Midpoint convexity from two recursive potentials

Proof of Proposition 2.2. The fields are positively homogeneous and subadditive. To verify subadditivity, induct upwards: the child Euclidean norms are monotone on nonnegative arrays and satisfy the triangle inequality. Each entry in either maximum in (2.1) is therefore subadditive, and a maximum of two subadditive functions is subadditive. Negation interchanges P_s and Q_s. This proves symmetry and the triangle inequality for each proposed norm. Moreover,

by following the child inequalities to the corresponding root. This proves definiteness and coordinate continuity. The coordinate vectors are linearly independent, so the spaces are infinite dimensional.

The fields of \Pi_Hv vanish outside H. Inside H its coordinates are unchanged, and each maximum is nondecreasing in the child fields; induction gives P_s(\Pi_Hv)\le P_s(v) and the same for Q_s. Thus \Pi_H is contractive. Ignore the fixed coordinate \rho when N=N_0. It extends to the completion and has finite-dimensional range. Applying I-\Pi_H to finite approximations proves tail density. Choose an increasing sequence (H_k) of finite initial sets covering the tree. If a finite vector v is supported in H_k, then

Density therefore gives \Pi_{H_k}x\to x for every completed vector x. In particular, the coordinates separate points. A completed vector whose coordinates vanish outside a finite initial set H equals \Pi_Hx, so support in H has its usual coordinate meaning. Subadditivity and sign exchange give

The fields therefore extend continuously, and the root norm formulas persist on the completions.

We prove reflexivity by describing the finite-height components. Set M_n(v)=\max\{P_{r_n}(v),Q_{r_n}(v)\} on c_{00}(T_n), and let E_n be its completion under this norm. Writing m_s=\max\{P_s,Q_s\} gives

The Euclidean norm of the m_s on one level is at most the Euclidean norm of the coordinates on that level plus the corresponding norm of m_c on the next level. Iteration and Cauchy–Schwarz yield M_n(v)\le\sqrt{n+1}\,\|v\|_2. Conversely the \ell_2 norm of either field on any level is at most its root value. Since v=P-Q,

Hence E_n is \ell_2(T_n) with an equivalent norm, and is reflexive. The recursion extends to its vectors by continuity or directly by the finitely many levels.

For v with zero joining coordinate, put L(v)=(\sum_n M_n(v|_{T_n})^2)^{1/2}. The two child arrays at \rho give

For a free root coordinate a=v_\rho, with the remaining coordinates written as w, let A=(\sum_nP_{r_n}(w)^2)^{1/2} and B=(\sum_nQ_{r_n}(w)^2)^{1/2}. Then

For the lower bound, N_J\ge |a| and N_J\ge A+B\ge L(w). These comparisons identify X_0 with the \ell_2 sum of the E_n under an equivalent norm, and X_J with that sum plus \R. The Hilbert comparisons for individual components need not be uniform in n: we take the outer sum in the norms M_n themselves.

For completeness, an \ell_2 sum of Banach spaces is complete: a Cauchy sequence converges in each component, and its Cauchy bounds pass to all finite component sums and then to their supremum. Its dual is the \ell_2 sum of the component duals. The upper norm bound is Cauchy–Schwarz; the lower bound follows by testing a functional on finitely many almost norming component vectors with scalar coefficients proportional to their dual norms. Finite component vectors are dense. Applying this dual identification twice proves reflexivity when each component is reflexive. Equivalent norms preserve reflexivity. This proves the assertion for both joined spaces.□

Proof. Put E_s=[d_s(2P_s+d_s)(2Q_s+d_s)]^{2/3}. We show that E_s\ge\sum_c E_c. Suppose first that P_s^2=\sum_cP_c^2, and put

Use the same definitions at the children. The upper inequality for P' and the equality for P give U_s\ge\sum_c U_c. The triangle inequality in \ell_2 applied to the child fields Q',Q gives V_s\ge\sum_c V_c. Consequently Hölder’s inequality yields

If the tight field at s is Q, interchange P,Q in this calculation. At a leaf in J, E_b=d_b^2. Iterating the inequality through the finite tree and discarding the nonnegative energies of the other leaves proves the assertion.□

Proof of Lemma 3.2 and the quantitative assertions of Theorem 2.1. We use the fields from (2.1). For a finite head vector x and a finite tail z vanishing on an initial set H containing \operatorname{supp}x, set

The coordinate difference identity and convexity give the same nonnegative defect d_s=P'_s-P_s=Q'_s-Q_s. The averaged fields satisfy the upper inequalities by the \ell_2 triangle inequality. At a first node b outside H, the head fields vanish on its whole subtree, so

Only finitely many such values are nonzero. Apply Lemma 3.1 to the finite relevant part of H and its first-exit boundary, retaining at each boundary node its already-computed full-subtree field values. Discard only subtrees where all four fields vanish. This gives the bound on \sum_b d_b^2. Figure 1 shows the finite part through which the inequality is propagated.

We next convert this energy into a norm estimate for each of the two root conventions. On H the coordinate of z is zero, so P_s(z)=Q_s(z) there. With r_s=\max\{P_s(z),Q_s(z)\} one has r_s^2\le\sum_c r_c^2 on H and r_b\le2d_b at its boundary. Thus r_o\le2(\sum_b d_b^2)^{1/2}.

The sum of the root fields. First use either root-sum space X_\Sigma or X_J, and write N=N_\Sigma or N=N_J, respectively. Assume N(x)=1 and put

Since N(z)=2r_o, the energy lemma gives

The last inequality uses P_o+Q_o=1. The field continuity and tail density from Proposition 2.2 extend this cubic inequality to every z\in F_H, with the finite unit center x fixed. This proves Lemma 3.2. If N(z)\ge t with 0<t<1, then d\ge t^3/256: for d\le1 use (1+d)^2\le4, and for d>1 the claim is immediate. Since the average gain is 2d, it is at least t^3/128.

Fix either root-sum space X\in\{X_\Sigma,X_J\} with its norm N. For any unit center x_0\in X and \varepsilon>0, choose a finite unit center x with N(x-x_0)<\varepsilon, then choose a finite initial H containing its support. For all y\in F_H with N(y)\ge1, put z=ty in the preceding estimate. Replacing x by x_0 loses at most \varepsilon in the average, uniformly in y. Thus the supremum over finite-codimensional subspaces at x_0 is at least t^3/128-\varepsilon. Letting \varepsilon\downarrow0 and taking the infimum over unit x_0 proves (2.3).

The common root field. Now use X_0 and write N=N_0, with o=\rho in the preceding calculation. By definition N(v)=P_o(v)=Q_o(v). Put

Now N(z)=r_o, and the same energy lemma yields

The triangle inequality gives d\le N(z). For z\ne0 this proves the homogeneous estimate

The tail density from Section 2 and continuity extend this inequality to every nonzero completed tail, proving (2.4).□

Proof. Set t_0=\alpha c/(2\beta M) and suppose 0<\gamma<\overline\delta_N(t_0). At a current vector u_s^{(h)}=u, choose a closed finite-codimensional subspace F such that

Convexity extends this inequality to every w\in F with N(w)\ge t_0N(u). More explicitly, for a unit center x and a unit direction y, convexity at the radii 0,t_0,r gives N(x+ry)-1\ge(r/t_0)(N(x+t_0y)-1) for r\ge t_0.

Weak nullity persists under an equivalent norm. In the finite-dimensional quotient X/F the increments converge to zero in norm, so there are w_k\in F with N(w_k-d_{s,k}^{(h)})\to0. For large k, N(w_k)\ge\alpha c/2\ge t_0N(u). Choose k with error less than \gamma N(u)/2. The preceding estimate then gives

Choose h such that \alpha m(1+\gamma/2)^h>\beta M and iterate this choice for h steps, a contradiction. Restricting an infinite tree to that height gives the same argument. Finally, \overline\delta_N(t_0)\ge0: intersect a finite-codimensional subspace with the kernel of a functional norming a given center. Both signs of every perturbation in that kernel have norm at least that of the center.□

Proof of the renorming assertion in Theorem 2.1. On the infinite tree define v_s=\sum_{r\preceq s}e_r. Along this path the fields of v_s are P=1,Q=0, so N_\Sigma(v_s)=1. Its child increments are e_{sk}, and for a finite scalar family (a_k) the recursion gives

Here a^+=\max\{a,0\} and a^-=\max\{-a,0\}. In particular N_\Sigma(e_{sk})=2. The map taking the unit vectors of \ell_2 to e_{sk} extends to a bounded linear map by (4.1). Since the unit vectors of \ell_2 are weakly null, so are these increments against every continuous linear functional on X_\Sigma. Lemma 4.1 applies with m=M=1,c=2.

For a node s\in T_n, let v_s be the indicator of the path from r_n to s, with joining coordinate zero. Its component fields are again P=1,Q=0 along the path. At \rho they both equal one. Thus

At a fixed non-leaf s in a component, the sibling sequence is weakly null: restricting any functional to that component gives a continuous functional on a space equivalent to \ell_2(T_n) by (2.5). Alternatively, the sibling computation in (4.1) gives a direct bounded map from \ell_2; for N_0 its factor two is replaced by one. The joined trees contain components of every finite height. Lemma 4.1 therefore applies to X_0 with m=M=c=1, and to X_J with m=M=c=2. In both cases the vanishing one-sided modulus occurs at t_0=\alpha/(2\beta); for X_\Sigma it occurs at t_0=\alpha/\beta.□

Proof. We first work with one pair x,z, writing H for its head and omitting n. Write C_s^1=P_s(x) and C_s^2=Q_s(x). For s\in H define

Every subtree rooted in H contains a nonzero coordinate of x. The child inequalities therefore imply w_s>0. At the root, 1/2\le w_o\le1, since P_o(x)+Q_o(x)=1. Also (u_s^1)^2+(u_s^2)^2=1. Let I_s be the children of s in H, and J_s the children outside H. Choose an index i_s\in\{1,2\} for which the child inequality of the base vector is tight at s.

For \sigma\in\{-1,1\} set

The two differences agree because z_s=0 on H. They are nonnegative: at exterior children the base fields are zero, and upward induction uses the monotonicity of the recursion in both child bounds. Thus

Here and below, C^1(v)=P(v) and C^2(v)=Q(v) when an argument is specified. Define the exterior energies by

Negation interchanges the fields, so the last sum is independent of our choices of i_s. Only finitely many exterior terms are nonzero. On H the fields of z agree. The recursion there gives

Telescoping over H proves

We will show that a small root defect forces this exterior energy to be small, independently of the head.

There is a second useful estimate, for every a\in H:

Choose one index i with u_a^i=\ell_a, and iterate the child inequality for the single field C^i(x+\sigma z) from a to all first exterior nodes below it. At those nodes the whole base subtree vanishes. Summing over signs makes the resulting boundary sum independent of i, and division by w_o gives (5.6).

The identity that propagates the defect. For a fixed sign abbreviate d_s=d_s^\sigma. Subtracting the tight base inequality for C^{i_s} from its endpoint inequality yields

Whenever u_s^{i_s}>0, put R_{sc}=u_c^{i_s}/u_s^{i_s}. Tightness gives \sum_{c\in I_s}\nu_cR_{sc}^2=\nu_s. Expanding the squares in the following display and using that identity rewrites (5.7) as

The last sum measures the failure of the defect to propagate by the ratios R_{sc}. Multiplying those ratios along a branch makes the left side telescope.

Fix 0<\lambda<1/2, 0<\zeta<1, and h>0. Inspect the root with Z_o=1. At an inspected node s, stop when

If none holds, call s a continuation node and inspect its children c\in I_s with Z_c=Z_sR_{sc}. The denominator in R_{sc} is at least \lambda at such a node. Let G and S denote the continuation and stop sets. Every leaf of H has \ell_s=0, so every inspected branch stops. The stops are incomparable, and the child inequalities give

Multiplying (5.8) at s\in G by Z_s/[2(u_s^{i_s}+d_s^\sigma)], which is at least \zeta/[2(1+h)], and summing gives

If the root itself stops, this is equality with G empty.

A nonroot stop T has a parent s\in G, and R_{sT}d_s^\sigma\le h/\lambda. Use (a+b)^2\le2a^2+2b^2, then (5.10) and (5.11), to obtain

The first term includes the possible root stop. Classify stops first by \ell_T<\lambda, then by \ell_T\ge\lambda and Z_T<\zeta. At every remaining stop, Z_T\ge\zeta and some d_T^\sigma>h. The boundary term in (5.11) therefore gives

Only the middle term still needs a bound uniform in the tree.

The probability of a small product. Run the following finite random process from o. At s\in H move to c\in I_s with probability w_c/w_s, and move to absorbing states labelled 1 and 0 with respective probabilities

These are nonnegative and sum with the child probabilities to one. Stay at an absorbing state and run to one step beyond the height of H. Reaching a particular node s has probability \nu_s. Give that node value p_s=P_s(x)^2/w_s, and an absorbing state its label. The conditional expectation of the next value is p_s, since

The successive values p_j thus form a martingale in [0,1] for the filtration generated by the path.

On [\lambda^2,1-\lambda^2] define H_1(p)=\tfrac12\log p and H_2(p)=\tfrac12\log(1-p). Their derivatives satisfy

At a step starting at s\in G take b_j=H_{i_s}'(p_s); at every other step take b_j=0. These coefficients are known before the step. With \Delta p_j=p_{j+1}-p_j, put

The sum M is a predictable martingale transform [4]. Here its second moment follows directly: conditional mean-zero increments make the cross terms vanish, giving

Suppose that the path reaches T\in S with \ell_T\ge\lambda and Z_T<\zeta. All preceding nodes are in G, and all values through T lie in [\lambda^2,1-\lambda^2]. On each such edge s\to c, Taylor’s inequality gives

After this stop the path meets no further node of G. Thus the later coefficients of M are zero, and allowing later nonnegative terms in V gives \log\zeta>\log Z_T\ge M-K_\lambda V. For L=\log(1/\zeta) this event is contained in \{M<-L/2\}\cup\{K_\lambda V>L/2\}. Chebyshev’s and Markov’s inequalities, followed by (5.15), show that

Different stops correspond to disjoint path events. For each fixed \lambda, the right side tends to zero as \zeta\downarrow0.

Finishing the estimate in the required order. The head consists of G and the disjoint subtrees beginning at its stops. By (5.6), the latter contribute at most

to \mathcal E. Write \alpha=\sum_\sigma d_o^\sigma and \beta_0=\sum_\sigma(d_o^\sigma)^2. Combining (5.11)–(5.13) with (5.16) proves the finite estimate

Restore the sequence index. The assumed convergence and (5.3), together with w_o\ge1/2, imply d_o^\sigma\to0 for both signs. Keep \lambda,\zeta,h fixed while taking this limit in (5.17). Then

Choose \lambda small, then choose \zeta small with that \lambda fixed, and finally choose h small. The right side becomes arbitrarily small. Therefore \mathcal E_n\to0, and (5.5) proves the lemma.□

Proof. Fix t>0. Lemma 5.1 gives an \eta>0, independent of the finite unit center x, such that the average endpoint norm is at least 1+\eta for every finite tail z vanishing on its predecessor closure and satisfying N_\Sigma(z)\ge t/2. Otherwise a sequence would contradict the lemma; convexity always bounds the average below by one.

Given a unit center x_0 in the completion, choose a finite unit vector x with N_\Sigma(x-x_0)<\eta/2, and let H be its predecessor closure. The finite tails are dense in \ker\Pi_H, as established in Section 2. A tail of norm at least t is the limit of finite tails whose norms are eventually at least t/2. Continuity therefore gives the endpoint estimate at x for this completed tail. Replacing x by x_0 loses less than \eta/2 in the average. Taking z=ty with y\in\ker\Pi_H and N_\Sigma(y)\ge1, then performing the defining infimum and supremum, gives \widehat\delta_{N_\Sigma}(t)\ge\eta/2>0.□

Proof. Put u=p+r/2 and v=q+r/2. These are upper fields by convexity. We first distribute probability along the tree. Set \mu_o=1, and choose masses recursively so that the masses of the children of every nonleaf sum to its mass and

At the root the right side is at most one. At every other step the sum of the children’s required lower bounds is at most the required lower bound at their parent. Assign those lower bounds first and distribute the remaining mass among the children. This constructs the masses. Nodes of zero mass have all three fields zero and can be omitted from the random process.

Start a path at the root, moving from s to c with probability \mu_c/\mu_s. Once it reaches a leaf, leave it there until the common terminal time l, the height of the tree. Relative to the filtration of the path, define

All three processes lie in [0,M], and U_i,V_i\ge R_i/2. The upper inequalities give \mathbb E_i U_{i+1}^2\le U_i^2 and \mathbb E_i V_{i+1}^2\le V_i^2. Conditional Cauchy–Schwarz also makes U,V supermartingales.

At each nonleaf choose U if the inequality for p is tight, and choose V otherwise. Write W_i^{(i)},W_{i+1}^{(i)} for the two consecutive values of the process chosen at time i; choose U once the path has stopped at a leaf. Subtracting the tight squared inequality for p from the one for p+r gives

The corresponding formula holds for v when q is tight. Consequently

Put L_i=W_{i+1}^{(i)}/W_i^{(i)} when the denominator is positive, and L_i=0 otherwise. A zero denominator forces R_i=0 and the conditional product on the right to vanish, so in every case R_i\ge\mathbb E_i[L_iR_{i+1}]. Iterating yields

The boundary hypothesis says \mathbb ER_l^2=\sum_b r_b^2\ge\beta^2. Since R_l\le M,

On this event U_l,V_l\ge\beta/4. For the first time U_i<c_*, condition on the path up to that time. The squared supermartingale inequality gives

This follows by summing the conditional estimate over the disjoint first-time events; the same argument applies to V. Our choice of c_* makes twice this bound at most h/2. Hence the event

has probability at least h/2.

We have retained a definite set of paths on which all ratios can be logged. It remains to prevent their product from becoming too small. Define the increasing concave function

The processes A_i=g(U_i)-g(0) and B_i=g(V_i)-g(0) are nonnegative supermartingales bounded by H_*. Indeed, conditional Jensen followed by the monotonicity of g proves their supermartingale inequalities. Let

For either process D=A or D=B, write d_i^D=D_i-\mathbb E_iD_{i+1}\ge0. Telescoping expectations gives \sum_i\mathbb E d_i^D\le H_*. Expanding squared increments gives

Split each selected increment in Z into its conditional mean and its centered part. The absolute expectation of the sum of the means is at most 2H_*. The centered parts are martingale differences, since the choice between A and B is made before the step. Their cross terms have zero expectation, and their total variance is at most 6H_*^2 by (6.5). Therefore

Markov’s inequality and (6.1) give \mathbb P(Z<-K)\le h/4, so \mathbb P(G\cap\{Z\ge-K\})\ge h/4. On G, the product of ratios in (6.3) is e^Z. Restricting its expectation to this last event proves r_o\ge (h/4)(\beta/2)e^{-K}=\gamma(M,\beta).□

Proof. Write o=\rho for the joining root. Use p=P(x) and q=Q(x), and define r by

The same r occurs because both differences of fields equal x. Convexity gives r\ge0. Take the finite tree consisting of H and the first vertices outside H on paths to the support of y. Call those exterior vertices B. Preserve the complete subtree values at them. The four restricted fields are upper fields; a base tight inequality remains tight at every internal node because omitted subtrees have zero base fields. At the root the endpoint average gives p_o+q_o+2r_o\le2. At b\in B,

On H the two fields of y agree. Upward induction therefore bounds both by the Euclidean norm of the boundary sums P_b(y)+Q_b(y) below the node. In particular,

Other leaves, if present, have nonnegative r_b^2 and cause no loss. Lemma 6.1 with M=2 and \beta=\zeta/4 gives r_o\ge\gamma(2,\zeta/4). The average norm equals N_J(x)+2r_o.□

Proof. The hypotheses imply N_J(z_j)\le1, so for \varepsilon>2 there is no such family and any \eta\in(0,1) works. Suppose 0<\varepsilon\le2 and fix 0<\delta<\min\{1/3,\varepsilon/8\}. Choose a finite vector x' with N_J(x-x')<\delta, and let H be the finite predecessor closure of \operatorname{supp}x' together with the root. The bounded sequence \Pi_H z_j lies in a finite-dimensional space. Hence we can choose i\ne j so that, for u=(z_i-z_j)/2, N_J(\Pi_Hu)<\delta. Then

For example, x+u is the average of x+z_i and x-z_j. The completed tail u-\Pi_Hu can be approximated, by the density statement of Section 2, by a finite tail y satisfying N_J(y-(u-\Pi_Hu))<\delta. Consequently

Apply Lemma 6.2 with \zeta=\varepsilon/4 to obtain

Let \delta\downarrow0. This proves the stated conclusion, with room to choose the displayed \eta strictly below one.□

Proof. Write p=p_h and M_s=\max\{P_s(v),Q_s(v)\}. The recursion gives

Taking the \ell_p norm on a level and proceeding upward proves that all the defining arrays have finite level norms and that

In particular, the recursion is defined for every v\in\ell_p(T_h). Conversely, iteration of the child inequalities gives, on every level j,

Since v=P(v)-Q(v), these estimates imply

They also give |v_s|\le\nu_h(v) for every coordinate s.

Negation exchanges the two arrays, and positive scaling scales them. If (P,Q) and (P',Q') are feasible pairs for v and w, then (P+P',Q+Q') is feasible for v+w, by the triangle inequality in \ell_p. Minimality therefore gives the triangle inequality for \nu_h. The lower bound in (7.2) gives definiteness. Norm equivalence proves completeness of E_h and density of its finitely supported vectors.

On a countable set, the usual dual of \ell_p, 1<p<\infty, is \ell_{p/(p-1)}: Hölder gives the pairing in one direction, and testing a functional on finitely supported signed powers of its coordinate values gives the converse and the dual norm. Applying this twice proves reflexivity, and finite support approximation proves dual separability. Equivalent norms preserve the continuous dual, bidual, and surjectivity of the canonical map into the bidual, so these properties hold for E_h.

The outer-sum argument in Proposition 2.2, applied to these component norms, proves completeness and reflexivity of X_{\mathrm v} and the dual identification

under the componentwise pairing. Finite component approximation and separability of each E_h^* give separability of this dual. Truncating the component sum and then approximating within the retained components proves finite total support density in X_{\mathrm v}. Coordinate evaluations are bounded by the norm; each finite coordinate projection is consequently a finite sum of bounded rank-one operators.□

Proof. For s\in T_h, let e_s be its coordinate vector and put b_s=\sum_{t\preceq s}e_t. The path indicator itself is a feasible P array for b_s, with Q=0, and its root coordinate supplies the reverse inequality. Hence \nu_h(b_s)=1. For a nonterminal s, the vector e_{si} has fields P=1,Q=0 at si and P=Q=1 at each proper ancestor. Thus \nu_h(e_{si})=2. The sequence of distinct child vectors (e_{si})_i is weakly null in E_h, by (7.2) and p_h>1, and hence in X_{\mathrm v}. Since b_{si}-b_s=e_{si} and the component heights are unbounded, Lemma 4.1 applies with m=M=1, c=2, \alpha=a, and \beta=b. It gives \overline\delta_N(a/b)=0, excluding an AUC norm.□

Proof. The total array R_s=P_s+Q_s of any vector satisfies \|(R_{si})_i\|_p\le R_s, by Minkowski. Consequently its pth powers sum to at most R_s^p on any antichain below s. This follows first for finite antichains by iterating the child inequality, and then for countable ones by increasing finite subsets.

There is a feasible pair for y which uses its minimal arrays off A and whose two values at s\in A both equal

Grouping boundary descendants by the first child proves the child inequalities. The antichain estimate just proved makes the sums finite. Minimality, (7.4), and Minkowski give

Choose a sign \sigma with \|(v_i^\sigma)_{i\in H}\|_p\ge Y/4. For this sign, put

Let L_s be the old value at s\in A of its chosen tight array. Subtract its old equality from the new child inequality. For children in A, use (a+d)^p-a^p\ge d^p, and for boundary children use that the old value was zero. It follows that

Here and below a node in A with no boundary descendants makes no contribution. Define z_s=d_s/R_s when R_s>0, and z_s=0 otherwise. All child values vanish when R_s=0; for positive R_s we have

The weights in this sum have total at most one.

To keep track of the boundary mass, define for s\in A\cup H

The antichain bound gives 0\le W_s\le1. At s\in H we have W_s=z_s^p, and at s\in A the quantity W_s is the weighted sum of the child quantities with the weights in (7.6). For a node with remaining height m=h-|s|, we claim

It holds at boundary nodes because z_s=W_s^{1/p}, 0\le W_s\le1, and C_m\le1. At nodes of A of depth h, W_s=0. For the induction step, insert the child bounds into (7.6) and apply Jensen to the power p^m, placing any missing weight at zero. This proves (7.7), including countably many children by the same Jensen inequality for a probability measure on a countable set.

At the root,

Indeed p^h\le\mathrm e and \log p\le p-1 bound the first geometric sum after taking logarithms. If Y=0, the conclusion is immediate from M_\pm\ge B. Otherwise let M=M_\sigma=R_o. The selected sign gives W_o\ge(Y/(4M))^p, so Y/(4M)\le1. Equation (7.7) therefore yields

Since M\le U,

The other squared endpoint norm is at least B^2, proving (7.5). If x=0, then U^2=2Y^2, and the same inequality holds directly because c_6\le8.□

Proof. Set Y_h=\nu_h(y^{(h)}) and U_h^2=\nu_h(x^{(h)}+y^{(h)})^2+ \nu_h(x^{(h)}-y^{(h)})^2. Lemma 7.3 applies also to components with x^{(h)}=0. Summing it reduces the assertion to

For S>0, this is Jensen for the cube with weights U_h^2/S and values Y_h^2/U_h^2. Components with U_h=0 have both vectors zero and can be omitted. The case S=0 is immediate.□

Proof. By homogeneity, assume M=1, so \nu_h(y)\le1. Use p=p_h, the common increments (7.3), and the same choice of tight label at each node as above. Denote that label by S in a calculation at its chosen node; the children in that calculation use the parent’s label. For a nonterminal s\in A, the subtraction that gave (7.6) gives

At an outside child, the two signs interchange P,Q. Thus the sum of the second term over signs is the sum of the pth powers of both arrays for y, independently of the parent’s label.

For 0\le j\le h, and for 0\le j<h respectively, define

An empty sum is zero. Summing (7.10) gives

The sum of S_s(x+\sigma y)^p on the right is at most two: for each sign, the sum of the pth powers of both arrays on a level is at most P_o(x+\sigma y)^p+Q_o(x+\sigma y)^p\le1. Moreover, the first-exit children form an antichain, so

The antichain argument includes infinitely many children by increasing finite subsets, as in Lemma 7.3.

We retain the exit mass from all preceding levels by setting

Add \sum_{i<j}H_i^p to (7.11), bound it by p\sum_{i<j}H_i^{p-1}H_i, and apply Hölder to this list of terms together with the terms indexed by (s,\sigma). The p/(p-1) powers of the first factors sum to at most 2+1=3, by the level bound and (7.12); the pth powers of the second factors sum to K_j^p. Therefore

Iterating gives

For the last inequality, use \sum_{i=1}^{h}p^{-i}\le1/(p-1) and p^{1/(p-1)}\le\mathrm e.

It remains to relate the accumulated exit mass to the norm of y. On A, the two arrays for y are equal, since y_s=0. Their common pth power at a nonterminal node s\in A is

At nodes of A of depth h it is zero. Bounding each maximum by the sum of its two entries and iterating upward yields P_o(y)^p\le\sum_{j<h}H_j^p. Consequently

Combining with (7.14), and using p^h\le\mathrm e<3 and 0\le\nu_h(y)/(6\mathrm e)\le1, gives

Finally,

This proves the normalized estimate and hence (7.9) by scaling. For x=0 its left side is \nu_h(y)=M, and the right side is c_3M\le M.□

Proof. In component h write B_h=\nu_h(x^{(h)}), Y_h=\nu_h(y^{(h)}), M_{h,\sigma}=\nu_h(x^{(h)}+\sigma y^{(h)}), M_h=\max_\sigma M_{h,\sigma}, and m_h=(M_{h,+}+M_{h,-})/2. Lemma 7.5, including its zero-center case, and m_h\ge M_h/2 give

whenever M_h>0. When M_h=0, both component vectors vanish. Hölder gives

Since \sum_hM_h^2\le2R^2, summing the component inequalities proves (7.15). The same calculation for finite partial sums and monotone convergence justifies all infinite sums.□

Proof. Fix \eta>0 and choose a vector x' of finite total support with \|x-x'\|_{\mathrm v}<\eta. In its nonzero components, take finite ancestral sets containing the support, and let P be the projection onto their union. Proposition 7.1 makes P bounded and finite rank, so weak nullity gives \|Py_j\|_{\mathrm v}\to0. Set y'_j=y_j-Py_j. The pair x',y'_j satisfies the hypotheses of Theorem 7.6, with R_j=1+\eta+\|Py_j\|_{\mathrm v}. Moreover, \liminf_j\|y'_j\|_{\mathrm v}\ge\varepsilon. First let j\to\infty in (7.15) to obtain

Then let \eta\downarrow0. This gives (7.16).□