Proof. Write \(t_0=\alpha c/(2\beta M)\) and suppose \(0<\gamma<\overline\delta_N(t_0)\). At any current node \(p_s\), the definition supplies a finite-codimensional \(F\) such that \[N(p_s+w)\ge(1+\gamma)N(p_s) \quad(w\in F,\ N(w)=t_0N(p_s)).\] The same inequality holds when \(N(w)\ge t_0N(p_s)\). To see this, write a shorter point on the ray as a convex combination of \(p_s\) and \(p_s+w\); convexity gives \(N(p_s+rw)-N(p_s)\ge r[N(p_s+w)-N(p_s)]\) for \(r\ge1\).

The quotient \(X/F\) is finite dimensional. Since \(d_{s,n}\) is weakly null, choose \(w_n\in F\) with \(N(w_n-d_{s,n})\to0\). Eventually \(N(w_n)\ge\alpha c/2\ge t_0N(p_s)\). Choosing the approximation error at most \(\gamma N(p_s)/2\) gives a child satisfying \[N(p_{s^\frown n})\ge(1+\gamma/2)N(p_s).\] Choose \(h\) large enough that \(\alpha m(1+\gamma/2)^h>\beta M\), and perform the selection in a tree of height \(h\). Repeating it \(h\) times yields a path whose last vector has \(N\)-norm at least \(\alpha m(1+\gamma/2)^h\). This exceeds the uniform bound \(\beta M\), a contradiction. Finally, \(\overline\delta_N(t_0)\ge0\): at a unit center intersect any candidate subspace with the kernel of a norming functional, on which the perturbed norms are at least one. This proves equality to zero. ◻

Completion of Theorem 1. The coordinate vectors are linearly independent, so each space is infinite dimensional. Finite rational coordinate combinations are dense, so each space is separable. For \(s\in I\), set \(p_s=\sum_{u\in I,\ u\preceq s}e_u\). Its norm is one: the path condition gives the upper bound, and a singleton test gives the lower bound. In the rootless model begin at any fixed nonroot node \(v\) and use the subtree \((p_{v^\frown s})_s\) so that the initial vector is nonzero. In all four models, \[\left\|\sum_{n\in A}a_ne_{s^\frown n}\right\|_K \le\left(\sum_{n\in A}|a_n|^2\right)^{1/2} \quad(A\subset\mathbb N\text{ finite}).\] This is the sibling constraint in the first two models. Siblings form an antichain in the third and are part of the globally constrained family in the fourth. If \(\phi\in X_K^*\), use \(a_n=\phi(e_{s^\frown n})\) in this inequality to obtain \[\sum_{n\in A}|\phi(e_{s^\frown n})|^2\le\|\phi\|^2.\] Thus the child increments are weakly null, and each has norm one. Lemma 3 applies with \(m=M=c=1\). In particular, for every equivalent norm as in that lemma, \(\overline\delta_N(\alpha/(2\beta))=0\). ◻

Proof. The choice \(h=x\), \(\mu=0\) proves the upper bound. Every representation of \(x\) satisfies \[\|x\|_2\le\|h\|_2+\sqrt{n+1}\|\mu\|_1 \le\sqrt{n+1}(\|h\|_2+\|\mu\|_1),\] which proves the lower bound. Scaling and adding representations, chosen within an arbitrary error of their infima, prove homogeneity and the triangle inequality. The lower bound gives definiteness; norm equivalence gives completeness and density of finite arrays.

For \(p_n(f)\le1\), absolute convergence allows us to write \[|\langle f,x\rangle| \le\|h\|_2+ \sum_s|\mu_s|\left|\sum_{r\preceq s}f_r\right| \le\|h\|_2+\|\mu\|_1.\] Thus the displayed supremum is at most \(L_n(x)\). Conversely, Hahn–Banach gives, for \(x\ne0\), a functional \(\varphi\) such that \(\varphi(x)=L_n(x)\) and \(|\varphi(z)|\le L_n(z)\). Since \(L_n(z)\le\|z\|_2\), it has the form \(\varphi(z)=\langle f,z\rangle\) with \(\|f\|_2\le1\). The representation \(b_s=B_ne_s\) gives \(L_n(b_s)\le1\), hence \(|\sum_{r\preceq s}f_r|=|\varphi(b_s)|\le1\). This proves equality, including the trivial case \(x=0\).

For an arbitrary \(f\in H_n\), the representation estimate gives dual norm at most \(p_n(f)\). The Hilbert unit ball and every \(\pm b_s\) belong to the \(L_n\) unit ball. Testing on these vectors gives both reverse bounds in (22), and therefore the exact dual norm. ◻

Proof. Each block is an equivalently normed Hilbert space and hence reflexive. For completeness of the sum, a Cauchy sequence has a limit in each block; its norm bounds and Cauchy estimates pass to the limit first on finite sets of blocks and then by increasing those sets. Cauchy–Schwarz shows that each displayed dual family defines a bounded functional. Conversely, restrict an arbitrary functional to finitely many blocks, choose almost norming vectors there, and weight them proportionally to the block dual norms. This proves that the sum of their squares is bounded by the square of the original functional norm. Density of finite-block vectors proves the representation and equality of norms. Applying the same argument to the dual sum proves surjectivity of the canonical map into the bidual. Finite arrays are dense in the dual by the same block truncation and Hilbert equivalence, so it is separable.

The cost formula gives \(L_n(e_s),L_n(b_s)\le1\). The dual vector \(e_s\) has \(p_n(e_s)=1\) and evaluates to one on both vectors, giving equality. The weak nullity assertion follows from square summability of every fixed dual array on the \(n\)th block.

For clarity we give the renorming argument with its scale and gain. Suppose \(\alpha\|\cdot\|\le N\le\beta\|\cdot\|\), where \(0<\alpha\le\beta\), and suppose \(d=\overline\delta_N(t)>0\) for some \(0<t\le\alpha/\beta\). Choose \(0<\eta<d/2\). For each unit center \(u\) there is a finite-codimensional closed subspace \(F\) such that \(N(u+tv)\ge1+2\eta\) on its unit sphere. Convexity on a ray gives \(N(u+v)\ge1+2\eta\) whenever \(v\in F\) and \(N(v)\ge t\).

At an internal vertex \(s\) take \(u=b_{n,s}/N(b_{n,s})\) and \(z_j=e_{n,s^\frown j}/N(b_{n,s})\). These vectors are weakly null and \(N(z_j)\ge\alpha/\beta\ge t\). Their quotient images in \(X/F\) tend to zero, since that quotient is finite-dimensional. Choose \(v_j\in F\) with \(N(v_j-z_j)\to0\). If \(N(v_j)<t\), replace \(v_j\) by \(tv_j/N(v_j)\); for large \(j\) this is defined, and the extra change is at most \(N(v_j-z_j)\). Thus \(z_j\) is approximated by vectors in \(F\) of norm at least \(t\). For a sufficiently large child index, \[N(b_{n,s^\frown j})\ge(1+\eta)N(b_{n,s}).\] Iterating to depth \(n\) gives \(\alpha(1+\eta)^n\le\beta\), impossible for all \(n\). The asymptotic modulus is nonnegative: the kernel of a norming functional at a unit center has \(N(u+tv)\ge1\). This proves the claimed zero value at every indicated scale. Taking \(\eta=d/4\) gives the often useful growth factor \(1+d/4\). ◻

Proof. In block \(n\) write \(A_n=A\cap T_n\) and set \[H(s)=\sum_{r\preceq s}h_{n,r},\qquad V(s)=\sum_{\substack{r\preceq s\\r\notin A_n}}g_{n,r}, \qquad a_n=p_n(h_n),\quad c_n=p_n(g_n).\] Here \(|H|\le a_n\). The sum \(V\) is a difference of two prefix sums of \(g_n\), or a single prefix sum if the path misses \(A_n\), so \(|V|\le2c_n\). It vanishes on \(A_n\). Define \[(V_1,V_2)= \begin{cases} (\min(V,0),-\max(V,0)),&H\ge0,\\ (\max(V,0),-\min(V,0)),&H<0. \end{cases}\] Let \(r_{l,n,s}=V_l(s)-V_l(s^-)\), with predecessor value zero at the root. These coefficients vanish on \(A_n\). Outside \(A_n\), the head sum \(H\) is constant along each remaining branch, and both scalar clipping maps are \(1\)-Lipschitz. Therefore \[|r_{l,n,s}|\le|g_{n,s}| \quad(s\notin A_n).\] At a first exit vertex this follows from the zero predecessor value of \(V_l\). Moreover the prefix sums of \(r_{l,n}\) are \(V_l\), and \(V_1-V_2=V\), proving the claimed difference.

The supports of \(h_n\) and \(r_{l,n}\) are disjoint. Hence their Hilbert norm satisfies \[\|\lambda h_n+\mu r_{l,n}\|_2^2 \le\lambda^2a_n^2+\mu^2c_n^2.\] The products \(HV_l\) are nonpositive, so each prefix sum satisfies \[|\lambda H+\mu V_l| \le\max\{\lambda a_n,2\mu c_n\}.\] These two estimates give \(p_n(\lambda h_n+\mu r_{l,n})^2 \le\lambda^2a_n^2+4\mu^2c_n^2\). Summing over \(n\) proves the asserted inequality; taking \(\lambda=0\), \(\mu=1\) proves \(r_l\in X^*\) and its norm bound. ◻

Proof. Choose norming dual vectors \(h,g\) for \(x,y\), with norms at most one, and replace \(h\) by \(P_Ah\). The projection preserves its evaluation at \(x\) and does not increase its norm. Lemma 7 gives \(r_1(y)-r_2(y)=\|y\|\), so at least one of \(|r_1(y)|,|r_2(y)|\) is at least \(\|y\|/2\). For this index choose a sign \(\varepsilon\) with \(\varepsilon r_l(y)\ge\|y\|/2\). Because \(h(y)=r_l(x)=0\), evaluation of \(h+cr_l\) on \(x+\varepsilon y\) proves (25). Its supremum over \(c>0\) is the displayed square root, by Cauchy–Schwarz in two dimensions; endpoint cases follow by a limit. ◻

Proof. Fix \(\eta>0\) and choose a finite head \(A\) with \(\|x-P_Ax\|\le\eta\), using the strong head exhaustion. Put \(z_j=(I-P_A)y_j\). Since \(P_A\) has finite rank and \(y_j\) is weakly null, \(\|P_Ay_j\|\to0\). In particular, \[\bigl|\|z_j\|-\|y_j\|\bigr|\le\|P_Ay_j\|\longrightarrow0.\] Apply Theorem 8 to \(P_Ax\) and \(z_j\). The triangle inequality then gives \[\max\{\|x+y_j\|,\|x-y_j\|\} \ge\sqrt{\|P_Ax\|^2+\|z_j\|^2/16} -\eta-\|P_Ay_j\|.\] Keep this head fixed while taking the lower limit in \(j\). The result is at least \(\sqrt{\|P_Ax\|^2+L^2/16}-\eta\). Finally let \(\eta\downarrow0\), choosing the heads along the strong exhaustion, so that \(\|P_Ax\|\to\|x\|\). This proves the stated bound. ◻

Proof. Choose a norming functional \(\phi\in\ell_2(U)\) for \(B_U\nu\). Its dual norm is at most one, hence \(\|\phi\|_2\le1\) and \(F_t:=\sum_{v\in U,\,v\preceq t}\phi(v)\) satisfies \(|F_t|\le1\). At a root set the value at its nonexistent parent equal to zero, and put \[\psi(v)=|F_v|-|F_{\operatorname{par}(v)}|.\] Then \(|\psi(v)|\le|\phi(v)|\), and the prefix sums of \(\psi\) are \(|F_t|\). Thus \(\psi\) also has dual norm at most one: pairing it with a decomposition \(h+B_U\alpha\) costs at most \(\|h\|_2+\|\alpha\|_1\). Absolute convergence of the coefficient sums gives \[L_U(B_U\nu)=\sum_t\nu(t)F_t \le\sum_t\mu(t)|F_t| =\langle B_U\mu,\psi\rangle\le L_U(B_U\mu).\] The claim concerns positive path coefficients; it asserts no general coordinatewise monotonicity of the norm. ◻

Proof. For the unit estimate set \(r=L_n(x)\) and choose a norming functional \(g=gP_A\), of norm at most one, with \(g(x)=r\); use zero if \(x=0\). For each sign choose \(x\pm y=h_\pm+B_n\alpha_\pm\) of cost at most \(1+\eta\), where \(\eta>0\). Put \(d=1+\eta-r\). Subtracting the value of \(g\) from the cost gives nonnegative terms with sum at most \(d\): \[\|h_\pm\|_2-g(h_\pm) +\sum_t\bigl(|\alpha_\pm(t)|-\alpha_\pm(t)g(b_t)\bigr)\le d.\] In particular, \[\|h_\pm|_U\|_2\le\sqrt{2(1+\eta)d}.\] For this bound, use \(|g(h_\pm)|\le\|P_Ah_\pm\|_2\) and factor the difference of squares. Its other factor is at most \(2(1+\eta)\). The coefficients satisfying \(\alpha_\pm(t)g(b_t)\le0\) have total mass at most \(d\).

Retain only the other coefficients ending in \(U\), and call their intrinsic forest contributions \(v_\pm\). Then \[y|_U=v_++w_+,\qquad -y|_U=v_-+w_-,\qquad L_U(w_\pm)\le K:=\sqrt{2(1+\eta)d}+d.\] Within each tree of \(U\), the value \(g(b_t)\) is constant, since the head of every such path is the same. If that value is zero, no coefficient is retained there. Otherwise both retained arrays have its sign. One isometric change of signs makes both arrays nonnegative. Lemma 10 now gives \[L_U(v_+)\le L_U(v_++v_-) =L_U(w_++w_-)\le2K.\] It follows that \(L_U(y|_U)\le3K\) and \(L_n(y)\le6K\). Letting \(\eta\downarrow0\), and using \(0\le1-r\le\sqrt{1-r}\), proves (27). Scaling gives the first bound in (28) when \(\beta>0\); when \(\beta=0\), both vectors are zero. Contractivity of \(P_A\) gives \(r\le\beta\) and the second bound. In each component both endpoint norms are at least the center norm. Thus the maximum of their squared differences from \(r^2\) is at most their sum. Summing over components proves (29), including components with empty heads. ◻

Proof. Only the finite ancestor hull of the supports matters. Write \(q(v)=\min\{G(v),W(v)\}\). Nonnegativity gives \[q(v)\le G(v),\qquad \sum_{w\text{ child of }v}q(w)\le q(v).\] Process vertices from the leaves towards the roots. At \(v\), the mass already marked in its child subtrees totals \(\sum_wq(w)\). Mark enough additional, previously unmarked mass of \(\mu\) below \(v\) to make the marked total there equal to \(q(v)\); fractions of an endpoint mass are allowed. The two displayed inequalities make this possible. Later steps can only increase a previously met quota. The total finally marked in a tree equals its root quota and is at most its total \(\nu\)-mass. Take \(\rho\) to be the marked mass. ◻

Proof. Choose \(g=gP_A\) norming \(x\), and finite path representations of both endpoints with cost at most \(1+\eta\). Put \(d=1+\eta-L_n(x)\le\varepsilon+\eta\). As in the preceding proof, their Hilbert tails have lengths \(K_\pm\le\sqrt{2(1+\eta)d}\), and their coefficient masses \(B_\pm\) with \(\alpha_\pm(t)g(b_t)\le0\) satisfy \(B_\pm\le d\). Call these coefficients bad and the others good.

In a complementary tree where the common value \(c=g(b_t)\) is zero, there are no good coefficients. Otherwise put \(\varsigma=\operatorname{sign}(c)\). All good coefficients have sign \(\varsigma\), and all bad nonzero coefficients have sign \(-\varsigma\). Apply Lemma 12 to the combined good mass \(\mu\) and combined bad mass \(\nu\) in each tree, and mark the selected portions with their originating endpoint representation. Denote the selected good mass by \(\rho\), and put \(G(v)=\mu(K_v)\), \(W(v)=\nu(K_v)\). At a vertex in a tree with \(c\ne0\), the vanishing sum of the two endpoint tails reads \[h_+(v)+h_-(v)+\varsigma\bigl(G(v)-W(v)\bigr)=0.\] Consequently, in every complementary tree, including those with no good mass, \[(G(v)-W(v))_+\le|h_+(v)+h_-(v)|\qquad(v\notin A).\] After removal, the good tail in either one of the representations has absolute coordinate at most \(G(v)-\rho(K_v)\le(G(v)-W(v))_+\). Its Hilbert norm is therefore at most \(K_++K_-\). The total removed mass is at most \(B_++B_-\), and each complementary path costs at most two in \(L_n\). Decomposing the plus endpoint tail into its Hilbert, bad, removed-good, and remaining-good parts gives \[L_n(y)\le2K_++K_-+2B_++2(B_++B_-) \le3\sqrt{2(1+\eta)d}+8d.\] Let \(\eta\downarrow0\). This proves (30), including its endpoint cases.

We give the outer-sum passage explicitly. For the \(j\)-th pair set \[\begin{align*} a_{j,n}^\pm&=L_n(x_{j,n}\pm y_{j,n}), &r_{j,n}&=L_n(x_{j,n}),\\ c_{j,n}&=\tfrac12(a_{j,n}^++a_{j,n}^-), &a_{j,n}&=\max(a_{j,n}^+,a_{j,n}^-). \end{align*}\] Convexity gives \(c_{j,n}\ge r_{j,n}\), and the endpoint assumptions give \(\|a_j^\pm\|_2\le1\). Hence \(\|c_j\|_2\to1\). The Hilbert parallelogram identity then gives \[\|a_j^+-a_j^-\|_2^2 =2\|a_j^+\|_2^2+2\|a_j^-\|_2^2-4\|c_j\|_2^2 \le4(1-\|x_j\|^2)\longrightarrow0.\] Since \(a_j=c_j+|a_j^+-a_j^-|/2\), we also have \(\|a_j\|_2\to1\). Ignore coordinates with \(a_{j,n}=0\), whose vectors are zero, and put \(\delta_{j,n}=1-r_{j,n}/a_{j,n}\in[0,1]\). Then \[\sum_na_{j,n}^2\delta_{j,n} \le\sum_n(a_{j,n}^2-r_{j,n}^2)\longrightarrow0.\] Applying the component estimate after division by \(a_{j,n}\) gives \(L_n(y_{j,n})/a_{j,n}\le3\sqrt{2\delta_{j,n}}+8\delta_{j,n}\); the same ratio is at most one by the triangle inequality. For fixed \(\tau>0\), the squared outer norm is therefore at most \[(3\sqrt{2\tau}+8\tau)^2\sum_na_{j,n}^2 +\tau^{-1}\sum_na_{j,n}^2\delta_{j,n}.\] First let \(j\to\infty\), then \(\tau\downarrow0\). ◻

Proof. Work first in one component. Set \(r=L_n(x)\) and \(M_\pm=L_n(x\pm y)\), so \(M_\pm\ge r\). Use a supported norming functional \(g\) and finite representations of costs at most \(M_\pm+\eta\). Put \(d_\pm=M_\pm+\eta-r\). Here a path coefficient is good when \(\operatorname{sign}(\alpha_\pm(t))g(b_t)>1/2\), and bad otherwise. The deficit calculation gives \[ B_\pm\le2d_\pm,\qquad K_\pm\le\sqrt{2(M_\pm+\eta)d_\pm}. \tag{32}\] Write \(c=g(b_t)\) for the common value on one complementary tree. If \(|c|\le1/2\), that tree has no good coefficients. Otherwise put \(\varsigma=\operatorname{sign}(c)\); good coefficients have sign \(\varsigma\), and bad nonzero coefficients have sign \(-\varsigma\). Let \(G_+(v),G_-(v)\) be the subtree masses of the good coefficients from the two representations, and let \(W(v)\) be the subtree mass of all bad coefficients. In a tree with \(|c|>1/2\), the endpoint tails satisfy \[h_+(v)+h_-(v)+\varsigma\bigl(G_+(v)+G_-(v)-W(v)\bigr)=0.\] For the quota selection take \(G=G_+\). Since \(G_-\ge0\), the identity gives the following bound, which is also immediate in a tree with no good mass: \[(G(v)-W(v))_+\le|h_+(v)+h_-(v)|.\] Lemma 12 removes at most \(B_++B_-\) from the plus good coefficients. The remaining plus good tail has Hilbert norm at most \(K_++K_-\); the removed tail has \(L_n\)-norm at most \(2(B_++B_-)\). Adding the plus Hilbert and bad tails yields \[\begin{split} L_n(y)&\le2(K_++K_-)+4(B_++B_-)\\ &\le11\sum_{\sigma\in\{+,-\}} \sqrt{(M_\sigma+\eta)d_\sigma}. \end{split}\] Here \(d_\sigma\le M_\sigma+\eta\) and \(2\sqrt2+8<11\). Squaring the last sum costs at most a factor two, and \((M_\sigma+\eta)d_\sigma\le(M_\sigma+\eta)^2-r^2\). Letting \(\eta\downarrow0\) proves the component estimate with \(2\cdot11^2=242\). Summation gives (31). Empty heads cause no exception; their centers are zero. ◻

Proof. Write \(x^s=x+sy\), \(s\in\{-1,1\}\), and denote averages over \(s\) by a bar. Fix \(\lambda>0\). In each nonzero component choose \[x_n^s=z_n^s+B_n\alpha_n^s,\qquad r_n^s=\|z_n^s\|_2+M_n^s\le(1+\lambda)L_n(x_n^s),\qquad M_n^s=\|\alpha_n^s\|_1.\] Use zero decompositions in zero components. Project to \(A'\): truncate the Hilbert terms and push each path endpoint to its last ancestor in \(A'\), discarding endpoints with no such ancestor. Both costs decrease, so all arrays may be taken supported in \(A'\).

In a fixed \(T_n\), let \(\pi_A(t)\) be the last ancestor of \(t\) in \(A\), or a dummy symbol \(\perp\) if none exists. For each fibre set \[L_b=\sum_{\pi_A(t)=b}\bar\alpha_n(t),\qquad l_b\in\{-1,1\},\quad l_bL_b=|L_b|.\] Either choice is allowed at zero. Introduce the mass on pairs \((s,t)\) given by \(m_n(s,t)=|\alpha_n^s(t)|/2\), and let \(h(s,t)\) be the coefficient’s sign, chosen arbitrarily at zero. Its wrong-sign mass is \[W_n=\sum_{s,t:\,h(s,t)\ne l_{\pi_A(t)}}m_n(s,t).\] Separating the two signs in each fibre, including the dummy fibre, gives \(\bar M_n-\sum_b|L_b|=2W_n\). Since \(x_n=P_{A\cap T_n}(\bar z_n+B_n\bar\alpha_n)\), its component norm is at most \(\overline{\|P_Az_n^s\|_2}+\sum_b|L_b|\); the dummy term is a harmless nonnegative addition to this cost. It follows that \[ \Delta_n:=\bar r_n-L_n(x_n) \ge2W_n+\frac12\overline{ \frac{\|P_Ez_n^s\|_2^2}{\|z_n^s\|_2}}\ge0, \tag{34}\] A quotient with zero denominator means zero. The Hilbert estimate used here is \[\|z\|_2-\|P_Az\|_2\ge\frac{\|P_Ez\|_2^2}{2\|z\|_2} \quad(z\ne0,\ P_{A'}z=z).\]

The quantity \(Q=\sum_n\bar r_n\Delta_n\) controls both Hilbert leakage and wrong-sign mass. For an array on the disjoint union of the trees, \(\|\cdot\|_2\) denotes its Hilbert norm, including summation over components. Since \(\|z_n^s\|_2\le2\bar r_n\), \[ \begin{split} \|P_E\bar z\|_2^2 &\le\overline{\|P_Ez^s\|_2^2}\le4Q,\qquad \sum_n\Delta_n^2\le Q,\\ Q&\le\sum_n(\bar r_n)^2-\|x\|^2 \le(1+\lambda)^2\overline{\|x^s\|^2}-\|x\|^2. \end{split} \tag{35}\] The first line follows from (34) and \(0\le\Delta_n\le\bar r_n\); the second uses \(L_n(x_n)\le\bar r_n\) and \((\bar r_n)^2\le\overline{(r_n^s)^2}\).

It remains to recover the norm of \(P_Ey\) from this budget. The zero case is immediate. Otherwise take a norming functional for \(P_Ey\) and precompose it with \(P_E=P_{A'}-P_A\). Its coefficient array \(\phi\), supported in \(E\), satisfies \[ \langle y,\phi\rangle=\|P_Ey\|,\qquad \|\phi\|_2\le2,\qquad \Bigl(\sum_nc_n^2\Bigr)^{1/2}\le2, \quad c_n=\sup_{t\in T_n}\left|\sum_{v\preceq t}\phi(v)\right|. \tag{36}\] The functional has norm at most two. Testing against Hilbert vectors gives the first coefficient bound. Testing against finite sums of one root path in each component, with arbitrary scalar multipliers, gives the second by the outer Hilbert norm. Approximate each supremum if it is not attained.

The Hilbert part of \(\langle y,\phi\rangle=\overline{s\langle x^s,\phi\rangle}\) has absolute value at most \(4\sqrt Q\), by (35) and (36). For the path part write \(g(t)=\sum_{v\preceq t}\phi(v)\) and \[I_n=\int s\,h(s,t)\,g(t)\,dm_n(s,t).\] For \(v\in E\cap T_n\), let \(m(v)\) be the mass of pairs with \(t\succeq v\), and \(w(v)\) its wrong-sign part. The entire subtree below \(v\) is in a single \(\pi_A\)-fibre. Its coefficients therefore satisfy \[\bigl(B_n\bar\alpha_n\bigr)(v) =l_{\pi_A(v)}\bigl(m(v)-2w(v)\bigr)=-\bar z(v),\] where the last equality uses \(x(v)=0\) on \(E\). Thus, whenever \(w(v)\le m(v)/4\), \[m(v)\le2|\bigl(B_n\bar\alpha_n\bigr)(v)|=2|\bar z(v)|.\]

Call a vertex \(v\in E\) retained if this inequality’s condition holds at \(v\) and at every preceding vertex of its path in \(E\). Split \(g(t)\) into its sum over retained vertices and the remainder. The remainder starts at the first vertex satisfying \(w(v)>m(v)/4\). Such first vertices form an antichain, so their disjoint descendant sets have total mass at most \(4W_n\). The remainder is the difference of two prefix sums and has absolute value at most \(2c_n\). It therefore contributes at most \(8c_nW_n\) to \(|I_n|\). The retained contribution is at most \(2\sum_{v\in E\cap T_n}|\phi(v)|\,|\bar z(v)|\). Summing over components and using \(2W_n\le\Delta_n\), we obtain \[\begin{split} \sum_n|I_n| &\le4\|P_E\bar z\|_2+4\sum_nc_n\Delta_n\\ &\le8\sqrt Q+8\sqrt Q=16\sqrt Q. \end{split}\] Together with the Hilbert part this proves \(\|P_Ey\|\le20\sqrt Q\). Apply (35), square, and let \(\lambda\downarrow0\) to obtain (33). ◻

Proof. There are only finitely many relevant coordinates, since there are finitely many sign histories and all the sets \(P_i\) are finite. We first represent the terminal value by one random atom in each component. The finite-head unit ball is exactly the convex hull in (23). This description provides finite random representations by a Hilbert-ball vector or a signed root path.

For each terminal history and each component, apply this description with \(P=P_m\cap T_n\), rescaled by \(L_n(x_m^n)\). On a finite extension of the probability space we may therefore choose \(h_n,c_n,u_n\) so that \[ \mathbb E[h_n+c_nb_{u_n}\mid\mathcal F_m]=x_m^n, \qquad \sum_n(\|h_n\|_2^2+|c_n|^2)\le C^2. \tag{40}\] For each component the sample is either a Hilbert-ball vector \(h_n\), with \(c_n=0\), or a signed root path with \(h_n=0\) and \(|c_n|=L_n(x_m^n)\). Zero components use both coefficients zero. An arbitrary root can be used for \(u_n\) when \(c_n=0\). The sigma-fields \(\mathcal F_i\) still record only the original signs. In particular \(x_i=\mathbb E_i(h_n+c_nb_{u_n})_n\).

Let \(v_i^n\) be the deepest vertex of the path to \(u_n\) in \(P_i\), or a symbol \(\partial\) if this intersection is empty; set \(b_\partial=0\). The path trace on the new layer is \[I_i^n=R_{\Delta_i\cap T_n}b_{u_n}=b_{v_i^n}-b_{v_{i-1}^n}, \qquad L_n(I_i^n)\le2.\] The sigma-fields \(\mathcal G_i^n=\sigma(\mathcal F_i,v_i^n)\) increase: from \(v_i^n\) and the previous sign history one recovers \(v_{i-1}^n\) by truncating its path to \(P_{i-1}\). Set \(a_i^n=\mathbb E[c_n\mid\mathcal G_i^n]\). Orthogonality of scalar martingale increments gives \[ \sum_n\sum_{i=1}^m\mathbb E|a_i^n-a_{i-1}^n|^2 \le\sum_n\mathbb E|c_n|^2\le C^2. \tag{41}\]

We split the new layer into its Hilbert contribution, its old predicted path coefficient, and the change in that prediction: \[\begin{align*} R_{\Delta_i}x_i&=H_i+M_i+W_i,\\ H_i&=\mathbb E_iR_{\Delta_i}(h_n)_n,\\ M_i&=\mathbb E_i(a_{i-1}^nI_i^n)_n,\\ W_i&=\mathbb E_i((a_i^n-a_{i-1}^n)I_i^n)_n. \end{align*}\] The identity follows because \(I_i^n\) is \(\mathcal G_i^n\)-measurable, so \(c_n\) may be replaced by \(a_i^n\) inside its conditional expectation. Conditional Jensen, the disjointness of the \(\Delta_i\) on each outcome, and (41) show that \[ \sum_i\mathbb E\|H_i\|^2\le C^2, \qquad \sum_i\mathbb E\|W_i\|^2\le4C^2. \tag{42}\] Here we use \(L_n\le\|\cdot\|_2\) for the first inequality. Denote conditional expectations given \(\mathcal F_{i-1}\) by primes. Predictability of \(\Delta_i\), together with the support of \(x_{i-1}\), gives \(H_i'+M_i'+W_i'=R_{\Delta_i}x_{i-1}=0\). Consequently \[ \sum_i\mathbb E\|M_i'\|^2\le10C^2. \tag{43}\] It remains to compare the old-coefficient contribution on a child history with its average on the parent history.

We prove \(\|M_i\|\le4\|M_i'\|\) pointwise. Fix a component and a history through time \(i-1\). For \(v\in \Delta_i\cap T_n\), let \(r(v)\) be the first vertex of \(\Delta_i\) on its root path. If \(v_i^n=v\), then \(v_{i-1}^n\) is the parent of \(r(v)\), or \(\partial\) when this first vertex is the root. Thus the value of \(a_{i-1}^n\) on these outcomes is a scalar \(q_{r(v)}\) depending only on this first vertex. It is the conditional mean on the old trace cell determined by the parent history and \(v_{i-1}^n=r(v)^-\), with the root case interpreted as \(\partial\). On a null old trace cell put \(q_{r(v)}=0\); all its parent and child trace probabilities then vanish. Let \(\mu(v)\) and \(\widetilde\mu(v)\) be the probabilities of \(v_i^n=v\) conditional on this parent history and on a specified child history, respectively. Since both children have probability \(1/2\), \(\widetilde\mu(v)\le2\mu(v)\).

Put \(I(v)=b_v-b_{r(v)^-}\), interpreting the second term as zero when \(r(v)\) is the root. Outcomes with \(v_i^n\notin\Delta_i\) have zero path trace. The two conditional mixtures are therefore \[\begin{aligned} M_i^n&=\sum_{v\in\Delta_i\cap T_n}q_{r(v)}\widetilde\mu(v)I(v),\\ (M_i')^n&=\sum_{v\in\Delta_i\cap T_n}q_{r(v)}\mu(v)I(v). \end{aligned}\]

For an admissible test \(a\), put \(A_v=\sum_{w\in[r(v),v]}a_w\). We have \[|\langle M_i^n,a\rangle| \le2\sum_{v\in \Delta_i\cap T_n}|q_{r(v)}|\mu(v)|A_v|.\] Define a test candidate \(f\), zero off \(\Delta_i\cap T_n\), by \[f_v=\operatorname{sign}(q_{r(v)})(|A_v|-|A_{v^-}|),\] where \(A_{v^-}=0\) at the first vertex of each interval and \(\operatorname{sign}(0)=0\). The reverse triangle inequality gives \(\|f\|_2\le\|a\|_2\). Each root path meets \(\Delta_i\) in at most one interval. Summing \(f\) on that interval telescopes to a signed \(|A_v|\), which is at most \(2\), since \(A_v\) is the difference of two root-path sums of \(a\). Hence \(f/2\) is admissible. Moreover \[\langle(M_i')^n,f\rangle =\sum_{v\in \Delta_i\cap T_n}|q_{r(v)}|\mu(v)|A_v|.\] Taking the supremum over \(a\) proves \(L_n(M_i^n)\le4L_n((M_i')^n)\). Summing the component squares proves the required comparison. By (43), the total energy of \(M_i\) is at most \(160C^2\). Finally, \(\|H_i+M_i+W_i\|^2\le3(\|H_i\|^2+\|M_i\|^2+\|W_i\|^2)\), giving \(3(1+160+4)C^2=495C^2\). ◻

Proof. If \(S=\varnothing\), then \(L_n(R_Tz^\pm)\le V_T(z^\pm)\le2g\), so the conclusion follows. Otherwise \(S\) contains the root. For \(u\in T\), write \(\pi(u)\) for its deepest ancestor in \(S\). Choose \(\sigma_s\in\{-1,1\}\) so that \(\sigma_sd_S(z,s)=|d_S(z,s)|\). Write \[d_T(z^\pm,u)=\sigma_{\pi(u)}(p_u^\pm-q_u^\pm), \qquad p_u^\pm,q_u^\pm\ge0,\] using the positive and negative parts. Collapsing the refined path coefficients to \(S\) shows that \[ g=2\sum_{u\in T}\tfrac12(q_u^++q_u^-)\ge0. \tag{46}\] For \(u\in U\), all descendants in \(T\) have the same \(\pi\)-label. Let \(P_u^\pm,Q_u^\pm\) be the respective sums of \(p_v^\pm,q_v^\pm\) over these descendants. Since \(z^\pm(u)=\sigma_{\pi(u)}(P_u^\pm-Q_u^\pm)\), \[ P_u^\pm\le2\operatorname{av}P_u^\pm \le2|z(u)|+2\operatorname{av}Q_u^\pm, \tag{47}\] where \(\operatorname{av}\) means the average of the two signs.

Fix one sign and apply Lemma 12 on the finite forest \(U\) to the endpoint masses \(\mu_u=p_u^\pm\) and \(\nu_u=q_u^++q_u^-=2\operatorname{av}q_u^\pm\). Initiality of \(S\) ensures that all descendants in \(T\) of a vertex in \(U\) remain in \(U\). The corresponding subtree masses are therefore \(P_u^\pm\) and \(2\operatorname{av}Q_u^\pm\). The lemma gives \(0\le r_u\le p_u^\pm\) with \[\sum_{v\succeq u}r_v \ge\min\{P_u^\pm,2\operatorname{av}Q_u^\pm\}, \qquad \sum_{u\in U}r_u\le2\sum_{u\in U}\operatorname{av}q_u^\pm\le g.\] By (47), the remaining mass satisfies \[P_u^\pm-\sum_{v\succeq u}r_v \le\bigl(P_u^\pm-2\operatorname{av}Q_u^\pm\bigr)_+ \le2|z(u)|.\] Also \(\sum_{u\in U}q_u^\pm\le g\), by (46). The remaining positive mass defines \[v(u)=\sigma_{\pi(u)}\sum_{w\succeq u}(p_w^\pm-r_w) \quad(u\in U),\qquad v=0\text{ off }U.\] Its coordinates satisfy \(|v(u)|\le2|z(u)|\). The exact decomposition \[R_Uz^\pm=v+\sum_{u\in U}\sigma_{\pi(u)}(r_u-q_u^\pm)R_Ub_u\] therefore costs at most \(2\|R_Uz\|_2+2(g+g)\), as required. ◻

Proof. For each terminal history choose, componentwise, \[x_m^n=h_n+B_nc_n, \qquad \|h_n\|_2+\|c_n\|_1\le2L_n(x_m^n),\] with both terms zero in a zero component. Only finitely many components are needed. Put \(h_i=\mathbb E_i(h_n)_n\), \(b_i=(B_n\mathbb E_i c_n)_n\); then \(x_i=h_i+b_i\). The terminal Hilbert and coefficient budgets are each at most \(4C^2\): \[\mathbb E\|(h_n)_n\|_{\ell_2(\coprod T_n)}^2\le4C^2, \qquad \mathbb E\sum_n\|c_n\|_1^2\le4C^2.\] At time \(i-1\), apply Lemma 17 in each component to \(S=P_{i-1}\cap T_n\), \(T=P_i\cap T_n\), and the two values of \(b_i^n\). These sets are fixed before the sign at time \(i\). If \(g_{i,n}\) is the variation gain in the lemma, then \[V_{P_{i-1}\cap T_n}(b_{i-1}^n)^2+g_{i,n}^2 \le\mathbb E_{i-1}V_{P_i\cap T_n}(b_i^n)^2.\] Indeed, the average variation equals the previous variation plus the nonnegative number \(g_{i,n}\); square and apply scalar Jensen. Telescoping and (44) give \[\sum_{i,n}\mathbb E g_{i,n}^2\le4C^2.\] Since \(R_{\Delta_i}x_{i-1}=0\), the lemma and the outer Euclidean norm imply \[\|R_{\Delta_i}x_i\| \le\|R_{\Delta_i}h_i\|_2+2\|R_{\Delta_i}h_{i-1}\|_2 +4\Big(\sum_ng_{i,n}^2\Big)^{1/2}.\] For \(j=i-1\) and \(j=i\), predictability gives \(R_{\Delta_i}h_j=\mathbb E_jR_{\Delta_i}(h_n)_n\). Conditional Jensen followed by pathwise disjointness of the layers bounds each of the two unscaled Hilbert energies by \(4C^2\). Squaring the last three-term inequality and summing now gives \(3(1+4+16)4C^2=252C^2\). ◻

Proof. For every terminal outcome and component choose \[x_m^n=h_n+B_n\lambda_n, \qquad \|h_n\|_2+\|\lambda_n\|_1\le2L_n(x_m^n).\] Write \(\lambda_n=\epsilon_n\mu_n\), where \(\mu_n=|\lambda_n|\), \(|\epsilon_n|\le1\), and \(M_n=\mu_n(T_n)\). Thus \(\mu_n\) is a positive measure on path endpoints, allowed to depend on the terminal outcome. The two budgets satisfy \[ \mathbb E\sum_n(\|h_n\|_2^2+M_n^2)\le4\mathbb E\|x_m\|^2. \tag{50}\] On outcome–endpoint pairs put the finite positive measure \[\nu_n(\{\omega\}\times\{u\})=2^{-m}\mu_n(\omega,\{u\}).\] This need not be a probability measure; conditional means on its positive-mass cells have their usual meaning. On null cells set them to zero. The operators \(\mathbb E_t\) still use the original uniform law on sign histories, whereas the conditional means of \(\epsilon_n\) below use \(\nu_n\), which need not give equal mass to the two sign children. Let \(a_{t,n}^-\) be the conditional mean of \(\epsilon_n\) given the first \(t\) signs and the truncation of the root path to \(A_t\). Let \(a_{t,n}^+\) be its conditional mean given the first \(t+1\) signs and its truncation to \(D_t\). Empty truncations are recorded by a dummy symbol. The conditioning partitions, ordered as \[(0,-),(0,+),(1,-),(1,+),\ldots,(m-1,-),(m-1,+),\] refine successively: the known sign history and truncation of a deeper trace to the earlier head recover the preceding data. Define \[q_{t,n}=\mathbb E_{t+1} \int|a_{t,n}^+-a_{t,n}^-|\,d\mu_n.\] We first establish the weighted sign estimate \[ \sum_{t<m}\mathbb E q_{t,n}^2\le4\mathbb E M_n^2. \tag{51}\]

For clarity fix \(n\) temporarily. Conditional Cauchy–Schwarz gives \[q_t^2\le(\mathbb E_{t+1}M)\, \mathbb E_{t+1}\int|a_t^+-a_t^-|^2\,d\mu.\] Under the original sign law the children are fair, so \(M\ge0\) gives \(\mathbb E_{t+1}M\le2\mathbb E_tM\). Put \[R_t=\max_{0\le j\le t}\mathbb E_jM,\qquad R_{-1}=0.\] The conversion from the sign law to the pair measure is \[\mathbb E\left[(\mathbb E_tM)\, \mathbb E_{t+1}\int|a_t^+-a_t^-|^2\,d\mu\right] =\int(\mathbb E_tM)|a_t^+-a_t^-|^2\,d\nu,\] because \(\mathbb E_tM\) is measurable at time \(t+1\). Hence the sum of the integrated right sides is at most \[2\sum_{t<m}\int R_t|a_t^+-a_t^-|^2\,d\nu.\] Expand \(R_t=\sum_{j\le t}(R_j-R_{j-1})\). Each nonnegative increment is constant on each cell \(C\) of the partition defining \(a_j^-\). Orthogonality of successive conditional-mean increments under \(\nu\) gives \[\begin{aligned} \sum_{t=j}^{m-1}\int_C|a_t^+-a_t^-|^2\,d\nu &\le\int_C|\epsilon-a_j^-|^2\,d\nu\\ &=\int_C|\epsilon|^2\,d\nu-\nu(C)|a_j^-|^2 \le\nu(C). \end{aligned}\] The left side uses only some increments of the refining sequence; omitting the others preserves the bound. Multiply by the cell-constant value of \(R_j-R_{j-1}\), then sum over cells and \(j\). The weighted sum \(2\sum_{t<m}\int R_t|a_t^+-a_t^-|^2\,d\nu\) is consequently at most \(2\int R_{m-1}\,d\nu=2\mathbb E(MR_{m-1})\).

The scalar input is the finite-horizon form of Doob’s \(L_2\) maximal inequality (Doob 1953). We include its first-crossing proof:

\[\Big\|\max_{j\le m}\mathbb E_jM\Big\|_2\le2\|M\|_2.\] Indeed, if \(R=\max_{j\le m}\mathbb E_jM\), partitioning the event \(\{R\ge u\}\) according to its first crossing gives \(u\mathbb P(R\ge u)\le\mathbb E[M\mathbf1_{\{R\ge u\}}]\). Integrating in \(u\) yields \(\mathbb ER^2\le2\mathbb E(MR)\); Cauchy–Schwarz proves the claim. Thus \(2\mathbb E(MR_{m-1})\le4\mathbb EM^2\), proving (51).

We next compare a child window with its parent window, keeping the component \(n\) fixed. For a vertex \(s\), write \(C_s=\{u:s\preceq u\}\) for the cone of endpoints whose paths pass through \(s\). Define the signed endpoint measures \[\lambda'_{t+1}=\mathbb E_{t+1}(a_t^-\mu_n), \qquad \rho_{t+1}=\mathbb E_{t+1}((a_t^+-a_t^-)\mu_n).\] For \(s\in U_t\), the child sign history together with membership in \(C_s\) is measurable for the partition defining \(a_t^+\). Hence \[\mathbb E_{t+1}\int_{C_s}\epsilon_n\,d\mu_n =\mathbb E_{t+1}\int_{C_s}a_t^+\,d\mu_n =\lambda'_{t+1}(C_s)+\rho_{t+1}(C_s).\] Thus their sum represents the path part of \(x_{t+1}^n\) on the window \(U_t\). The error measure satisfies \(\|\rho_{t+1}\|_1\le q_{t,n}\). Let \(h_{t,n}=\mathbb E_th_n\), \(\lambda'_t=\mathbb E_t\lambda'_{t+1}\), and \(\rho_t=\mathbb E_t\rho_{t+1}\). Averaging gives, at each \(s\in U_t\), \[ x_t^n(s)=h_{t,n}(s)+\lambda'_t(C_s)+\rho_t(C_s). \tag{52}\]

Fix a history through time \(t\). On every cone \(C_s\) with \(s\in U_t\), truncation to \(A_t\) is fixed. Thus \(a_t^-\) is constant on that cone, including all completions of the fixed sign history. There is no cancellation within the cone in either \(\lambda'_t\) or \(\lambda'_{t+1}\). With \(\phi_s=|\lambda'_t(C_s)|\), fair binary conditioning therefore gives \[ |\lambda'_{t+1}|(C_s)\le2\phi_s. \tag{53}\] Choose another path decomposition, now only for the parent window: \[R_{U_t}x_t^n=d+B_n\theta, \qquad \|d\|_2+\|\theta\|_1\le2r_{t,n}, \qquad r_{t,n}=L_n(R_{U_t}x_t^n).\] If \(\beta=|\theta|+|\rho_t|\), then (52) gives \[ \phi_s\le|d(s)-h_{t,n}(s)|+\beta(C_s), \qquad \|\beta\|_1\le\|\theta\|_1+\mathbb E_tq_{t,n}. \tag{54}\]

Stop along each descending line at the first vertex \(s\in U_t\) where \(\beta(C_s)\ge\phi_s/2\). These first vertices form an antichain, so their endpoint cones are disjoint. At window vertices neither stopped nor below a stop, \(\phi_s\le2|d(s)-h_{t,n}(s)|\). By (53), the vector formed by the child’s path coordinates at these vertices has Hilbert norm at most \(4(\|d\|_2+\|R_{U_t}h_{t,n}\|_2)\). On the remaining part of the window, every contributing endpoint lies in a stopping cone. Its total \(|\lambda'_{t+1}|\)-mass is at most \(4\|\beta\|_1\), using the stopping inequality, the factor \(2\) in (53), and disjointness of the cones. Each path contributes there on an interval, of \(L_n\)-norm at most \(2\). This part therefore costs at most \(8\|\beta\|_1\). The error measure \(\rho_{t+1}\) costs at most \(2q_{t,n}\). Combining these estimates gives the explicit component bound \[ \begin{split} L_n(R_{U_t}x_{t+1}^n)\le{}& \|R_{U_t}h_{t+1,n}\|_2+4\|R_{U_t}h_{t,n}\|_2\\ &+16r_{t,n}+8\mathbb E_tq_{t,n}+2q_{t,n}. \end{split} \tag{55}\] Here \(4\|d\|_2+8\|\theta\|_1\le16r_{t,n}\).

We have reduced the vector estimate to square summation. Conditional Jensen and predictability of \(U_t\) give, for \(j=t,t+1\), \[\mathbb E\|R_{U_t}h_{j,n}\|_2^2 \le\mathbb E\|R_{U_t}h_n\|_2^2.\] The windows are disjoint on every outcome, so summing over \(t,n\) bounds either Hilbert energy by \(\mathbb E\sum_n\|h_n\|_2^2\). The same summation for \(q_{t,n}\) and \(\mathbb E_tq_{t,n}\) is bounded by \(4\mathbb E\sum_nM_n^2\), by (51) and Jensen. Square the five terms in (55), bounding their sum’s square by five times the sum of squares, and use (50). This proves (49) with an absolute constant. All sums over components involve nonnegative terms; Tonelli’s theorem justifies them even when the terminal vectors have infinitely many nonzero components. ◻

Proof. We first work in one component, suppressing \(n\) and restricting each set to its tree. At each terminal outcome choose \[W_m=H+B_nc, \qquad \|H\|_2+\|c\|_1\le2L_n(W_m),\] and let \(H_i=\mathbb E_iH\). For an endpoint \(s\), let \(\pi_i(s)\) be its deepest ancestor in \(A_i\), or \(\partial\) if none exists. Set \(\pi_i(\partial)=\partial\) and \(b_\partial=0\). Define \[c_r^i=\mathbb E_i\sum_{\pi_i(s)=r}c_s, \qquad v_i=\sum_{r\in A_i\cup\{\partial\}}|c_r^i|.\] The dummy coefficient is kept in this last sum even though its path vector vanishes. This ensures the exact refinement identity \[ \mathbb E_{i-1}\sum_{\pi_{i-1}(r)=q}c_r^i=c_q^{i-1} \quad(q\in A_{i-1}\cup\{\partial\}). \tag{57}\] The child window itself has the representation \[R_{\Delta_i}W_i=R_{\Delta_i}H_i+\sum_rc_r^iR_{\Delta_i}b_r.\]

Orient each refined coefficient by its old collapsed coefficient: \[\sigma_r=\operatorname{sign}(c^{i-1}_{\pi_{i-1}(r)}), \qquad c_r^i=\sigma_r(g_r-q_r),\qquad g_r,q_r\ge0,\] where sign at zero is chosen to be \(+1\), and \(g_r,q_r\) are the positive and negative parts of \(\sigma_rc_r^i\). The signs are known on the parent history. By (57), \[\beta_i:=\mathbb E_{i-1}\sum_rq_r =\tfrac12(\mathbb E_{i-1}v_i-v_{i-1})\ge0.\] The elementary inequality \((\mathbb E_{i-1}v_i-v_{i-1})^2 \le\mathbb E_{i-1}v_i^2-v_{i-1}^2\) therefore telescopes to \[ \sum_i\mathbb E\beta_i^2\le\tfrac14\mathbb Ev_m^2 \le\tfrac14\mathbb E\|c\|_1^2. \tag{58}\]

Fix a step and a parent history. For \(t\in \Delta_i\), all its descendant labels in \(A_i\) have the same old ancestor and therefore the same sign \(\sigma_t\). Set \(g(t)=\sum_{r\succeq t}g_r\) and \(q(t)=\sum_{r\succeq t}q_r\), with the dummy omitted. Averaging the coordinate formula gives \[W_{i-1}(t)=H_{i-1}(t)+ \sigma_t(\mathbb E_{i-1}g(t)-\mathbb E_{i-1}q(t)).\] Decompose the leaked parent window once more: \[Y=R_{\Delta_i}W_{i-1}=L+B_nd, \qquad \|L\|_2+\|d\|_1\le2L_n(Y).\] There are two equally likely children and \(g(t)\ge0\), so \[g(t)\le2\mathbb E_{i-1}g(t) \le2\left(|L(t)|+|H_{i-1}(t)|+ \sum_{s\succeq t}|d_s|+\mathbb E_{i-1}q(t)\right).\] Call \(t\in \Delta_i\) large when \(g(t)>4(|L(t)|+|H_{i-1}(t)|)\). At a large node, \[g(t)\le4\left(\sum_{s\succeq t}|d_s|+ \mathbb E_{i-1}q(t)\right).\] The minimal large nodes form an antichain. Mark every coefficient label in \(A_i\) below one of them. Their descendant sets are disjoint, so the total marked positive mass is at most \(4(\|d\|_1+\beta_i)\). The total negative mass satisfies \(\sum_rq_r\le2\beta_i\), again by fair binary averaging. Using \(L_n(R_{\Delta_i}b_r)\le2\), all marked positive terms together with all negative terms cost at most \(8\|d\|_1+12\beta_i\).

It remains to estimate the unmarked positive part. Its coordinate is zero at every large vertex, since all descendant labels there are marked. At other vertices its absolute value is at most \(g(t)\le4(|L(t)|+|H_{i-1}(t)|)\). Its Hilbert norm is consequently at most \(4(\|L\|_2+\|R_{\Delta_i}H_{i-1}\|_2)\). Including the child Hilbert term yields \[ \begin{split} L_n(R_{\Delta_i}W_i)\le{}&\|R_{\Delta_i}H_i\|_2 +4\|R_{\Delta_i}H_{i-1}\|_2+12\beta_i\\ &+16L_n(R_{\Delta_i}W_{i-1}). \end{split} \tag{59}\] The last coefficient follows from \(4\|L\|_2+8\|d\|_1\le16L_n(Y)\).

For \(j=i-1,i\), conditional Jensen and the fixed, disjoint windows give \(\sum_i\mathbb E\|R_{\Delta_i}H_j\|_2^2\le\mathbb E\|H\|_2^2\). Squaring the four terms in (59), using four times their squared sum, and applying (58), gives \[\sum_i\mathbb E L_n(R_{\Delta_i}W_i)^2 \le68\mathbb E\|H\|_2^2+144\mathbb E\|c\|_1^2 +1024\sum_i\mathbb E L_n(R_{\Delta_i}W_{i-1})^2.\] The first two terms are at most \(144\mathbb E(\|H\|_2+\|c\|_1)^2 \le576\mathbb E L_n(W_m)^2\). Finally sum over the components and replace \(576\) by \(1024\) in the terminal term. This proves (56). ◻

Proof. Scaling and adding representations in (60) proves homogeneity and the triangle inequality. Every representation satisfies \[\|z\|_2\le\|h\|_2+\sqrt{n+1}\|\lambda\|_1 \le\sqrt{n+2}\bigl(\|h\|_2^2+\|\lambda\|_1^2\bigr)^{1/2};\] the choice \(h=z\) gives the upper bound. This proves positive definiteness, completeness, and equivalence to the Hilbert norm. The dual and bidual therefore have the same underlying continuous linear functionals as for \(H_n\), so \(Z_n\) is reflexive. For each coordinate, \(|z_t|\le\|h\|_2+\|\lambda\|_1\), which gives the second assertion of (61). A single atomic mass represents \(p_t\) with cost one. For \(e_t\) use its Hilbert representation. At the root, the representation \(e_o=e_o/2+J_n(\delta_o/2)\) attains the matching cost \(1/\sqrt2\).

The quotient map \((h,\lambda)\mapsto h+J_n\lambda\) has domain \(H_n\oplus_2\ell_1(T_n)\). Pulling the functional \(f\) back gives \((f,J_n^*f)\), whose norm is \(\bigl(\|f\|_2^2+\|J_n^*f\|_\infty^2\bigr)^{1/2}\). Indeed the upper bound is Cauchy–Schwarz, and the reverse bound follows by choosing the Hilbert vector in the direction of \(f\), an atomic coordinate arbitrarily close to the supremum for \(J_n^*f\), and then optimizing their two nonnegative lengths. The pullback and quotient functional norms agree: a representation bounds the functional by its cost, and representations can have cost arbitrarily close to \(|z|_n\). Since \((J_n^*f)(t)=\sum_{s\preceq t}f_s\), this proves (62) without requiring a maximizing endpoint.

For completeness of the outer sum, a Cauchy sequence converges in each \(Z_n\); its Cauchy bounds pass first to every finite sum and then to the whole sum, proving membership and convergence of the limit. A bounded functional on this sum restricts to functionals \(f_n\in Z_n^*\) with \(\sum_n\|f_n\|^2<\infty\). To prove the bound on this sum, test on finite collections of almost norming unit vectors multiplied by arbitrary Euclidean unit scalar vectors. Conversely such \((f_n)\) defines a functional by Cauchy–Schwarz. Density of finite component sums shows that these constructions are inverse and isometric. Applying the same description to the bidual and using component reflexivity proves reflexivity of \(X_{\mathrm q}\). Finally truncate the outer sum and then the Hilbert coordinates in each remaining component to obtain finite node support. ◻

Proof. The triangle inequality bounds the absolute sum in every fiber of \(r\), so \(R_A\) contracts each \(\ell_1\) component and hence \(L\). Coordinate restriction gives \(P_Ap_t=p_{r(t)}\) and thus \(P_AJ=JR_A\). Projecting \(h\) orthogonally in \(H\) and folding \(\lambda\) in (63) proves contractivity of \(P_A\). The remaining bounds follow from \(Q_A=I-P_A\) and \(\|J\lambda\|_{\mathrm q}\le\|\lambda\|_L\). ◻

Proof. For each fixed nonterminal \(s\in T_n\), the coordinate vectors \(e_{s^\frown j}\) are weakly null in \(Z_n\), by its equivalence to \(H_n\), and therefore in \(X_{\mathrm q}\). Their norms and the norms of \(p_s\) are at least \(1/\sqrt2\), while \(\|p_s\|_{\mathrm q}\le1\). Suppose \(\overline\delta_N(t_0)>0\) and choose \(c>0\) with \(2c<\overline\delta_N(t_0)\). At \(u=p_s/N(p_s)\) choose a closed finite-codimensional subspace \(F\) such that \(N(u+t_0v)>1+2c\) for every \(v\in F\) with \(N(v)=1\). The normalized sequence \(v_j=e_{s^\frown j}/N(e_{s^\frown j})\) is weakly null, because its denominators are bounded below. Its image in \(X_{\mathrm q}/F\) is norm null. Approximation in \(F\), followed by normalization, gives vectors of \(F\cap S_N\) whose \(N\)-distance from \(v_j\) tends to zero. Consequently \(N(u+t_0v_j)>1+c\) for all sufficiently large \(j\).

For a convex function \(r\mapsto N(u+rv_j)\) with value one at zero, a lower bound greater than one at \(t_0\) persists for \(r\ge t_0\). Apply this with \(r=N(e_{s^\frown j})/N(p_s)\ge a/(\sqrt2 b)\) to obtain \(N(p_{s^\frown j})\ge(1+c)N(p_s)\). Choosing successors for \(n\) steps gives \(b\ge (1+c)^n a/\sqrt2\), impossible for arbitrarily large \(n\). The asymptotic modulus is nonnegative: for each unit center one may restrict to the kernel of a norming functional. It is therefore zero. ◻

Proof. The Hilbert parallelogram identity and the scalar square inequality give \(C\ge\|\bar h\|_H^2+\|\widetilde h\|_H^2+\sum_nM_n^2\). On the other hand the displayed representation of \(x\) has squared cost \(\|P\bar h\|_H^2+\sum_nq_n^2\ge\|x\|_{\mathrm q}^2\). Subtract, using the orthogonal decomposition in \(H\), to prove (64). Now \(M_n^2-q_n^2\ge(M_n-q_n)^2\) and \[\tfrac12(\|Qh^+\|_H^2+\|Qh^-\|_H^2) =\|Q\bar h\|_H^2+\|Q\widetilde h\|_H^2.\] These identities imply (65). ◻

Proof. Set \(\alpha_t=V(t)-\sum_{c\text{ child of }t}V(c)\ge0\). Induction from the last level, using nonnegative sums, gives \(V(t)=\sum_{s\succeq t}\alpha_s\) and \(\sum_{t\in O}\alpha_t=B\). This induction is over the finite height, not over a finite set of children. Hence the required vector is \[\sum_{t\in O}\varepsilon_t\alpha_t (p_t-p_{r(t)}).\] The series is absolutely convergent in \(Z_n\) by Lemma 22 and has norm at most \(2B\). Continuity of the coordinates identifies its sum with the stated array. Finally \[\sum_c\min(V(c),W(c)) \le\min\Bigl(\sum_cV(c),\sum_cW(c)\Bigr) \le\min(V(t),W(t)),\] which proves the assertion about minima. ◻

Proof. Convexity gives \(\|x\|_{\mathrm q}\le R\), so the right side is defined. The case \(R=0\) is immediate. Choose the representations above with each squared cost at most \(R^2+\eta\), where \(\eta>0\). Then \(E\le D:=R^2+\eta-\|x\|_{\mathrm q}^2\).

Fix a component. For every fiber of \(r(t)\) choose its majority sign \(\varepsilon_r\) among the coefficients of both labeled arrays \(\lambda^+,\lambda^-\). If the positive and negative masses in a nondiscarded fiber are \(a_r,b_r\), its contribution to \(d_n\) is \(\min(a_r,b_r)\); for a discarded fiber its contribution is \((a_r+b_r)/2\ge\min(a_r,b_r)\). Thus the total minority mass over the two arrays is at most \(d_n\).

Outside \(A_n\), each descendant has the same \(r(t)\) as its ancestor. For \(\epsilon\in\{+,-\}\) define the majority and minority occupancies \[U^\epsilon(t)=\sum_{s\succeq t}(\varepsilon_{r(s)} \lambda_n^\epsilon(s))_+, \qquad V^\epsilon(t)=\sum_{s\succeq t}(-\varepsilon_{r(s)} \lambda_n^\epsilon(s))_+, \qquad V=V^++V^-.\] Here \(a_+=\max(a,0)\). All these arrays satisfy the child inequality of Lemma 25, and the total initial mass of \(V\) is at most \(d_n\). Since \(x\) vanishes outside the head, \[\varepsilon_{r(t)}(U^+(t)+U^-(t)-V(t)) =-h_n^+(t)-h_n^-(t).\] Let \(S(t)=\min(U^+(t),V(t))\). It is a stopped flow with initial mass at most \(d_n\), and \[0\le U^+(t)-S(t) \le (U^+(t)+U^-(t)-V(t))_+ \le |h_n^+(t)+h_n^-(t)|.\] Consequently the tail of \(y_n\), which agrees with that of \(x_n+y_n\), is the sum of the signed flows \(S,-V^+\) and the Hilbert vector \(h_n^++\varepsilon_r(U^+-S)\), all restricted to the outside. Lemma 25 gives \[|Q_Ay_n|_n\le4d_n+\|Q_Ah_n^+\|_2+ \|Q_A(h_n^++h_n^-)\|_2.\] Take the outer Euclidean norm and use (65). The three terms are at most \(4\sqrt D\), \(\sqrt{2D}\), and \(2\sqrt D\), respectively. Thus \(\|Q_Ay\|_{\mathrm q}\le(6+\sqrt2)\sqrt D\le8\sqrt D\). Let \(\eta\downarrow0\). No minimizing representation was assumed. ◻

Proof. Approximate \(x\) within \(\rho>0\) by \(x_0\) supported in a finite ancestral head \(A\). Theorem 26 bounds each \(Q_Ay_j\) by \(8\sqrt{(R+\rho)^2-\|x_0\|_{\mathrm q}^2}\). The sequence \((y_j)\) is bounded, and \(P_A\) has finite rank, so two distinct head projections can be arbitrarily close. Their difference and the two tail bounds give \(\varepsilon\le16\sqrt{(R+\rho)^2-\|x_0\|_{\mathrm q}^2}\). Let \(\rho\downarrow0\). ◻

Proof. It suffices to prove the assertion in one component, since the result can then be summed. Choose endpoint representations whose squared costs exceed \(|x\pm y|_n^2\) by at most \(\eta>0\), and use the notation of Lemma 24, omitting the component index. Put \(E=(|x+y|_n^2+|x-y|_n^2)/2+\eta-|x|_n^2\). Then \[\|Q\bar h\|_2^2+\|\widetilde h\|_2^2+M^2-q^2\le E.\] Orient each nondiscarded fiber by the sign of its folded mean coefficient \(\nu_{r(t)}\), choosing \(+1\) when it is zero, and choose either sign in \(\{-1,1\}\) on a discarded fiber. Define half-weighted positive and negative endpoint masses by \[\alpha_t^\epsilon=\tfrac12(\varepsilon_{r(t)} \lambda_t^\epsilon)_+, \qquad \beta_t^\epsilon=\tfrac12(-\varepsilon_{r(t)} \lambda_t^\epsilon)_+, \quad\epsilon\in\{+,-\}.\] Let \(\beta=\beta^++\beta^-\). In a nondiscarded fiber the positive mass minus negative mass is the absolute folded coefficient. The discarded fiber contributes zero to \(q\). It follows that \(\|\beta\|_1\le M-q\le\sqrt E\).

For \(t\notin A_n\) put \(A^\epsilon(t)=\sum_{s\succeq t}\alpha_s^\epsilon\) and \(B^\epsilon(t)=\sum_{s\succeq t}\beta_s^\epsilon\), and let \(A_{\mathrm{tot}}=A^++A^-\), \(B_{\mathrm{tot}}=B^++B^-\). The endpoint identities give \[0=\bar h(t)+\varepsilon_{r(t)}(A_{\mathrm{tot}}(t)-B_{\mathrm{tot}}(t)), \qquad y(t)=\widetilde h(t)+\varepsilon_{r(t)} (A^+(t)-A^-(t)-B^+(t)+B^-(t)).\] For each sign put \(S^\epsilon=\min(A^\epsilon,B_{\mathrm{tot}})\) and \(V^\epsilon=A^\epsilon-S^\epsilon\). Then \(0\le V^\epsilon\le(A_{\mathrm{tot}}-B_{\mathrm{tot}})_+ \le|\bar h|\) outside the head. Each \(S^\epsilon\) is a stopped flow of initial mass at most \(\|\beta\|_1\), and the same bound applies to each \(B^\epsilon\). Lemma 25 estimates their four signed vectors by \(8\|\beta\|_1\) in total. The remaining terms have Hilbert norm at most \(\|Q\widetilde h\|_2+2\|Q\bar h\|_2\). Since \(y\) vanishes on the head, \[|y|_n\le\|Q\widetilde h\|_2+2\|Q\bar h\|_2 +8\|\beta\|_1\le12\sqrt E.\] Let \(\eta\downarrow0\), and sum the resulting component inequalities. Empty heads cause no exception: their entire mass is in the discarded fiber, and the same bounds hold. ◻

Proof. Choose \(x_0\) of finite node support with \(\|x-x_0\|_{\mathrm q}<\epsilon\), and a finite ancestral head \(A\) supporting it. Finite rank and boundedness give distinct \(i,j\) with \(\|P_Az\|_{\mathrm q}<\epsilon\) for \(z=(y_i-y_j)/2\). Convexity gives \(\|x\pm z\|_{\mathrm q}\le r\), while \(\|z\|_{\mathrm q}\ge d\). Apply (67) to \(x_0,Q_Az\). These vectors have endpoint norms at most \(r+2\epsilon\), and \(\|Q_Az\|_{\mathrm q}\ge d-\epsilon\). Let \(\epsilon\downarrow0\). ◻

Proof. Choose the endpoint representations with squared costs at most \((R+\eta)^2\). In the notation preceding Lemma 24 define the nonnegative endpoint measure \(m_n=(|\lambda_n^+|+|\lambda_n^-|)/2\). Both \(|\sigma_n|\) and \(|\tau_n|\) are bounded pointwise by \(m_n\), and \(\|m_n\|_1=M_n\). The lemma gives \[ \|Q\bar h\|_H^2+\|\widetilde h\|_H^2+\sum_nd_n^2 \le (R+\eta)^2-\|x\|_{\mathrm q}^2=:E. \tag{68}\]

Write \(S_t=\{s:s\succeq t\}\) for a descendant set. Mark a node \(t\notin A_n\) when \(|\sigma_n(S_t)|<m_n(S_t)/2\), and retain only the first marked nodes on each path outside the head. Every marked node has such a first ancestor, since it has finitely many ancestors. The retained descendant sets are pairwise disjoint.

Each retained set lies in a single fiber of \(r\). If disjoint subsets \(S_j\) lie in a nondiscarded fiber \(C\), then \[\sum_j\bigl(m_n(S_j)-|\sigma_n(S_j)|\bigr) \le m_n(C)-|\sigma_n(C)|.\] Indeed \(|\sigma_n(C)|\) is at most the sum of the displayed absolute sums plus the \(m_n\)-mass of the remaining set. This argument also holds for countably many subsets by absolute convergence. Discarded fibers are bounded by their whole \(m_n\)-mass. Summing over fibers yields \[\sum_{t\text{ first marked}} \bigl(m_n(S_t)-|\sigma_n(S_t)|\bigr)\le d_n, \qquad \sum_{t\text{ first marked}}m_n(S_t)\le2d_n.\] Split \(\tau_n=\tau_n^{\mathrm r}+\tau_n^{\mathrm k}\) by restricting the first part to the union of these descendant sets. Then \(\|\tau_n^{\mathrm r}\|_1\le2d_n\) and \[\|QJ\tau^{\mathrm r}\|_{\mathrm q} \le4\Bigl(\sum_nd_n^2\Bigr)^{1/2}.\] If an outside node has a first marked ancestor, the occupancy of \(\tau_n^{\mathrm k}\) there is zero. Otherwise it is unmarked, and \[|(J_n\tau_n^{\mathrm k})(t)| \le m_n(S_t)\le2|\sigma_n(S_t)|=2|\bar h_n(t)|.\] The last equality uses \(Qx=0\) in \(x=\bar h+J\sigma\). Thus \(QJ\tau^{\mathrm k}\in H\) and has Hilbert norm at most \(2\|Q\bar h\|_H\). Since \(y=\widetilde h+J\tau\), \[\|Qy\|_{\mathrm q} \le\|Q\widetilde h\|_H+2\|Q\bar h\|_H +4\Bigl(\sum_nd_n^2\Bigr)^{1/2} \le7\sqrt E.\] Let \(\eta\downarrow0\). ◻

Proof. If \(\|x\|_{\mathrm q}>1-\delta\), choose a finite-support \(x_0\) within \(\delta\) of \(x\), and an ancestral head \(A\) for \(x_0\). Then \(\|x_0\|_{\mathrm q}>1-2\delta\) and \(\|x_0\pm y_j\|_{\mathrm q}\le1+\delta\). Proposition 30 gives \[\|Q_Ay_j\|_{\mathrm q} <7\sqrt{(1+\delta)^2-(1-2\delta)^2} \le7\sqrt{6\delta}<c/2.\] Two bounded head projections are less than \(c\) apart, contradicting the assumed separation of the full vectors. ◻

Proof. Use representations with squared costs at most \(R^2+\eta\) and write \(E=R^2+\eta-\|x\|_{\mathrm q}^2\). In each nondiscarded fiber \(C_r\), orient the coefficients by the sign of the folded coefficient \(\nu_n(r)\), choosing \(+1\) when it is zero; use either sign in \(\{-1,1\}\) in a discarded fiber. Let \(g_n,b_n\) be half the total absolute masses of coefficients agreeing and disagreeing with this sign, respectively, counting both labeled endpoint representations. Then \(\|b_n\|_1\le d_n\). More precisely, if \(A_n\ne\varnothing\), all nodes have an ancestor in \(A_n\) and \(2\|b_n\|_1=d_n\); when \(A_n=\varnothing\), simply \(d_n=M_n\).

On the outside put \(G(t)=g_n(S_t)\) and \(B(t)=b_n(S_t)\). Constancy of the orientation along each outside subtree gives \[|(J_n\sigma_n)(t)|=|G(t)-B(t)|=|\bar h_n(t)|.\] In the labeled half-difference \(\tau_n=(\lambda_n^+-\lambda_n^-)/2\), remove every minority term and every majority term whose endpoint has an outside ancestor \(t\) with \(G(t)\le2B(t)\). Denote the removed and kept arrays by \(\tau_n^{\mathrm r},\tau_n^{\mathrm k}\). The first such stopping nodes have disjoint descendant sets. The removed majority mass is at most \(2\|b_n\|_1\); hence \(\|\tau_n^{\mathrm r}\|_1\le3d_n\). This accounting is on the labeled endpoint terms, before any cancellation in their half-difference, so it bounds that difference as well.

At a node having a stopping ancestor the kept occupancy vanishes. At every other outside node, \(G(t)>2B(t)\) and \[|(J_n\tau_n^{\mathrm k})(t)| \le G(t)\le2|G(t)-B(t)|=2|\bar h_n(t)|.\] Therefore \[|Qy_n|_n\le\|Q\widetilde h_n\|_2 +2\|Q\bar h_n\|_2+6d_n.\] Lemma 24 and the triangle inequality in the outer Euclidean sum give \(\|Qy\|_{\mathrm q}\le9\sqrt E\). Let \(\eta\downarrow0\). ◻

Proof. Approximate \(x\) within \(\gamma\) by a finite-support \(x_0\), and choose a finite ancestral head for it. The tail of every point in the lens has norm less than \(\rho/3\) by Proposition 32. The points in the lens have norm at most one, so in an infinite sequence two head projections are less than \(\rho/3\) apart. Their full distance is less than \(\rho\). ◻

Proof. Scale to \(\|x\|_{\mathrm q}=1\). Choose \(x_0\) of finite node support with \(\|x-x_0\|_{\mathrm q}\le\delta\) and an ancestral head \(A\) containing its support. Then \(\|x_0\pm u\|_{\mathrm q}\le1+2\delta\). Choose representations \(x_0\pm u=h^\pm+J\lambda^\pm\) with each squared cost at most \((1+3\delta)^2\). Using the notation above for these representations, let \(C\) be their average squared cost and set \[\mathcal D=C-\left(\|P\bar h\|_H^2+\sum_nq_n^2\right).\] The parallelogram identity and the scalar square inequality used in Lemma 24 give, before comparing the center representation with its norm, \[\begin{aligned} \|Q\bar h\|_H^2+\|\widetilde h\|_H^2+ \sum_n(M_n^2-q_n^2)&\le\mathcal D,\\ \tfrac12(\|Qh^+\|_H^2+\|Qh^-\|_H^2)+ \sum_nd_n^2&\le\mathcal D. \end{aligned}\] The second inequality uses \(M_n^2-q_n^2\ge d_n^2\) and the parallelogram identity for \(Qh^\pm\). The center representation has cost at least \(\|x_0\|_{\mathrm q}\), so \[ 0\le\mathcal D\le(1+3\delta)^2-\|x_0\|_{\mathrm q}^2 \le(1+3\delta)^2-(1-\delta)^2\le16\delta. \tag{69}\]

Orient each nondiscarded fiber \(C_r\) by the sign of \(\nu_n(r)\), using \(+1\) for a zero folded coefficient or a discarded fiber. This time use the full, rather than half, outside absolute masses: let \(p_n^\pm\) agree with the orientation and \(b_n^\pm\) disagree with it, and put \(B_n=\|b_n^++b_n^-\|_1\). The outside minority mass is at most the total cancellation bound, giving \(B_n\le2d_n\) and \[ \sum_nB_n^2\le4\mathcal D. \tag{70}\] For outside nodes let \(G(t)=(p_n^++p_n^-)(S_t)\) and \(V(t)=(b_n^++b_n^-)(S_t)\). Since the averaged represented vector vanishes there, \[|G(t)-V(t)|=|h_n^+(t)+h_n^-(t)|.\] Stop at the first outside nodes satisfying \(G(t)>0\) and \(V(t)\ge G(t)/2\). Their descendant sets are disjoint, and the total majority mass in them is at most \(2B_n\). Remove all these majority terms and all minority terms from the outside part of \(\lambda_n^+\); denote what remains by \(\lambda_n^{\mathrm k}\) and what was removed by \(\lambda_n^{\mathrm r}\). Then \(\|\lambda_n^{\mathrm r}\|_1\le3B_n\) and, by (70), \[\|\lambda^{\mathrm r}\|_L\le6\sqrt{\mathcal D}, \qquad \|QJ\lambda^{\mathrm r}\|_{\mathrm q}\le12\sqrt{\mathcal D}.\] Every outside node either has a stopping ancestor, has \(G(t)=0\), or satisfies \(V(t)<G(t)/2\). In all cases the kept occupancy is at most \(2|G(t)-V(t)|\) in absolute value. Thus \[\|QJ\lambda^{\mathrm k}\|_H \le2\|Q(h^++h^-)\|_H\le4\sqrt{\mathcal D}.\] The atomic terms whose endpoints belong to \(A\) have zero tail. The representation of \(x_0+u\), together with \(\|Qh^+\|_H\le\sqrt{2\mathcal D}\), now gives \[\|Qu\|_{\mathrm q}\le(12+4+\sqrt2)\sqrt{\mathcal D} \le18\sqrt{\mathcal D}\le72\sqrt\delta.\] All stopping decisions use a first ancestor, so infinite coefficient support and countably many children cause no change. The choice of \(A\) preceded the choice of \(u\), as required. ◻

Proof. Apply the proof of Theorem 26 in this single component, retaining the coefficient \(6+\sqrt2\) before its final rounding to \(8\). Here \(Q_Ay=y\). Explicitly, approximate endpoint representations of cost below \(1+e+\varepsilon\) give \(\kappa=(1+e+\varepsilon)^2-1\), Hilbert tails at most \(\sqrt{2\kappa}\), combined Hilbert tail at most \(2\sqrt\kappa\), and folded-mass loss at most \(\sqrt\kappa\) by Lemma 24. Because the head contains the root, no fiber is discarded: the combined minority mass is exactly this folded-mass loss. Grouping by the common truncated endpoint and clipping one majority occupancy against the combined minority occupancy gives two signed stopped flows of norm at most \(2\sqrt\kappa\) each, by Lemma 25. The remaining majority occupancy is bounded in Hilbert norm by \(2\sqrt\kappa\). Adding the original Hilbert tail gives \((6+\sqrt2)\sqrt\kappa\). These are the same decomposition and signed path representations used in that theorem, so no invariance of \(|\cdot|_n\) under arbitrary coordinate sign changes is needed. Let \(\varepsilon\downarrow0\). Finally \((1+e)^2-1\le3e\) and \(3(6+\sqrt2)^2=114+36\sqrt2<169\). ◻

Proof. The conditions defining \(Y_x\) involve only finitely many continuous coordinate functionals, so the assertion about codimension follows. Put \[a_n=|x_n|_n,\quad b_n=|x_n+y_n|_n,\quad c_n=|x_n-y_n|_n, \quad v_n=\frac{b_n^2+c_n^2}{2},\quad d_n=v_n-a_n^2.\] The triangle inequality gives \(a_n,|y_n|_n\le(b_n+c_n)/2\). Therefore \[ d_n\ge0,\qquad |y_n|_n^2\le v_n, \qquad \sum_na_n^2=1, \qquad \sum_nd_n\le(1+e)^2-1\le3e. \tag{73}\] Fix \(0<\eta\le1\). Write \[E_\eta=\{n:d_n>\eta a_n^2\}.\] On these components \(v_n=a_n^2+d_n\le(1+1/\eta)d_n\), and hence \[ \sum_{n\in E_\eta}|y_n|_n^2 \le3e(1+1/\eta). \tag{74}\] If \(a_n=0\) and \(y_n\ne0\), then \(d_n=v_n=|y_n|_n^2>0\), so \(n\in E_\eta\). If both vectors vanish in that component, its contribution is zero. Thus every remaining nonzero contribution has \(a_n>0\).

Now assume additionally that \(\eta\le1/4\). For \(n\notin E_\eta\) with \(a_n>0\), let \(m_n=(b_n+c_n)/2\). We have \(m_n\ge a_n\) and \[m_n^2+\left(\frac{b_n-c_n}{2}\right)^2=v_n \le(1+\eta)a_n^2.\] It follows that \[\frac{\max(b_n,c_n)}{a_n} \le\sqrt{1+\eta}+\sqrt\eta=1+\delta_\eta, \qquad \delta_\eta:=\sqrt{1+\eta}+\sqrt\eta-1 \le\frac32\sqrt\eta\le1.\] Apply Lemma 35 to \(x_n/a_n\) and \(y_n/a_n\). Summing its squared estimate gives \[ \sum_{n\notin E_\eta}|y_n|_n^2 \le169\delta_\eta\sum_{n\notin E_\eta}a_n^2 \le\frac{507}{2}\sqrt\eta. \tag{75}\] Take \(\eta=e^{2/3}\), which is at most \(1/4\) when \(e\le1/8\). Combining (74) and (75), \[\|y\|_{\mathrm q}^2 \le\frac{507}{2}e^{1/3}+3e+3e^{1/3} \le\frac{519}{2}e^{1/3}.\] The second assertion follows from \(519/2<17^2\). ◻

Proof. By finite-coordinate density, choose a finitely supported unit vector \(x^0\) with \(\|x-x^0\|_{\mathrm q}<e\). Such unit vectors are dense: approximate \(x\) by a nonzero finitely supported vector and normalize it. Use the finite coordinate exclusions for \(x^0\) from Proposition 36. If \(y\) satisfies the hypotheses of (76), then \[\|x^0+y\|_{\mathrm q},\|x^0-y\|_{\mathrm q}\le1+2e.\] Since \(2e\le1/8\), Proposition 36 proves (76). Given \(t>0\), choose \(0<e_t\le1/16\) with \(17(2e_t)^{1/6}<t\). For \(u\in S_Y\), both endpoint norms at \(tu\) cannot be at most \(1+e_t\), by (76). Each corresponding maximum is therefore greater than \(1+e_t\), giving the displayed infimum bound. ◻

Proof. Monotonicity and homogeneity follow directly by induction through the levels. The same induction and the triangle inequality in \(\ell_2\) give subadditivity. Squaring (77) and discarding its nonnegative cross term proves the lower bound in (78). The \(\ell_2\) norm of the \(q_t\) on level \(d\) is at most the \(\ell_2\) norm of the \(m_t\) on that level plus the \(\ell_2\) norm of the \(q_s\) on level \(d+1\). Iterating gives the middle bound; Cauchy–Schwarz gives the last one. Keeping only the child on the path to \(v\) at each step gives the path bound.

For the energy inequality, induction proves \(q_t(m+r)\ge(q_t(m)^2+q_t(r)^2)^{1/2}\). Indeed, let \(A\) and \(B\) be the \(\ell_2\) norms of the child values for \(m\) and \(r\). The induction hypothesis gives \[q_t(m+r)\ge m_t+r_t+\sqrt{A^2+B^2} \ge\sqrt{m_t^2+r_t^2}+\sqrt{A^2+B^2} \ge\sqrt{(m_t+A)^2+(r_t+B)^2}.\] The last inequality is the triangle inequality in \(\mathbb R^2\). Summing its square over the roots proves (79). These arguments apply to countably many children: all sums are nonnegative, and the \(\ell_2\) inequalities hold for finite partial sums and then by monotone convergence. This also shows that each cost commutes with increasing limits of nonnegative arrays. ◻

Proof. First we bound the output of a representation. Fix an output level \(d\) and a start level \(h\le d\). Segments starting at a single node \(t\) of level \(h\) contribute total absolute mass at most \(m_t(a)\) to level \(d\); starts at different nodes of level \(h\) have disjoint descendants. Therefore the \(\ell_2\) norm of this part of the output is at most \(\|m(a)|_{\mathbb N^h}\|_2\). Summing over \(h\le d\) bounds the norm on level \(d\) by \(\sqrt{n+1}\,M_n(m(a))\). Summing the squares over \(d\) gives \[ \|Pa\|_2\le(n+1)M_n(m(a)). \tag{82}\] Conversely, the singleton representation \(a_{tt}=x_t\) has cost at most \(\sqrt{n+1}\|x\|_2\). This proves finiteness and the comparisons. Homogeneity is immediate. Adding representations and using monotonicity and subadditivity of the cost proves the triangle inequality; the lower comparison proves definiteness. The norm is equivalent to the Hilbert norm, and hence is complete and reflexive.

The countable \(\ell_2\) sum is complete. Its dual is isometrically the \(\ell_2\) sum of the component duals: restriction of a functional gives the component functionals, finite collections of approximate norming vectors give the lower bound for their \(\ell_2\) norm, and Cauchy–Schwarz gives the reverse bound. Finite-component vectors are dense, so restrictions determine the functional. Applying this identification twice and using the reflexivity of every component proves reflexivity of \(X_S\).

A forest representation bounds each component norm by its component cost, and therefore bounds \(\|x\|_S\) by its total cost. Conversely, for \(\eta>0\) choose a representation in component \(n\) with cost at most \(\|x|_{\mathcal T_n}\|_{S,n}+\eta2^{-n}\). Minkowski’s inequality makes its total cost less than \(\|x\|_S+\eta\). Finally, finite support is dense in each component by (81), and finite-component vectors are dense in the sum. ◻

Proof. The absolute value of the output at \(v\) is at most the sum of start masses on the ancestor path to \(v\), which is bounded by \(M(m(a))\). Take the infimum over representations. A singleton or a root-to-node path has a representation of cost one, and its nonzero coordinate gives the matching lower bound.

Restrict each representing segment to \(C\). Since \(C\) contains every ancestor of its nodes, this restriction is empty or a segment with the same start. The start masses cannot increase. Taking infima proves \(\|P_Cx\|_S\le\|x\|_S\); the bound for \(I-P_C\) follows by the triangle inequality. The child vectors lie in a fixed component and are weakly null there, because its norm is equivalent to the Hilbert norm. They remain weakly null under its inclusion in \(X_S\). ◻

Proof. Fix \(\eta>0\) and put \(R=\max\{\|p+y\|_S,\|p-y\|_S\}+\eta\). Choose representations of \(p+y\) and \(p-y\) of cost at most \(R\). Halve every coefficient and concatenate the two lists, retaining the labels \(+\) and \(-\) that record their origins. These labels are independent of the signs of the coefficients. The combined list has output \(p\) and start mass \(m\) with \(M(m)\le R\). Figure 1 illustrates the two uses of a matched pair.

A first node outside \(C\) means a node \(s\notin C\) whose parent lies in \(C\), or a root outside \(C\). Every segment is then of exactly one of three kinds: it lies wholly in \(C\); it contains one first node outside \(C\); or its start lies strictly below a first node outside \(C\). The second kind includes segments starting at that first node. At each such \(s\), match the positive and negative weights of all segments containing \(s\) in pieces of equal absolute weight. Their total positive and total negative masses are finite and equal, since their sum is the output \(p_s=0\). For completeness, place the positive weights consecutively on an interval of their total length, do the same with the absolute negative weights, and use the intersections of these intervals as matching pieces. This produces a countable matching that uses every weight. A segment contains at most one first node outside \(C\), so the matchings do not interfere.

In each matched pair, both starts are ancestors of its first outside node \(s\). Call the piece with deeper start the later piece, breaking a tie arbitrarily. Define \(r_t\) to be the sum of the sizes of later pieces starting at \(t\), together with the sizes of all original pieces starting strictly below a first node outside \(C\). These are disjoint portions of the list; hence \(0\le r\le m\).

We first recover \(p\) at cost at most \(M(m-r)\). Delete all pieces starting strictly below a first outside node. For a matched pair with coincident starts, discard both pieces. Otherwise discard the later piece and shorten the earlier piece to end immediately before the later start. The remaining part lies in \(C\). On \(C\) the removed overlapping parts had equal opposite coefficients and therefore canceled. Segments wholly inside \(C\) are unchanged. The modified list consequently represents \(p\), and its start masses are at most \(m-r\). By (79), \[ \|p\|_S^2+M(r)^2\le M(m-r)^2+M(r)^2\le M(m)^2. \tag{84}\]

It remains to control \(y\) by the saved mass. The output of the \(+\) list outside \(C\) is \(y/2\). We represent it using start masses at most \(4r\). Include each \(+\) piece starting strictly below a first outside node without change. For a \(+\) piece in a pair at \(s\), let \(t\) be the pair’s later start and \(v\) the endpoint of this piece. Replace its exterior part by the segment \([t,v]\) with its original coefficient. If \(t\in C\), also include the segment from \(t\) to the parent of \(s\) with the opposite coefficient. These two segments leave precisely the desired output from \(s\) to \(v\). If \(t=s\), no subtraction is needed. Each pair has at most two \(+\) pieces, and each uses at most two segments of the matched size, all starting at \(t\). Their total start mass is therefore at most four times the later piece’s mass. No segment wholly in \(C\) is needed. Thus \[ \|y\|_S\le2M(4r)=8M(r). \tag{85}\]

All the modified lists are legitimate: their absolute start masses have finite cost, and each coordinate has only finitely many ancestors. Combining (84) and (85) gives \(\|p\|_S^2+\|y\|_S^2/64\le R^2\). Finally let \(\eta\downarrow0\). ◻

Proof. Choose a finite-support \(p\) arbitrarily close to \(x\), let \(C\) be its finite ancestral closure and set \(F=\ker P_C\). Theorem 41 gives, uniformly for \(y\in F\) with \(\|y\|_S\ge1\), \[\max_{\pm}\|x\pm ty\|_S \ge\sqrt{\|p\|_S^2+t^2/64}-\|x-p\|_S.\] As \(\|x-p\|_S\to0\), the right side tends to \(\sqrt{1+t^2/64}\). Coordinate continuity makes \(F\) closed and finite-codimensional, proving the assertion. ◻

Proof. At each fixed level of \(T_h\), the singleton segments form an incomparable family. Hence the squared norm of \(g\) dominates the sum of \(|g(v)|^2\) over that level. Summing over the \(h+1\) levels gives the lower bound in (87). For the upper bound, every segment has at most \(h+1\) nodes, so Cauchy–Schwarz and disjointness give \[\sum_i\left|\sum_{v\in S_i}g(v)\right|^2 \leq(h+1)\sum_i\sum_{v\in S_i}|g(v)|^2 \leq(h+1)\|g\|_2^2.\] Thus \(Y_{\mathrm I,h}\) is isomorphic to the Hilbert space \(\ell_2(T_h)\) and is reflexive.

A test family separates into families on the individual trees. Conversely, the union of finitely many such families, one in each of distinct trees, is incomparable. Choosing each component family arbitrarily close to its supremum proves \[\|g\|_{Y_{\mathrm I}}^2 =\sum_h\|g|_{T_h}\|_{Y_{\mathrm I,h}}^2\] for finitely supported \(g\), and then the asserted sum decomposition follows by completion. For completeness, the dual of an \(\ell_2\)-sum of Banach spaces is the \(\ell_2\)-sum of their duals: Cauchy–Schwarz bounds the sum of the pairings, while finitely many nearly norming component vectors with suitable scalar weights give the opposite norm inequality. Restrictions to the components determine a functional by density. Applying this description twice, componentwise reflexivity shows that the canonical map into the bidual is onto. Thus \(Y_{\mathrm I}\) is reflexive, as is its dual \(X_{\mathrm I}\). ◻

Proof. Every atom has norm at most one, by Cauchy–Schwarz and (86), and for every \(g\in Y_{\mathrm I}\), \[\sup_{A\text{ atom}} A(g)=\|g\|_{Y_{\mathrm I}}.\] If the closed convex hull of the atoms missed a point of the dual unit ball, Hahn–Banach separation would give a separating functional in \(X_{\mathrm I}^*=Y_{\mathrm I}\), using Proposition 43. The displayed identity rules out such a separation. Each atom has finite support, which gives the density assertion. ◻

Proof. The trees (89) exist at every finite height and satisfy Lemma 3 with path bounds and increment lower bound all equal to one. More explicitly, for an equivalent norm \(N\) with \(a\|u\|_{\mathrm I}\leq N(u)\leq b\|u\|_{\mathrm I}\), that lemma uses \(t_0=a/(2b)\). If \(\delta=\overline\delta_N(t_0)>0\), its finite-codimensional approximation and convexity argument selects a child at every nonterminal node for which \[N(v_{t_j})\geq (1+\delta/4)N(v_s).\] Iteration to height \(h\) gives \(a(1+\delta/4)^h\leq b\), a contradiction for sufficiently large \(h\). ◻

Proof. The nonempty intersections of an incomparable segment family with \(E\) are again incomparable segments; the same is true of the portions outside \(E\). Here prefix closure ensures that a segment is cut into an initial part and a terminal part. Applying these restricted tests to the original array proves both contraction inequalities on \(Y_{\mathrm I}\), first on finite arrays and then on the completion. The adjoints are contractions and act as the stated coordinate projections on each \(e_v\), hence on \(X_{\mathrm I}\) by Lemma 44. The range of \(P_E\) is contained in the span of the finitely many \(e_v\) with \(v\in E\). Finally, approximate \(x\) by a finitely supported vector and let \(E\) contain its support and all its ancestors. The contraction bound for \(Q_E\) gives the last assertion. ◻

Proof. Convexity gives \(\|x\|_{\mathrm I}\leq1\). Choose \(f\in Y_{\mathrm I}\) supported on \(E\) such that \[\|f\|_{Y_{\mathrm I}}\leq1, \qquad x(f)=\|x\|_{\mathrm I}.\] Indeed, Hahn–Banach gives a norming element of \(X_{\mathrm I}^*\); reflexivity identifies it with an element of \(Y_{\mathrm I}\), and applying \(P_E\) preserves its pairing with \(x\) without increasing its norm. If \(x=0\), we may simply take \(f=0\).

For an atom with representation (88), define \[p_i^A=z_{S_i}(f),\qquad d(A)=1-A(f)=1-\sum_i a_i p_i^A\geq0.\] Both coefficient vectors \((a_i)\) and \((p_i^A)\) have Euclidean norm at most one. The identity \[2d(A)=\sum_i(a_i-p_i^A)^2 +\left(1-\sum_i a_i^2\right) +\left(1-\sum_i(p_i^A)^2\right)\] therefore gives \[ \sum_i(a_i-p_i^A)^2\leq2d(A), \qquad 1-\sum_i(p_i^A)^2\leq2d(A). \tag{91}\] Replace the coefficients on the terminal pieces by setting \[\widehat A=\sum_i p_i^A z_{S_i\setminus E}.\] Those nonempty terminal pieces remain incomparable. The atom norm bound and (91) imply \[ \|\widehat A-Q_EA\|_{\mathrm I}\leq\sqrt{2d(A)}. \tag{92}\]

Let \(\mathcal R\) be the minimal nodes of \(\mathcal T\setminus E\). They form an antichain and their descendant subtrees partition the complement. If \(p_i^A\ne0\) and \(S_i\setminus E\) is nonempty, the original segment meets \(E\) and exits it through exactly one \(q\in\mathcal R\). For a fixed \(q\), at most one segment in an incomparable family can do this, since two such segments would both contain \(q\). We may write \[ \widehat A=\sum_{q\in\mathcal R}r_q^A z_{[q,t_q^A]}. \tag{93}\] Only finitely many coefficients are nonzero. When \(r_q^A\ne0\), let \(s_q^A\in E\) be the starting node of the original segment and let \(t_q^A\succeq q\) be its terminal node. Then \(r_q^A\) is the sum of \(f\) from \(s_q^A\) to the predecessor of \(q\). For zero coefficients choose \(t_q^A=q\); no starting node is needed.

We next show that opposite coefficients in two atoms consume their norming defects: \[ \sum_{q:\,r_q^Ar_q^{A'}<0}|r_q^Ar_q^{A'}| \leq \frac12\left(1-\sum_i(p_i^A)^2 +1-\sum_i(p_i^{A'})^2\right) \leq d(A)+d(A'). \tag{94}\] For any exit occurring in this sum, the two starting nodes lie on the same ancestor path of \(q\). They are distinct, since identical starts would give identical sums of \(f\), rather than opposite signs. Consider first the exits for which \(s_q^A\prec s_q^{A'}\). Shorten the corresponding segment of \(A\) to the interval from \(s_q^A\) to the predecessor of \(s_q^{A'}\). This interval is nonempty and is a subsegment of the original segment. Distinct exits correspond to distinct segments, so all these shortenings can be made simultaneously. Every modified family is still incomparable. Its corresponding squared evaluation changes from \((r_q^A)^2\) to \((r_q^A-r_q^{A'})^2\), an increase of \[(r_q^{A'})^2+2|r_q^Ar_q^{A'}| \geq2|r_q^Ar_q^{A'}|.\] All its squared evaluations sum to at most one. Thus the sum of these products is at most \(\tfrac12(1-\sum_i(p_i^A)^2)\). For the exits with the opposite ordering, shorten the segments of \(A'\) instead and obtain the other half of (94). The final inequality follows from (91).

We have controlled the replacement error for one atom and opposite exit coefficients for two atoms. It remains to apply these facts to convex combinations approximating the two endpoints \(x+y\) and \(x-y\). Fix \(\varepsilon>0\). By Lemma 44, choose a finite probability distribution of atoms for each endpoint whose mean differs from that endpoint in norm by at most \(\varepsilon\). Choose \(\sigma\in\{-1,1\}\) with equal probabilities, then choose \(A\) from the distribution for \(x+\sigma y\). Consequently \[\|\mathbb E A-x\|_{\mathrm I}\leq\varepsilon, \qquad \|\mathbb E(\sigma A)-y\|_{\mathrm I}\leq\varepsilon, \qquad 0\leq\bar d:=\mathbb E d(A) \leq1-\|x\|_{\mathrm I}+\varepsilon.\] Contractivity of \(Q_E\), the identity \(Q_Ex=0\), and Cauchy–Schwarz applied to (92) give \[ \|\mathbb E\widehat A\|_{\mathrm I} \leq\varepsilon+\sqrt{2\bar d}, \qquad \|\mathbb E(\sigma\widehat A)-Q_Ey\|_{\mathrm I} \leq\varepsilon+\sqrt{2\bar d}. \tag{95}\]

Put \(m_q=\mathbb E r_q^A\) and \(b_q=\mathbb E|r_q^A|\). Apply (94) to independent copies \(A,A'\) of the unconditional atom distribution. With \(r_+=\max(r,0)\) and \(r_-=\max(-r,0)\), independence gives \[2\sum_q\mathbb E[(r_q^A)_+]\,\mathbb E[(r_q^A)_-] \leq2\bar d.\] Since \(b_q^2-m_q^2=4\mathbb E[(r_q^A)_+]\,\mathbb E[(r_q^A)_-]\), \[ \sum_q b_q^2\leq\sum_qm_q^2+4\bar d. \tag{96}\] At this fixed \(\varepsilon\), the distributions are finite and their atoms have finite support. Thus all sums of \(m_q\) and \(b_q\) above are finite, even though \(\mathcal R\) can be infinite.

An array supported on the antichain \(\mathcal R\) has \(Y_{\mathrm I}\)-norm equal to its Euclidean norm: a segment meets at most one of its support nodes, and singleton segments realize the Euclidean test. The coefficient of \(\mathbb E\widehat A\) at \(q\) is \(m_q\), so duality gives \[ \left(\sum_qm_q^2\right)^{1/2} \leq\|\mathbb E\widehat A\|_{\mathrm I}. \tag{97}\] For \(g\in Y_{\mathrm I}\) with \(\|g\|_{Y_{\mathrm I}}\leq1\), define \[L_q=\sup_{t\succeq q}|z_{[q,t]}(g)|.\] For each finite subset of exits, choose segments approaching these suprema. The chosen segments are incomparable, so their squared evaluations sum to at most one. Passing to the suprema and then taking the supremum over finite subsets proves \(\sum_qL_q^2\leq1\); no supremum need be attained. It follows that \[|\mathbb E(\sigma\widehat A)(g)| \leq\sum_qb_qL_q \leq\left(\sum_qb_q^2\right)^{1/2}.\] Taking the supremum over \(g\), and using (95)–(97), yields \[\|Q_Ey\|_{\mathrm I} \leq\varepsilon+\sqrt{2\bar d} +\left((\varepsilon+\sqrt{2\bar d})^2+4\bar d\right)^{1/2}.\] The right-hand side is increasing in \(\bar d\). Substitute \(\bar d\leq1-\|x\|_{\mathrm I}+\varepsilon\) and let \(\varepsilon\downarrow0\). The two limiting contributions are \(\sqrt{2(1-\|x\|_{\mathrm I})}\) and \(\sqrt{6(1-\|x\|_{\mathrm I})}\), proving (90). ◻

Proof. Fix \(0<\eta<c\) and use Lemma 46 to choose \(E\) with \(\|Q_Ex\|_{\mathrm I}<\eta\). The sequence \((u_j)\) is bounded, since \(u_j=((x+u_j)-(x-u_j))/2\). The range of \(P_E\) is finite dimensional, so there are distinct \(j,k\) such that, for \(w=(u_j-u_k)/2\), one has \(\|P_Ew\|_{\mathrm I}<\eta\). The separation assumption gives \(\|w\|_{\mathrm I}\geq c\). Moreover, \(x+w\) is the midpoint of \(x+u_j\) and \(x-u_k\), and \(x-w\) is the midpoint of the other two endpoints. Hence \(\|x\pm w\|_{\mathrm I}\leq1\). Therefore \[\|P_Ex\pm Q_Ew\|_{\mathrm I}\leq1+2\eta, \qquad \|Q_Ew\|_{\mathrm I}\geq c-\eta.\] Apply Theorem 47 after dividing both vectors by \(1+2\eta\): \[\frac{c-\eta}{1+2\eta} \leq K\sqrt{1-\frac{\|P_Ex\|_{\mathrm I}}{1+2\eta}}.\] Since \(\|P_Ex-x\|_{\mathrm I}<\eta\), letting \(\eta\downarrow0\) gives \(c\leq K\sqrt{1-\|x\|_{\mathrm I}}\), which is (98). ◻