Midpoint lenses in segment spaces

Proof. Singleton tests give \|u\|_2\le\rho(u). On a component of height h, each segment has at most h+1 vertices. Thus Cauchy–Schwarz and disjointness give

Write J_h for the completion on that component. Segment families from different components combine freely, so

Each J_h is isomorphic to Hilbert space. The dual of an \ell_2 sum is isometrically the \ell_2 sum of the duals: Cauchy–Schwarz proves one inequality, and testing on almost norming unit vectors in finitely many components, with proportional weights, proves the other. Applying this identity twice proves reflexivity. One may also pass from reflexivity of J to that of its dual through the canonical maps. Write \iota_J:J\to J^{**} and \iota_X:X\to X^{**}. For \Phi\in X^{**}, define x=\Phi\circ\iota_J\in J^*=X. Every f\in X^*=J^{**} is \iota_J(p) for some p\in J, so

Thus \iota_X is onto. In particular X^*=J canonically. If the coordinate functionals did not have dense span in X, Hahn–Banach would give a nonzero element of X^*=J annihilating them. All its coordinates would vanish, a contradiction. Countability of \mathcal F now proves separability; the independent coordinates prove infinite dimensionality.

A segment meets H in a prefix and its complement in a suffix. Restricting disjoint segment tests to either set therefore proves contractivity on J. The adjoints give precisely the coordinate projections on X, and remain contractive. Finally, approximate x by a finite coordinate combination. Once H contains its support, contractivity bounds the discarded tail of x by that approximation error. This proves convergence along an ancestral exhaustion.□

Proof. On a vector supported outside H, every segment test reduces to a segment in one cone. Grouping a disjoint family by its gates gives the upper estimate for the first identity; combining almost maximizing families from finitely many distinct cones gives the reverse estimate. Pass to the completion. A functional supported outside H has exactly the norm of its restriction to Q_HJ, because Q_H is contractive. Duality for the \ell_2 sum therefore proves the second identity.

A segment contains at most one gate, so \rho(u)=\|u\|_2 for vectors supported on G_H. Testing x on such finite vectors proves the gate-coordinate contraction.□

Proof. A segment intersects H in a prefix and its complement in a suffix. Restricting every member of a disjoint testing family proves the primal contractions. Their adjoints preserve the coordinate span, hence its closure, and give the asserted dual contractions. Applying Q_H to finite approximants proves density in its range. An exhaustion eventually fixes each finite coefficient vector; contractivity extends convergence to the completion. Finally the norm of x=P_Hx can be computed using only P_HJ_\infty, whose unit ball is compact. A maximizer is a unit test when x\ne0.□

Proof. There are finitely many disjoint segment families in H. For each family, its atoms form the continuous image of a Euclidean unit ball. Their union is compact, as is its finite-dimensional convex hull C. For a test f=P_Hf, the support function of C is precisely \rho(f), by optimizing the coefficients in each family. Finite-dimensional separation identifies C with the dual unit ball of P_HJ_\infty; contractivity identifies this with the inherited ball of P_HB_\infty. Carathéodory’s theorem gives |H|+1 atoms. Equivalently, its elementary proof repeatedly removes a term from any longer convex combination using an affine dependence among its points, keeping the coefficients nonnegative and their sum one.

For the gauge assertion, every displayed representation bounds the norm by \sum_j\lambda_j. Conversely, write x=r(x/r), where r=\|x\|>0, and represent x/r\in C; x=0 is immediate. For the global gauge, choose an ancestral H containing the support of x. Projection P_H sends every global atom to an atom in H: intersect its disjoint segments with H and discard empty intersections. Thus every global representation projects to a local representation with no larger total weight. Local representations are already global ones, so the two gauges agree on every finite coefficient vector. The completion identification follows from (3.1).

For a completed unit vector, Lemma 3.1 gives P_Hx\to x with each P_Hx in the finite atomic ball. Thus the norm-closed convex hull of all finite atoms is the completed unit ball. This last assertion is an approximation statement; the finite exact representations are used only after a finite projection.□

Proof. Each exterior part of a segment lies in one cone. Grouping tests by cones gives one inequality in (3.2); joining nearly maximizing disjoint families from the finitely many nonzero cones gives the other. Completion gives the displayed primal sum. The norm of Q_Hx is exactly that of its restriction to Q_HJ_\infty, by contractivity. The dual norm of an \ell_2 sum is the \ell_2 sum of the restriction norms: Cauchy–Schwarz proves the upper bound, and finitely many almost norming tests with proportional weights give the lower bound. This proves (3.3) without any assertion that B_\infty is the full dual. Finally a segment meets the antichain G_H in at most one point, so tests supported there have precisely their Euclidean norm. Testing x on them gives the contraction. The same argument shows that x\mapsto(x(e_v))_{v\in A} is a contraction to \ell_2(A) for every antichain A, not only for a gate set.□

Proof. There are finitely many disjoint segment families in K. For each, its atoms are the continuous image of a Euclidean unit ball, so the union is compact. Its convex hull is compact in the finite-dimensional space P_KX. For a test p supported in K, optimizing the atom coefficients shows that the support function of this convex hull is \rho(p). In computing \rho(p), all segment families can be restricted to K. Finite-dimensional separation therefore identifies the hull with the dual unit ball on that coordinate space. By contractivity of P_K, this is precisely the inherited unit ball of P_KX.

To obtain the stated bound on the number of atoms, consider a convex representation with more than |K|+1 positive weights. Its atoms are affinely dependent, so there are real numbers \theta_i, not all zero, with \sum_i\theta_i=0 and \sum_i\theta_i A_i=0. Vary the weights along this direction until the first weight reaches zero; all weights remain nonnegative, and both their sum and their represented vector are unchanged. Repeating leaves at most |K|+1 terms.

Likewise, for x=P_Hx the dual pairing may be restricted to P_HJ without altering its norm. That space is finite dimensional, so its compact unit ball contains a norming test, of norm one when x\ne0.□

Proof. The coefficient estimate puts every atom in B_X. Conversely, for x\in B_X, choose an increasing finite ancestral exhaustion (K_n). Contractivity gives P_{K_n}x\in B_X, and Lemma 4.1 puts each P_{K_n}x in \operatorname{conv}\mathcal A. The convergence P_{K_n}x\to x proves the reverse inclusion. For a nonempty segment S, its functional has norm at most one by the coefficient estimate, and evaluation on e_v, for any v\in S, gives equality: the singleton vector e_v has \rho(e_v)=1.

There is also a direct separation proof of the completed hull identity. Optimizing the atom coefficients gives \sup_{A\in\mathcal A}A(p)=\rho(p) for every p\in J. Since X^*=J, separation of any point of B_X from the norm-closed convex hull would give x(p)>\rho(p), contradicting \|x\|\le1. This second proof uses reflexivity; the finite-projection proof only uses the coordinate completion and its contractive exhaustion.□

Proof. At a gate with opposite nonzero sums b,b', the two heads end at the same parent v. They are therefore nested and cannot coincide. Suppose the unprimed head [a,v] is longer than the primed head [a',v]. Split the unprimed segment immediately before a', retaining its own endpoint in the second piece. Since p vanishes outside H, the two new sums are b-b' and b'. The gain in its squared test is

Figure 1 distinguishes the two original segments from the two pieces created by the split.

Do this at every opposite-sign gate, in the family with the longer head. Within each family each original segment is cut at most once, because it crosses only one gate. Disjointness is preserved, so the total gain in each family is at most its deficit L. Summing proves (5.1).□

Proof of Theorem 1.1. Put r=\|x\|. If r=0, contractivity gives \|Q_Hy\|\le\|y\|\le R, which proves the assertion. Assume r>0. First suppose that y has finite support, and enlarge H to a finite ancestral set K containing that support. Choose a head test p of norm one such that x(p)=r. By the finite atomic-ball formula, choose finite laws of represented vectors

with means x+y and x-y. Choose a sign \sigma with equal probabilities and then the corresponding law. Thus \mathbb EA=x and \mathbb E(\sigma A)=y.

For each outcome put

Both V and L are nonnegative. Since \mathbb E\sum_i a_i b_i=r\mathbb EA(p)=r^2, expansion gives

The atom coefficient estimate also gives \|Q_H(A-A^\circ)\|\le\sqrt V.

Let G be the gates outside H which occur in the finite representations. At a gate g, write b(g) for the coefficient of the crossing segment in A^\circ, and put it equal to zero when there is no crossing segment. Every segment wholly outside H has replaced coefficient zero. If W=Q_H(A-A^\circ), then

Indeed, the gates are incomparable, so each segment contains at most one gate, and disjointness assigns different gates to different coefficients. Since \mathbb EA(e_g)=x(e_g)=0,

The simultaneous opposite-head splitting argument, applied to the test rp, gives for two independent outcomes

Here the available squared-test budget is r^2; this is the only change from the norm-one version. Consequently, with m(g)=\mathbb E|b(g)|,

Equations (6.2) and (6.3) therefore give \|m\|_2\le\sqrt{\eta+2\theta}.

On the cone rooted at g, each replaced outcome is b(g)d_{J(g)}, where J(g) is its crossing tail starting at g. The term is zero when there is no crossing segment. Thus the signed mean decomposes as

Each cone term has norm at most \mathbb E|b(g)|=m(g), since |\sigma|=1 and a segment indicator has norm one. The cone identity therefore gives

Lemma 15.5 gives a direct convex-product proof of the same reconstruction, allowing any multiplier of absolute value at most one in place of \sigma.

The signed mean of the original atoms is y, so its tail satisfies

For the last inequality, square both sides and use \sqrt{\eta(\eta+2\theta)}\le\eta+\theta. Now apply (6.1).

For an arbitrary completed vector y, take an increasing finite ancestral exhaustion K_n\supset H. Contractivity gives \|x\pm P_{K_n}y\|\le R, and P_{K_n}y\to y in norm. The finite estimate passes to the limit. No bound on the height or the number of children entered the calculation.□

Proof. For a unit x=P_Hx and y\in\ker P_H with \|y\|\ge1, Theorem 1.1 applied to ty gives

Both endpoint norms are at least one, by applying the contractive projection P_H. Their average therefore exceeds one by at least the right side of (7.4).

For an arbitrary unit center x, choose a finitely supported unit vector x_0 with \|x-x_0\|<\varepsilon, and let H contain its ancestral support. The estimate for x_0 applies to every y\in\ker P_H of norm at least one. Replacing x_0 by x loses at most \varepsilon in the average. Take the infimum over these y, then the supremum over finite-codimensional subspaces, and finally let \varepsilon\downarrow0. The resulting bound is uniform over the original unit center.□

Proof. For a finite vector u, each segment sum of P_Eu is the sum of u over the intersection of that segment with E. Nonempty intersections are disjoint segments when the original segments are disjoint. Taking the supremum proves \rho(P_Eu)\le\rho(u). Extend by density and take the adjoint to obtain the assertion on X. Contractivity and density identify P_EJ with the indicated completion. Since P_E maps the unit ball of J onto the unit ball of its range, restriction has exactly the asserted dual norm. A segment meets an antichain in at most one vertex; singleton tests then give the \ell_2 norm on its primal range. The restriction identity and \ell_2 duality give the dual assertion.□

Proof. Every segment belongs to one component. Grouping a test family by component gives one inequality in the primal sum formula. For the other, choose nearly maximizing finite families on each of the finitely many components meeting the support and combine them. Completion gives the displayed primal identity, and duality for \ell_2 sums gives the dual identity. On component i,

Thus each component and its dual are reflexive Hilbert space renormings with dense coordinates. The sum identities give reflexivity and coordinate density on the entire forest. No bound uniform in i on the Hilbert space comparison is needed. Projection contractivity uses only segment intersection, as in Lemma 8.1; the complementary-subtree identity uses independent segment tests in its incomparable subtrees.

For the last assertion, put h=n-1. The component n=1 is the one-dimensional singleton tree with the absolute-value norm. The remaining components have heights 1,2,\ldots, and the coordinate map (n,\sigma)\mapsto(n-1,\sigma) preserves their segments and all their tests. Splitting off the singleton proves both isometries. The component bound and the number of extension steps follow from the maximal chain cardinality n.□

Proof. Choose \eta with \gamma<\eta<\overline\delta_N(t) and then F\in\operatorname{cof}(Z) such that N(x+tv)>1+\eta for every v\in F\cap S_N. Put u_j=z_j/N(z_j). These vectors are weakly null because 1/N(z_j)\le1/t. Their images in Z/F converge to zero weakly, hence in norm, since the quotient is finite dimensional. Choose v_j\in F with N(v_j-u_j)\to0. Then N(v_j)\to1, so w_j=v_j/N(v_j) is defined for large j, belongs to F\cap S_N, and satisfies N(w_j-u_j)\to0. Consequently

for large j. For r\ge t, convexity gives

Apply this at r=N(z_j) to finish the proof.□

Proof. Suppose the modulus at t=ac/(bM) is positive and choose 0<\gamma<\overline\delta_N(t). At a nonterminal s, put

The equivalent norms have the same continuous linear functionals, so (z_j) is weakly null, and N(z_j)\ge ac/(bM)=t. Lemma 8.3 gives a child for which N(u_{s^\frown j})>(1+\gamma)N(u_s). Starting at the root and repeating k times gives am(1+\gamma)^k<bM, which is impossible for sufficiently large k. Thus the modulus at t is zero. Applying the same argument at t/2, using either the lemma or the direct approximation form above, proves the stated smaller-radius conclusion.□

Proof. For a vertex s in a component with root o, put u_s=\sum_{o\preceq v\preceq s}e_v^*. A single segment test gives \|u_s\|\le1, while evaluation on e_s gives equality. Likewise \|e_s^*\|=1, and u_{s^\frown j}-u_s=e_{s^\frown j}^*. These increments are weakly null: under the canonical identification X_{\boldsymbol h}^*=J_{\boldsymbol h}, their evaluations on g are the coordinates g(s^\frown j), which tend to zero since \|g\|_2\le\rho(g). Equivalently, the coordinate map on this fixed component is a Hilbert space isomorphism, so every continuous functional on the full space restricts to a continuous Hilbert space functional there. Components of unbounded finite height give the trees required by Proposition 8.4, with m=M=c=1. For the convention of n levels the same iteration makes n-1 steps. Both step counts tend to infinity.□

Proof. Singleton tests and Cauchy–Schwarz give

For coefficients supported on an antichain, equality holds: the test with proportional coefficients on that antichain has its Euclidean norm, since a segment meets it in at most one point. In particular sibling coordinate vectors span an isometric \ell_2. Thus the map sending the unit vectors of \ell_2(T_\infty) to e_v^* extends boundedly into B_\infty. For any L\in B_\infty^*, finite Euclidean duality gives \sum_v|L(e_v^*)|^2\le\|L\|^2. Hence distinct coordinates are weakly null against the whole continuous dual. A root-path functional z_v=d_{[o,v]} has norm one, by the single segment upper bound and any coordinate test on that path. Its child is z_{v^\frown n}=z_v+e_{v^\frown n}^*.

Use the one-sided modulus defined in (8.1). It is nonnegative: for each x, the kernel of a norming functional makes every displayed norm at least one. Suppose its value is positive at a radius r_0\le a/b. Choose 0<d<\overline\delta_N(r_0). For a path z, put x=z/N(z) and choose F so that N(x+r_0w)\ge1+d on the N-unit sphere of F. The normalized child directions v_n=e_{v^\frown n}^*/N(e_{v^\frown n}^*) are weakly null. Their images in the finite-dimensional quotient by F converge to zero in norm. Approximate them in F and normalize those approximants to obtain N-unit vectors w_n\in F with N(v_n-w_n)\to0. Choose N(v_n-w_n)<d/(2r_0). Then N(x+r_0v_n)>1+d/2 for large n. The actual child step has radius

Convexity along this ray, including the case r_n=r_0, gives

Choosing a child repeatedly contradicts a\le N(z_v)\le b. For the forest, choose the component height in advance large enough that a(1+d/2)^h>b and carry out only h such selections.□

Proof. The path norms, weak nullity against the whole dual and vanishing one-sided modulus are Proposition 9.1. For the table, choose respectively a cofinite gain \gamma=d, \gamma=\varepsilon or \gamma=d/2. The uniform raw-child error bound N(e_v^*-f)<\min(a/2,a\gamma/2) from (9.2) gives growth at least 1+\gamma/2. These are exactly the three rows. In the last row the selected gain d/2 lies strictly below the named modulus, as required by the supremum over finite-codimensional subspaces.□

Proof. Take a disjoint segment family. Write A_i and B_i for its head f-sum and exterior w-sum. Then \sum_iB_i^2\le\rho(v)^2. A positive product A_iB_i occurs only at a crossing, and different crossings use different cone indices (or different last head vertices in the grouped construction). If |A_i|<sc_j, its cross term is bounded by 2s^2c_j^2. Their total is at most 2s^2\rho(v)^2.

For a remaining positive product, the head qualifies for the suffix selection. It contains the selected suffix, whose sum C_i has the opposite sign to A_i and magnitude at least sc_j. The suffix is proper, so splitting it off leaves two nonempty head segments. The increase in their squared f-sums is

These cuts are simultaneous cuts in distinct original segments. Their pieces and the other nonempty heads are a valid disjoint test family for f. Thus the original head energy plus all large positive cross terms is at most \rho(f)^2. Drop nonpositive cross terms, add the small-crossing budget 2s^2\rho(v)^2, and add the pure exterior budget s^2\rho(v)^2. This proves the result. For a strict threshold, use |A_i|\le sc_j in the small case and |A_i|>sc_j in the large case; the proof is unchanged.□

Proof. Fix any one of the budgets above and choose its signs by Lemma 10.1. For each group let \Phi_j(r)=\sigma_j\max\{\sigma_j r,0\}. On an exterior cone in that group, set the cumulative potentials of h_1,h_2 equal to \Phi_j(V_g(w)) and \Phi_j(-V_g(w)), respectively, and take successive differences down the edges from initial value zero. Since \Phi_j(r)-\Phi_j(-r)=r, the difference of the two tests is v. If v(w)=0, its cumulative values at w and its parent are equal, so both new increments vanish. Thus no new support appears, even along infinite descendant paths.

On any segment in a cone, its h_j-sum is a difference of two values of a single 1-Lipschitz map. Its magnitude is therefore at most the absolute v-sum on that same segment. The same assertion holds for a crossing after deleting its zero head portion. In a last-head group, both endpoints use the same grouping index, including the artificial zero at the head predecessor. Hence \rho(h_j)\le\rho(v). Root prefixes have the chosen sign and the chosen budget bounds their magnitudes. Lemma 10.1 completes the proof.□

Proof. Use the paired test procedure (10.8), let the tail test approach its norm, and put s=t/6. At a finite unit center the average is at least

Arbitrarily close normalized approximation of the center preserves this bound. The construction applies to untouched forest components because they have no crossing terms. It needs no bound on the component height.□

Proof. In (10.9), put t=1 and s=\|y\|/(6\|x\|) when both vectors are nonzero. This gives (10.10). If y=0 the assertion is equality; if x=0, it follows from \|y\|\ge\|y\|/\sqrt{12}.

For the second assertion, given \varepsilon>0, choose H with \|w-P_Hw\|\le\varepsilon. The bounded sequence P_HZ_j has two distinct terms for which h=(Z_j-Z_l)/2 satisfies \|P_Hh\|\le\varepsilon. Then \|h\|\ge\sigma and \|w\pm h\|\le1 by convexity. Formula (10.9) at P_Hw,Q_Hh, again with t=1, gives

Let \varepsilon\downarrow0, put s=\sigma/6, and use \sqrt{1+3s^2}\le1+3s^2/2. This retains the stated 1/24 loss.□

Proof. The exterior-test requirement loses no norming accuracy: replacing any finite test v by Q_Hv preserves y(v) and cannot increase \rho(v). Let v^+,v^- be the successive differences of \max(V_g,0) and \max(-V_g,0) on each cone. Both have nonnegative root prefixes, v=v^+-v^-, and each segment sum is dominated in absolute value by the corresponding v-sum.

For the first construction choose q\in\{v^+,v^-\} with |y(q)|>\|y\|/4. Partition its cones by \sigma_g=1 and \sigma_g=-1, writing q=q_++q_-. One part has pairing of absolute value greater than \|y\|/8. For that sign \sigma put w=\sigma q_\sigma. This last sign makes its root prefixes inward; the eventual endpoint sign is chosen from the sign of y(w). These two choices must not be confused.

For the second construction form w_{\sigma,+}=\sigma v^+\mathbf1_{\{\sigma_g=\sigma\}} and w_{\sigma,-}=\sigma v^-\mathbf1_{\{\sigma_g=\sigma\}} for \sigma\in\{-1,1\}. Then v=\sum_\sigma\sigma(w_{\sigma,+}-w_{\sigma,-}). The triangle inequality for their four pairings gives the claimed 1/8 bound for one of them. Each of the four was fixed before the endpoint was selected.

For the third construction write v_g=v\mathbf1_{U_g}. In each nonzero cone select q_g' from the two cumulative parts so that |y(q_g')|\ge|y(v_g)|/2. Reverse the selected part if necessary to make its root prefixes inward. Call the oriented test r_g. Then

Keep either the positive-pairing cones or the negative-pairing cones, whichever has larger total absolute pairing, and sum their r_g. This loses another factor two. The selected half in each cone can differ; this procedure is not the choice of one of two fixed global clips.

In all three constructions, masking cones and reversing complete cone pieces preserves the same-segment absolute-sum domination. Thus \rho(w)\le\rho(v) and all prefix signs are inward. Lemma 10.1 proves the test-cost bound.□

Proof. Put q_g=\rho(v\mathbf1_{U_g}), so \sum_gq_g^2\le1. At each attached gate with q_g>0, choose the shortest head suffix of magnitude at least 2sq_g, if one exists, and let \tau_g be its sign. Otherwise use \tau_g=1. Set a(g)=-\tau_gq_g at the gates and zero elsewhere. The gates form an antichain, so \rho(a)^2=\sum_gq_g^2\le1. For w=a\pm v, every root prefix at g has sign -\tau_g or is zero, and magnitude at most 2q_g. Also \rho(w)\le2.

In a disjoint testing family write the head and tail sums as A_i,B_i. The pure tail contribution s^2\sum_iB_i^2 is at most 4s^2. A positive cross term with |A_i|<2sq_g costs at most 8s^2q_g^2. Their total is at most 8s^2. For a larger positive term the selected suffix has sum C_i opposite to A_i and magnitude at least 2sq_g. Splitting it off gains

The simultaneous cuts are disjoint, so all these larger terms are paid by the head norm. This proves (11.1). Unattached roots contribute only to the pure tail budget.

For a finite unit center choose its head normer f and a half-norming exterior test v, so |y(v)|\ge1/2 for \|y\|\ge1. One of a+v,a-v has pairing of absolute value at least |y(v)|, because their difference is 2v. Selecting that test and then the endpoint sign gives

Take s=t/24. The ratio equals \sqrt{1+t^2/48}; testing on f makes the other endpoint at least one. Finally approximate an arbitrary unit center within one quarter of this ratio’s excess. The average retains precisely the lower bound in (11.2).□

Proof. Since the gates form an antichain, \rho(a)^2=\sum_ga_g^2\le1. Each root prefix of h has sign -\tau_g or is zero and magnitude at most 2a_g, and \rho(h)\le2. Zero weights give zero cross terms; unattached roots have no head crossing.

For a disjoint test family write A_i,B_i for head and tail sums and L=\sum_iA_i^2\le1. Small positive products, with |A_i|\le\alpha a_g, have total at most 2\alpha. For every other positive product, the simultaneous suffix cuts in (10.4) release at least 2|A_i|\alpha a_g. Their total release is at most 1-L, whereas |B_i|\le2a_g. Thus the large positive products total at most (1-L)/\alpha. Since \sum_iB_i^2\le4,

This uses precisely 2c\le\alpha. Strict thresholds use the same small class, including equality. Formula (10.1) verifies both proposed choices of weights.□

Proof. First choose c>0, depending only on D, with D(\sqrt{1+16c^2}-1)<c/16, and then choose 0<\eta<c/[64(1+c)]. Choose a head H, a head test f, and a center tail error so that \rho(f)\le1, x(f)>\|x\|-\eta, and \|Q_Hx\|<\eta. Boundedness in the finite-dimensional head gives distinct indices with \|P_H(z_i-z_j)\|<1/4, hence their tail difference has norm greater than 3/4. A finite tail test v, \rho(v)\le1, has |(z_i-z_j)(v)|>1/2. Rename one index so |z_i(v)|>1/4.

Use prefix weights a_g=m_g and strict threshold 2c m_g in Lemma 11.2. The two perturbations h_0=a, h_1=a+v differ by v, have norms at most one and two, and have the same inward prefix signs with magnitudes at most 2m_g. One choice h has |z_i(h)|>1/8. We retain the following direct whole-crossing estimate:

For a crossing write its entire sum as A+B, where B already includes the factor c, so |B|\le2cm_g. If |A|\le2cm_g, its square is at most A^2+12c^2m_g^2. If |A|>2cm_g and AB\le0, it is at most A^2. Otherwise the chosen suffix has sum C opposite to A, with |C|>2cm_g\ge|B|, and

The simultaneous head cuts remain disjoint. Crossings and wholly head segments therefore cost at most 1+12c^2; wholly tail segments add at most c^2\rho(h)^2\le4c^2. This proves (11.5).

Put M=\max_{\pm}\|x\pm z_i\|, and choose \epsilon\in\{-1,1\} so that \epsilon z_i(h)=|z_i(h)|. Test x+\epsilon z_i on f+ch, and x-\epsilon z_i on f, and add. The head evaluations of z_i cancel. With L=\sqrt{1+16c^2}, this gives

Here x(h)>-2\eta. It follows that

Thus c_D=c/[32(1+L)] works. The parameter choice depends on the fixed bound D; no unit-tail hypothesis on the selected z_i was used.□

Proof. If R=0, both vectors vanish. If x=0, contractivity of Q_H gives \|Q_Hy\|\le R, which implies the theorem. In the remaining case, scale to R=1 and put d=1-\|x\|. By Lemma 4.1, choose a norming test p=P_Hp with \rho(p)=1 and x(p)=\|x\|.

Fix a finite ancestral K\supset H. Represent each of P_K(x+y) and P_K(x-y) as a finite convex combination of atoms, as supplied by Lemma 4.1. On a finite probability space choose a uniform sign \sigma\in\{-1,1\} and conditionally choose a represented atom A=\sum_i a_i d_{S_i} from the corresponding combination. Then

For each outcome set b_i=p(S_i), A^\circ=\sum_i b_i d_{S_i}, and L=1-\sum_i b_i^2. Replacing the original coefficients by these segment sums has the joint error estimate

Taking expectations and applying (4.1) gives

We have paid for coefficient replacement. It remains to control the tail of the replaced signed mean. At a gate g, let b(g) be the crossing sum of the represented family, or zero. Segments entirely outside H have replaced coefficient zero. Thus A^\circ(e_g)=b(g) and its restriction to J(U_g) has norm at most |b(g)|. Define finite gate vectors

Because x(e_g)=0, Lemma 2.2 and (12.2) give \|h\|_2\le\sqrt{2d}. Take two independent copies of the atom data. Lemma 5.1 and independence yield

Consequently \|m\|_2\le\sqrt{6d}. By the cone identity (2.1),

Its distance to Q_HP_Ky is at most \sqrt{2d} by (12.2). Let K increase through an ancestral exhaustion. Lemma 2.1 gives P_Ky\to y, proving (12.1) for R=1. Scaling gives the stated theorem.□

Proof. For a finite-head unit vector u=P_Hu and unit y\in\ker P_H, put R=\max\{\|u+ty\|,\|u-ty\|\}. Convexity gives R\ge1. Since (\sqrt2+\sqrt6)^2<16, Proposition 12.1 implies t^2\le16R(R-1). Solving this quadratic inequality gives the asserted lower bound for R-1.

For an arbitrary unit x, choose an ancestral head H with \|Q_Hx\|<\eta<1. Then u=P_Hx/\|P_Hx\| is well defined and \|x-u\|<2\eta. The preceding bound for R-1, diminished by 2\eta, therefore holds for all unit y\in\ker P_H with center x. Take the infimum over these y, the supremum over finite-codimensional subspaces, and then let \eta\downarrow0. The bound is independent of x, so the final infimum proves the corollary.□

Proof. The test exists by Lemma 4.1, since \|x\|\ge1-d>0. Both endpoint evaluations on p are at most one, so |y(p)|\le1-\|x\|\le d and (x\pm y)(p)\ge1-2d. The coefficient identity from the proof of Proposition 12.1 gives

Equation (4.1) and Cauchy–Schwarz give the second bound in (12.3).

Combine the laws with equal probabilities. Their mean is x, whose pairing with p is at least 1-d. The coefficient and opposite-crossing calculation in the proof of Proposition 12.1 therefore gives \|m\|_2\le\sqrt{6d}. The coordinate inequality in (12.4) follows from m=(m_1+m_{-1})/2.

These separate-law bounds also give a tail estimate directly. The cone identity implies

Thus (12.3) gives \|Q_HP_Ky\|\le2\sqrt d+2\sqrt{6d}. Ancestral exhaustion yields

The larger constant records the cost of retaining the two conditional laws separately; Proposition 12.1 uses their signed mixture.□

Proof. The head support of p and P_Hy=0 imply P_K(x+\tau y)(p)=r for both signs. Expanding exactly as in (6.1) gives the first assertion of (13.1), and Cauchy–Schwarz gives the second. For the equal mixture, put \eta=\mathbb EV and \theta=\mathbb EL. Then \eta+\theta\le\Delta, so \eta+2\theta\le2\Delta. Thus (6.3) bounds \|m\|_2 as asserted. Writing m_\tau(g)=\mathbb E_\tau|b(g)|, positivity gives m_\tau\le2m coordinatewise. Hence

The original conditional tail is \tau P_Ky. Adding its replacement error gives \|P_Ky\|\le(1+2\sqrt2)\sqrt\Delta for every finite ancestral K\supset H. Along an ancestral exhaustion, P_Ky\to y in norm, proving (13.2) for the completed vector.

Alternatively, use only \eta\le\Delta and \theta\le\Delta. Then \eta+2\theta\le3\Delta, and the identical endpoint recovery proves (13.3). This calculation is useful when starting from the gauge definition. First let x and y have finite coordinate support. For every \lambda>\max\{\|x+y\|,\|x-y\|\} there are finite convex representations with coefficient radii at most \lambda. Carrying out the calculation with \lambda gives

Letting \lambda\downarrow R gives (13.3), without assuming that the infimum in the gauge definition is attained. For a completed tail y, first apply this finite-coordinate argument to P_Ky, where K\supset H is finite and ancestral. Contractivity bounds both projected endpoint norms by R, and P_Kx=x. The resulting estimate for P_Ky passes to y along an ancestral exhaustion, as in the preceding proof.□

Proof. The convex hull of \mathcal A is symmetric and absorbing. Disjointness bounds every atom’s coordinates by one, while singleton segments make x/\|x\|_2 an atom for x\ne0. Hence its gauge is a norm and

For every \lambda>a(x) there is a finite convex representation

Indeed, the defining infimum gives such a representation at some radius less than \lambda; adjoining the zero atom permits any larger radius. No attainment of the gauge infimum is used here.

If every segment meets E\subset\mathcal F in a segment or the empty set, then P_E preserves atoms: intersect their disjoint segment families with E and discard empty terms. Thus P_E is contractive for a and extends to X_a. This applies to ancestral sets, descendant subtrees, and unions of incomparable descendant subtrees. Along a finite ancestral exhaustion, P_Hx\to x in X_a, because these projections are contractive and eventually fix each finite coordinate vector.

Density of c_{00}(\mathcal F) in J gives q(x)=\|x\|; tests approaching the unit ball may be rescaled to have norm at most one. Every atom belongs to B_X, so \|x\|\le a(x). Conversely, place the support of a nonzero x in a finite ancestral set H. Lemma 4.1 expresses x/\|x\| as a finite convex combination of atoms supported in H, giving a(x)\le\|x\|. The zero vector is immediate. This proves equality of the three norms. Coordinate density in Lemma 2.1 makes both completion maps onto X.

In particular the component duality already proved in that lemma applies to the finite polar construction without a change of norm: q_h=q|_{c_{00}(T_h)} is the norm inherited from J_h^*, and

On each component the inherited bounds are

Thus the polar completion has the asserted component decomposition and is reflexive.□

Proof. First take a finite family and x\in c_{00}(\mathcal F). A segment meets at most one U_j. Consequently the squared coefficient norms of an atom’s restrictions sum to at most one. The seminorm z\mapsto(\sum_j a(P_{U_j}z)^2)^{1/2} is at most one on every atom, so it is at most a(z) by convexity and passage through radii \lambda>a(z).

For the reverse inequality, choose \lambda_j>a(P_{U_j}x). Write each P_{U_j}x as a finite convex average \sum_{k_j}\theta_{j,k_j}A_{j,k_j} of disjoint-segment sums with coefficient norm at most \lambda_j. Project these representations onto U_j, preserving that bound. For each tuple (k_1,\ldots,k_m), the sum \sum_j A_{j,k_j} still uses disjoint segments and has coefficient norm at most (\sum_j\lambda_j^2)^{1/2}. Give this sum the product weight \prod_j\theta_{j,k_j}. These weights are nonnegative and sum to one, and their average is \sum_jP_{U_j}x=x. Therefore

Let each \lambda_j decrease to a(P_{U_j}x), including when this value is zero. This proves the finite-coordinate identity.

Projection onto the union of the subtrees preserves atoms. Apply it to finite-coordinate approximants of x to obtain the finite-family identity in the completion. For a countable family, the finite-subtree truncations converge: a finite-coordinate approximant in the union meets only finitely many subtrees, and projection onto the remaining subtrees has norm at most one. The finite-family formula passes to the limit and proves the asserted sum.

Taking the component roots gives the atomic construction’s outer \ell_2 sum directly. Project atomic representations onto T_h. Every resulting atom has Euclidean norm at most \sqrt{h+1}, since its segments are disjoint and each has at most h+1 vertices. Convexity and the strict-radius limit therefore give \|x\|_2\le\sqrt{h+1}\,a(x); the reverse Hilbert bound is (14.1). The components are thus Hilbert-isomorphic, and the \ell_2-sum reflexivity argument of Lemma 2.1 applies to this direct construction.□

Proof. The identity is an expansion of the square; its right-hand inequality uses \sum_i c_i^2\le1. Applying (15.1) to the coefficients c_i-s_i and using contractivity of Q_H proves the replacement bound. For the exterior segments s_i=0, so the coefficient norm bound for disjoint segment functionals gives the first inequality in (15.6). The crossing indices form a subset of the full coefficient array, giving the second. A crossing segment contains its gate, and all other contributions there come from O(A), proving the coordinate identity. On any finite array supported on the gate antichain, \rho equals the Euclidean norm: a segment meets that antichain at most once, and singleton tests give the reverse inequality. Dual testing proves the final contraction, also for an infinite gate set by taking finite subsets.□

Proof by truncation. At an opposite-sign gate, both head segments end at the parent of that gate. Their starting vertices lie on the same ancestor chain, so the heads are nested. They are unequal because equal heads have equal sums on p. Suppose A has the longer head, with sum s, and the shorter head of B has sum s'. Replace the segment of A by its prefix ending just before the shorter head starts. This prefix is nonempty, and its sum is s-s'. Its squared sum increases by

Perform these truncations at every opposite-sign gate where A has the longer head. Each original segment is used at most once, since it has only one gate. The new segments are subsets of the old ones, so they remain disjoint. Their total squared test sum is at most one, and hence the total increase is at most D(A). The other nesting direction has budget D(B). Adding proves the assertion.

This comparison can also be indexed first by the last interior vertex: at most one crossing segment of an atom uses any given such vertex, and the same nesting proof applies. Before using a cancellation of coordinates, however, one must reindex by the first exterior vertex. Different children of one interior vertex are different coordinates. Applying the same comparison after this reindexing gives (15.8) in precisely the form needed below.□

Proof by splitting. Apply Lemma 5.1 to the two underlying segment families. In its notation, L_{\mathcal S}=D(A), L_{\mathcal S'}=D(B), and the crossing sums are s_g(A),s_g(B). Its simultaneous two-piece cuts retain the longer segment’s original endpoint and give exactly (15.8). The proof uses only nesting of finite paths and the bound on their squared p-sums, so it applies unchanged in either coordinate model. Gates that are untouched component roots have zero crossing sums and contribute nothing.□

Proof. The map r\mapsto(-r)_+ is one-Lipschitz. The difference between the product sums formed with c and with s is therefore bounded by

The crossing arrays c(B) and s(A) have Euclidean norms at most one. Cauchy–Schwarz and Lemma 15.1 bound this error by \sqrt{2L(A)}+\sqrt{2L(B)}. Lemma 15.2 bounds the remaining sum by (D(A)+D(B))/2\le L(A)+L(B).□

Proof. The mean of \|c-s\|_2 is at most \sqrt{2k}, which proves (15.10). Put a_g=\mathbb E(s_g)_+ and b_g=\mathbb E(-s_g)_+. For independent copies A,B, the expected opposite-sign product sum is 2\sum_g a_gb_g. Thus Lemma 15.2 and \mathbb E D\le2k give

Since M_s=a+b and \mathbb E s=a-b, the same expansion gives

Together with (15.10), this proves (15.11). The coordinate inequality ||c_g|-|s_g||\le|c_g-s_g| gives \|M_c-M_s\|_2\le\sqrt{2k}, proving (15.12).

For the other route put u_g=\mathbb E(c_g)_+ and w_g=\mathbb E(-c_g)_+. Lemma 15.3 and Cauchy–Schwarz in the probability space give

Now M_c=u+w and \mathbb E c=u-w, so \|M_c\|_2^2=\beta^2+4\sum_g u_gw_g gives (15.13).□

Proof by convex products. Ignore roots with M_g=0. At each remaining root, select j with probability |a_{g,j}|/M_g and choose the signed segment \operatorname{sgn}(a_{g,j})d_{[g,w_{g,j}]}. Make these choices independently at the finitely many roots. Put r=\|(M_g)_g\|_2. If r=0, all mixtures vanish. Otherwise multiply the selected segment at g by M_g/r. The selected segments lie in incomparable subtrees and are disjoint, and their squared coefficients sum to one. Every resulting functional is therefore an atom. Its expectation is (\sum_g f_g)/r, which proves (15.15).

For the crossing expectation in (15.17), the contribution at g divided by M_c(g)>0 is a convex combination of signed segments starting at g, with zero allowed to account for |\theta|<1. The same finite product construction applies; choosing zero simply omits that segment. Apply it separately to either conditional law. Since \mathbb E_\pm|c_g|\le2\mathbb E|c_g| coordinatewise, it also proves (15.18).

Under the one-sign assumption, M_g=|f_g(e_g)|. Antichain testing gives the reverse inequality in (15.16). Adding same-sign coefficients increases every absolute root coordinate, proving the asserted monotonicity. This conclusion concerns these rooted mixtures; it does not assert lattice monotonicity of the norm on Z.

For the stated extensions, apply the finite inequalities first. Their upper bounds show that truncating absolutely summable coefficients at each root and then square-summable root masses gives a Cauchy sequence in Z. Root coordinates converge by continuity, so the equalities pass to the limit as well. No infinite product is needed to produce an atom.□

Proof by disjoint test segments. For v\in c_{00}, let m_g(v) be the largest |d_R(v)| among the rooted segments appearing at g, or zero if there are none. Choose a maximizing segment for every nonzero m_g(v). They lie in incomparable subtrees and are disjoint, so \sum_gm_g(v)^2\le\rho(v)^2. Consequently,

Density extends this bound to J. For a finite atom law use the crossing tails in its support in the same calculation:

The conditional argument is identical, and the root tests used in the first proof again give the one-sign equality.

One may maximize over all rooted segments in a fixed finite ancestral set, or take the supremum over all such segments in a subtree. In the latter case, choose almost maximizing segments for finitely many roots before taking the limit. This proves the same estimate without assuming global attainment. The norm-limit statements then follow by the finite truncations described above.□

Proof. First suppose each endpoint has a finite law of represented atoms. Write \mathbb E_i for its law and \mathbb E=\sum_i\lambda_i\mathbb E_i for their mixture. The pointwise budget of Lemma 15.1 gives

The same argument under the ith law bounds its replacement error by \sqrt{2\delta_i}.

Let s_g(A) denote the head sum of the crossing at g, with value zero when no segment crosses there. A segment wholly outside H has replacement coefficient zero. Thus Q_HA^\circ consists of segments starting at the gates, with coefficients s_g(A). Put v=\mathbb E s and m=\mathbb E|s|, coordinatewise. Gate-coordinate contraction and (16.3) give \|v\|_2\le\eta+\sqrt{2\delta}. Applying the signed–unsigned identity (15.14) to the mixture yields

Equivalently, its positive and negative mean masses l_g,n_g satisfy \sum_gl_gn_g\le\delta.

The conditional absolute masses m_i=\mathbb E_i|s| satisfy m_i\le m/\lambda_i coordinatewise. Rooted-mass recovery, Lemma 15.5, therefore gives

This comparison concerns nonnegative Euclidean mass vectors; it does not require lattice monotonicity of the coordinate-space norm. Adding the conditional replacement error proves (16.1).

For completed endpoints, choose finite coordinate approximants and divide each by the maximum of one and its norm. They converge in norm inside B_Z and have finite atom laws by the finite-hull theorem for the chosen model. Evaluation and both coordinate projections are continuous, so the inequality passes to the limit. Finally, nonnegativity of the individual deficits gives \delta_i\le\delta/\lambda_i and hence (16.2).□

Proof. Apply (15.14) to the head-sum arrays of A. The pointwise replacement budget gives \|\mathbb Es\|_2\le\eta+\sqrt{2\delta}, and hence

The bounded-multiplier case of Lemma 15.5, applied to the replaced atoms A^\circ, bounds \|Q_H\mathbb E(\nu A^\circ)\| by this absolute mass. The remaining term satisfies

Adding proves the first inequality. The second follows from \sqrt{a^2+b^2}\le a+b for a,b\ge0. No independence between \nu and A was used.

For finite convex representations of x+u and x-u, take their equal mixture and let \nu be its endpoint sign. Then \mathbb EA=x and \mathbb E(\nu A)=u. For completed endpoints, approximate both inside their finite atomic balls and pass to the limit, as in Theorem 16.1.□

Proof. For the first assertion choose a norming test in the finite-dimensional head. Then \delta\le\varepsilon and \eta=0, so (16.2) proves the claim. At the level of finite laws the proof retains the separate estimates

For the second assertion apply Theorem 16.1 to z_\pm=h_\pm/R. Both individual deficits and the mixed deficit are at most d, while \eta=0. The conditional replacement cost is therefore \sqrt{2d} and its crossing cost is at most 2\sqrt{6d}. Multiplication by R proves the assertion, including d=0. Alternatively, projecting both endpoints onto increasing finite ancestral sets containing H preserves their sum and their evaluations on p, and gives the same limit.□

Proof. Use finite atom laws first, with equal mixture \mathbb E. The decomposition Q_HA=C(A)+O(A) from the toolkit separates crossing and wholly exterior segments. Let c_g(A) and s_g(A) be the original crossing coefficient and its head sum. The pointwise estimates (15.6) and the gate identity (15.7) imply

By (15.14), \|\mathbb E|s|\|_2\le\|\mathbb Es\|_2+2\sqrt\delta. The coordinate inequality \big||c_g|-|s_g|\big|\le|c_g-s_g| and Jensen’s inequality therefore give

Lemma 15.5 bounds the crossing mean of either conditional law by 2\|\mathbb E|c|\|_2. Its exterior mean costs at most

Their sum proves (16.8). For a norming head test, \delta=1-\|P_Hz\|; gate contraction gives \kappa\le\|Q_Hz\|, and 6\sqrt2+6<15.

Keeping the weaker conditional exterior estimate \mathbb E_i\|O(A)\|\le2\sqrt{2\delta} instead records the alternative budget

For \delta=0 the last inequality is non-strict. Either the one-piece truncation or the two-piece comparison in Lemma 15.2 supplies the required opposite-sign estimate. Thus this original-coefficient calculation is compatible with both surgeries. Normalized finite approximants and norm continuity give the result for completed endpoints.□

Proof. Fix a finite ancestral K\supset H and represent P_K(x\pm u)/(1+e) by finite atom laws. Under each law the expectation on p is at least (1-e)/(1+e)\ge1-2e. The pointwise budget gives

Let l_g,n_g be the positive and negative crossing masses of the equal mixture after replacement. Its mean before replacement is P_Kx/(1+e), so gate contraction gives \|l-n\|_2\le2\sqrt e+e.

Take independent outcomes of this mixed law. At each gate compare only a positive head sum in the first outcome with a negative head sum in the second. If the sums of the longer and shorter nested heads are b,b', truncation leaves the sum b-b' and gains (b-b')^2-b^2=(b')^2-2bb'\ge2|bb'| in the squared test. As in Lemma 15.2, different gates require different original segments, so the gains total at most D(A)+D(A'). Independence now gives the one-orientation estimate 2\sum_gl_gn_g\le\mathbb E(D(A)+D(A'))\le8e. Hence

Rooted-mass recovery bounds each conditional replaced tail by 2\|l+n\|_2. Either proof in Lemma 15.5 applies; in its finite-test proof all maximizing segments may be taken in the finite set K.

Using the sufficient bound \|I-P_H\|\le2, restriction of each replacement error costs at most 4\sqrt e. Half the difference of the conditional tails is P_Ku/(1+e), since the center cancels. Consequently

Here e\le\sqrt e/2. Coordinate exhaustion proves the claim.□

Proof. For the first assertion set d=t^2/256. Let x_0 be a finitely supported unit vector and choose H containing its support. For y\in\ker P_H, \|y\|\ge1, both endpoint norms are at least one. If both were at most 1+d, their normalized midpoint x_0/(1+d) would have norm at least 1-d. Equation (16.6) would give

whereas d<1 makes the left side greater than t/2. The average therefore exceeds 1+d/2. For an arbitrary unit x, choose a finitely supported unit x_0 with \|x-x_0\|<d/4 and use its head subspace. The average loses less than d/4, leaving gain d/4=t^2/1024.

For the second assertion choose a head with \|Q_Hx\|\le e and a norming head test with x(p)\ge1-e. Intersect \ker P_H with the kernel of a norming functional for x on Z. Both actual endpoints have norm at least one. For u=ty, \|y\|\ge1, Proposition 16.5 forces one endpoint to exceed 1+e, so their average exceeds 1+e/2. Taking the modulus proves the claim.□

Proof. Projection onto A gives L_+,L_-\ge r. Thus L<r is impossible. If r=0, then x=0 and L_+=L_-=L=\|z\|, and (17.1) follows because C\ge1. This includes L=0. If an endpoint norm is zero, projection forces r=0 and then z=0, so no zero denominator is needed in what follows. For r>0, divide x and z by r; it is enough to prove the assertion with r=1. In this case L_+,L_-\ge1.

First suppose z is finitely supported, and choose a finite ancestral set B containing A and the support of z. By Lemma 3.1, choose a head test f=P_Af with \rho(f)=1 and x(f)=1. The segment test norm gives

for every disjoint segment family.

Choose a sign \sigma with probabilities

Conditionally on that sign, choose a finite distribution of represented atoms with mean (x+\sigma z)/L_\sigma. Such distributions exist by Lemma 3.2; the random data retain a chosen coefficient list and segment family for every atom. These probabilities give both identities

The first identity cancels the tail. The second recovers the desired displacement without passing to one conditional distribution.

Apply Corollary 16.2 to this finite law, with multiplier \nu=\sigma. Its mean x/L is supported in A, so the center-tail parameter in that corollary is zero. Its evaluation deficit under f is

The first inequality in (16.4), together with the two identities in (17.4), therefore gives

Multiplying by L proves (17.1) when r=1. The multiplier may depend on the represented atom; the corollary requires only |\sigma|\le1, which holds for the endpoint sign used here.

For general z\in\ker P_A, use the finite tail approximants from Lemma 3.1. Their endpoint norms and averages converge, so the finite-support inequality passes to the limit. Finally, rescaling from r=1 gives rC\sqrt{(L/r)(L/r-1)}=C\sqrt{L(L-r)}.□

Proof. For a finite-support unit center x, choose a finite ancestral set A containing its support and take F=\ker P_A. If y\in F and \|y\|\ge1, Theorem 17.1 applied to z=ty yields t^2\le C^2 L(L-1). Since L\ge1, solving this quadratic gives L-1\ge\omega(t). The projection P_A has finite rank, so F is closed and has finite codimension.

For an arbitrary unit center x, density and normalization provide a finite-support unit vector x_0 with \|x-x_0\|<\omega(t)/2. Choose the subspace associated with x_0 before choosing y. Each endpoint norm changes by at most \|x-x_0\|, uniformly for all such y, so their average exceeds 1+\omega(t)/2. Taking the infimum over y, the supremum over F, and then the infimum over unit centers gives the assertion. This argument imposes no upper bound on t.□

Proof. By Lemma 4.1, choose p=P_Hp with \rho(p)=1 and x(p)=\|x\|\ge1-d. The two ball inclusions give x(p)\pm h(p)\le1, and hence |h(p)|\le d. Fix \eta>0. Corollary 4.2 provides finite represented-atom mixtures approximating the two endpoints within \eta. Write \mathbb E_+, \mathbb E_- for their expectations and \mathbb E=(\mathbb E_++\mathbb E_-)/2 for the equal mixture. For each atom Z=\sum_i a_i d_{S_i} set b_i=p(S_i), \widehat Z=\sum_i b_i d_{S_i}, D=\sum_i(a_i-b_i)^2, and L=1-\sum_i b_i^2. Pointwise D+L\le2(1-Z(p)). The mixture pairs with p by at least 1-d-\eta, whereas each conditional law pairs with p by at least 1-2d-\eta. Therefore

In particular the conditional replacement error is at most 2\sqrt{d+\eta}.

Write z_\pm=\mathbb E_\pm Z and z=(z_++z_-)/2. These finite atom means lie in B_X. Their approximation to the intended endpoints gives

Apply Theorem 16.1 to z_+,z_- with equal weights and this same head test. It gives

The conditional boundary-mass factor is two, while the conditional replacement term is at most 2\sqrt{d+\eta}. Restoring the endpoint approximation costs at most \eta, since Q_Hx=0 and Q_H is contractive. This proves (17.6). Letting \eta\downarrow0 gives the asserted limit.□

Proof. Choose a head test p=P_Hp with \rho(p)\le1 and x(p)=\|x\|. Approximate x\pm y within \varepsilon by finite represented-atom means z_\pm, using Corollary 4.2. On their equal mixture, let \sigma=1 on the plus law and \sigma=-1 on the minus law. Then \|\mathbb EA-x\|\le\varepsilon and \|\mathbb E(\sigma A)-y\|\le\varepsilon. Its mean test deficit is at most d+\varepsilon, and its mean tail has norm at most \varepsilon. Corollary 16.2 applies to this finite law with multiplier \sigma, and hence gives

There is no conditional boundary-mass factor: the signed expectation recovers the difference of the two approximating endpoints directly. Since Q_Hy=y, adding the endpoint approximation error \varepsilon proves (17.7) with the stated e_\varepsilon. Let \varepsilon decrease to zero. For endpoints of norm at most 1+\lambda about a supported unit center, first divide by 1+\lambda and take d=\lambda/(1+\lambda) in the same calculation.□

Proof. Coordinate density and contractivity give a finite ancestral H with \|Q_Hx\|\le\delta. Under the identification X^*=J in Proposition 14.1, choose a norming test f\in J with \rho(f)=1 and x(f)=1. Set p=P_Hf. Then \rho(p)\le1 and x(p)=1-(Q_Hx)(f)\ge1-\delta, proving (17.8). This choice uses only x and \delta.

Fix any permitted y. The radius-(1+\delta) version of Corollary 4.2 provides two finite laws of represented functionals

whose respective means approximate x+y and x-y within \delta. Write \mathbb E_+ and \mathbb E_- for these laws and \mathbb E=(\mathbb E_++\mathbb E_-)/2 for their equal mixture. Thus

The last bound uses \rho(p)\le1 and (17.8).

In each represented outcome define

Both L and E are nonnegative. Expanding the square and taking expectations gives the unscaled joint budget

The disjoint-segment coefficient estimate (15.1) gives \|V-W\|\le\sqrt E. Cauchy–Schwarz, nonnegativity, and the equal mixture therefore yield

Let G_H be the first exterior vertices. A segment wholly outside H has replacement coefficient zero. At each g\in G_H there is at most one crossing segment in an outcome; let q_g be its head sum b_i, or zero if it is absent. Thus W(e_g)=q_g and the restriction of Q_HW to the cone at g is either zero or q_g times a segment functional starting at g. Put m_g=\mathbb E|q_g| and v_g=\mathbb E q_g. These vectors have finite support. Antichain contraction, followed by (17.11), gives

Apply Lemma 5.1 to two independent copies of the represented segment families, both tested against p. Its simultaneous two-piece splits retain the longer segment’s exterior endpoint, and their available squared-test budgets are L and L'. Consequently

Independence factors every product expectation. In combination with (17.10), this gives

For either conditional law, the absolute mass in the cone at g is at most \mathbb E_\pm|q_g|\le2m_g. The rooted-mixture estimate in Lemma 15.5, equivalently the subtree decomposition of Proposition 14.3, therefore gives

Restoring the conditional coefficient error and the endpoint approximation yields

Taking half the difference of the two endpoint tails proves the same bound for Q_Hy. Finally F(\delta)\to0 as \delta\downarrow0. Choosing F(\delta)<\varepsilon before choosing x, and then choosing H as above before y, proves the uniform assertion.□

Proof. First suppose that x,z are supported in a finite ancestral K\supset H. Choose a head test p with \rho(p)\le1 and x(p)=\|x\|; when x\ne0 it can be chosen of norm one, and when x=0 take p=0. The finite atomic-ball identity represents each endpoint as a finite law of represented atoms. Let \tau be the endpoint sign and use the equal mixture of the laws. Then

Here the last equality uses z(p)=0. The gate coordinates of the mixed mean vanish. Lemma 15.1 therefore gives

Thus (15.11) and (15.12) yield

The signed mixture in (18.3), followed by Lemma 15.5, gives

This signed reconstruction has no conditional factor of two.

For the second route use the original-product estimate (15.13). If u_g=\mathbb E(c_g)_+ and w_g=\mathbb E(-c_g)_+, it says

Let \mathbf1_+ indicate the plus law in the mixture. Now \mathbb E(\mathbf1_+Q_HA)=z/2. Applying the test-segment proof of Lemma 15.5 to \mathbf1_+C(A), and bounding the exterior term by its pooled mean, gives

The factor of two comes from this indicator reconstruction. This proves (18.2). The fourth-root order comes from the product perturbation, while the leading factor two comes from the indicator recovery; the two losses have separate causes.

The same loss can be recorded as two separate conditional bounds. If d\le d_0, positivity of the conditional weights and (15.18) give

Indeed each conditional mean of a nonnegative quantity is at most twice its pooled mean. Since z=\mathbb E_+Q_HA, adding these bounds gives \omega_{\mathrm{coef}}(d_0) directly, with the exterior and crossing errors still distinguished.

For a general z, let K increase through finite ancestral sets containing H. Contractivity gives \|x\pm P_Kz\|\le1, and P_HP_Kz=0. Apply the finite calculation and use P_Kz\to z in Z. The function \omega_{\mathrm{coef}} is nondecreasing, so d\le d_0 proves the last assertion. This also handles nonattainment in the completed space without a representation of its endpoints.□

Proof. For finite y, use the exact laws of (18.3), with z=y and d=s=1-\|x\|. The original crossing coefficients obey (18.6), so

The signed reconstruction y=\mathbb E(\tau Q_HA) and Lemma 15.5 give

There is no conditional factor of two in this calculation. It keeps the original-product transfer error, which is responsible for the fourth-root order. For general y, apply this finite estimate to P_Ky along a finite ancestral exhaustion containing H. Contractivity preserves both endpoint bounds, and norm convergence proves the assertion. When x=0, the head test used in the exact-law calculation is p=0, as before.□

Proof. Apply (18.1) to x/R,y/R, whose deficit is h=1-1/R, and multiply by R. In particular, both the mean coefficient error squared and the mean head slack are at most 2h; the wholly exterior term has mean norm at most \sqrt{2h}. The crossing means have norms at most \sqrt{2h} and 2\sqrt{2h} before absolute values, and the unsigned head mean has squared norm at most 12h. These are the terms retained in the stated constant.□

Proof. Choose a norming head test p as above. Equivalently, in the gauge presentation choose a norm-one functional norming x and compose it with P_H; its coordinate values define p, since optimizing over disjoint-segment atoms gives \sum_i p(S_i)^2\le1. For \epsilon>0, choose finite convex combinations w^\pm of atoms with

Such approximants follow from the closed atomic-ball description, not an exact convex representation in the completion. This closure also follows directly from the strict-gauge property: approximate a completed unit-ball vector u by finite u_n with \|u_n-u\|<1/n. Then u_n/(1+2/n) has gauge strictly below one and belongs to the convex hull of atoms, while still converging to u. Use the two endpoint approximants’ finite laws and their equal mixture, and put k=d+\epsilon. Since y(p)=0,

The gate restriction is contractive and vanishes on x. Thus (15.7) yields

Use the splitting proof of Lemma 15.2. Independence gives the same conflict budget as in Lemma 15.4, so

For the second line, first bound the square root in the first line by \epsilon+2\sqrt{2k}+2\sqrt k, and then use the mean coefficient error \sqrt{2k}.

The exterior term under the plus law has norm at most \sqrt{2k} by (18.10). Apply the convex-product proof of Lemma 15.5 to its crossing part. The conditional absolute coefficient mean is bounded coordinatewise by twice the pooled mean. Hence

Since Q_H is contractive and Q_H(x+y)=y, add \epsilon to this bound for \|y\|/(1+d). Let \epsilon\downarrow0. This proves (18.9). Only when d\le1 may one use 1+d\le2 to obtain 2(7\sqrt2+4)\sqrt d<32\sqrt d.□

Proof. Choose H so that \|Q_Hx\|\le\delta, and choose a primal unit test p=P_Hp with x(p)\ge1-\delta. Such a test norms P_Hx, whose norm is at least 1-\delta>0. Both choices depend only on x,\delta, before y is given.

Approximate each scaled endpoint (x\pm y)/(1+\delta) within \delta by a finite atom mean, and mix the two laws equally. With k=\mathbb E(1-A(p)), the pooled evaluation and tail satisfy

Thus the pooled gate norm \kappa is at most 2\delta. The product-transfer estimate used here is

proved in Lemma 15.3 by the one-Lipschitz property of the positive part and the head-conflict bound. Applying the displayed mass estimate above gives

For either conditional law, the crossing tail has norm at most 2B and the wholly exterior part has norm at most 2\sqrt{2k}. Restore the endpoint approximation error \delta, take half the difference of the two endpoint tails, and multiply by 1+\delta. The bound is therefore

Using k\le3\delta gives exactly the stated \eta(\delta). Its leading order is \delta^{1/4}, inherited from the original-product estimate.□

Proof. By density of finitely supported vectors in J, choose p\in B_J with finite support and x(p)\ge\|x\|-\lambda. Use the contractive exhaustion in Lemma 2.1 to choose a finite ancestral H containing this support and satisfying \|Q_Hx\|\le\lambda. Make these choices before selecting y.

By Corollary 4.2, choose finite represented atom laws with means z_+,z_- such that \|z_\pm-(x\pm y)\|\le\lambda. Write \mathbb E_\pm for their expectations and \mathbb E=(\mathbb E_++\mathbb E_-)/2. For z=(z_++z_-)/2 one has

Use the notation of Section 15. For an atom A=\sum_i c_i d_{S_i}, put s_i=p(S_i) and L=1-A(p). Its wholly exterior part is O(A), and c_g,s_g are its original and corrected crossing coefficients at g\in G_H. The pointwise budgets in Lemma 15.1 give

Here the last two vectors are the crossing arrays. In particular \mathbb EL\le3\lambda and \mathbb E\sqrt{2L}\le\sqrt{6\lambda}.

For two independent represented atoms, the original-product estimate of Lemma 15.3 reads

Its head comparison may use either the truncation or the two-piece proof in Lemma 15.2; both give the same transfer bound. This is the estimate underlying the common mass bound above. Thus, for a_g=\mathbb E(c_g)_+ and b_g=\mathbb E(-c_g)_+, independence yields

The gate-coordinate identity (15.7) also gives

Consequently the original absolute mass m=a+b=\mathbb E|c| satisfies

For either endpoint law, positivity of the mixture weights gives

Lemma 15.5, or its proof by maximizing one test segment in each cone, therefore bounds the whole crossing tail by 2\|m\|_2. The segments selected in different cones are disjoint, so their primal test sums have squared sum at most one; this is what turns the absolute masses into the whole-tail bound. Thus \|Q_Hz_\pm\|\le2\sqrt{6\lambda}+2\|m\|_2. Finally \|y-(z_+-z_-)/2\|\le\lambda, and hence

Substitute (18.19). All choices of H and p preceded y, as required.□

Proof. For the first assertion, approximate any unit x by a unit finite head x_0=P_Hx_0 with \|x-x_0\|<e/4, and use F=\ker P_H. For y\in F with \|y\|\ge1, Corollary 18.3 rules out both \|x_0\pm ty\|\le1+e, since their lens bound is less than t/2<t\|y\|. Both endpoint norms are at least one by projection. Their average is therefore greater than 1+e/2, and replacing x_0 by x leaves it greater than 1+e/4.

For the second assertion choose H with \|x-P_Hx\|\le c and again let F=\ker P_H. Both \|x\pm ty\|\ge\|P_Hx\|\ge1-c. If their average were at most 1+c, each would be at most 1+3c. It would follow that \|P_Hx\pm ty\|\le1+4c=R. For every finite ancestral K\supset H, apply (18.2) to P_Hx/R and tP_Ky/R. Its deficit is at most 1-(1-c)/R=5c/R, so

Exhaustion gives t\|y\|<t, a contradiction. Notice that the definition of \omega_{\mathrm{coef}} for all nonnegative arguments makes this parameter substitution meaningful even before c is made small.

For the third assertion choose H with \|x-P_Hx\|<d, and let f\in Z^* have norm one and f(x)=1. Put F=\ker P_H\cap\ker f. Then \|x\pm ty\|\ge1 for y\in F. If both were at most 1+d, then \|P_Hx\pm ty\|\le1+2d, and the normalized head would have norm at least

Proposition 18.1 would give t\le(1+2d)\omega_{\mathrm{coef}}(3d) for \|y\|\ge1, contrary to the chosen parameter. Thus the average is greater than 1+d/2.

Finally set d=(t/64)^2 for the fourth assertion. Choose H so that \|x-P_Hx\|<d/8. The normalized head x_0=P_Hx/\|P_Hx\| then exists and satisfies \|x-x_0\|<d/4, because normalization costs at most the projection error again. For y\in\ker P_H of norm at least one, Proposition 18.4 and 32\sqrt d=t/2<t show that one endpoint x_0\pm ty has norm greater than 1+d. Both have norm at least one. Their average exceeds 1+d/2, and replacing x_0 by x leaves more than 1+d/4=1+t^2/16384.□

Proof. If (18.20) failed, represent the two scaled endpoints by exact finite atom laws and use their equal mixture as in (18.3). Put \eta=\delta/(1+\delta); this is the deficit of the scaled unit head. The conditional and pooled deficits all equal \eta. Using the truncation proof of Lemma 15.2, the calculations (18.4)–(18.5) give

In the plus law, \mathbb E_+\|O(A)\|\le\sqrt{2\eta} because its own mean deficit is \eta. For the crossing part use the disjoint-test-segment proof of Lemma 15.5 and the bound of its conditional absolute mean by twice the pooled one. Therefore

Indeed 2\sqrt{12}+3\sqrt2<12 and \eta\le\delta. But \delta<1/625 and

contradicting (18.22).

For a completed unit tail y, take finite ancestral projections P_Ky\to y with K\supset H. For all sufficiently large K the projections are nonzero. Normalize them to unit vectors; they still vanish on H and converge to y. Passing to the limit in the finite strict inequality yields the stated non-strict one.

Given an arbitrary unit center x, choose H so that its normalized head x_0 satisfies \|x-x_0\|<\delta/2. Choose a norm-one f\in Z^* with f(x)=1, and set F=\ker P_H\cap\ker f. For unit y\in F, both actual endpoint norms are at least one, and the maximum is at least 1+\delta/2. Their average is at least 1+\delta/4. For a fixed direction, the average s\mapsto(\|x+sy\|+\|x-sy\|)/2 is even and convex, hence nondecreasing for s\ge0. This extends the bound to all y\in F of norm at least one and gives \widehat\delta_Z(t)\ge\delta/4=t^2/2500.□

Proof. Only crossing segments contribute to B_\omega. At any gate an outcome supplies at most one crossing, with signed coefficient b_{\omega g}. Regard its absolute mass p_\omega|b_{\omega g}| as a supply. Match positive and negative supplies at this gate by repeatedly assigning the smaller of the two remaining supplies to a pair and deleting every exhausted supply. This finite procedure produces pair masses m\ge0 whose total is \min(P_g,M_g). The total mass using any one outcome at this gate is at most its supply.

In a pair of positive mass, the two head intervals end at the parent of the gate. They are strictly nested: equality would give equal primal evaluations, whereas the signs are opposite. Denote the absolute evaluations of the longer and shorter heads by u>0 and v>0, respectively. Splitting the longer crossing immediately before the shorter head gives a squared-test gain

Charge (m/u)D to the outcome containing the longer crossing. This charge is at least mv.

Here is the capacity argument for all these charges. Fix an outcome \omega and one of its crossing segments I. Its absolute evaluation u is fixed. The sum of the fractions m/u over pairs in which this segment supplies the longer head is at most p_\omega. Consequently its total charge is at most p_\omega times the largest proposed gain on this segment, or zero if there is no proposal. Choose one split giving this largest gain on every affected segment of the outcome. These segments were disjoint, so all chosen splits can be made together. Their total gain is at most L_\omega, because the new family is a legal test on the primal unit vector. Summing gives a bound p_\omega L_\omega for the charge to this outcome. Thus

At a fixed gate the shorter member of a pair contributes the quantity m/v. The supply constraint for each outcome implies that the sum of these quantities over pairs having that outcome as shorter member is at most its probability. There is at most one crossing per outcome at the gate, hence \sum m/v\le\sum_\omega p_\omega=1. Cauchy–Schwarz now gives

Sum over gates to obtain (19.1).□

Proof. Choose a finitely supported primal unit vector z with x(z)>1-h, and enlarge its ancestor closure so that \|Qx\|<h. Approximate each (x\pm y)/(1+h) within h by a finite mean of atoms. Let \mathbb E be their equal mixture and \mathbb E_\pm the two conditional means. The identity

has three nonnegative terms on the right. Also \mathbb EA(z)\ge(1-h)/(1+h)-h\ge1-3h. Therefore

The corresponding conditional replacement error is at most 2\sqrt{6h}. Since \mathbb EA approximates x/(1+h) within h, gate-coordinate contraction gives

By Lemma 19.1 and the coordinatewise identity P+M=|P-M|+2\min(P,M), \|P+M\|_2\le2h+3\sqrt{6h}. Each conditional absolute coefficient mass is at most twice the mixed mass, whence \|\mathbb E_+B\|\le2(2h+3\sqrt{6h}). Approximation and replacement bound the tail of (x+y)/(1+h) by the bracket in (19.2); subtracting Qx proves the result. The displayed bound tends to zero with h uniformly in the unit center.□

Proof. Let R be the minimal vertices of \mathcal F\setminus A, and let U_r be the descendant subtree rooted at r\in R. Lemma 2.1 supplies the contractive projections, and Lemma 2.2 supplies both the exact subtree norm identity (2.1) and the contractive root-coordinate map.

If x=0, the conclusion follows from \|y\|\le1. Otherwise, Lemma 4.1 gives a head-supported test a\in J with \rho(a)=1 and x(a)=\|x\|. Since P_Ay=0, both endpoints z^\pm=x\pm y have value 1-d on a.

Write each represented atom as

with its disjoint segment family included in the data. Corollary 4.2 supplies the norm approximations used below.

Fix \eta>0. Choose two finite laws on represented atoms, with expectations E_+,E_-, such that \|E_\pm m-z^\pm\|\le\eta. For each atom put

The coefficient identity gives

Since each endpoint has the same evaluation on a, each law separately satisfies

Consequently, for e=\eta+\sqrt{2(d+\eta)},

The estimates for separate laws here use y(a)=0; they do not follow from a bound on their mixture alone.

Only segments crossing from A into its complement contribute to Q_Am_a. In each U_r there is at most one such segment in a represented atom, because it must pass through r. Let b_r(m) be its coefficient, or zero if it is absent. The restriction of m_a to J(U_r) has norm at most |b_r(m)| and value b_r(m) on the unit vector at r. Segments wholly outside A, including those in components disjoint from A, have corrected coefficient zero; their original coefficients have already been included in (19.4).

The truncation comparison in Lemma 15.2, applied to the test a and the two represented families, gives

The important choice here is the law of each sample: we compare one sample from each endpoint, instead of two samples from their mixture.

Now take m from the plus law and m' independently from the minus law. Set

All these arrays have finite support. Independence and (19.4)–(19.6) yield

The endpoint sum z^++z^-=2x vanishes outside A. By (19.5) and the contractive root-coordinate map, \|S^++S^-\|_2\le2e. Therefore

and hence

Let y_r=y|_{J(U_r)} and let e_r^\pm be the norm of the restriction of z^\pm-E_\pm m_a to that subtree. Equation (2.1) and (19.5) give \|(e_r^\pm)_r\|_2\le e. The restrictions of z^\pm are \pm y_r, so

Combining (2.1), the Euclidean triangle inequality, and (19.8), we obtain the complete error estimate

Let \eta decrease to zero. This gives (19.3), since \sqrt6+2\sqrt2<6.□

Proof. First suppose that y is finitely supported and choose a finite ancestral set B\supseteq A containing its support. Choose a test a supported in A such that J(a)=1 and \langle x,a\rangle=\|x\|. Lemma 3.1 supplies such a test when x\ne0; when x=0, any J-unit test supported in the nonempty set A has the required equality.

Choose a uniform sign \sigma\in\{-1,1\} and, conditionally on that sign, a finitely supported law of represented molecules m=\sum_i\lambda_i\mathbf1_{S_i}, with S_i\subset B and \sum_i\lambda_i^2\le1, as supplied by Lemma 3.2, such that

The representation includes the segment family and its coefficients. Thus \mathbb Em=x and \mathbb E(\sigma m)=y. For each outcome define

Both D and H are nonnegative, and

The nonempty suffixes S_i\setminus A remain disjoint segments. Therefore, with Q=I-P_A,

Let R consist of the minimal nodes of B\setminus A. These nodes are incomparable, and their descendant subtrees intersected with B partition B\setminus A. For each r\in R, let c_r be the coefficient u_i of the segment crossing from A through r, or zero if there is no crossing. A segment entirely outside A has u_i=0, and each segment meets at most one member of R. Consequently (Q\widetilde m)(r)=c_r and

Since \mathbb Em=x vanishes on R, Jensen’s inequality gives

Compare two represented molecules, using primes for the second. Write c^+=\max(c,0) and c^-=\max(-c,0) coordinatewise. Retain only conflicts with c_r>0>c'_r in the nonnegative sum of Lemma 15.2. Its truncation proof, applied to these families and the test a, gives

Thus the ordered estimate uses only one of the two opposite-sign orientations controlled by the common boundary comparison. Take two independent copies of the mixture law. Independence and (19.13) imply

Thus the unsigned means satisfy

The coefficient 10 results from retaining just one ordered conflict and then using the separate bounds on \mathbb EH and \mathbb ED in (19.10). Using both orientations in the common boundary comparison gives a smaller unsigned-mass bound; the present calculation records the cost of the ordered choice.

Only crossing segments contribute to Q\widetilde m. For a finitely supported test b, put

The segments attaining these finitely many maxima lie in disjoint subtrees, so \sum_r L_r(b)^2\le J(b)^2. Hence

Taking the supremum over J(b)\le1 and using (19.11) proves the finite-support estimate. If R is empty, all these tail sums are zero and the same conclusion holds.

For general y\in B_\infty, apply the finite estimate to x\pm P_By=P_B(x\pm y) along the exhaustion in Lemma 3.1. These pairs remain in the unit ball, and P_By\to y in norm, so the same bound passes to the limit.□

Proof. The case x=0 in (20.1) follows from \|y\|\le1. Otherwise choose a primal unit vector z supported on H with x(z)=\|x\|. Set

If h_g>0 and g has parent p_g\in H, inspect the terminal head intervals [q,p_g] with absolute z-sum at least \theta h_g. When there are any, choose the shortest such interval and let d_g\in\{-1,1\} be the sign of its sum. In all other cases put d_g=1. The signs depend on z,y,H, and will be the same for both endpoint approximations.

For an atom A=\sum_Ia_I\mathbf1_I, put b_I=z(I), \widetilde A=\sum_Ib_I\mathbf1_I, and L_A=1-\sum_Ib_I^2. Let b_g be its crossing coefficient, zero if there is no crossing. We claim

Call a gate bad if h_g>0, d_gb_g<0, and |b_g|\ge\theta h_g. Its crossing head qualifies for the selection above; it strictly contains the selected shortest suffix, since the signs disagree. Delete that suffix and the tail from the crossing, retaining its nonempty initial part. The squared-test gain is at least 2|b_g|\theta h_g. All bad gates use distinct segments, so the deletions preserve a disjoint test family and

At remaining gates with d_gb_g<0, the sum of h_g|b_g| is at most \theta\sum_gh_g^2=\theta R^2. The identity |b_g|=d_gb_g+2|b_g|\mathbf1_{\{d_gb_g<0\}} proves (20.3).

Fix \rho>0, and approximate each endpoint x+\sigma y, \sigma=\pm1, separately within \rho by a finite atom mean. For its expectation, the usual coefficient identity gives

Here each law has this bound because y(z)=0. If r_g=\|[\mathbb E\widetilde A-(x+\sigma y)]|_{T_g}\|, then \|(r_g)\|_2\le E_\rho. Thus

Multiply by h_g, sum, and apply (20.3) to obtain

All sums converge: (h_g)\in\ell_2, |y(g)|\le h_g, and the error-cone norms are square summable. Both signs yield this inequality with the same d_g. Select the sign for which the first term is nonpositive and let \rho\downarrow0. With \theta=1/4,

Solving this quadratic proves (20.1), including \Delta=0.

For the separated-sequence consequence, fix \rho>0 and choose a finite ancestor-closed H with \|Qp\|\le\rho. The head projections of the bounded sequence have a pair with \|P(u_i-u_j)\|\le2\rho. Let v=(u_i-u_j)/2. Convexity gives \|p\pm v\|\le1, while \|Pv\|\le\rho and \|v\|\ge\varepsilon/2. Therefore

Apply (20.1) with its constant 8 after division by 1+2\rho, then let \rho\downarrow0. This gives \varepsilon/2\le8\sqrt{1-\|p\|}, which is (20.2).□

Proof. First suppose u has finite support. The finite-hull theorem for Z gives two finite laws of represented vectors A=\sum_i c_i d_{S_i} with disjoint segments and \sum_i c_i^2\le L^2, such that \mathbb E_+A=x+u and \mathbb E_-A=x-u. In B_\infty these laws also follow directly from the strict gauge representation, since both endpoint norms are strictly less than L. Choose a head test p=P_Hp with \rho(p)=1 and x(p)=1. For each represented outcome set

Every such disjoint family is a test for \rho, so D\ge0. Both conditional expectations of \sum_i c_is_i equal one. Expansion and Cauchy–Schwarz give, separately for both laws,

For the second inequality, use 1\le L(\mathbb E_\pm\sum_i s_i^2)^{1/2}.

Pool the laws with probabilities 1/2, and write \mathbb E for their mixture. At a vertex v\in H there is at most one segment of an outcome that crosses with last head vertex v: two such segments would both contain v. Let C_v be its s_i value, or zero if it is absent. Define

Take an independent second outcome, denoted by primes. If C_vC'_v<0, their head portions are distinct nested segments ending at v. Split the longer immediately before the shorter starts. If its old sum is C_v, the new sums are C_v-C'_v,C'_v; the gain in squared sums is

Make every such split in the outcome having the longer head. Different v require different original segments, so these splits preserve disjointness in both outcomes. Their total gain is at most D+D'. Independence and (21.2) yield

Choose sign + at v if p_v\ge n_v and sign - otherwise. The minority masses m_v=\min(p_v,n_v) satisfy

In Q_HA, replace each coefficient c_i by s_i. In the expectation of either law this changes the vector by at most \sqrt{L^2-1}, by (15.1), contractivity of Q_H, and (21.2). Segments wholly outside H disappear because s_i=0. Split the remaining expected crossing terms into G_\pm+B_\pm, retaining in G_\pm the preferred parent sign and putting the other terms in B_\pm.

For children g\in G_H of a fixed v, the sum of absolute coefficients in B_\pm is at most 2m_v: a nonnegative conditional expectation is at most twice its mixture expectation. Put b_g=|B_\pm(e_g)|. Formula (15.16), followed by \sum_g b_g^2\le(\sum_g b_g)^2 within each parent group, gives

Thus \pm u=G_\pm+E_\pm, where

The two G vectors have the same sign at each exit vertex. Lemma 15.5 therefore gives \|G_+\|\le\|G_++G_-\|\le2b_L, and hence \|u\|\le3b_L.

For a completed tail u, approximate it by Q_Hu_j with finitely supported u_j. The strict endpoint hypothesis persists for all sufficiently large j. Passing to the limit proves the proposition.□

Proof. Put \delta=(L-1)/8. Given a unit x, choose a finite-head unit x_0 with \|x-x_0\|<\delta, supported on an ancestral H. If y\in\ker P_H, \|y\|\ge1, and the average of \|x\pm ty\| were at most 1+\delta, the average at x_0 would be less than 1+2\delta. Each endpoint at x_0 has norm at least one by P_H-contractivity. Their maximum would therefore be less than 1+4\delta<L. Proposition 21.1 would give t\le\|ty\|\le3b_L<t, a contradiction.□

Proof. If h=0, then \|u\|\le1 proves the assertion. Otherwise fix a norming head test p, so \rho(p)=1 and h(p)=\|h\|. For \epsilon>0, approximate h\pm u within \epsilon by finite convex combinations V_\pm=\mathbb E_\pm A of atoms. Put d_\epsilon=1-\|h\|+\epsilon and mix the two laws equally. The individual evaluation deficits obey \mathbb E_\pm L(A)\le d_\epsilon. Therefore Lemma 15.1 gives the prelimit budgets

Group only the crossings by their last head vertex v, and let p_v,n_v be their positive and negative mean head-sum masses in the mixture. The simultaneous parent-indexed splitting in the proof of Proposition 21.1 gives \sum_vp_vn_v\le d_\epsilon. This step compares crossings that may leave through different children of the same parent. The proof applies in each forest component separately; untouched roots have no crossing term.

Choose the same preferred parent sign for both laws. Each law’s wrong-sign mass at v is at most 2\min(p_v,n_v), so the sum of its squared parent masses is at most 4d_\epsilon. Splitting each parent mass among its child gates cannot increase that square sum. Thus the discarded replaced tail has norm at most 2\sqrt{d_\epsilon} by (15.16). Replacement itself costs at most \sqrt{2d_\epsilon} under each law. If M_\pm are the retained replaced tails, contractivity and the approximation errors give

The signs of M_+ and M_- agree at every gate. Hence their rooted norm formula gives \|M_+\|\le\|M_++M_-\|, at most twice the right side of (21.7). Adding the remaining error and letting \epsilon\downarrow0 proves (21.5).□

Proof. First suppose y is supported on K\setminus H for a finite ancestral K\supset H. By Lemma 3.2, represent (x\pm y)/(1+\delta) as finite expectations \mathbb E_\pm A of atoms A=\sum_i c_id_{S_i} supported in K, where \sum_i c_i^2\le1. Choose a norming head test p for x, set s_i=p(S_i), and put D_A=1-\sum_i s_i^2. The shared identity (15.3) gives

Both terms on the left are nonnegative. Since \mathbb E_\pm A(p)=1/(1+\delta), both conditional laws and their equal mixture satisfy

For a crossing at g\in G_H\cap K, denote its coefficients by c_g=c_i and s_g=s_i, using zero for an absent crossing. Take independent outcomes from the mixture. The full opposite-sign comparison (15.8), applied to their head sums, gives

Its expectation is 4\sum_g\mathbb E(s_g)_+\,\mathbb E(-s_g)_+ on the left and at most 4\eta on the right. Consequently

Choose \sigma_g\in\{-1,1\} to be a majority sign for s_g. The Euclidean norm of the vector of expected wrong-sign s_g masses is at most \sqrt\eta, because \min(P,Q)^2\le PQ. Moreover, Jensen’s inequality in Euclidean space and (21.9) imply

Using (-\sigma_gc_g)_+\le(-\sigma_gs_g)_++|c_g-s_g|, we obtain

For either conditional law the bound is twice this large.

We now recover the tail vectors from these masses. Wholly exterior segments have s_i=0. Their expected vector in either law has norm at most \sqrt{2\eta} by (21.9) and (15.1). A crossing at g has exterior segment L_g=[g,w]. Its retained and discarded conditional means are

with zero terms for absent crossings. Since \|d_{L_g}\|=1, (15.2) and (21.11) bound \|\sum_gh_{\pm,g}\| by 2(1+\sqrt2)\sqrt\eta. Thus, with g_\pm=\sum_gg_{\pm,g},

To verify the required comparison explicitly, put C_{\pm,g}=\mathbb E_\pm(\sigma_gc_g)_+. Then \|g_{\pm,g}\|\le C_{\pm,g} and g_{\pm,g}(e_g)=\sigma_gC_{\pm,g}. The upper and lower estimates in (15.2) give

Adding the remainder in (21.12) proves the claim for finite support.

For a general completed tail, let K increase through a finite ancestral exhaustion containing H. Contractivity gives \|x\pm P_Ky\|\le1+\delta, and P_Ky\to y in norm by coordinate-span density. The finite-support estimate passes to the limit with the same \delta. This uses exact finite closed-ball representations, rather than a strict-radius approximation.□

Proof. Set \delta_t=(t/(12C_{\mathrm{ex}}))^2<1. For a unit finite-head x_0 and y\in\ker P_H with \|y\|\ge1, the two inequalities \|x_0\pm ty\|\le1+\delta_t would imply

which is impossible. Each endpoint has norm at least one, so their average exceeds 1+\delta_t/2. For an arbitrary unit x, choose a finite-head unit x_0 with \|x-x_0\|<\delta_t/4. The resulting average at x exceeds 1+\delta_t/4, uniformly on this tail subspace. Formula (7.2) gives the conclusion.□

Proof. We first prove the finite-support assertion. Choose radii R_n>\max\{\|x_n+z_n\|,\|x_n-z_n\|\} with R_n\to1. The gauge representation supplies two finite laws of atoms with means (x_n\pm z_n)/R_n. Pool them equally. Suppress the index n and choose a norming head test p for x. For an outcome A=\sum_i c_id_{S_i}, put s_i=p(S_i),

Since \mathbb E_\pm A(p)=1/R, the same calculation as in (21.9) gives

All expectations in what follows are finite, even though the tree has infinitely many children.

For a crossing at g, let c_g,s_g be its coefficient and head sum. Its head is a suffix [v,\operatorname{parent}(g)] of the ancestral path. Define d_g\ge0 as the maximum gain in the sum of squared p-sums obtainable by splitting this head at one place, leaving the exterior portion attached to the later piece. Include the option of no split, with gain zero. The head is finite, so this maximum exists. Define all three quantities to be zero if there is no crossing at g. Maximizing splits for different crossings act on distinct disjoint segments. Therefore

Let W be the part of A consisting of wholly exterior segments. These have s_i=0, so

Also the Euclidean norm of the expected root-coordinate vector (\mathbb EW(e_g))_g is o(1). Indeed, exit vertices are incomparable, so a wholly exterior segment meets at most one of them; disjointness then bounds the root-coordinate norm by the Euclidean coefficient norm of the exterior segments.

Put m_g=\mathbb E|c_g|. We will show \|(m_g)\|_2=o(1). Indices with m_g=0 can be omitted. Discard an observation at g if any one of the following holds:

The first condition removes an expected absolute mass at most m_g/4. For the union B_g of the last two conditions, put p_g=\mathbb P(B_g) and b_g=\mathbb E[|c_g|1_{B_g}]. Markov’s inequality, (22.1), and (22.2) imply

Write the last quantity as \varepsilon_n\to0.

To bound actual mass removal, fix \eta>0 and separate the components according to p_g>\eta or p_g\le\eta. On the first set, b_g\le m_g, hence

On the second set, Cauchy–Schwarz gives b_g^2\le p_g\mathbb E[c_g^21_{B_g}]; therefore

Choose \eta=\eta_n\to0 slowly enough that \varepsilon_n/\eta_n\to0 (for example \eta_n=\sqrt{\varepsilon_n}+1/n). Thus \|b\|_2=o(1).

Every surviving observation has c_g and s_g of the same sign, |s_g|\ge m_g/8, and d_g<(m_g/8)^2. If two survivors at one exit vertex had opposite signs, their head suffixes would be distinct and nested. Splitting the longer at the shorter would increase its squared sum by at least 2(m_g/8)^2, contradicting the definition and bound of its maximal gain. Hence the survivors have one common sign at each g.

Let a_g be the mass discarded by the first condition; thus a_g\le m_g/4. The total discarded mass is at most a_g+b_g, so the single-sign conclusion implies

The mixture has mean x/R and therefore zero exterior coordinates. At the exit roots its crossing part has values \mathbb Ec_g; the remaining values come from \mathbb EW. By (22.3), \|(\mathbb Ec_g)_g\|_2=o(1). Equation (22.7) consequently yields

The expected absolute crossing mass at each g in either conditional law is at most 2m_g. By (15.15), its expected crossing tail has norm at most 2\|m\|_2=o(1). Add the wholly exterior part from (22.3). Since the conditional tail means are \pm z/R, we obtain \|z/R\|=o(1), the desired contradiction.

For completed tails, choose finitely supported approximations and apply Q_{H_n}, with errors tending to zero. Contractivity preserves the tail condition and the endpoint limits, while the lower norm bound persists. The finite-support contradiction applies.□

Proof. The modulus is nonnegative by convexity. If it vanished at t, there would be unit vectors v_n for which the supremum over finite-codimensional subspaces in (7.2) tends to zero. Choose unit finite-head x_n with \|x_n-v_n\|<1/n and support contained in an ancestral H_n. Using \ker P_{H_n} in that supremum, choose z_n there with \|z_n\|\ge t and

The choice of z_n uses approximate infima; attainment is unnecessary. Both endpoint norms are at least one by projection, and therefore both converge to one. Approximation by finite tails gives, if desired, \|z_n\|\ge t-o(1) with the same endpoint limit. This contradicts Proposition 22.1.□

Proof. The bounded set of images Pu_j is totally bounded in the finite-dimensional range of P. Cover it by finitely many open balls of radius \varepsilon/2. Two images with distinct indices belong to the same ball. The norm bounds for their half-difference follow immediately, and x+v=((x+u_i)+(x-u_j))/2 and x-v=((x-u_i)+(x+u_j))/2 belong to R B by convexity.□

Proof. By Lemma 4.1, choose p_n=P_{H_n}p_n with \rho(p_n)\le1 and x_n(p_n)=\|x_n\|; take p_n=0 when x_n=0. Let \eta_n>0 tend to zero. Corollary 4.2 provides approximations to x_n+y_n and x_n-y_n within \eta_n by finite convex combinations of represented atoms Z=\sum_i a_i d_{S_i}, where the segments in each represented atom are pairwise disjoint and \sum_i a_i^2\le1. Sample the two combinations with equal probabilities, recorded by a sign \sigma. Thus

The probability space may depend on n, but it is finite for each n. Put b_i=p_n(S_i), \widehat Z=\sum_i b_i d_{S_i}, L=1-\sum_i b_i^2, and D=\sum_i(a_i-b_i)^2. The defining segment test gives L\ge0, and expansion of the square gives

Consequently, with d_n=1-\|x_n\|+\eta_n\longrightarrow0,

Thus both corrected means approximate their targets with error e_n=\eta_n+\sqrt{2d_n}\longrightarrow0.

Let G_n be the minimal nodes outside H_n. For a represented atom, write A_r for the corrected coefficient of its crossing segment at r\in G_n, or zero if no such segment exists. There is at most one such segment at each r, and segments wholly outside H_n have corrected coefficient zero. Hence \sum_r A_r^2\le1 pointwise. The gate-coordinate contraction and x_n(e_r)=0 imply

Only finitely many gates occur for each n.

Take an independent copy of the represented data, with crossing coefficients B_r and deficit L'. Whenever A_rB_r<0, the two head portions end at the parent of r and are properly nested. Split the longer head at the start of the shorter, in the family to which the longer head belongs. If the two sums are a,b, with a corresponding to the longer head, the increase in the squared segment test is

Splitting the actual segment into two retains its original endpoint; p_n vanishes on its tail, so these are precisely the two new sums. Each segment crosses at only one gate. All assigned splits can therefore be performed simultaneously in each family and preserve disjointness. Their total increases are bounded by L and L'. It follows that

Cauchy–Schwarz on the product probability space and the finite set of gates now supplies the step from smaller squares to products:

For m_r=\mathbb E|A_r| and h_r=\mathbb EA_r, independence gives

Together with the signed-mean estimate this proves \|(m_r)\|_2\to0.

In the cone rooted at r, the corrected tail is a segment functional with coefficient A_r, or zero. Its signed expectation has norm at most m_r. Propositions 14.1 and 14.3 therefore give

Restoring the error e_n and using Q_{H_n}y_n=y_n proves the lemma.□

Proof. If the first statement failed, taking \delta=1/n would produce a sequence contradicting Lemma 23.2. This contradiction also proves that the choice is independent of the head and its height.

Fix t>0 and apply the first statement with \varepsilon=t/2. Choose 0<e<1 so that e/(1+e)<\delta. If u=P_Hu is a unit vector and v=Q_Hv has norm t, the inequality \max\{\|u+v\|,\|u-v\|\}\le1+e would allow its endpoints to be divided by 1+e. The normalized center has norm greater than 1-\delta, whereas the normalized displacement has norm t/(1+e)>t/2, a contradiction. Thus the maximum is greater than 1+e for every such u,v.

For any unit x\in X, choose a finite ancestor-closed H such that u=P_Hx/\|P_Hx\| is defined and \|x-u\|<e/2, using the contractive exhaustion in Proposition 14.1. On the unit sphere of the finite-codimensional space \ker P_H, apply the preceding result to v=ty. It gives

Taking the infimum over that sphere, the supremum over finite-codimensional spaces, and the infimum over unit centers yields a midpoint modulus at least e/2>0.□

Proof. Otherwise choose such configurations indexed by n, with \|x_n\|\to1. Choose finite ancestor-closed H_n with \|Q_{H_n}x_n\|<1/n. The sequence (P_{H_n}v_j) is bounded in a finite-dimensional space. Hence two different indices i,j have \|P_{H_n}(v_i-v_j)/2\|<1/n. Put w_n=(v_i-v_j)/2. Separation gives \|w_n\|\ge\varepsilon/2, and convexity of the two opposite endpoint pairs gives x_n\pm w_n\in B_X. The head center a_n=P_{H_n}x_n and tail displacement b_n=Q_{H_n}w_n satisfy

Dividing both by 1+2/n contradicts Lemma 23.2.□

Proof. First recall the separate-endpoint bound for a finite ancestral head: if a=P_Ha, P_Hb=0, and \|a\pm b\|\le R, then

For a\ne0 choose a norming head test by Lemma 4.1 and apply (16.7) with d=1-\|a\|/R. Both endpoints have the same test value \|a\|, since the test vanishes on b; thus the individual replacement cost is \sqrt{2d} and the conditional crossing cost is 2\sqrt{6d}. Scaling gives the displayed inequality. If a=0 or R=0, it follows directly from \|b\|\le R.

Fix \varepsilon>0. Lemma 2.1 supplies a finite ancestral H with \|Q_Hx\|\le\varepsilon. Coordinatewise convergence on this fixed finite set implies \|P_Hz_j\|\le\varepsilon for all sufficiently large j. For those indices,

Applying (23.3) with R=1+2\varepsilon gives

Take the upper limit in j with this head fixed. Since \|P_Hx\|\ge\|x\|-\varepsilon, the right side is at most \varepsilon+C\sqrt{(1+2\varepsilon)(1-\|x\|+3\varepsilon)}. Letting \varepsilon\downarrow0 proves the result. The argument uses only convergence of the finitely many coordinates on each chosen head.□

Proof. Suppose instead that \widehat\delta_{B_\infty}(t)=0. There are unit vectors x_n for which the supremum over finite-codimensional subspaces is less than 1/n. Approximate x_n within 1/n by a finitely supported unit vector a_n, and let A_n be a finite nonempty initial set containing its support. Taking the subspace \ker P_{A_n} in the definition, choose y_n there with \|y_n\|\ge1 and

By Lemma 3.1, choose a finite vector z_n\in\ker P_{A_n} within 1/n of ty_n. It follows that

Put r_n^\pm=\|a_n\pm z_n\|. Projection onto the head shows that r_n^\pm\ge1, so both sequences tend to 1. Moreover,

This boundedness is needed when the two normalizations are compared.

For each n, take a finite initial set B_n\supseteq A_n containing both supports. Choose a head test p_n=P_{A_n}p_n with \rho(p_n)=a_n(p_n)=1. Lemma 3.2 gives two finite laws of represented molecules G=\sum_i\lambda_i\mathbf1_{S_i} satisfying

Write w_{n,\pm} for these means. Their deficits on p_n are \delta_{n,\pm}=1-1/r_n^\pm\to0. With D,E computed using p_n, each law retains the budget

Write Q_n=I-P_{A_n} and let \mathbb E_n denote the equally weighted mixture. Its tail is

It need not vanish exactly. Its norm \eta_n tends to zero because r_n^\pm\to1 and (24.3) bounds \|z_n\|. Apply (24.2) with the head A_n, the test p_n, the deficits \delta_{n,\pm}, and this \eta_n. It gives \|Q_nw_{n,\pm}\|\to0. These norms are \|z_n\|/r_n^\pm, which cannot tend to zero because \|z_n\|\ge t-1/n and r_n^\pm\to1. This contradiction proves the proposition.□