§ 1The contest, on one page
Two players fight over two rounds. Player $i$ values winning at $v_i;$ player 1 starts round one with a head start $\Delta$ on the score. Effort costs $c(e),$ the same convex function for both, with $c(0)=c'(0)=0.$ The number that turns value into effort is the reach,
$$ \begin{align*} w_i &:= c^{-1}(v_i), \qquad\text{the effort at which player } i \text{ has burned the whole prize,}\\ f &:= w_2 - w_1 - \Delta, \qquad\text{the effective lead, once reaches are netted out.} \end{align*} $$Round two is a static contest fought under whatever lead $\Delta_2$ entered it, and its value is known in closed form (the paper’s Lemma 1). It is piecewise, and the pieces are the whole story:
$$ \begin{align*} U_1^{T}(\Delta_2) &= \begin{cases} 0, & \Delta_2 \le w_2 - w_1,\\ v_1 - c(w_2 - \Delta_2), & w_2 - w_1 \lt \Delta_2 \lt w_2,\\ v_1, & \Delta_2 \ge w_2, \end{cases} \\[6pt] U_2^{T}(\Delta_2) &= \begin{cases} v_2, & \Delta_2 \le -w_1,\\ v_2 - c(w_1 + \Delta_2), & -w_1 \lt \Delta_2 \lt w_2 - w_1,\\ 0, & \Delta_2 \ge w_2 - w_1. \end{cases} \end{align*} $$Read the middle branch as economics: a player who enters the last round behind by a margin must spend the cost of catching up to the rival’s reach, and keeps the value net of that. The outer branches are saturation: a lead so large that the last round is not fought at all. On the interior, note, it is the rival’s reach that appears. That is what makes the asymmetric case subtle and the symmetric case deceptively simple.
In round one player $i$ picks $e_i,$ and the entering lead of round two is $\Delta_2 = \Delta + e_1 - e_2.$ Player 1’s total payoff is $U_1^T(\Delta + e_1 - e_2) - c(e_1).$ The rival’s effort enters through the lead itself, inside every $c(w \pm \Delta_2),$ not merely through a win probability. So round one is not a standard all-pay auction, and the usual closed forms do not apply. Throughout, the primitives are competitive, $-w_1 \lt \Delta \lt w_2$: outside that range the head start is decisive and nothing is contested.
§ 2Combs: what an equilibrium looks like
Equilibria are mixed, and they are combs: player 1 spreads probability over finitely many efforts $x_1 \lt \cdots \lt x_{n_1}$ with masses $p,$ player 2 over $y_1 \lt \cdots \lt y_{n_2}$ with masses $q.$ Only $n_2 \in \{n_1, n_1+1\}$ occurs, and the pair is written $\sigma(n_1,n_2).$ One orientation is fixed once and for all: the player who earns zero rent is called player 2, so $u_2 = 0$ and her lowest atom sits at zero effort. In the shifted coordinate $\tilde y_\ell = y_\ell + f$ the two supports interleave,
$$ \tilde y_1 = f \;\lt\; x_1 \;\lt\; \tilde y_2 \;\lt\; x_2 \;\lt\; \cdots $$and the lead $f$ appears in exactly two places, both on player 2’s own side. Geometrically, each atom of one player sits strictly between two atoms of the other. Economically, the two supports are teeth that mesh: each effort is chosen to be exactly indifferent against the rival’s comb.
Two equations per atom
An atom is placed by two conditions: the player is stationary there (a tangency, using $c'$) and indifferent there (a level, using $c$). Written against the rival’s atoms, with the truncation that only pairs on the interior branch of Lemma 1 contribute,
$$ \begin{align*} \text{player 1 at } x_k: && \sum_{\ell \le k} q_\ell\, c'\!\bigl(w_1 + \tilde y_\ell - x_k\bigr) &= c'(x_k),\\ && \sum_{\ell \le k} q_\ell\,\bigl[v_1 - c(w_1 + \tilde y_\ell - x_k)\bigr] - c(x_k) &= u_1,\\[8pt] \text{player 2 at } \tilde y_\ell: && \sum_{k \lt \ell} p_k\, c'\!\bigl(w_2 + x_k - \tilde y_\ell\bigr) &= c'(\tilde y_\ell - f),\\ && \sum_{k \lt \ell} p_k\,\bigl[v_2 - c(w_2 + x_k - \tilde y_\ell)\bigr] - c(\tilde y_\ell - f) &= 0 . \end{align*} $$The bracket $v_i - c(\text{margin})$ is Lemma 1 again: the value net of the cost of catching up from that margin in round two. The index sets are asymmetric on purpose, $\ell \le k$ on one side and $k \lt \ell$ on the other: an atom only “sees” the rival atoms it can still catch. The boundary atom $\tilde y_1 = f$ has no tangency condition; it is a pin.
The dial
Everything above is a forward recursion in one free number,
$$ t := q_1,\qquad \text{player 2’s mass on her zero-effort atom.} $$Economically, $t$ is the probability that the underdog sits round one out. Given $t,$ player 1’s first atom solves the base equation $t\,c'(w_1 + f - x_1) = c'(x_1),$ which fixes her rent $u_1 = t\,[v_1 - c(w_1 + f - x_1)] - c(x_1),$ and then the two players take turns: player 2’s pair of equations yields $(p_k, \tilde y_{k+1}),$ player 1’s pair yields $(q_{k+1}, x_{k+1}).$ The owner of the top atom, the apex player, takes whatever probability is left. So exactly one condition remains to be met, the closing residual:
$$ G(t) = \begin{cases} \sum_{\ell \le n_1} q_\ell(t) - 1, & \text{squares } \sigma(n_1,n_1),\\[2pt] \sum_{k \le n_1} p_k(t) - 1, & \text{rectangles } \sigma(n_1,n_1+1). \end{cases} $$The non-apex player’s masses must sum to one. A candidate equilibrium at this depth is a zero of $G.$ Read $G$ as a budget: negative means the recursion has not yet spent that player’s probability, so the comb could grow another tooth; zero means exactly spent; positive means over-spent, and the rung is infeasible at that dial.
§ 3Rungs, residuals, and the Nash gate
Index the rungs by their chain index $j$: $\sigma(n_1,n_1)$ is $j = 2n_1-1$ and $\sigma(n_1,n_1+1)$ is $j = 2n_1,$ so the chain reads
$$ \sigma(1,1) \to \sigma(1,2) \to \sigma(2,2) \to \sigma(2,3) \to \sigma(3,3) \to \cdots \qquad G_1,\, G_2,\, G_3,\, G_4,\, G_5,\, \ldots $$and note that all the residuals live on the same recursion: $G_{j+1}$ is what the march says after one more tooth than $G_j.$ The first is explicit, $G_1(t) = t - 1.$ Rung $j$ can only be reached where every shallower rung has budget to spare,
$$ U_j := \{\, t : G_i(t) \lt 0 \ \text{ for all } i \lt j \,\}, $$because the recursion needs each earlier player’s probability not yet exhausted in order to place the next tooth. A zero of $G_j$ on $U_j$ is a comb of rung $j;$ it is a Nash equilibrium exactly when three inequalities hold:
$$ \begin{align*} G_{j-1}(t^\ast) &\le 0 &&\text{(NN)}\quad\text{the apex mass } {-G_{j-1}} \text{ is nonnegative,}\\ G_j(t^\ast) &= 0 &&\text{(CL)}\quad\text{the rung closes,}\\ G_{j+1}(t^\ast) &\ge 0 &&\text{(NJ)}\quad\text{the non-apex player cannot profitably jump above the top.} \end{align*} $$together with positive interior masses and $u_1 \ge 0.$ The third line is the one that distinguishes a comb from an equilibrium. Indifference on one’s own support rules out every deviation below the top of the combined support; what it cannot rule out is an overshoot strictly above it, a fresh concave hump carrying no own atom. It turns out that this overshoot is unprofitable precisely when the successor residual is nonnegative: the march would place the next tooth with nonpositive budget, so there is no room for it. That is why the whole Nash layer is a sign pattern on one family of curves.
§ 4Chain monotonicity, and its reduction
The claim. Along the chain, player 1’s equilibrium rent $u_1^\ast$ strictly decreases between all distinct Nash profiles. Deeper rungs pay the leader less. Economically: the finer the mixing, the more of the leader’s advantage is competed away. A rung with more teeth is a more contested game.
Step one: rent rises with the dial
Along the march, at any cost, the rent is a function of the dial alone, $u_1(t) = t\,[v_1 - c(m_0)] - c(x_1)$ with $m_0 = w_1 + f - x_1(t),$ the margin player 1 still has to make up in round two when the underdog sits out. The bracket is what that round is worth to her, and individual rationality signs it: $u_1 \ge 0$ gives $t\,[v_1 - c(m_0)] \ge c(x_1) > 0,$ so the bracket is positive on the admissible window. Differentiate. The base equation is exactly the statement that $x_1$ is stationary, so by the envelope argument the $x_1'$ terms cancel:
$$ \begin{align*} u_1'(t) &= \bigl[v_1 - c(m_0)\bigr] + \underbrace{t\,c'(m_0)\,x_1' - c'(x_1)\,x_1'}_{=\,0\ \text{by the base equation}} = v_1 - c(m_0) \;>\; 0, \\[4pt] x_1'(t) &= \frac{c'(m_0)}{c''(x_1) + t\,c''(m_0)} \;>\; 0 \qquad\text{(differentiate the base equation),} \end{align*} $$so $m_0$ falls as $t$ rises and $u_1'$ is increasing: $u_1$ is convex, and strictly increasing in $t$ on the admissible window, above the individual-rationality floor $u_1 \ge 0.$ In words: the more likely the underdog is to sit out, the more the leader keeps. This is the general, cost-free half of the theorem.
Step two: the dials fall
So the theorem reduces to “the closing dials descend along the chain.” Here is where the residual curves do the work. Suppose every $G_j$ has only up-crossing zeros on $U_j$: wherever $G_j(t^\ast) = 0,$ the residual is negative just to the left and positive just to the right. Then three things follow at once.
- Each $U_j$ is an initial segment $(t_0, \tau_{j-1}),$ ending at the closing of rung $j-1.$ Rung $j$ can only live below where rung $j-1$ closed.
- On that interval $G_j$ changes sign at most once, from $-$ to $+.$ So it has at most one zero, and above its Nash zero it is nonnegative. A distinct deeper rung cannot close at or above that dial. This is the ordering; the same sentence also says each rung has at most one Nash dial.
- With step one, deeper rungs pay less.
This is the $k$-free reduction of the record: it uses no cost class at all. The entire burden is the hypothesis, which has a name.
Note the quantifier: every zero, including the non-Nash ones. The proof cannot look at the gate first and then at the slope; the sign-change argument needs the residual’s behaviour on the whole interval. A gated proof would be weaker to state and, on record, has no mechanism behind it.
§ 5Why single crossing, and not widening
The cubic was already known to satisfy chain monotonicity, by a machine-verified proof of a much stronger property: mass widening. A mass drift is the derivative of a mass with respect to the dial, $p_k'$ or $q_\ell',$ and a gap drift is the derivative of the distance between two neighbouring teeth, $(x_k - \tilde y_k)'$ or $(\tilde y_{k+1} - x_k)'.$ Mass widening is the statement that no mass drift is negative: raising the dial never takes probability away from a tooth. The proof gives more: along the march every mass drift and every gap drift is nonnegative, at every node, at every dial on the window, and single crossing follows. The trouble is that widening is false on the general class $X$ of costs with $c', c'', c''' > 0$ (already the polynomial $e^2 + e^{12}$ has a falling mass at a legal dial), so nothing in that proof ports.
Single crossing is the weaker, $k$-free hypothesis that the reduction of §4 actually consumes, and it is a shape of statement that may survive on a larger class of cost functions: a statement about the sum of the drifts at a zero, not about each drift everywhere. On $X$ itself, single crossing fails at a knee-shaped cost at $f = 0,$ and the theorem as stated fails at another knee cost at $f > 0.$ So the pure cubic is the ground test of the mechanism: the case with no knees, $c''' \equiv 6$ and $c'''' \equiv 0,$ where every coefficient is explicit and where the question is not whether single crossing holds, but whether it can be proved from the objects that exist at a closing. If it can, the proof’s ledger says which cubic facts each step used, and hence what a general class must supply.
The rule of the game, then: no drift sign may be assumed; nothing from the widening proof may be cited; the target is the sign of $G_j'$ at zeros, and only there.
§ 6The closing identity, and the theorem
Call the non-apex player’s side the closing side: its masses, the $q_\ell$ on a square and the $p_k$ on a rectangle, are the ones that must sum to one. Write $\mu_1, \ldots, \mu_i$ for the masses of its teeth below the last one, and $\mu_{\rm new}$ for the mass of the last. The probability the closing side has left before its last tooth is its unused budget $B,$ and the last tooth takes a share $\beta$ of it:
$$ B := 1 - \sum_{j \le i} \mu_j , \qquad \beta := \frac{\mu_{\rm new}}{B} . $$What is left after the last tooth follows by subtraction,
$$ B_{\rm new} := 1 - \sum_{j \le i} \mu_j - \mu_{\rm new} = B - \beta\,B = B\,(1 - \beta), $$and it holds at every dial, not only at a closing, because it is the definition of $\beta$ rearranged. What is left is also the closing residual of §2 with its sign flipped, $G = \sum_{j \le i} \mu_j + \mu_{\rm new} - 1 = -B_{\rm new}.$
A rung closes when the last tooth takes all of the remaining budget: $\beta = 1$ at the terminal contact. That is what $G = 0$ means, and it gives the derivative of the residual for free. Differentiate $B_{\rm new} = B\,(1 - \beta)$ first, and impose $\beta = 1$ afterwards:
$$ \begin{align*} G' &= -B_{\rm new}' = B\,\beta' - (1 - \beta)\,B' \bigr|_{\beta = 1} = B\,\beta' \;>\; 0 . \tag{CR-7} \end{align*} $$Here $\beta' > 0$ follows from (CR-5) in §7, and $B > 0$ because the dial lies in the reachable set $U_j$ of §3: before its last tooth the closing side still has budget to spare, $B = -G_{j-2} > 0;$ on the first two rungs no tooth precedes the last, and $B = 1.$ The rate at which the last share fills is positive: the rung closes transversally, from deficit to surplus. That is single crossing at this closing, Nash or not.
The proof is §4 applied to (CR-7): the $k$-free reduction makes the reachability sets initial segments, the sign-change count orders the dials, and rent rises with the dial. The same count says that each rung of the cubic chain has at most one Nash dial at $f \ge 0,$ a byproduct recorded as cor:cubic-sc-rung-unique.
§ 7The cubic proof: the contact state
§6 left one sign open: the share of the last atom rises with the dial, $\beta' > 0.$ This section proves it for the cubic, at a nonnegative lead $f \ge 0,$ at the closing of a square or a rectangle of any depth. Subsections 7.1 to 7.3 hold for any cost of §1: they rewrite the two equations per atom of §2 as two contact equations and reduce $\beta' > 0$ to a few sign conditions. Subsection 7.4 shows that the cubic meets them, and 7.5 runs the induction along the comb.
The proof treats the two players with one set of coordinates. A side is one player’s atoms and masses as the other player, the rival, sees them: player 1’s side carries the masses $p_k,$ which close the rectangles, and player 2’s side carries the $q_\ell,$ which close the squares. On a side, write $\mu_i$ for the masses and $d_i$ for the atoms, placed so that the rival’s round-two margin against atom $i$ at her own effort $y$ is $d_i - y.$ Write $z_i$ for the rival’s atom just above atom $i,$ measured in her own effort, $w$ for her reach and $u$ for her level, which is her rent:
| side | closes | rival | $\mu_i$ | $d_i$ | $z_i$ | $w$ | $u$ |
|---|---|---|---|---|---|---|---|
| player 1’s side | rectangles | player 2 | $p_i$ | $w_2 + x_i - f$ | $\tilde y_{i+1} - f$ | $w_2$ | $0$ |
| player 2’s side | squares | player 1 | $q_i$ | $w_1 + \tilde y_i$ | $x_i$ | $w_1$ | $u_1$ |
At $f \ge 0$ every margin that appears is strictly positive, so the raw cost and the zero-extended cost agree wherever the march evaluates them: this is the one place the sign of the lead enters, and it is where a negative lead would force a saturation remainder into every formula below.
7.1 The hump, and its double contact
The first step compresses everything the placed atoms do to the rival into one function of one variable. Take a prefix: the atoms placed so far on a side, masses $\mu_1, \ldots, \mu_i$ at $d_1, \ldots, d_i,$ and the contact $z = z_i.$ The rival’s payoff from an own effort $y,$ less her level, is the hump
$$ \Phi(y) := \sum_{j \le i} \mu_j\bigl[c(w) - c(d_j - y)\bigr] - c(y) - u . $$Each term is mass times value net of catch-up cost, less own cost, less the level; the rival’s value is written $c(w),$ by the definition of her reach. From here on a prime is the derivative in the dial $t,$ and a subscript $y$ or $\eta$ is a partial derivative at a fixed dial, as in $\Phi_y := \partial\Phi/\partial y.$ The rival’s atom at $z$ is stationary and indifferent, which are §2’s two equations for that atom:
$$ \Phi(z) = \Phi_y(z) = 0 . $$The hump is tangent to its level at the contact: a double contact. The next contact lies beyond $z,$ so measure the hump from the contact, per unit of the prefix’s unused budget $B,$ as in §6:
$$ B := 1 - \sum_{j \le i}\mu_j , \qquad F(\eta) := -\frac{\Phi(z + \eta)}{B}, \qquad F(0) = F_\eta(0) = 0 . $$Call $F$ the shortfall of the hump below its level, at a distance $\eta$ past the contact. It depends on the dial through the prefix and the contact.
7.2 One contact more
Each step of the march adds one atom to a side and one contact for the rival; this subsection writes that step as two equations, the contact equations. The march alternates between the sides. Call each use of one player’s pair of equations from §2 a visit: player 2’s pair, which yields $(p_k, \tilde y_{k+1}),$ is a visit to player 1’s side, and player 1’s pair, which yields $(q_{k+1}, x_{k+1}),$ is a visit to player 2’s side. A visit finds the side’s next atom already placed, at $d = d_{i+1}$: the previous visit, to the other side, placed it as its new contact. The visit returns two numbers. One is the mass on that atom, $\mu_{\rm new} = \beta B,$ a share $\beta$ of the unused budget as in §6. The other is the next contact, a step $s$ past the old one, at $z + s.$ Between the new contact and the new atom lies the margin
$$ m := d - z - s . $$The new hump is $\Phi(y) + \beta B\,[c(w) - c(d - y)],$ and it must be tangent to its level at $z + s$:
$$ \begin{align*} \Phi(z + s) + \beta B\,\bigl[c(w) - c(m)\bigr] &= 0, \\ \Phi_y(z + s) + \beta B\,c'(m) &= 0 . \end{align*} $$Since $\Phi(z + \eta) = -B\,F(\eta)$ and $\Phi_y(z + \eta) = -B\,F_\eta(\eta),$ dividing by $B$ leaves $\beta\,[c(w) - c(m)] = F(s)$ and $\beta\,c'(m) = F_\eta(s).$ Divide one by the other and the share drops out; what remains is the ratio $\psi := F/F_\eta,$ which we call the contact price. These are the contact equations:
$$ \begin{align*} \frac{c(w) - c(m)}{c'(m)} &= \psi(s), \tag{CT1}\\ \beta &= \frac{F_\eta(s)}{c'(m)} . \tag{CT2} \end{align*} $$Two equations in the two unknowns, $s$ and $\beta,$ since $d - z$ is given and the margin is $m = d - z - s.$ Read the first as an accounting identity for the prize, $c(w) = c(m) + \psi(s)\,c'(m)$: the whole prize is spent on the margin and on the contact price, charged at the marginal cost of the margin. These identities use the two tangencies and nothing else; in particular no derivative of any mass has been taken.
7.3 Moving the dial
Everything is now a function of $t,$ and this subsection finds when the share rises with it. Write $\rho := d' - z'$ for the incoming difference, the rate at which the new atom pulls away from the old contact, so that $m' = \rho - s'.$ A subscript $t$ is a derivative along the dial at a fixed step, as in $\psi_t := \partial\psi/\partial t$ at fixed $\eta.$ Differentiate the first contact equation (CT1) with respect to the dial $t.$ The reach is fixed, and by the equation itself the left side has $m$-derivative
$$ \frac{d}{dm}\,\frac{c(w) - c(m)}{c'(m)} = -1 - \frac{c(w) - c(m)}{c'(m)}\,\frac{c''(m)}{c'(m)}, \qquad\text{and define}\qquad \lambda := 1 + \psi(s)\,\frac{c''(m)}{c'(m)} . $$So $-\lambda\,m' = \psi_\eta\,s' + \psi_t,$ and substituting $m' = \rho - s'$:
$$ s' = \frac{\lambda\,\rho + \psi_t}{\lambda - \psi_\eta}, \qquad s' - \rho = \frac{\psi_\eta\,\rho + \psi_t}{\lambda - \psi_\eta} . $$The share follows. Differentiate the second contact equation (CT2):
$$ \beta' = \frac{F_{\eta\eta}(s)\,s' + F_{\eta t}(s)}{c'(m)} - \beta\,\frac{c''(m)}{c'(m)}\,m' . $$Under the one-step conclusion the last term is positive, since $\beta > 0$ and $m' \lt 0,$ so $\beta' > 0$ as soon as $F_{\eta\eta}(s) \ge 0$ and $F_{\eta t}(s) \ge 0.$ The sign of $\beta'$ is derived, from the rates of the contact and the margin. It is not a hypothesis on a mass drift, which is exactly what the rules forbade; this is the point at which the proof pays its own way rather than borrowing from widening.
7.4 What the cubic supplies
Now take $c(e) = e^3.$ The cubic meets every sign condition that 7.3 puts on the contact price and the shortfall, and it turns the two that involve the dial, $\psi_t \ge 0$ and $F_{\eta t} \ge 0,$ into one condition on one number. Expand the hump: since $-(d_j - y)^3 = (y - d_j)^3 = y^3 - 3d_j y^2 + \cdots,$
$$ \begin{align*} \Phi(y) &= -\Bigl(1 - \sum_j \mu_j\Bigr) y^3 \;-\; 3\Bigl(\sum_j \mu_j d_j\Bigr) y^2 \;+\; (\text{lower order}) \;=\; -B\,y^3 - 3M\,y^2 + \cdots,\\[4pt] M &:= \sum_{j \le i}\mu_j d_j . \end{align*} $$A cubic with a double root at $z$ is $-(y-z)^2$ times an affine factor. Its leading coefficient is $-B,$ and matching the $y^2$ coefficient fixes the factor’s value at the contact:
$$ \begin{align*} \Phi(y) &= -(y - z)^2\bigl[B(y - z) + 3N\bigr], \qquad N := M + B z , \qquad R := \frac{N}{B},\\ F(\eta) &= \eta^2\,(\eta + 3R) . \tag{CR-1} \end{align*} $$That is the whole memory of the prefix, and the level $u$ has dropped out: the prefix and the contact fix it. Whatever the atoms below did, the rival sees only two numbers, $B$ and $N,$ and the shortfall sees only one: the state $R.$ Its numerator $N$ is a centre of mass, the probability-weighted location of the placed atoms with the unused budget parked at the contact, measured in the same effort units. Geometrically: a cubic tangent to a line is a parabola-shaped touch times a straight line, and the straight line has slope $B$ and height $3N$ at the contact.
From (CR-1), $F_\eta = 3\eta\,(\eta + 2R),$ so the contact price at the step is a third of a number $P,$ and with $c(m) = m^3$ the contact equations become explicit:
$$ \begin{align*} \psi(s) &= \frac{P}{3}, & P &:= \frac{s\,(3R + s)}{2R + s}, \\ w^3 &= m^2\,(m + P), & \beta &= \frac{s\,(2R + s)}{m^2}, \\ R_{\rm new} &= \frac{R + s + \beta m}{1 - \beta} \quad (\beta \lt 1) . \tag{CR-2} \end{align*} $$In the cubic the accounting identity reads $w^3 = m^3 + P\,m^2$: the whole prize, in effort units cubed, is spent on the margin and on $P,$ three times the contact price. The last line is bookkeeping, and the cost does not enter it: the new atom adds $\beta B\,d$ to $M$ and leaves ${B_{\rm new} = B(1 - \beta)}$ unused, now parked at the new contact $z + s,$ so
$$ N_{\rm new} = M + \beta B\,d + B(1 - \beta)(z + s) = B\,(R + s + \beta m), \qquad R_{\rm new} = \frac{N_{\rm new}}{B_{\rm new}} . $$Two elementary bounds on $B$ carry the whole argument. With $P_s$ and $P_R$ the partial derivatives of $P$ in $s$ and in $R,$ they sign the two derivatives of the contact price:
$$ \begin{align*} \psi_\eta &= \frac{P_s}{3}, & P_s &= 1 + \frac{2R^2}{(2R + s)^2} \;\in\; \bigl[1, \tfrac32\bigr), \\ \psi_t &= \frac{P_R\,R'}{3}, & P_R &= \frac{s^2}{(2R + s)^2} \;>\; 0 . \tag{CR-3} \end{align*} $$The other two quantities of 7.3 are as explicit. Since $c''(m)/c'(m) = 2/m,$ $\lambda = 1 + 2P/(3m),$ and differentiating $F_\eta = 3\eta^2 + 6R\eta$ once more gives the second line:
$$ \begin{align*} \lambda - \psi_\eta &= \frac{C}{3m}, & C &:= 3m + 2P - m\,P_s \;>\; 0, \\ F_{\eta\eta} &= 6\,(R + \eta), & F_{\eta t} &= 6\,\eta\,R' . \tag{CR-4} \end{align*} $$$C$ is positive because $P_s \lt \tfrac32$ gives $m P_s \lt \tfrac32 m \lt 3m + 2P.$ The march meets only positive steps, since the contacts climb the comb, and nonnegative states, since $N$ averages nonnegative locations. There every sign condition that 7.3 puts on the contact price and the shortfall holds, except the two that involve the dial, and those reduce to one: the state is not falling, $R' \ge 0.$ With it and a positive incoming difference, $s' > \rho > 0,$ $m' \lt 0,$ and
$$ \begin{align*} \beta' &= \frac{2(R + s)\,s' + 2s\,R'}{m^2} - \frac{2\beta\,m'}{m} \;>\; 0, \\[4pt] R_{\rm new}' &= \frac{(1 + 2s/m)\,R' + (1 + 2(R + s)/m)\,s' - \beta\,m' + R_{\rm new}\,\beta'}{1 - \beta} \;>\; 0 . \tag{CR-5} \end{align*} $$The first line is the share formula of 7.3 with the cubic’s $F_{\eta\eta},$ $F_{\eta t},$ $c'$ and $c''.$ For the second, differentiate the last line of (CR-2), written as $R_{\rm new}(1 - \beta) = R + s + \beta m,$ and use the first line to replace $\beta' m.$ Every term on the right is nonnegative and $s' > 0,$ so a state that is not falling is handed on to the next visit to the same side as a rising one.
7.5 The base, and the alternation
It remains to start the induction on each side and to run it along the comb. Both sides start cleanly. Player 2’s side, after its first atom, carries $q_1 = t$ at $d_1 = w_1 + f,$ with contact $x_1$ and unused budget $B = 1 - t,$ so $N = t\,(w_1 + f) + (1 - t)\,x_1$ and its state after one atom is
$$ \begin{align*} R_1 &= x_1 + \frac{t\,(w_1 + f)}{1 - t}, \qquad R_1' = x_1' + \frac{w_1 + f}{(1 - t)^2} \;>\; 0 , \tag{CR-6} \end{align*} $$using $x_1' > 0$ from §4. Player 1’s side starts from the empty prefix: $z = R = R' = 0,$ $B = 1,$ and the hump $-y^3,$ which is tangent to its level $u = 0$ at zero because $c(0) = c'(0) = 0.$ Its first visit receives the atom $d = w_2 + x_1 - f,$ so $\rho = x_1' > 0,$ and returns $\tilde y_2' > x_1'.$
Now alternate. A visit to player 1’s side receives $\rho = x_k' - \tilde y_k' > 0$ and returns $\tilde y_{k+1}' > x_k';$ a visit to player 2’s side receives $\rho = \tilde y_{k+1}' - x_k' > 0$ and returns $x_{k+1}' > \tilde y_{k+1}';$ (CR-5) carries a rising state to the next visit to the same side. Every state that a later visit needs has $B > 0$: the masses still to come on that side are positive, and all of its masses sum to at most one. Induction on the merged comb gives, at every legal closing and every depth,
$$ 0 = \tilde y_1' \;\lt\; x_1' \;\lt\; \tilde y_2' \;\lt\; x_2' \;\lt\; \tilde y_3' \;\lt\; \cdots \qquad\text{(the record’s } \texttt{prop:cubic-rate-order-sc}\text{).} $$Geometrically every atom but the bottom one moves to the right as the dial rises, and the higher the atom the faster. Economically, when the underdog concedes a little more, every effort above her zero rises, and the top of the contest rises fastest. At the closing the last visit has $\beta = 1,$ and its $\beta' > 0$ is the sign that (CR-7) of §6 needed.
§ 8What the cubic paid, and what comes next
The whole skeleton of the proof, from the reduction of §4 to the terminal identity of §6, is cost-general. The cubic enters in one object only: the double-contact factorization (CR-1), which compresses the prefix into a one-dimensional state, and the exact rational contact map that gives the two bounds (CR-3). Everything else is derivation.
| step | cubic property consumed | what a general class must supply |
|---|---|---|
| (CR-1) factorization | $c'''' = 0$: the remainder after a double contact is affine | a finite, or at least controlled, description of the prefix hump |
| (CR-3) bounds | the exact rational contact map | the same two signs for the implicit node map |
| (CR-5) invariance | the state is one number, $R$ | an invariant positive cone in whatever the state becomes |
| margins positive | $f \ge 0,$ budgets, increasing cost; not cubic | the same, plus a saturation remainder at $f \lt 0$ |
| (CR-7) terminal identity | none: $B_{\rm new} = B(1-\beta),$ $\beta = 1$ | nothing |
| §4 reduction | none | nothing |
Three extensions, as of 27 September 2026
- A linear term. $c(e) = e^3 + b\,e$ with $b > 0$ drops $c'(0) = 0.$ The factorization survives verbatim, since the linear term only touches coefficients the double zero already fixes; $b$ enters the first contact equation (CT1), which becomes $3[w^3 - m^3 + b(w - m)] = (3m^2 + b)P,$ and the underdog’s zero-effort atom becomes a strict corner whose first contact is a boundary pin. Rebuilt as one node and pushed through the same recursion: thm:cubicb-chain-sc, at $f \ge 0.$
- Negative lead. At $f \lt 0$ player 1’s prefix can saturate against the underdog’s zero atom, and the raw cubic acquires a zero-extension remainder. The reduced state past the saturated atom is still exact, and when player 1 values at least as much as player 2 the crossing step’s new mass still rises: thm:cubic-chain-sc-fneg. The case $v_1 \lt v_2$ at $f \lt 0$ is open.
- Pure powers. For $c = e^\theta$ with $\theta$ not an integer there is no finite state: the hump is the state, and the next contact reads only its value and slope at one point. The honest target is a dial-monotonicity of the normalized hump at the contacts, with a base that is cost-general. For $\theta = n$ integer the state is $(n-2)$-dimensional and $e^4$ is the natural next rung. None of this is proved; it is what the ledger says the proof would need.
Reading list, in the record
| label | what it states |
|---|---|
| lem:payoff-dial | $u_1$ strictly increasing in the dial, general cost |
| prop:nash-signs, prop:nash-layer | the Nash set as the sign pattern $G_{j-1} \le 0 = G_j \le G_{j+1}$ |
| cor:kfree-reduction | single crossing at every rung implies the chain ordering, no cost class |
| lem:pred-orders | a residual that does not fall back below its Nash zero excludes deeper profiles above it |
| prop:cubic-rate-order-sc | (CR-1) to (CR-6): the merged location-rate order at every cubic closing |
| prop:cubic-closing-race | (CR-7), (CR-8): the terminal identity and the square rent compensation |
| thm:cubic-chain-sc | the theorem of §6 |
| cor:cubic-sc-rung-unique | at most one Nash dial per rung, pure cubic, $f \ge 0$ |
| thm:cubicb-chain-sc, thm:cubic-chain-sc-fneg | the two extensions above |