complex Hilbert Space Containing a Dense Isometric Image, and Bounded Complex-Linear Maps Extend to It §extension-linear with $V=\mathcal{F}$, the form $h$, $K=\mathcal{H}$ and $C=\beta$ (the completion being $\mathcal{H}$ and $J_{h}=J$ by The Amalgamated Free Product Space of Two Noncommutative Laws over a Common Marginal, Its Vacuum Vector and Vacuum State §space), there is exactly one continuous map $\Lambda_{\varepsilon}(b):\mathcal{H}\to\mathcal{H}$ with $\Lambda_{\varepsilon}(b)J\xi=J\ell_{\varepsilon}(b)\xi$ for every $\xi\in\mathcal{F}$, and it belongs to $\mathcal{L}(\mathcal{H})$ with bound $\beta$; hence $\lVert\Lambda_{\varepsilon}(b)\rVert_{\mathrm{op}}\le\beta$ by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §least-bound. For uniqueness within $\mathcal{L}(\mathcal{H})$, let $S\in\mathcal{L}(\mathcal{H})$ satisfy $SJ\xi=J\ell_{\varepsilon}(b)\xi$ for every $\xi$. For $w,w'\in\mathcal{H}$, linearity of $S$ and the fact that $\lVert S\rVert_{\mathrm{op}}$ is a bound for $S$ (Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §least-bound) give $\lVert Sw-Sw'\rVert_{\mathcal{H}}=\lVert S(w-w')\rVert_{\mathcal{H}}\le\lVert S\rVert_{\mathrm{op}}\lVert w-w'\rVert_{\mathcal{H}}$. The metric of $\mathcal{H}$ is $(w,w')\mapsto\lVert w-w'\rVert_{\mathcal{H}}$ (Complex Hilbert Spaces and Bounded Linear Maps: Standing Notation §spaces, claim 3 of The Induced Norm is a Norm, and Induces a Metric), so $S$ is Lipschitz with constant $\lVert S\rVert_{\mathrm{op}}$ and therefore continuous by A Lipschitz Map is Uniformly Continuous. By the uniqueness in The Complex Hilbert Completion is a Complex Hilbert Space Containing a Dense Isometric Image, and Bounded Complex-Linear Maps Extend to It §extension-linear, $S=\Lambda_{\varepsilon}(b)$.