Midpoint convexity from bounded tree potentials and path costs

Proof. Choose a head test f with f(x)=\|x\|_K and an arbitrary test g. We construct two inward increment families from g. On each component of T\setminus D, let r be its last ancestor in D, put c=P_f(r), and use the relative potential

Let \sigma=-1 if c\ge0 and \sigma=1 if c<0, and write \xi_+=\max\{\xi,0\} for \xi\in\mathbb R. Define

and set U=V=0 on D. Thus both U and V point inward from c, and U-V=R. Let u,v be their edge increments, with root coefficient zero if the root is a coordinate.

Taking a signed positive part is 1-Lipschitz and fixes zero. Along an edge within a component the reference ancestor is unchanged; on an edge leaving D its parent has relative potential zero. It follows in both cases that

Each quadratic constraint on u or v is therefore at most one, using the same coordinate groups as in the chosen row of the table. Their potentials are bounded by two, so u/2,v/2\in K. Thus u and v extend to continuous functionals on X_K with norm at most two. Coefficient domination also preserves finite support for K_{\rm a} and square summability for K_{\rm g}. The coefficient identity gives (u-v)(y)=g(y) first on finite tails and then, by continuity, on F_D.

Put a=\sqrt{1-\theta^2}. We claim af+\theta u,af+\theta v\in K. On a tail component their potentials are ac+\theta U(s) and ac+\theta V(s). If c\ge0, each lies between ac-2\theta and ac; if c<0, each lies between ac and ac+2\theta. Both intervals lie in [-1,1] because a\le1 and 2\theta\le1. On D the potentials are simply those of af. For any quadratic group B, the head and tail supports are disjoint, so

The same calculation applies to v; all support requirements remain satisfied. This proves the claim for all four rows without identifying any two of their norming sets.

Evaluate x+ty against af+\theta u, and x-ty against af+\theta v. Since f vanishes on tails and u,v vanish on the head, adding gives

Take the supremum over g\in K to obtain (4).□

Proof of the midpoint estimate. Fix t\in(0,1). For a finite unit center x and any y\in F_D with \|y\|_K\ge1, choose \theta=t/\sqrt{4+t^2}<1/2 in Lemma 2.2. Its right side becomes \sqrt{1+t^2/4}. Now let x_0 be any unit vector in the completion. For every \varepsilon>0, choose a finite unit vector x with \|x-x_0\|_K<\varepsilon, then choose D containing its support. The triangle inequality loses at most \varepsilon in the average, uniformly over all y\in F_D of norm at least one. Therefore

The subspace is allowed to depend on \varepsilon. Letting \varepsilon\downarrow0 and then taking the infimum over x_0 proves (3).□

Proof. Write t_0=\alpha c/(2\beta M) and suppose 0<\gamma<\overline\delta_N(t_0). At any current node p_s, the definition supplies a finite-codimensional F such that

The same inequality holds when N(w)\ge t_0N(p_s). To see this, write a shorter point on the ray as a convex combination of p_s and p_s+w; convexity gives N(p_s+rw)-N(p_s)\ge r[N(p_s+w)-N(p_s)] for r\ge1.

The quotient X/F is finite dimensional. Since d_{s,n} is weakly null, choose w_n\in F with N(w_n-d_{s,n})\to0. Eventually N(w_n)\ge\alpha c/2\ge t_0N(p_s). Choosing the approximation error at most \gamma N(p_s)/2 gives a child satisfying

Choose h large enough that \alpha m(1+\gamma/2)^h>\beta M, and perform the selection in a tree of height h. Repeating it h times yields a path whose last vector has N-norm at least \alpha m(1+\gamma/2)^h. This exceeds the uniform bound \beta M, a contradiction. Finally, \overline\delta_N(t_0)\ge0: at a unit center intersect any candidate subspace with the kernel of a norming functional, on which the perturbed norms are at least one. This proves equality to zero.□

Completion of Theorem 2.1. The coordinate vectors are linearly independent, so each space is infinite dimensional. Finite rational coordinate combinations are dense, so each space is separable. For s\in I, set p_s=\sum_{u\in I,\ u\preceq s}e_u. Its norm is one: the path condition gives the upper bound, and a singleton test gives the lower bound. In the rootless model begin at any fixed nonroot node v and use the subtree (p_{v^\frown s})_s so that the initial vector is nonzero. In all four models,

This is the sibling constraint in the first two models. Siblings form an antichain in the third and are part of the globally constrained family in the fourth. If \phi\in X_K^*, use a_n=\phi(e_{s^\frown n}) in this inequality to obtain

Thus the child increments are weakly null, and each has norm one. Lemma 2.3 applies with m=M=c=1. In particular, for every equivalent norm as in that lemma, \overline\delta_N(\alpha/(2\beta))=0.□

Proof. Fix 0<t<1. Let x be a finite unit vector supported in an initial subtree D, and choose a norming potential b^0 frozen outside D. For every y\in F_D,

Suppose first that y\in F_D is finite and \|y\|_{\mathrm e}=1. Let E=\{r\notin D:r^-\in D\} be the first-exit nodes. Split y=y_E+y_L into its coordinates on E and on the deeper edges.

First-exit case. Suppose some c\in\mathcal B satisfies |f_c(y_E)|\ge1/2. For each v\in D set

Cauchy–Schwarz in each sibling group gives \sum_{v\in D}L_v\ge1/2. Choose the inward sign \sigma_v=-1 when b^0_v\ge0 and \sigma_v=1 when b^0_v<0. For \varepsilon\in\{-1,1\} define

Since L_{v,1}+L_{v,-1}\ge L_v, one common sign \varepsilon has \sum_vL_{v,\varepsilon}\ge1/4.

For 0<h<1, multiply all head labels by 1-h^2. At a selected exit r from v, assign the jump

assign zero at the other exits, omit groups with L_{v,\varepsilon}=0, and freeze each exit label throughout its descendant subtree. Every nonzero jump has sign \sigma_v and magnitude at most h. It points toward zero from the head label; if it crosses zero, its resulting magnitude is at most h<1. Thus all labels remain in [-1,1]. At a parent in D the internal and exit increments have total square sum at most

and outside D all further increments vanish. This is an admissible potential. Its evaluations on x and y are respectively 1-h^2 and \varepsilon h\sum_vL_{v,\varepsilon}. Hence

Taking h=t/8 gives the lower bound 1+t^2/64.

Deeper-edge case. If the preceding case fails, every admissible test has |f_c(y_E)|<1/2. Choose c\in\mathcal B with f_c(y)>5/6; then f_c(y_L)>1/3. We shall preserve the norming head on the horizontal axis and use the vertical coordinate to detect this deeper contribution.

Consider labels B_v\in\mathbb R^2 satisfying

Every scalar projection of B onto a unit vector belongs to \mathcal B. Therefore the linear map

satisfies |A_Bz|_2\le\|z\|_{\mathrm e} for finite z, and extends contractively to X_{\mathrm e}.

Set B_v=(b^0_v,0) on D. For each first-exit node r, put u=b^0_{r^-} and, throughout the subtree rooted at r, define

The first-exit increment is zero because z_r=0. Also |z_v|\le2k=1/4, and |B_v|_2^2=u^2(1-z_v^2)+z_v^2\le1. For fixed u and c_r, the map

has

It is consequently 1-Lipschitz. At a parent outside D this same map applies to all its children, so the child-increment constraint follows from that for c. At a parent in D, the original head increments remain admissible and all first-exit increments are zero. This verifies (14) at the boundary as well as in every descendant subtree.

Now A_Bx=(1,0), whereas the second coordinate of A_By is k f_c(y_L)>1/24. The first exits contribute zero to that coordinate. The parallelogram identity gives

Completion and arbitrary centers. Put G(t)=\sqrt{1+t^2/576}-1. Since G(t)\le t^2/1152<t^2/64, both cases give a maximum endpoint norm at least 1+G(t). Combining this with (12) yields an average at least 1+G(t)/2. Finite unit vectors are dense in the unit sphere of F_D, by (10) followed by normalization, so continuity gives this same estimate for every unit y\in F_D. For a unit u\in F_D, the function

is convex and even, hence nondecreasing on [0,\infty). The estimate therefore also holds for every y\in F_D with \|y\|_{\mathrm e}\ge1. Given an arbitrary unit center, choose a finite unit center within G(t)/4 and use its subspace F_D. The reverse triangle inequality loses at most G(t)/4 in the average. Taking the infimum over these tail directions, the supremum over finite-codimensional subspaces, and then the infimum over unit centers proves (11), with the direction convention of (1).

The renorming obstruction. For nonroot s, let p_s be the sum of the edges on the path from o to s. Telescoping gives f_b(p_s)=b_s-b_o, and the singleton increment test gives the lower bound in

For every fixed parent s and finite scalar family (a_n),

This follows directly from the sibling constraint in (8). If \varphi\in X_{\mathrm e}^*, applying (20) with a_n=\varphi(e_{s^\frown n}) shows

thus e_{s^\frown n}\to0 weakly against the whole continuous dual. These increments have norm one, and p_{s^\frown n}=p_s+e_{s^\frown n}. Start at any fixed nonroot v and use the tree (p_{v^\frown s})_{s\in T}. By (19) and (20), Lemma 2.3 applies with m=1, M=2 and c=1. Its obstruction scale is exactly \alpha c/(2\beta M)=\alpha/(4\beta), proving the final assertion.□

Proof. The choice h=x, \mu=0 proves the upper bound. Every representation of x satisfies

which proves the lower bound. Scaling and adding representations, chosen within an arbitrary error of their infima, prove homogeneity and the triangle inequality. The lower bound gives definiteness; norm equivalence gives completeness and density of finite arrays.

For p_n(f)\le1, absolute convergence allows us to write

Thus the displayed supremum is at most L_n(x). Conversely, Hahn–Banach gives, for x\ne0, a functional \varphi such that \varphi(x)=L_n(x) and |\varphi(z)|\le L_n(z). Since L_n(z)\le\|z\|_2, it has the form \varphi(z)=\langle f,z\rangle with \|f\|_2\le1. The representation b_s=B_ne_s gives L_n(b_s)\le1, hence |\sum_{r\preceq s}f_r|=|\varphi(b_s)|\le1. This proves equality, including the trivial case x=0.

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

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

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

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

At an internal vertex s take u=b_{n,s}/N(b_{n,s}) and z_j=e_{n,s^\frown j}/N(b_{n,s}). These vectors are weakly null and N(z_j)\ge\alpha/\beta\ge t. Their quotient images in X/F tend to zero, since that quotient is finite-dimensional. Choose v_j\in F with N(v_j-z_j)\to0. If N(v_j)<t, replace v_j by tv_j/N(v_j); for large j this is defined, and the extra change is at most N(v_j-z_j). Thus z_j is approximated by vectors in F of norm at least t. For a sufficiently large child index,

Iterating to depth n gives \alpha(1+\eta)^n\le\beta, impossible for all n. The asymptotic modulus is nonnegative: the kernel of a norming functional at a unit center has N(u+tv)\ge1. This proves the claimed zero value at every indicated scale. Taking \eta=d/4 gives the often useful growth factor 1+d/4.□

Proof. In block n write A_n=A\cap T_n and set

Here |H|\le a_n. The sum V is a difference of two prefix sums of g_n, or a single prefix sum if the path misses A_n, so |V|\le2c_n. It vanishes on A_n. Define

Let r_{l,n,s}=V_l(s)-V_l(s^-), with predecessor value zero at the root. These coefficients vanish on A_n. Outside A_n, the head sum H is constant along each remaining branch, and both scalar clipping maps are 1-Lipschitz. Therefore

At a first exit vertex this follows from the zero predecessor value of V_l. Moreover the prefix sums of r_{l,n} are V_l, and V_1-V_2=V, proving the claimed difference.

The supports of h_n and r_{l,n} are disjoint. Hence their Hilbert norm satisfies

The products HV_l are nonpositive, so each prefix sum satisfies

These two estimates give p_n(\lambda h_n+\mu r_{l,n})^2 \le\lambda^2a_n^2+4\mu^2c_n^2. Summing over n proves the asserted inequality; taking \lambda=0, \mu=1 proves r_l\in X^* and its norm bound.□

Proof. Choose norming dual vectors h,g for x,y, with norms at most one, and replace h by P_Ah. The projection preserves its evaluation at x and does not increase its norm. Lemma 5.3 gives r_1(y)-r_2(y)=\|y\|, so at least one of |r_1(y)|,|r_2(y)| is at least \|y\|/2. For this index choose a sign \varepsilon with \varepsilon r_l(y)\ge\|y\|/2. Because h(y)=r_l(x)=0, evaluation of h+cr_l on x+\varepsilon y proves (25). Its supremum over c>0 is the displayed square root, by Cauchy–Schwarz in two dimensions; endpoint cases follow by a limit.□

Proof. Fix \eta>0 and choose a finite head A with \|x-P_Ax\|\le\eta, using the strong head exhaustion. Put z_j=(I-P_A)y_j. Since P_A has finite rank and y_j is weakly null, \|P_Ay_j\|\to0. In particular,

Apply Theorem 5.4 to P_Ax and z_j. The triangle inequality then gives

Keep this head fixed while taking the lower limit in j. The result is at least \sqrt{\|P_Ax\|^2+L^2/16}-\eta. Finally let \eta\downarrow0, choosing the heads along the strong exhaustion, so that \|P_Ax\|\to\|x\|. This proves the stated bound.□

Proof. Choose a norming functional \phi\in\ell_2(U) for B_U\nu. Its dual norm is at most one, hence \|\phi\|_2\le1 and F_t:=\sum_{v\in U,\,v\preceq t}\phi(v) satisfies |F_t|\le1. At a root set the value at its nonexistent parent equal to zero, and put

Then |\psi(v)|\le|\phi(v)|, and the prefix sums of \psi are |F_t|. Thus \psi also has dual norm at most one: pairing it with a decomposition h+B_U\alpha costs at most \|h\|_2+\|\alpha\|_1. Absolute convergence of the coefficient sums gives

The claim concerns positive path coefficients; it asserts no general coordinatewise monotonicity of the norm.□

Proof. For the unit estimate set r=L_n(x) and choose a norming functional g=gP_A, of norm at most one, with g(x)=r; use zero if x=0. For each sign choose x\pm y=h_\pm+B_n\alpha_\pm of cost at most 1+\eta, where \eta>0. Put d=1+\eta-r. Subtracting the value of g from the cost gives nonnegative terms with sum at most d:

In particular,

For this bound, use |g(h_\pm)|\le\|P_Ah_\pm\|_2 and factor the difference of squares. Its other factor is at most 2(1+\eta). The coefficients satisfying \alpha_\pm(t)g(b_t)\le0 have total mass at most d.

Retain only the other coefficients ending in U, and call their intrinsic forest contributions v_\pm. Then

Within each tree of U, the value g(b_t) is constant, since the head of every such path is the same. If that value is zero, no coefficient is retained there. Otherwise both retained arrays have its sign. One isometric change of signs makes both arrays nonnegative. Lemma 5.6 now gives

It follows that L_U(y|_U)\le3K and L_n(y)\le6K. Letting \eta\downarrow0, and using 0\le1-r\le\sqrt{1-r}, proves (27). Scaling gives the first bound in (28) when \beta>0; when \beta=0, both vectors are zero. Contractivity of P_A gives r\le\beta and the second bound. In each component both endpoint norms are at least the center norm. Thus the maximum of their squared differences from r^2 is at most their sum. Summing over components proves (29), including components with empty heads.□

Proof. Only the finite ancestor hull of the supports matters. Write q(v)=\min\{G(v),W(v)\}. Nonnegativity gives

Process vertices from the leaves towards the roots. At v, the mass already marked in its child subtrees totals \sum_wq(w). Mark enough additional, previously unmarked mass of \mu below v to make the marked total there equal to q(v); fractions of an endpoint mass are allowed. The two displayed inequalities make this possible. Later steps can only increase a previously met quota. The total finally marked in a tree equals its root quota and is at most its total \nu-mass. Take \rho to be the marked mass.□

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

In a complementary tree where the common value c=g(b_t) is zero, there are no good coefficients. Otherwise put \varsigma=\operatorname{sign}(c). All good coefficients have sign \varsigma, and all bad nonzero coefficients have sign -\varsigma. Apply Lemma 5.8 to the combined good mass \mu and combined bad mass \nu in each tree, and mark the selected portions with their originating endpoint representation. Denote the selected good mass by \rho, and put G(v)=\mu(K_v), W(v)=\nu(K_v). At a vertex in a tree with c\ne0, the vanishing sum of the two endpoint tails reads

Consequently, in every complementary tree, including those with no good mass,

After removal, the good tail in either one of the representations has absolute coordinate at most G(v)-\rho(K_v)\le(G(v)-W(v))_+. Its Hilbert norm is therefore at most K_++K_-. The total removed mass is at most B_++B_-, and each complementary path costs at most two in L_n. Decomposing the plus endpoint tail into its Hilbert, bad, removed-good, and remaining-good parts gives

Let \eta\downarrow0. This proves (30), including its endpoint cases.

We give the outer-sum passage explicitly. For the j-th pair set

Convexity gives c_{j,n}\ge r_{j,n}, and the endpoint assumptions give \|a_j^\pm\|_2\le1. Hence \|c_j\|_2\to1. The Hilbert parallelogram identity then gives

Since a_j=c_j+|a_j^+-a_j^-|/2, we also have \|a_j\|_2\to1. Ignore coordinates with a_{j,n}=0, whose vectors are zero, and put \delta_{j,n}=1-r_{j,n}/a_{j,n}\in[0,1]. Then

Applying the component estimate after division by a_{j,n} gives L_n(y_{j,n})/a_{j,n}\le3\sqrt{2\delta_{j,n}}+8\delta_{j,n}; the same ratio is at most one by the triangle inequality. For fixed \tau>0, the squared outer norm is therefore at most

First let j\to\infty, then \tau\downarrow0.□

Proof. Work first in one component. Set r=L_n(x) and M_\pm=L_n(x\pm y), so M_\pm\ge r. Use a supported norming functional g and finite representations of costs at most M_\pm+\eta. Put d_\pm=M_\pm+\eta-r. Here a path coefficient is good when \operatorname{sign}(\alpha_\pm(t))g(b_t)>1/2, and bad otherwise. The deficit calculation gives

Write c=g(b_t) for the common value on one complementary tree. If |c|\le1/2, that tree has no good coefficients. Otherwise put \varsigma=\operatorname{sign}(c); good coefficients have sign \varsigma, and bad nonzero coefficients have sign -\varsigma. Let G_+(v),G_-(v) be the subtree masses of the good coefficients from the two representations, and let W(v) be the subtree mass of all bad coefficients. In a tree with |c|>1/2, the endpoint tails satisfy

For the quota selection take G=G_+. Since G_-\ge0, the identity gives the following bound, which is also immediate in a tree with no good mass:

Lemma 5.8 removes at most B_++B_- from the plus good coefficients. The remaining plus good tail has Hilbert norm at most K_++K_-; the removed tail has L_n-norm at most 2(B_++B_-). Adding the plus Hilbert and bad tails yields

Here d_\sigma\le M_\sigma+\eta and 2\sqrt2+8<11. Squaring the last sum costs at most a factor two, and (M_\sigma+\eta)d_\sigma\le(M_\sigma+\eta)^2-r^2. Letting \eta\downarrow0 proves the component estimate with 2\cdot11^2=242. Summation gives (31). Empty heads cause no exception; their centers are zero.□

Proof. Write x^s=x+sy, s\in\{-1,1\}, and denote averages over s by a bar. Fix \lambda>0. In each nonzero component choose

Use zero decompositions in zero components. Project to A': truncate the Hilbert terms and push each path endpoint to its last ancestor in A', discarding endpoints with no such ancestor. Both costs decrease, so all arrays may be taken supported in A'.

In a fixed T_n, let \pi_A(t) be the last ancestor of t in A, or a dummy symbol \perp if none exists. For each fibre set

Either choice is allowed at zero. Introduce the mass on pairs (s,t) given by m_n(s,t)=|\alpha_n^s(t)|/2, and let h(s,t) be the coefficient’s sign, chosen arbitrarily at zero. Its wrong-sign mass is

Separating the two signs in each fibre, including the dummy fibre, gives \bar M_n-\sum_b|L_b|=2W_n. Since x_n=P_{A\cap T_n}(\bar z_n+B_n\bar\alpha_n), its component norm is at most \overline{\|P_Az_n^s\|_2}+\sum_b|L_b|; the dummy term is a harmless nonnegative addition to this cost. It follows that

A quotient with zero denominator means zero. The Hilbert estimate used here is

The quantity Q=\sum_n\bar r_n\Delta_n controls both Hilbert leakage and wrong-sign mass. For an array on the disjoint union of the trees, \|\cdot\|_2 denotes its Hilbert norm, including summation over components. Since \|z_n^s\|_2\le2\bar r_n,

The first line follows from (34) and 0\le\Delta_n\le\bar r_n; the second uses L_n(x_n)\le\bar r_n and (\bar r_n)^2\le\overline{(r_n^s)^2}.

It remains to recover the norm of P_Ey from this budget. The zero case is immediate. Otherwise take a norming functional for P_Ey and precompose it with P_E=P_{A'}-P_A. Its coefficient array \phi, supported in E, satisfies

The functional has norm at most two. Testing against Hilbert vectors gives the first coefficient bound. Testing against finite sums of one root path in each component, with arbitrary scalar multipliers, gives the second by the outer Hilbert norm. Approximate each supremum if it is not attained.

The Hilbert part of \langle y,\phi\rangle=\overline{s\langle x^s,\phi\rangle} has absolute value at most 4\sqrt Q, by (35) and (36). For the path part write g(t)=\sum_{v\preceq t}\phi(v) and

For v\in E\cap T_n, let m(v) be the mass of pairs with t\succeq v, and w(v) its wrong-sign part. The entire subtree below v is in a single \pi_A-fibre. Its coefficients therefore satisfy

where the last equality uses x(v)=0 on E. Thus, whenever w(v)\le m(v)/4,

Call a vertex v\in E retained if this inequality’s condition holds at v and at every preceding vertex of its path in E. Split g(t) into its sum over retained vertices and the remainder. The remainder starts at the first vertex satisfying w(v)>m(v)/4. Such first vertices form an antichain, so their disjoint descendant sets have total mass at most 4W_n. The remainder is the difference of two prefix sums and has absolute value at most 2c_n. It therefore contributes at most 8c_nW_n to |I_n|. The retained contribution is at most 2\sum_{v\in E\cap T_n}|\phi(v)|\,|\bar z(v)|. Summing over components and using 2W_n\le\Delta_n, we obtain

Together with the Hilbert part this proves \|P_Ey\|\le20\sqrt Q. Apply (35), square, and let \lambda\downarrow0 to obtain (33).□

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

For each terminal history and each component, apply this description with P=P_m\cap T_n, rescaled by L_n(x_m^n). On a finite extension of the probability space we may therefore choose h_n,c_n,u_n so that

For each component the sample is either a Hilbert-ball vector h_n, with c_n=0, or a signed root path with h_n=0 and |c_n|=L_n(x_m^n). Zero components use both coefficients zero. An arbitrary root can be used for u_n when c_n=0. The sigma-fields \mathcal F_i still record only the original signs. In particular x_i=\mathbb E_i(h_n+c_nb_{u_n})_n.

Let v_i^n be the deepest vertex of the path to u_n in P_i, or a symbol \partial if this intersection is empty; set b_\partial=0. The path trace on the new layer is

The sigma-fields \mathcal G_i^n=\sigma(\mathcal F_i,v_i^n) increase: from v_i^n and the previous sign history one recovers v_{i-1}^n by truncating its path to P_{i-1}. Set a_i^n=\mathbb E[c_n\mid\mathcal G_i^n]. Orthogonality of scalar martingale increments gives

We split the new layer into its Hilbert contribution, its old predicted path coefficient, and the change in that prediction:

The identity follows because I_i^n is \mathcal G_i^n-measurable, so c_n may be replaced by a_i^n inside its conditional expectation. Conditional Jensen, the disjointness of the \Delta_i on each outcome, and (41) show that

Here we use L_n\le\|\cdot\|_2 for the first inequality. Denote conditional expectations given \mathcal F_{i-1} by primes. Predictability of \Delta_i, together with the support of x_{i-1}, gives H_i'+M_i'+W_i'=R_{\Delta_i}x_{i-1}=0. Consequently

It remains to compare the old-coefficient contribution on a child history with its average on the parent history.

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

Put I(v)=b_v-b_{r(v)^-}, interpreting the second term as zero when r(v) is the root. Outcomes with v_i^n\notin\Delta_i have zero path trace. The two conditional mixtures are therefore

For an admissible test a, put A_v=\sum_{w\in[r(v),v]}a_w. We have

Define a test candidate f, zero off \Delta_i\cap T_n, by

where A_{v^-}=0 at the first vertex of each interval and \operatorname{sign}(0)=0. The reverse triangle inequality gives \|f\|_2\le\|a\|_2. Each root path meets \Delta_i in at most one interval. Summing f on that interval telescopes to a signed |A_v|, which is at most 2, since A_v is the difference of two root-path sums of a. Hence f/2 is admissible. Moreover

Taking the supremum over a proves L_n(M_i^n)\le4L_n((M_i')^n). Summing the component squares proves the required comparison. By (43), the total energy of M_i is at most 160C^2. Finally, \|H_i+M_i+W_i\|^2\le3(\|H_i\|^2+\|M_i\|^2+\|W_i\|^2), giving 3(1+160+4)C^2=495C^2.□

Proof. If S=\varnothing, then L_n(R_Tz^\pm)\le V_T(z^\pm)\le2g, so the conclusion follows. Otherwise S contains the root. For u\in T, write \pi(u) for its deepest ancestor in S. Choose \sigma_s\in\{-1,1\} so that \sigma_sd_S(z,s)=|d_S(z,s)|. Write

using the positive and negative parts. Collapsing the refined path coefficients to S shows that

For u\in U, all descendants in T have the same \pi-label. Let P_u^\pm,Q_u^\pm be the respective sums of p_v^\pm,q_v^\pm over these descendants. Since z^\pm(u)=\sigma_{\pi(u)}(P_u^\pm-Q_u^\pm),

where \operatorname{av} means the average of the two signs.

Fix one sign and apply Lemma 5.8 on the finite forest U to the endpoint masses \mu_u=p_u^\pm and \nu_u=q_u^++q_u^-=2\operatorname{av}q_u^\pm. Initiality of S ensures that all descendants in T of a vertex in U remain in U. The corresponding subtree masses are therefore P_u^\pm and 2\operatorname{av}Q_u^\pm. The lemma gives 0\le r_u\le p_u^\pm with

By (47), the remaining mass satisfies

Also \sum_{u\in U}q_u^\pm\le g, by (46). The remaining positive mass defines

Its coordinates satisfy |v(u)|\le2|z(u)|. The exact decomposition

therefore costs at most 2\|R_Uz\|_2+2(g+g), as required.□

Proof. For each terminal history choose, componentwise,

with both terms zero in a zero component. Only finitely many components are needed. Put h_i=\mathbb E_i(h_n)_n, b_i=(B_n\mathbb E_i c_n)_n; then x_i=h_i+b_i. The terminal Hilbert and coefficient budgets are each at most 4C^2:

At time i-1, apply Lemma 6.2 in each component to S=P_{i-1}\cap T_n, T=P_i\cap T_n, and the two values of b_i^n. These sets are fixed before the sign at time i. If g_{i,n} is the variation gain in the lemma, then

Indeed, the average variation equals the previous variation plus the nonnegative number g_{i,n}; square and apply scalar Jensen. Telescoping and (44) give

Since R_{\Delta_i}x_{i-1}=0, the lemma and the outer Euclidean norm imply

For j=i-1 and j=i, predictability gives R_{\Delta_i}h_j=\mathbb E_jR_{\Delta_i}(h_n)_n. Conditional Jensen followed by pathwise disjointness of the layers bounds each of the two unscaled Hilbert energies by 4C^2. Squaring the last three-term inequality and summing now gives 3(1+4+16)4C^2=252C^2.□

Proof. For every terminal outcome and component choose

Write \lambda_n=\epsilon_n\mu_n, where \mu_n=|\lambda_n|, |\epsilon_n|\le1, and M_n=\mu_n(T_n). Thus \mu_n is a positive measure on path endpoints, allowed to depend on the terminal outcome. The two budgets satisfy

On outcome–endpoint pairs put the finite positive measure

This need not be a probability measure; conditional means on its positive-mass cells have their usual meaning. On null cells set them to zero. The operators \mathbb E_t still use the original uniform law on sign histories, whereas the conditional means of \epsilon_n below use \nu_n, which need not give equal mass to the two sign children. Let a_{t,n}^- be the conditional mean of \epsilon_n given the first t signs and the truncation of the root path to A_t. Let a_{t,n}^+ be its conditional mean given the first t+1 signs and its truncation to D_t. Empty truncations are recorded by a dummy symbol. The conditioning partitions, ordered as

refine successively: the known sign history and truncation of a deeper trace to the earlier head recover the preceding data. Define

We first establish the weighted sign estimate

For clarity fix n temporarily. Conditional Cauchy–Schwarz gives

Under the original sign law the children are fair, so M\ge0 gives \mathbb E_{t+1}M\le2\mathbb E_tM. Put

The conversion from the sign law to the pair measure is

because \mathbb E_tM is measurable at time t+1. Hence the sum of the integrated right sides is at most

Expand R_t=\sum_{j\le t}(R_j-R_{j-1}). Each nonnegative increment is constant on each cell C of the partition defining a_j^-. Orthogonality of successive conditional-mean increments under \nu gives

The left side uses only some increments of the refining sequence; omitting the others preserves the bound. Multiply by the cell-constant value of R_j-R_{j-1}, then sum over cells and j. The weighted sum 2\sum_{t<m}\int R_t|a_t^+-a_t^-|^2\,d\nu is consequently at most 2\int R_{m-1}\,d\nu=2\mathbb E(MR_{m-1}).

The scalar input is the finite-horizon form of Doob’s L_2 maximal inequality [6]. We include its first-crossing proof:

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

We next compare a child window with its parent window, keeping the component n fixed. For a vertex s, write C_s=\{u:s\preceq u\} for the cone of endpoints whose paths pass through s. Define the signed endpoint measures

For s\in U_t, the child sign history together with membership in C_s is measurable for the partition defining a_t^+. Hence

Thus their sum represents the path part of x_{t+1}^n on the window U_t. The error measure satisfies \|\rho_{t+1}\|_1\le q_{t,n}. Let h_{t,n}=\mathbb E_th_n, \lambda'_t=\mathbb E_t\lambda'_{t+1}, and \rho_t=\mathbb E_t\rho_{t+1}. Averaging gives, at each s\in U_t,

Fix a history through time t. On every cone C_s with s\in U_t, truncation to A_t is fixed. Thus a_t^- is constant on that cone, including all completions of the fixed sign history. There is no cancellation within the cone in either \lambda'_t or \lambda'_{t+1}. With \phi_s=|\lambda'_t(C_s)|, fair binary conditioning therefore gives

Choose another path decomposition, now only for the parent window:

If \beta=|\theta|+|\rho_t|, then (52) gives

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

Here 4\|d\|_2+8\|\theta\|_1\le16r_{t,n}.

We have reduced the vector estimate to square summation. Conditional Jensen and predictability of U_t give, for j=t,t+1,

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

Proof. We first work in one component, suppressing n and restricting each set to its tree. At each terminal outcome choose

and let H_i=\mathbb E_iH. For an endpoint s, let \pi_i(s) be its deepest ancestor in A_i, or \partial if none exists. Set \pi_i(\partial)=\partial and b_\partial=0. Define

The dummy coefficient is kept in this last sum even though its path vector vanishes. This ensures the exact refinement identity

The child window itself has the representation

Orient each refined coefficient by its old collapsed coefficient:

where sign at zero is chosen to be +1, and g_r,q_r are the positive and negative parts of \sigma_rc_r^i. The signs are known on the parent history. By (57),

The elementary inequality (\mathbb E_{i-1}v_i-v_{i-1})^2 \le\mathbb E_{i-1}v_i^2-v_{i-1}^2 therefore telescopes to

Fix a step and a parent history. For t\in \Delta_i, all its descendant labels in A_i have the same old ancestor and therefore the same sign \sigma_t. Set g(t)=\sum_{r\succeq t}g_r and q(t)=\sum_{r\succeq t}q_r, with the dummy omitted. Averaging the coordinate formula gives

Decompose the leaked parent window once more:

There are two equally likely children and g(t)\ge0, so

Call t\in \Delta_i large when g(t)>4(|L(t)|+|H_{i-1}(t)|). At a large node,

The minimal large nodes form an antichain. Mark every coefficient label in A_i below one of them. Their descendant sets are disjoint, so the total marked positive mass is at most 4(\|d\|_1+\beta_i). The total negative mass satisfies \sum_rq_r\le2\beta_i, again by fair binary averaging. Using L_n(R_{\Delta_i}b_r)\le2, all marked positive terms together with all negative terms cost at most 8\|d\|_1+12\beta_i.

It remains to estimate the unmarked positive part. Its coordinate is zero at every large vertex, since all descendant labels there are marked. At other vertices its absolute value is at most g(t)\le4(|L(t)|+|H_{i-1}(t)|). Its Hilbert norm is consequently at most 4(\|L\|_2+\|R_{\Delta_i}H_{i-1}\|_2). Including the child Hilbert term yields

The last coefficient follows from 4\|L\|_2+8\|d\|_1\le16L_n(Y).

For j=i-1,i, conditional Jensen and the fixed, disjoint windows give \sum_i\mathbb E\|R_{\Delta_i}H_j\|_2^2\le\mathbb E\|H\|_2^2. Squaring the four terms in (59), using four times their squared sum, and applying (58), gives

The first two terms are at most 144\mathbb E(\|H\|_2+\|c\|_1)^2 \le576\mathbb E L_n(W_m)^2. Finally sum over the components and replace 576 by 1024 in the terminal term. This proves (56).□

Proof. Scaling and adding representations in (60) proves homogeneity and the triangle inequality. Every representation satisfies

the choice h=z gives the upper bound. This proves positive definiteness, completeness, and equivalence to the Hilbert norm. The dual and bidual therefore have the same underlying continuous linear functionals as for H_n, so Z_n is reflexive. For each coordinate, |z_t|\le\|h\|_2+\|\lambda\|_1, which gives the second assertion of (61). A single atomic mass represents p_t with cost one. For e_t use its Hilbert representation. At the root, the representation e_o=e_o/2+J_n(\delta_o/2) attains the matching cost 1/\sqrt2.

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

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

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

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

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

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

These identities imply (65).□

Proof. Set \alpha_t=V(t)-\sum_{c\text{ child of }t}V(c)\ge0. Induction from the last level, using nonnegative sums, gives V(t)=\sum_{s\succeq t}\alpha_s and \sum_{t\in O}\alpha_t=B. This induction is over the finite height, not over a finite set of children. Hence the required vector is

The series is absolutely convergent in Z_n by Lemma 7.2 and has norm at most 2B. Continuity of the coordinates identifies its sum with the stated array. Finally

which proves the assertion about minima.□

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

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

Outside A_n, each descendant has the same r(t) as its ancestor. For \epsilon\in\{+,-\} define the majority and minority occupancies

Here a_+=\max(a,0). All these arrays satisfy the child inequality of Lemma 7.5, and the total initial mass of V is at most d_n. Since x vanishes outside the head,

Let S(t)=\min(U^+(t),V(t)). It is a stopped flow with initial mass at most d_n, and

Consequently the tail of y_n, which agrees with that of x_n+y_n, is the sum of the signed flows S,-V^+ and the Hilbert vector h_n^++\varepsilon_r(U^+-S), all restricted to the outside. Lemma 7.5 gives

Take the outer Euclidean norm and use (65). The three terms are at most 4\sqrt D, \sqrt{2D}, and 2\sqrt D, respectively. Thus \|Q_Ay\|_{\mathrm q}\le(6+\sqrt2)\sqrt D\le8\sqrt D. Let \eta\downarrow0. No minimizing representation was assumed.□

Proof. Approximate x within \rho>0 by x_0 supported in a finite ancestral head A. Theorem 7.6 bounds each Q_Ay_j by 8\sqrt{(R+\rho)^2-\|x_0\|_{\mathrm q}^2}. The sequence (y_j) is bounded, and P_A has finite rank, so two distinct head projections can be arbitrarily close. Their difference and the two tail bounds give \varepsilon\le16\sqrt{(R+\rho)^2-\|x_0\|_{\mathrm q}^2}. Let \rho\downarrow0.□

Proof. It suffices to prove the assertion in one component, since the result can then be summed. Choose endpoint representations whose squared costs exceed |x\pm y|_n^2 by at most \eta>0, and use the notation of Lemma 7.4, omitting the component index. Put E=(|x+y|_n^2+|x-y|_n^2)/2+\eta-|x|_n^2. Then

Orient each nondiscarded fiber by the sign of its folded mean coefficient \nu_{r(t)}, choosing +1 when it is zero, and choose either sign in \{-1,1\} on a discarded fiber. Define half-weighted positive and negative endpoint masses by

Let \beta=\beta^++\beta^-. In a nondiscarded fiber the positive mass minus negative mass is the absolute folded coefficient. The discarded fiber contributes zero to q. It follows that \|\beta\|_1\le M-q\le\sqrt E.

For t\notin A_n put A^\epsilon(t)=\sum_{s\succeq t}\alpha_s^\epsilon and B^\epsilon(t)=\sum_{s\succeq t}\beta_s^\epsilon, and let A_{\mathrm{tot}}=A^++A^-, B_{\mathrm{tot}}=B^++B^-. The endpoint identities give

For each sign put S^\epsilon=\min(A^\epsilon,B_{\mathrm{tot}}) and V^\epsilon=A^\epsilon-S^\epsilon. Then 0\le V^\epsilon\le(A_{\mathrm{tot}}-B_{\mathrm{tot}})_+ \le|\bar h| outside the head. Each S^\epsilon is a stopped flow of initial mass at most \|\beta\|_1, and the same bound applies to each B^\epsilon. Lemma 7.5 estimates their four signed vectors by 8\|\beta\|_1 in total. The remaining terms have Hilbert norm at most \|Q\widetilde h\|_2+2\|Q\bar h\|_2. Since y vanishes on the head,

Let \eta\downarrow0, and sum the resulting component inequalities. Empty heads cause no exception: their entire mass is in the discarded fiber, and the same bounds hold.□

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

Proof. Choose the endpoint representations with squared costs at most (R+\eta)^2. In the notation preceding Lemma 7.4 define the nonnegative endpoint measure m_n=(|\lambda_n^+|+|\lambda_n^-|)/2. Both |\sigma_n| and |\tau_n| are bounded pointwise by m_n, and \|m_n\|_1=M_n. The lemma gives

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

Each retained set lies in a single fiber of r. If disjoint subsets S_j lie in a nondiscarded fiber C, then

Indeed |\sigma_n(C)| is at most the sum of the displayed absolute sums plus the m_n-mass of the remaining set. This argument also holds for countably many subsets by absolute convergence. Discarded fibers are bounded by their whole m_n-mass. Summing over fibers yields

Split \tau_n=\tau_n^{\mathrm r}+\tau_n^{\mathrm k} by restricting the first part to the union of these descendant sets. Then \|\tau_n^{\mathrm r}\|_1\le2d_n and

If an outside node has a first marked ancestor, the occupancy of \tau_n^{\mathrm k} there is zero. Otherwise it is unmarked, and

The last equality uses Qx=0 in x=\bar h+J\sigma. Thus QJ\tau^{\mathrm k}\in H and has Hilbert norm at most 2\|Q\bar h\|_H. Since y=\widetilde h+J\tau,

Let \eta\downarrow0.□

Proof. If \|x\|_{\mathrm q}>1-\delta, choose a finite-support x_0 within \delta of x, and an ancestral head A for x_0. Then \|x_0\|_{\mathrm q}>1-2\delta and \|x_0\pm y_j\|_{\mathrm q}\le1+\delta. Proposition 7.10 gives

Two bounded head projections are less than c apart, contradicting the assumed separation of the full vectors.□

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

On the outside put G(t)=g_n(S_t) and B(t)=b_n(S_t). Constancy of the orientation along each outside subtree gives

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

At a node having a stopping ancestor the kept occupancy vanishes. At every other outside node, G(t)>2B(t) and

Therefore

Lemma 7.4 and the triangle inequality in the outer Euclidean sum give \|Qy\|_{\mathrm q}\le9\sqrt E. Let \eta\downarrow0.□

Proof. Approximate x within \gamma by a finite-support x_0, and choose a finite ancestral head for it. The tail of every point in the lens has norm less than \rho/3 by Proposition 7.12. The points in the lens have norm at most one, so in an infinite sequence two head projections are less than \rho/3 apart. Their full distance is less than \rho.□

Proof. Scale to \|x\|_{\mathrm q}=1. Choose x_0 of finite node support with \|x-x_0\|_{\mathrm q}\le\delta and an ancestral head A containing its support. Then \|x_0\pm u\|_{\mathrm q}\le1+2\delta. Choose representations x_0\pm u=h^\pm+J\lambda^\pm with each squared cost at most (1+3\delta)^2. Using the notation above for these representations, let C be their average squared cost and set

The parallelogram identity and the scalar square inequality used in Lemma 7.4 give, before comparing the center representation with its norm,

The second inequality uses M_n^2-q_n^2\ge d_n^2 and the parallelogram identity for Qh^\pm. The center representation has cost at least \|x_0\|_{\mathrm q}, so

Orient each nondiscarded fiber C_r by the sign of \nu_n(r), using +1 for a zero folded coefficient or a discarded fiber. This time use the full, rather than half, outside absolute masses: let p_n^\pm agree with the orientation and b_n^\pm disagree with it, and put B_n=\|b_n^++b_n^-\|_1. The outside minority mass is at most the total cancellation bound, giving B_n\le2d_n and

For outside nodes let G(t)=(p_n^++p_n^-)(S_t) and V(t)=(b_n^++b_n^-)(S_t). Since the averaged represented vector vanishes there,

Stop at the first outside nodes satisfying G(t)>0 and V(t)\ge G(t)/2. Their descendant sets are disjoint, and the total majority mass in them is at most 2B_n. Remove all these majority terms and all minority terms from the outside part of \lambda_n^+; denote what remains by \lambda_n^{\mathrm k} and what was removed by \lambda_n^{\mathrm r}. Then \|\lambda_n^{\mathrm r}\|_1\le3B_n and, by (70),

Every outside node either has a stopping ancestor, has G(t)=0, or satisfies V(t)<G(t)/2. In all cases the kept occupancy is at most 2|G(t)-V(t)| in absolute value. Thus

The atomic terms whose endpoints belong to A have zero tail. The representation of x_0+u, together with \|Qh^+\|_H\le\sqrt{2\mathcal D}, now gives

All stopping decisions use a first ancestor, so infinite coefficient support and countably many children cause no change. The choice of A preceded the choice of u, as required.□

Proof. Apply the proof of Theorem 7.6 in this single component, retaining the coefficient 6+\sqrt2 before its final rounding to 8. Here Q_Ay=y. Explicitly, approximate endpoint representations of cost below 1+e+\varepsilon give \kappa=(1+e+\varepsilon)^2-1, Hilbert tails at most \sqrt{2\kappa}, combined Hilbert tail at most 2\sqrt\kappa, and folded-mass loss at most \sqrt\kappa by Lemma 7.4. Because the head contains the root, no fiber is discarded: the combined minority mass is exactly this folded-mass loss. Grouping by the common truncated endpoint and clipping one majority occupancy against the combined minority occupancy gives two signed stopped flows of norm at most 2\sqrt\kappa each, by Lemma 7.5. The remaining majority occupancy is bounded in Hilbert norm by 2\sqrt\kappa. Adding the original Hilbert tail gives (6+\sqrt2)\sqrt\kappa. These are the same decomposition and signed path representations used in that theorem, so no invariance of |\cdot|_n under arbitrary coordinate sign changes is needed. Let \varepsilon\downarrow0. Finally (1+e)^2-1\le3e and 3(6+\sqrt2)^2=114+36\sqrt2<169.□

Proof. The conditions defining Y_x involve only finitely many continuous coordinate functionals, so the assertion about codimension follows. Put

The triangle inequality gives a_n,|y_n|_n\le(b_n+c_n)/2. Therefore

Fix 0<\eta\le1. Write

On these components v_n=a_n^2+d_n\le(1+1/\eta)d_n, and hence

If a_n=0 and y_n\ne0, then d_n=v_n=|y_n|_n^2>0, so n\in E_\eta. If both vectors vanish in that component, its contribution is zero. Thus every remaining nonzero contribution has a_n>0.

Now assume additionally that \eta\le1/4. For n\notin E_\eta with a_n>0, let m_n=(b_n+c_n)/2. We have m_n\ge a_n and

It follows that

Apply Lemma 7.15 to x_n/a_n and y_n/a_n. Summing its squared estimate gives

Take \eta=e^{2/3}, which is at most 1/4 when e\le1/8. Combining (74) and (75),

The second assertion follows from 519/2<17^2.□

Proof. By finite-coordinate density, choose a finitely supported unit vector x^0 with \|x-x^0\|_{\mathrm q}<e. Such unit vectors are dense: approximate x by a nonzero finitely supported vector and normalize it. Use the finite coordinate exclusions for x^0 from Proposition 7.16. If y satisfies the hypotheses of (76), then

Since 2e\le1/8, Proposition 7.16 proves (76). Given t>0, choose 0<e_t\le1/16 with 17(2e_t)^{1/6}<t. For u\in S_Y, both endpoint norms at tu cannot be at most 1+e_t, by (76). Each corresponding maximum is therefore greater than 1+e_t, giving the displayed infimum bound.□

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

For the energy inequality, induction proves q_t(m+r)\ge(q_t(m)^2+q_t(r)^2)^{1/2}. Indeed, let A and B be the \ell_2 norms of the child values for m and r. The induction hypothesis gives

The last inequality is the triangle inequality in \mathbb R^2. Summing its square over the roots proves (79). These arguments apply to countably many children: all sums are nonnegative, and the \ell_2 inequalities hold for finite partial sums and then by monotone convergence. This also shows that each cost commutes with increasing limits of nonnegative arrays.□

Proof. First we bound the output of a representation. Fix an output level d and a start level h\le d. Segments starting at a single node t of level h contribute total absolute mass at most m_t(a) to level d; starts at different nodes of level h have disjoint descendants. Therefore the \ell_2 norm of this part of the output is at most \|m(a)|_{\mathbb N^h}\|_2. Summing over h\le d bounds the norm on level d by \sqrt{n+1}\,M_n(m(a)). Summing the squares over d gives

Conversely, the singleton representation a_{tt}=x_t has cost at most \sqrt{n+1}\|x\|_2. This proves finiteness and the comparisons. Homogeneity is immediate. Adding representations and using monotonicity and subadditivity of the cost proves the triangle inequality; the lower comparison proves definiteness. The norm is equivalent to the Hilbert norm, and hence is complete and reflexive.

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

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

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

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

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

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

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

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

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

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

Proof. Choose a finite-support p arbitrarily close to x, let C be its finite ancestral closure and set F=\ker P_C. Theorem 8.4 gives, uniformly for y\in F with \|y\|_S\ge1,

As \|x-p\|_S\to0, the right side tends to \sqrt{1+t^2/64}. Coordinate continuity makes F closed and finite-codimensional, proving the assertion.□

Proof. At each fixed level of T_h, the singleton segments form an incomparable family. Hence the squared norm of g dominates the sum of |g(v)|^2 over that level. Summing over the h+1 levels gives the lower bound in (87). For the upper bound, every segment has at most h+1 nodes, so Cauchy–Schwarz and disjointness give

Thus Y_{\mathrm I,h} is isomorphic to the Hilbert space \ell_2(T_h) and is reflexive.

A test family separates into families on the individual trees. Conversely, the union of finitely many such families, one in each of distinct trees, is incomparable. Choosing each component family arbitrarily close to its supremum proves

for finitely supported g, and then the asserted sum decomposition follows by completion. For completeness, the dual of an \ell_2-sum of Banach spaces is the \ell_2-sum of their duals: Cauchy–Schwarz bounds the sum of the pairings, while finitely many nearly norming component vectors with suitable scalar weights give the opposite norm inequality. Restrictions to the components determine a functional by density. Applying this description twice, componentwise reflexivity shows that the canonical map into the bidual is onto. Thus Y_{\mathrm I} is reflexive, as is its dual X_{\mathrm I}.□

Proof. Every atom has norm at most one, by Cauchy–Schwarz and (86), and for every g\in Y_{\mathrm I},

If the closed convex hull of the atoms missed a point of the dual unit ball, Hahn–Banach separation would give a separating functional in X_{\mathrm I}^*=Y_{\mathrm I}, using Proposition 9.1. The displayed identity rules out such a separation. Each atom has finite support, which gives the density assertion.□

Proof. The trees (89) exist at every finite height and satisfy Lemma 2.3 with path bounds and increment lower bound all equal to one. More explicitly, for an equivalent norm N with a\|u\|_{\mathrm I}\leq N(u)\leq b\|u\|_{\mathrm I}, that lemma uses t_0=a/(2b). If \delta=\overline\delta_N(t_0)>0, its finite-codimensional approximation and convexity argument selects a child at every nonterminal node for which

Iteration to height h gives a(1+\delta/4)^h\leq b, a contradiction for sufficiently large h.□

Proof. The nonempty intersections of an incomparable segment family with E are again incomparable segments; the same is true of the portions outside E. Here prefix closure ensures that a segment is cut into an initial part and a terminal part. Applying these restricted tests to the original array proves both contraction inequalities on Y_{\mathrm I}, first on finite arrays and then on the completion. The adjoints are contractions and act as the stated coordinate projections on each e_v, hence on X_{\mathrm I} by Lemma 9.2. The range of P_E is contained in the span of the finitely many e_v with v\in E. Finally, approximate x by a finitely supported vector and let E contain its support and all its ancestors. The contraction bound for Q_E gives the last assertion.□

Proof. Convexity gives \|x\|_{\mathrm I}\leq1. Choose f\in Y_{\mathrm I} supported on E such that

Indeed, Hahn–Banach gives a norming element of X_{\mathrm I}^*; reflexivity identifies it with an element of Y_{\mathrm I}, and applying P_E preserves its pairing with x without increasing its norm. If x=0, we may simply take f=0.

For an atom with representation (88), define

Both coefficient vectors (a_i) and (p_i^A) have Euclidean norm at most one. The identity

therefore gives

Replace the coefficients on the terminal pieces by setting

Those nonempty terminal pieces remain incomparable. The atom norm bound and (91) imply

Let \mathcal R be the minimal nodes of \mathcal T\setminus E. They form an antichain and their descendant subtrees partition the complement. If p_i^A\ne0 and S_i\setminus E is nonempty, the original segment meets E and exits it through exactly one q\in\mathcal R. For a fixed q, at most one segment in an incomparable family can do this, since two such segments would both contain q. We may write

Only finitely many coefficients are nonzero. When r_q^A\ne0, let s_q^A\in E be the starting node of the original segment and let t_q^A\succeq q be its terminal node. Then r_q^A is the sum of f from s_q^A to the predecessor of q. For zero coefficients choose t_q^A=q; no starting node is needed.

We next show that opposite coefficients in two atoms consume their norming defects:

For any exit occurring in this sum, the two starting nodes lie on the same ancestor path of q. They are distinct, since identical starts would give identical sums of f, rather than opposite signs. Consider first the exits for which s_q^A\prec s_q^{A'}. Shorten the corresponding segment of A to the interval from s_q^A to the predecessor of s_q^{A'}. This interval is nonempty and is a subsegment of the original segment. Distinct exits correspond to distinct segments, so all these shortenings can be made simultaneously. Every modified family is still incomparable. Its corresponding squared evaluation changes from (r_q^A)^2 to (r_q^A-r_q^{A'})^2, an increase of

All its squared evaluations sum to at most one. Thus the sum of these products is at most \tfrac12(1-\sum_i(p_i^A)^2). For the exits with the opposite ordering, shorten the segments of A' instead and obtain the other half of (94). The final inequality follows from (91).

We have controlled the replacement error for one atom and opposite exit coefficients for two atoms. It remains to apply these facts to convex combinations approximating the two endpoints x+y and x-y. Fix \varepsilon>0. By Lemma 9.2, choose a finite probability distribution of atoms for each endpoint whose mean differs from that endpoint in norm by at most \varepsilon. Choose \sigma\in\{-1,1\} with equal probabilities, then choose A from the distribution for x+\sigma y. Consequently

Contractivity of Q_E, the identity Q_Ex=0, and Cauchy–Schwarz applied to (92) give

Put m_q=\mathbb E r_q^A and b_q=\mathbb E|r_q^A|. Apply (94) to independent copies A,A' of the unconditional atom distribution. With r_+=\max(r,0) and r_-=\max(-r,0), independence gives

Since b_q^2-m_q^2=4\mathbb E[(r_q^A)_+]\,\mathbb E[(r_q^A)_-],

At this fixed \varepsilon, the distributions are finite and their atoms have finite support. Thus all sums of m_q and b_q above are finite, even though \mathcal R can be infinite.

An array supported on the antichain \mathcal R has Y_{\mathrm I}-norm equal to its Euclidean norm: a segment meets at most one of its support nodes, and singleton segments realize the Euclidean test. The coefficient of \mathbb E\widehat A at q is m_q, so duality gives

For g\in Y_{\mathrm I} with \|g\|_{Y_{\mathrm I}}\leq1, define

For each finite subset of exits, choose segments approaching these suprema. The chosen segments are incomparable, so their squared evaluations sum to at most one. Passing to the suprema and then taking the supremum over finite subsets proves \sum_qL_q^2\leq1; no supremum need be attained. It follows that

Taking the supremum over g, and using (95)–(97), yields

The right-hand side is increasing in \bar d. Substitute \bar d\leq1-\|x\|_{\mathrm I}+\varepsilon and let \varepsilon\downarrow0. The two limiting contributions are \sqrt{2(1-\|x\|_{\mathrm I})} and \sqrt{6(1-\|x\|_{\mathrm I})}, proving (90).□

Proof. Fix 0<\eta<c and use Lemma 9.4 to choose E with \|Q_Ex\|_{\mathrm I}<\eta. The sequence (u_j) is bounded, since u_j=((x+u_j)-(x-u_j))/2. The range of P_E is finite dimensional, so there are distinct j,k such that, for w=(u_j-u_k)/2, one has \|P_Ew\|_{\mathrm I}<\eta. The separation assumption gives \|w\|_{\mathrm I}\geq c. Moreover, x+w is the midpoint of x+u_j and x-u_k, and x-w is the midpoint of the other two endpoints. Hence \|x\pm w\|_{\mathrm I}\leq1. Therefore

Apply Theorem 9.5 after dividing both vectors by 1+2\eta:

Since \|P_Ex-x\|_{\mathrm I}<\eta, letting \eta\downarrow0 gives c\leq K\sqrt{1-\|x\|_{\mathrm I}}, which is (98).□