The chain runs out of probability

§ 1The question, and the answer in one line

Fix a cost $c$ and primitives $(v_1, v_2, \Delta)$ with $f = w_2 - w_1 - \Delta \ne 0$ (every symbol is recalled in § 2). The rungs $\sigma(1,1), \sigma(1,2), \sigma(2,2), \ldots$ are candidate equilibria with one more atom each. Finiteness of the chain says that beyond some depth no rung carries a Nash equilibrium; a rung is Nash-live at a dial if its closed comb there is a Nash equilibrium, and a cutoff is a number $K$ such that every Nash-live rung has chain index at most $K.$ The two results on this page produce such a $K,$ in two ways.

The mechanism. A comb is paid for in probability. Each atom the march places pins a mass on the atom below it, and each player has exactly one unit to pin. The chain could only be infinite if the atoms could crowd: infinitely many inside a bounded stretch of effort, with pins shrinking fast enough to fit under a budget of one. They cannot. While neither player has overspent, every atom lies below the reach, and the gaps between consecutive atoms cannot shrink by more than a fixed factor in total, because the curvature of each player's payoff is bounded below and of bounded variation. Only finitely many atoms fit, so before the march can place one more, some budget is overspent.

Where the lead enters. That count is made at one dial, and it bounds the atoms by the room divided by the first gap. It becomes a cutoff only if the first gap has a positive floor over every dial that matters, and that is what $f \ne 0$ buys. At $f < 0$ the underdog must spend effort $-f > 0$ before a leader who spends nothing is even matched, so at the floor dial, the lowest dial with nonnegative rent (§ 3), her atom $Y_2,$ the mass $p_1$ it forces onto $x_1,$ and the gap above $Y_2$ all have a definite size; at $f > 0$ the floor dial is itself positive, and the comb there has a definite size as well. At $f = 0$ nothing sets such a size. At the floor dial the leader spends nothing, the underdog matches her at no cost, nothing is pinned, and the march is a single point at zero effort. Just above that dial every gap and every pin shrinks with the dial, so no fixed count of atoms is forced; in the quadratic case they shrink in exact proportion to it, and lowering the dial makes room for any number of atoms under the same budget of one. That is how the chain is infinite at the symmetric point $v_1 = v_2,$ $\Delta = 0,$ which forces $f = 0$⁠: the rungs close at dials falling to zero, and their combs fill in toward a distribution with no atoms at all. § 5 does the arithmetic, exactly in the quadratic case.

That gives two statements, for costs in the class (CBV): $c$ is $C^2$ and strictly convex with $c(0)=c'(0)=0,$ and its curvature $g := c''$ is positive on $(0, W]$ and of bounded variation on $[0, W],$ where $W = \max(w_1, w_2)$ is the larger reach. The curvature $g$ may be zero at the boundary, $g(0)=0.$

The barrier is the new work, §§ 3–5. The cutoff is a short set-theoretic argument once the barrier is in hand, § 6.

§ 2The contest, the comb, and the budgets

Two players fight over two rounds. Player $i$ values winning at $v_i;$ player 1 enters round one with a head start $\Delta$ on the score. Effort costs $c(e),$ the same strictly convex $C^2$ function for both, with $c(0) = c'(0) = 0;$ its curvature is $g := c''.$ The reach $w_i := c^{-1}(v_i)$ is the effort at which player $i$ has spent the whole prize, and the lead is $f := w_2 - w_1 - \Delta,$ the underdog’s reach advantage net of the head start. Primitives are competitive, $-w_1 < \Delta < w_2,$ and this page assumes $f \ne 0.$ One orientation is fixed throughout: the player with zero rent is called player 2, the underdog; player 1, the leader, has rent $u_1 > 0.$ Write $M := w_1 + f = w_2 - \Delta.$

What a round-one effort earns

Round two is a static contest whose value is known in closed form, and the only piece of it this page needs is the continuation kernel: what player $i$ earns from the last round when her round-one effort exceeds a rival atom by $s,$ measured in the shifted coordinate in which the underdog's efforts are moved up by $f$⁠:

$$ \psi_i(s) = \begin{cases} 0, & s \le 0,\\ v_i - c(w_i - s), & 0 < s < w_i,\\ v_i, & s \ge w_i. \end{cases} $$

Behind, she abandons the last round and earns nothing; ahead by less than her reach, she keeps the prize net of the cost of defending that margin; ahead by her reach or more, the last round is not fought and she keeps the whole prize (saturation). Against the rival measure $\mu_{-i}$ with atoms $a,$ player $i$'s payoff at shifted effort $z$ is

$$ U_i(z) = \sum_a \mu_{-i}(\{a\})\,\psi_i(z - a) \;-\; c(z - d_i),\qquad d_1 = 0,\quad d_2 = f, $$

her own effort being $z - d_i.$ Note that $\psi_i$ is continuous with $\psi_i(0) = 0,$ and $\psi_i'(s) = c'(w_i - s)$ on $(0, w_i)$⁠: the slope jumps from $0$ to $c'(w_i)$ where a rival atom activates and returns continuously to $0$ at saturation, since $c'(0) = 0.$

Combs and the march

In this class at $f \ne 0$ every Nash equilibrium is a comb $\sigma(m,n)$ (a classification theorem of the record, taken as given here): the leader spreads mass over atoms $x_1 < \cdots < x_m,$ the underdog over $Y_1 < \cdots < Y_n$ in the shifted coordinate, with $n \in \{m, m+1\}$ and the supports interleaving,

$$ Y_1 = f \;<\; x_1 \;<\; Y_2 \;<\; x_2 \;<\; \cdots . $$

The underdog's bottom atom is at zero effort, $Y_1 = f,$ and carries the one free number, the dial $t := q_1,$ her probability of sitting round one out. Everything else is a forward recursion in $t,$ the march. Each atom is placed by two conditions on its owner's payoff: she is stationary there and she earns her target, $u_1$ for the leader and $0$ for the underdog. The first atom solves the base equation $t\,c'(M - x_1) = c'(x_1)$ and fixes the rent $u_1(t) = t\,[v_1 - c(M - x_1)] - c(x_1).$ Then the players alternate: the underdog's two conditions at her next atom determine the mass $p_1$ the leader puts on $x_1$ together with the atom $Y_2$⁠; the leader's two conditions at her next atom determine $q_2$ on $Y_2$ together with $x_2$⁠; and so on. Name the atoms in the order the march creates them,

$$ z_1 = x_1,\quad z_2 = Y_2,\quad z_3 = x_2,\quad z_4 = Y_3,\quad \ldots $$

and call $z_n$ the $n$-th response. Step $n$ of the march pins a mass on the previous response $z_{n-1}$ (with $z_0 := Y_1$) and creates $z_n.$ The newest response carries no mass yet; its own mass is the next pin. Its stationarity and target conditions already hold, against every rival atom pinned so far, and they stay true after every later step, because every later rival atom lies above it and contributes nothing to its owner's payoff there.

Budgets

Let $P_m := p_1 + \cdots + p_m$ and $Q_m := q_1 + \cdots + q_m$ be the pinned masses. The budget after step $n$ is the spending of the player just pinned, minus one:

$$ G_{2m-1} := Q_m - 1,\qquad G_{2m} := P_m - 1,\qquad\text{so } G_1(t) = t - 1 . $$

Read $G_n < 0$ as “not yet spent, the comb can grow,” $G_n = 0$ as “exactly spent,” $G_n > 0$ as “overspent.” Index the rungs by the same counter: $\sigma(m,m)$ has chain index $j = 2m-1$ and $\sigma(m, m+1)$ has $j = 2m.$ A rung of index $j$ closes at dial $t$ when $G_j(t) = 0$⁠: the player pinned at step $j$ has spent exactly one. Her rival's top atom, the apex $z_j,$ takes the leftover mass $-G_{j-1}(t),$ which must be nonnegative; a zero apex makes the comb a copy of the rung below it. A closed comb with nonnegative apex at index $j$ therefore has

$$ G_j(t) = 0,\qquad G_{j-1}(t) \le 0,\qquad G_i(t) < 0 \quad (i \le j-2), $$

the last because the next pin of the same player is strictly positive, so $G_i < G_{i+2} \le 0.$ A Nash equilibrium of rung $j$ needs more than closing — the rival must have no profitable jump above the apex — but this page never uses the extra condition. It uses only the direction

$$ \begin{gathered} \text{Nash-live at } t \ \Longrightarrow\ \text{closed with nonnegative apex at } t \ \Longrightarrow\ t \in E_j,\\[4pt] E_j := \{\,t : G_i(t) \le 0 \text{ for } 1 \le i \le j\,\}. \end{gathered} $$

$E_n$ is the budget-feasible set of depth $n$⁠: the dials at which the first $n$ pins fit under the budgets. The sets are nested, $E_1 \supseteq E_2 \supseteq \cdots,$ and the barrier is the statement that one of them is empty.

§ 3The floor, and a march that does not stop

This section delivers the domain: a closed dial interval $T$ on which every finite prefix of the march is defined and continuous, even after a budget has failed. Compactness of the sets $E_n$ is what turns the pointwise count of § 4 into the uniform cutoff of § 5.

The rent along the dial

The base problem is $u_1(t) = \max_x\{\,t\,[v_1 - c(M - x)] - c(x)\,\},$ attained at $x_1(t).$ Three values are immediate. At $t = 0$ the objective is $-c(x),$ so $x_1(0) = 0$ and $u_1(0) = 0.$ By the envelope argument the derivative is the bracket at the maximizer, $u_1'(t) = v_1 - c(M - x_1(t)),$ so the right slope at zero is $v_1 - c(M).$ At $t = 1$ the objective is $v_1 - c(M-x) - c(x),$ and the two costs are minimized together at $x = M/2$ by convexity, so $u_1(1) = v_1 - 2c(M/2).$ The function $u_1$ is strictly convex in $t$⁠: its derivative is the bracket, and $x_1$ strictly increases with $t$ by the base equation, whose left side rises with $t$ and falls with $x$ while its right side rises with $x.$ The sign of $f$ decides the shape.

• $f < 0.$ Then $M < w_1,$ so $c(M) < v_1$ and the initial slope is positive: $u_1 > 0$ on all of $(0, 1].$ Set $t_* := 0.$ At $t = 0$ itself the march has $q_1 = x_1 = u_1 = 0$⁠: an analytical endpoint, not a profile with positive rent.
• $f > 0.$ Then $M > w_1$ and the initial slope is negative: the rent dips below zero. If $u_1(1) \le 0$ it never returns and no dial has positive rent; there is nothing to bound and the cutoff is $K = 0.$ Otherwise strict convexity gives exactly one zero $t_* \in (0,1),$ the floor, and the rent is positive precisely above it.

Write $D := (t_*, 1]$ for the positive-rent dials and $T := [t_*, 1]$ for the closed interval. One more fact about the base atom: $x_1 > f$ on $T.$ At $f < 0$ this is $x_1 \ge 0 > f.$ At $f > 0,$ if $x_1 \le f$ then $M - x_1 \ge w_1,$ the bracket $v_1 - c(M - x_1)$ is nonpositive and $u_1 = t\,[\,\cdot\,] - c(x_1) < 0,$ against $u_1 \ge 0$ on $T.$ By compactness $x_1 - f$ has a positive minimum on $T.$ This is the only place the sign of the lead is used, and at $f = 0$ it fails exactly at the floor, where $x_1(0) = 0 = f$ (§ 5).

Each step has exactly one answer

Consider step $n \ge 2.$ The responding player faces a new rival atom at $b := z_{n-1},$ above her own previous atom, and must receive her target $u$ ($u_1$ or $0$) at a new atom, by placing a mass $r$ on $b.$ Let $U_0$ be her payoff from all rival atoms pinned before this step; above the last of them it is strictly concave and peaks at her previous atom, where it equals $u,$ so $U_0(b) < u.$ For a trial mass $r \ge 0$ let

$$ \Phi(r) := \max_{\,b \le e \le b + w_i}\bigl\{\,U_0(e) + r\,\psi_i(e - b)\,\bigr\}. $$

The objective is strictly concave in $e,$ so the maximizer is unique; $\Phi$ is continuous and nondecreasing in $r,$ starts at $U_0(b) < u,$ and tends to infinity (evaluate at any fixed $e$ strictly inside the interval, where $\psi_i > 0$). At a root $\Phi(r) = u$ the maximizer lies strictly above $b,$ because the value at $b$ itself is $U_0(b) < u$ whatever $r$ is; so $\psi_i(e-b) > 0$ there and $\Phi$ strictly increases through the root. The mass $r > 0$ is therefore unique, and so is the new atom. The atom is interior: at $e = b + w_i$ every kernel has saturated and is flat, so the payoff is falling at the rate $c'$ of her own effort. Thus the new atom satisfies both the stationarity and the target condition exactly, and it does so whether or not a budget has already failed.

The construction is continuous in the data it consumes. A strict bracket around the unique root persists under small changes of $U_0,$ so $r$ is continuous; the maximizer of a continuous objective on a compact interval, when unique, is continuous as well. Induction over the steps gives:

The first underdog step at the floor $t = 0$ (when $f<0$) is covered too: her payoff from the rival is then zero, but the new rival atom sits at raw effort $x_1 - f \ge -f > 0,$ which is all the argument needs. Consequently each $E_n = \{t \in T : G_i(t) \le 0,\ i \le n\}$ is a closed subset of the compact interval $T$⁠: compact, and nested in $n.$

§ 4Why the atoms cannot crowd

This section proves the pointwise count: at a dial where the first $n$ budgets are all within one, $n$ is at most three plus the length of the stretch the atoms share divided by the smallest gap the curvature allows. That quantity, called $H(t),$ is assembled at the end of the section from three parts: budget-feasible atoms live below the reach; the curvature of a payoff is bounded below and of bounded variation; and equal payoffs at consecutive own atoms tie the two gaps around a rival atom to each other.

Below the reach

Fix $t \in E_n.$ Each response $z_k,$ $k \le n,$ satisfies its target condition against the rival masses pinned so far, whose total is at most one because the corresponding budget $G_k$ is at most zero. The kernel lies between $0$ and $v_i$ and the target is nonnegative, so for the leader's atom $x_k$ and the underdog's atom $Y_{k+1}$⁠:

$$ \begin{gathered} c(x_k) = \sum_{\ell \le k} q_\ell\,\psi_1(x_k - Y_\ell) - u_1 \le v_1 Q_k \le v_1,\\[4pt] c(Y_{k+1} - f) = \sum_{k' \le k} p_{k'}\,\psi_2(Y_{k+1} - x_{k'}) \le v_2 P_k \le v_2 . \end{gathered} $$

Hence $x_k \le w_1$ and $Y_{k+1} \le f + w_2$ wherever created. For $n \ge 4$ the last two responses belong to different players and $z_{n-1} < z_n,$ so $z_{n-1}$ is below both bounds:

$$ z_{n-1} \;\le\; R := \min(w_1,\ f + w_2). $$

The newest response $z_n$ is only known to be below its owner's reach; that is why the count below stops at the penultimate atom and carries the constant $3.$ Likewise $Y_2 \le R$ on $E_3$⁠: $Y_2 \le f + w_2$ by its own bound and $Y_2 < x_2 \le w_1.$ All responses from $Y_2$ on have positive own effort for both players. On the stretch $I := [Y_2, z_{n-1}]$ the leader's own effort is at least $\ell_1(t) := Y_2(t)$ and the underdog's at least $\ell_2(t) := Y_2(t) - f,$ both positive since $Y_2 > x_1 > f.$

The curvature of a payoff

Take the two measures made of the pins placed through step $n,$ each of total mass at most one. Between consecutive rival atoms, $U_i'$ is absolutely continuous and $U_i'' = -\kappa_i$ almost everywhere, where, differentiating the payoff display of § 2 twice and using $\psi_i''(s) = -g(w_i - s)$ on $(0, w_i)$ and $0$ outside,

$$ \kappa_i(z) = g(z - d_i) + \sum_a \mu_{-i}(\{a\})\,\gamma_i(z - a), \qquad \gamma_i(s) := g(w_i - s)\,\mathbf 1_{[0, w_i)}(s). $$

Both terms are nonnegative, so on the stretch $I,$ where player $i$'s own effort runs in $[\ell_i(t), w_i],$

$$ \kappa_i \;\ge\; m_i(t) := \min_{[\ell_i(t),\, w_i]} g \;>\; 0, $$

positive because $g$ is continuous and positive on the compact interval $[\ell_i(t), w_i] \subset (0, W].$ For the variation, let $V_i$ be the total variation of $g$ on $[0, w_i].$ The rival kernel $\gamma_i$ has variation at most $V_i + g(0) + g(w_i)$ on the line: the variation of $g$ itself, plus the jump $g(w_i)$ where an atom activates and the jump $g(0)$ where it saturates. Rival mass is at most one, so

$$ \operatorname{TV}_I \kappa_i \;\le\; V_i + 1\cdot\bigl[V_i + g(0) + g(w_i)\bigr] \;=:\; B_i . $$

Write $\rho_i$ for the variation measure of $\kappa_i$ on $I$⁠: $\rho_i(I) \le B_i.$

Equal losses on both sides of a rival atom

Take three consecutive responses $a < b < d$ in $I,$ the outer two belonging to player $i$ and $b$ to her rival. Player $i$ is stationary at $a$ and at $d$ and earns her target at both: $U_i'(a) = U_i'(d) = 0$ and $U_i(a) = U_i(d).$ Her payoff is continuous at $b$ (the kernel starts at zero) although its slope jumps there, so integrate from each stationary end toward $b$ separately. On $(a, b)$ there is no rival atom, $U_i'(s) = -\int_a^s \kappa_i,$ and

$$ \begin{gathered} U_i(b) - U_i(a) = \int_a^b U_i'(s)\,ds = -\int_a^b (b - s)\,\kappa_i(s)\,ds;\\[4pt] \text{likewise}\qquad U_i(b) - U_i(d) = -\int_b^d (s - b)\,\kappa_i(s)\,ds . \end{gathered} $$

Equal payoffs at $a$ and $d$ give the equal-loss identity

$$ \int_a^b (b - s)\,\kappa_i(s)\,ds \;=\; \int_b^d (s - b)\,\kappa_i(s)\,ds . $$

Moving from either own atom toward the rival's costs the same loss; Figure 1 draws it. Now put $L := b - a$ and $L' := d - b,$ and define the weighted curvature averages $\bar\kappa_L := (2/L^2)\int_a^b (b-s)\,\kappa_i$ and $\bar\kappa_{L'} := (2/L'^2)\int_b^d (s-b)\,\kappa_i,$ so that the identity reads $\bar\kappa_L\,L^2 = \bar\kappa_{L'}\,L'^2$ and

$$ \Bigl|\log\frac{L'}{L}\Bigr| = \tfrac12\,\bigl|\log \bar\kappa_L - \log \bar\kappa_{L'}\bigr| \;\le\; \frac{\rho_i((a,d))}{2\,m_i(t)} . $$

The inequality: both averages lie between the infimum and the supremum of $\kappa_i$ on $(a, d),$ hence are at least $m_i(t),$ and $\log$ is $1/m_i(t)$-Lipschitz on $[m_i(t), \infty),$ so the difference of the logs is at most the oscillation of $\kappa_i$ on $(a,d)$ divided by $m_i(t),$ and the oscillation is at most $\rho_i((a,d)).$ In a world of constant curvature the two gaps would be equal; curvature can tilt them only by as much as it varies.

Summing over the stretch

For a fixed player the open intervals $(a, d)$ of her consecutive triples are pairwise disjoint, so summing the bound over all triples inside $I,$ for both players,

$$ \sum_{\text{triples in } I} \Bigl|\log\frac{L'}{L}\Bigr| \;\le\; \frac{\rho_1(I)}{2m_1(t)} + \frac{\rho_2(I)}{2m_2(t)} \;\le\; C(t) := \frac{B_1}{2m_1(t)} + \frac{B_2}{2m_2(t)} . $$

The gaps on the stretch are $z_3 - z_2, \ldots, z_{n-1} - z_{n-2},$ $n - 3$ of them, and the first is $h(t) := x_2(t) - Y_2(t) > 0.$ Each gap is the next one's $L$ and the previous one's $L',$ so every gap is the first times a product of ratios whose logs sum, in absolute value, to at most $C(t)$⁠: every gap is at least $h(t)\,e^{-C(t)}.$ They fit inside the stretch,

$$ (n - 3)\,h(t)\,e^{-C(t)} \;\le\; z_{n-1} - Y_2(t) \;\le\; R - Y_2(t), $$

that is,

$$ n \;\le\; H(t) \;:=\; 3 + \frac{R - Y_2(t)}{h(t)}\,e^{C(t)} . $$

For $n = 3$ the bound holds because $R \ge Y_2.$ Nothing in this argument normalized the two measures to probability or tested any deviation: it is a statement about an unfinished march, which is what a budget-feasible prefix is.

§ 5The barrier: a cutoff with no single crossing

Everything so far was at one dial. Compactness makes it uniform.

If $E_3$ is empty, no closed comb has index $3$ or more, and $K_B := 2.$ Otherwise $E_3$ is compact and nonempty, and on it the functions $\ell_1(t) = Y_2(t),$ $\ell_2(t) = Y_2(t) - f$ and $h(t) = x_2(t) - Y_2(t)$ are continuous and positive, so they have positive minima $\ell_1, \ell_2, h.$ Put

$$ m_i := \min_{[\ell_i,\, w_i]} g,\qquad C := \frac{B_1}{2m_1} + \frac{B_2}{2m_2},\qquad H := 3 + \frac{R - \ell_1}{h}\,e^{C},\qquad K_B := \lfloor H \rfloor . $$

Since $\ell_i \le \ell_i(t)$ the minimum $m_i$ is taken over a larger interval, so $m_i \le m_i(t)$ and $C(t) \le C;$ also $R - Y_2(t) \le R - \ell_1$ and $h(t) \ge h.$ Hence $H(t) \le H$ on $E_3,$ and since $E_n \subseteq E_3$ for $n \ge 3,$ a dial in $E_n$ would need $n \le H.$ Therefore

$$ E_{\lfloor H\rfloor + 1} = \varnothing . $$

The floor index

The floor $t_*$ is a point of $T,$ so the march runs there too, and its budgets must overrun: the floor index

$$ J := \min\{\,j \ge 1 : G_j(t_*) > 0\,\} $$

is finite. If $t_* \notin E_3$ then $J \le 3$⁠; otherwise $t_* \in E_{J-1}$ gives $J - 1 \le H(t_*)$ and $J \le \lfloor H(t_*)\rfloor + 1.$ A budget met exactly, $G_j(t_*) = 0,$ is not a violation; $J$ is the first strict one. At $f < 0$ the floor is the analytical endpoint $t = 0$⁠; at $f > 0$ it is the zero-rent dial, whose closed combs of index $j \ge J$ the same argument excludes.

Three controls

The quadratic $c = e^2$ with $v_1 = v_2 = 1$ has $g \equiv 2$⁠: $V_i = 0,$ $B_i = 4,$ $m_i = 2,$ $C = 2.$ At $\Delta = 1/10$ the lead is $f = -1/10,$ the floor is $t_* = 0,$ and the floor march gives $Y_2 = 1/11,$ $h = 1/12,$ $R = 9/10,$ so $H(t_*) = 3 + (534/55)\,e^2 \approx 74.7.$ The march itself overruns at step $10$⁠: the leader's fifth pin brings $P_5$ to $1092974185/1040449536 > 1,$ all nine earlier budgets being strictly negative. At $\Delta = -1/10$ the lead is $f = 1/10,$ the floor is $t_* = 21/100,$ and the overrun comes at step $9.$ The cubic $c = e^3$ at $(v_1, v_2, \Delta) = (9/10, 1, 1/5)$ has $f = 1 - (9/10)^{1/3} - 1/5 \approx -0.166$ and its floor overrun at step $46.$ The two quadratic controls were checked in exact rational arithmetic against the full payoff and stationarity equations of every atom; the cubic index is read from the ball-arithmetic certificate of the separately proved cubic result. A control is a worked case computed independently of the proof, to check its conventions and indices against known numbers.

cost and $(v_1, v_2, \Delta)$	$f$	floor $t_*$	floor index $J$	$\lfloor H(t_*)\rfloor + 1$	cutoff $J+1$ under (SC0), § 6	cutoff $J-1$ under (SC+), § 6
$e^2,$ $(1,\, 1,\, +\tfrac1{10})$	$-\tfrac1{10}$	$0$	10	75	11	9
$e^2,$ $(1,\, 1,\, -\tfrac1{10})$	$+\tfrac1{10}$	$\tfrac{21}{100}$	9	73	10	8
$e^3,$ $(\tfrac9{10},\, 1,\, \tfrac15)$	$\approx -0.166$	$0$	46	not evaluated	47	45

The analytic bound is loose by a factor of seven or eight, and the reason is visible in its two factors. Without the allowance $e^{C},$ the count $3 + (R - Y_2)/h = 3 + 534/55 \approx 12.7$ says how many atoms fit in the stretch at the first gap's spacing, close to the ten that occur. The factor $e^{C} = e^2 \approx 7.4$ is the price of letting the gaps shrink by all the curvature's variation at once, and for the quadratic the only variation is the two kernel jumps $g(0) + g(w_i) = 4$ per player, while the actual gaps shrink gently: a gap $\delta$ is followed by a gap $\delta/(1+\delta)$ (Figure 2). The bound is a device for finiteness, not a count of rungs.

Why the lead must not vanish

The sign of the lead was used once, in § 3, and what it buys is the positive minimum $h$ of the first gap on $E_3.$ Without it the count $H(t)$ is finite at every dial yet unbounded as the dial falls, no $E_n$ is empty, and the chain can be infinite. The pointwise count of § 4 never used the sign. The compactness step above did: the positive minimum of $x_1 - f$ on $T$ is what makes the mass $p_1$ that the underdog's first step puts on $x_1$ positive at the floor, hence $h$ continuous and positive on the compact set $E_3.$

Put $f = 0.$ Then $M = w_1,$ the initial rent slope $v_1 - c(M)$ is zero, and strict convexity still gives $u_1 > 0$ on $(0, 1]$⁠: the floor is $t_* = 0,$ as at $f < 0.$ But at the floor the base atom is $x_1(0) = 0 = f.$ The underdog's first problem then has the rival atom at her raw effort $x_1 - f = 0,$ and her payoff $U_0(e) = -c(e)$ already equals her target $0$ at $e = 0$⁠: the strict inequality $U_0(b) < u$ of § 3, which made the pin positive, is an equality. The mass $p_1 = 0$ on $x_1$ and the response $Y_2 = 0$ satisfy her target and stationarity conditions, since $c'(0) = 0$⁠; every later step inherits a rival atom at $0,$ a target $0$ and a payoff $-c(e),$ and pins nothing. The march at the floor is the zero comb: every atom at effort $0,$ every pin $0,$ every budget equal to $-1.$ Its first gap is $h(0) = 0.$

Nothing in § 4 fails at a dial $t > 0$⁠: there $x_1(t) > 0 = f,$ every pin is positive, and $n \le H(t)$ holds on $E_n.$ What fails is uniformity. As $t \downarrow 0$ the base equation gives $c'(x_1) = t\,c'(w_1 - x_1) \le t\,c'(w_1),$ so $x_1(t) \to 0.$ The next two steps follow it. Each pin solves a target condition whose value at zero pin, $U_0(b) = -c(b)$ for the underdog and $t\,\psi_1(Y_2) - c(Y_2)$ for the leader, tends to zero as $b \to 0$ and $t \to 0,$ while the target is $0$ or $u_1(t) \to 0.$ For any fixed pin $r > 0$ the value $\Phi(r)$ of § 3 then exceeds the target once $b$ and $t$ are small, because at a small effort $s$ above $b$ the kernel pays $r\,\psi_i(s) \approx r\,c'(w_i)\,s$ against an own cost $c(s)$ of smaller order; so the root lies below $r,$ and the pin tends to zero. Each response $z$ then obeys, by stationarity and $\psi_i' \le c'(w_i),$ the bound $c'(z - d_i) \le (\text{rival mass pinned so far}) \cdot c'(w_i),$ which tends to zero, so $z \to 0.$ Hence $Y_2(t), x_2(t) \to 0,$ the first gap $h(t) \to 0,$ and

$$ H(t) \;\ge\; 3 + \frac{R - Y_2(t)}{h(t)} \;\longrightarrow\; \infty . $$

The same two bounds, applied at each later step, send every pin and every response of a fixed prefix to zero as $t \downarrow 0,$ so each budget $G_i(t)$ tends to $-1$ and every $E_n$ contains an interval $(0, \varepsilon_n]$ above the floor: none is empty. When $g(0) = 0,$ as for every pure power $e^\theta$ with $\theta > 2,$ the lower efforts $\ell_i(t) \to 0$ drag $m_i(t)$ to zero too, and the allowance $e^{C(t)}$ diverges as well.

At the symmetric point the quadratic march is explicit and shows the whole picture; the two displays below are its closed form (prop:quad-sym-closed-form). Take $c = e^2,$ $v_1 = v_2 = w^2$ and $\Delta = 0,$ so $w_1 = w_2 = w$ and $Y_1 = f = 0.$ Consecutive responses are spaced by

$$ z_n - z_{n-1} = \frac{w\,t}{1 + n\,t}\qquad (n \ge 1,\ z_0 = Y_1 = 0), $$

so every gap is below $wt$ and the first gap of the stretch is $h(t) = wt/(1 + 3t).$ The pins start at $q_1 = t,$ and the cumulative spendings are

$$ \begin{gathered} 1 + Q_m = (1 + t)\prod_{j=1}^{m-1}\Bigl(\frac{1 + (2j+1)\,t}{1 + 2j\,t}\Bigr)^{2},\\[4pt] 1 + P_m = \prod_{j=1}^{m}\Bigl(\frac{1 + 2j\,t}{1 + (2j-1)\,t}\Bigr)^{2}. \end{gathered} $$

Every factor is $(1 + t/(1 + kt))^2 \le (1+t)^2$ for some $k \ge 1,$ so after step $n$ of either parity $G_n(t) \le (1+t)^n - 2,$ and

$$ E_n \;\supseteq\; \bigl(0,\ 2^{1/n} - 1\bigr] \qquad\text{for every } n, $$

an interval of length about $(\ln 2)/n$ reaching down to the floor. No $E_n$ is empty, and the barrier says nothing. At this cost $g \equiv 2,$ $C = 2$ and $R = w,$ so the count is $H(t) \approx e^{2}/t,$ while the depth the budgets actually allow is about $1/t$⁠: the two scale alike, and both are unbounded as the dial falls. The rungs do close. The rung $\sigma(m,m)$ closes where $G_{2m-1} = Q_m - 1 = 0,$ at a dial $t_m$ with ${1/t_m = 2m - \tfrac32 + O(1/m)}$ (lem:quad-sym-dial-asymptotics), and every one of these closed combs is a Nash equilibrium (cor:quad-sym-gate). The chain is infinite; the rents ${u_1 = w^2 t_m^2\big/(1 + t_m)}$ fall to zero; and the combs converge to the equilibrium in which both players mix continuously, with no atoms at all, within $43\,t_m$ in the cumulative distribution (prop:quad-sym-convergence). The atoms do crowd, as the mechanism of § 1 said they would have to, but not at one dial: along the chain, with the dial falling and the whole comb shrinking at its base.

Two cautions. First, $f = 0$ is a surface of primitives, not the symmetric point alone: at $c = e^2$ it is $\Delta = \sqrt{v_2} - \sqrt{v_1},$ nonzero whenever the prizes differ. The march at the floor is the zero comb on the whole surface, so the budget argument fails everywhere on it; but off the symmetric point the chain is finite nonetheless (prop:quad-finite-f0 at $c = e^2,$ thm:general-chain-finite in its class), for instance at $(v_1, v_2, \Delta) = (\tfrac{81}{100},\, 1,\, \tfrac1{10}),$ where $w_1 = \tfrac9{10},$ $w_2 = 1,$ $f = 0$ exactly, and the last Nash-live rung is $\sigma(4,4).$ What ends those chains is the deviation test that decides Nash liveness, which this page never uses. Second, the barrier does not pass to the limit $f \to 0$⁠: its constant $h$ is the minimum over $E_3$ of a first gap which, on the surface, vanishes at the floor.

§ 6With single crossing, the floor decides

The barrier bound $K_B$ is uniform but bulky: it needs the minima of $\ell_i$ and $h$ over the compact set $E_3.$ The floor index $J$ is one march, run at one dial until the first strict overrun. This section shows that a single-crossing property of the budgets, assumed only on the reachable dials and only through index $J,$ makes $J$ a cutoff for every dial in $D.$ Two conventions are in use, and they give different numbers.

Reachable sets and the exhaustion index

$$ \begin{gathered} S_i := \{\,t \in D : G_k(t) < 0 \text{ for all } k < i\,\},\\[4pt] \overline S_i := \{\,t \in D : G_k(t) \le 0 \text{ for all } k < i\,\} = E_{i-1} \cap D \end{gathered} $$

are the dials at which the march reaches index $i$ with every earlier budget strictly, respectively weakly, unspent. The bar is a notation for the weak version, not a topological closure. The exhaustion index $N(t) := \min\{\,i : G_i(t) \ge 0\,\}$ is the first index at which a budget is met or exceeded, so $S_i = \{N \ge i\}.$ At a closed comb of index $j \ge 2$ the display of § 2 gives $N(t) \in \{j - 1, j\},$ the first case being exactly the zero apex.

Convention one: a met budget stays met

$$ \text{(SC0)}\qquad s < t,\ \ s, t \in S_i,\ \ G_i(s) \ge 0 \ \Longrightarrow\ G_i(t) \ge 0 \qquad (1 \le i \le J). $$

On the dials where index $i$ is reachable, once the $i$-th budget is met it stays met as the dial rises. Call a subset $A$ of $D$ an up-set if $s \in A$ and $s < t \in D$ force $t \in A.$ Then $\{N \le i\}$ is an up-set of $D$ for every $i \le J,$ by induction: $\{N \le 0\}$ is empty, and

$$ \{N \le i\} = \{N \le i-1\} \;\cup\; \bigl(\{G_i \ge 0\} \cap S_i\bigr), $$

where the first set is an up-set of $D$ by induction, $S_i$ is its complement in $D,$ and the second set is an up-set of $S_i$ by (SC0). A dial above a member of the union is either in the first set, or in $S_i$ and above a member of the second set, hence in it. Now $G_J(t_*) > 0$ and $G_J$ is continuous on $T,$ so $G_J > 0,$ and therefore $N \le J,$ at every dial in $D$ close enough to the floor. The up-set property spreads this to all of $D$⁠: $N(t) \le J$ everywhere. A closed comb of index $j$ then has $j \le N(t) + 1 \le J + 1.$

Convention two: an overspent budget stays overspent

$$ \text{(SC+)}\qquad s < t,\ \ s, t \in \overline S_i,\ \ G_i(s) > 0 \ \Longrightarrow\ G_i(t) > 0 \qquad (1 \le i \le J). $$

Call $A \subseteq D$ a down-set if $t \in A$ and $s < t$ in $D$ force $s \in A.$ Then $E_i \cap D$ is a down-set for every $i \le J$⁠: it equals $\overline S_i \cap \{G_i \le 0\},$ where $\overline S_i = E_{i-1} \cap D$ is a down-set by induction (with $E_0 := T$), and within $\overline S_i$ the set $\{G_i \le 0\}$ is a down-set by (SC+), since a dial below one with $G_i \le 0$ cannot have $G_i > 0.$ If $E_J \cap D$ contained a dial $t,$ it would contain every dial between $t_*$ and $t,$ so $G_J \le 0$ arbitrarily close to the floor and, by continuity, $G_J(t_*) \le 0$⁠: against the definition of $J.$ Hence $E_J \cap D = \varnothing,$ and every closed comb, whose dial lies in $E_j \cap D,$ has $j \le J - 1.$

Why two conventions

Neither implies the other. (SC0) preserves “$\ge 0$” and so forbids a met budget from becoming unspent again, but lets an overspent budget come back down to exactly zero; (SC+) preserves “$> 0$” and so forbids that touch, but lets a budget met exactly at one dial become unspent at a higher one. (SC+) is also demanded on the larger weak-budget sets $\overline S_i.$ The two rungs it buys, $J - 1$ against $J + 1,$ are paid for by a different hypothesis, not a weaker one, so a crossing theorem must be read for which convention it delivers before the sharper cutoff is quoted. At $c = e^2$ the budgets rise strictly with the dial on their reachable sets (a separately proved quadratic property; Figure 3 shows the first ten), so both hold; Figure 3 shows the resulting picture at $\Delta = 1/10$⁠: the first ten budgets against the dial, the closed combs at their zeros, the feasible sets shrinking to empty at $E_{10},$ and the deepest closed comb at index $8 = J - 2,$ one below the (SC+) cutoff.

In either version single crossing is assumed only up to $J,$ never at $J + 1,$ and the floor overrun that anchors both arguments is proved, not assumed. What is assumed is the crossing itself: nothing on this page shows that a given cost in (CBV) has single-crossing budgets.

§ 7What the two results say, and what they do not

The class. Costs $c \in C^2[0,\infty),$ strictly convex, $c(0) = c'(0) = 0,$ with curvature positive on $(0, W]$ and of bounded variation on $[0, W].$ Zero curvature at the origin is allowed: pure powers $e^\theta$ with $\theta \ge 2$ and positive sums of them are in. Bounded variation is automatic for every $C^3$ cost and every cost analytic on $[0,W]$⁠; for those only the positivity of the curvature on $(0, W]$ remains to be checked. Outside the proof: a cost whose curvature vanishes somewhere inside $(0, W],$ or has infinite variation there; a positive initial marginal cost $c'(0) > 0$⁠; and the surface $f = 0,$ at any asymmetry, where the march at the floor is the zero comb (§ 5). Both signs of $f \ne 0$ are covered, in the orientation $u_1 > 0,$ $u_2 = 0$⁠; the mirror orientation is a second application with the players renamed, which negates $\Delta$ and hence $f.$

What is bounded. Chain indices of closed combs with nonnegative apex, hence of Nash-live rungs, zero-apex copies included. Not the number of Nash dials at a given rung, not which closed comb is Nash, and not the exact last live rung: in the quadratic control of Figure 3 the deepest closed comb sits two below $J,$ in the cubic control the separately proved cubic result places the only Nash-live rung at index $3$ against $J = 46,$ and the analytic bound $\lfloor H(t_*)\rfloor + 1$ is seven to eight times $J.$

Where single crossing stands. The barrier and the finiteness of $J$ need none. The cutoffs $J + 1$ and $J - 1$ assume the stated convention on the reachable dials through $J$ and prove nothing about it.

What kind of number $K_B$ is. A finite expression in the first three atoms of the march, two compact minima, and curvature bounds: finite for every cost in the class, but not a closed form in $(v_1, v_2, \Delta)$ and the cost's parameters, and not an algorithm for an arbitrary cost. Evaluating it needs a representation of $c$ in which roots, extrema and variation can be certified. The three controls are exact because the quadratic march has a closed form and the cubic floor was certified separately.

Against the earlier finiteness theorem. The older result that the Nash-live chain is finite away from the fully symmetric primitive $v_1 = v_2,$ $\Delta = 0$ was proved on a more restrictive cost class, by compactness of equilibria in the weak topology, and it names no index. At $f \ne 0$ that class satisfies (CBV), so its $f \ne 0$ part is recovered under weaker hypotheses, with the floor index $J$ attached. Its $f = 0$ part, finiteness off the symmetric point, is untouched; at the symmetric point itself the chain is infinite (§ 5).

Numbers on this page: the two quadratic controls are exact rationals from the control script; the fan in Figure 3 is the closed-form quadratic march evaluated at each dial and checked against the full payoff and stationarity equations to $10^{-15}$⁠; the cubic index is read from its separate certificate.