§ 1The hypothesis as the proposition states it
Two players, two rounds, one convex cost $c$ of effort, at the symmetric primitive $(v_1,v_2,\Delta) = (v,v,0)$: equal prizes, no lead, so both reaches equal $w = c^{-1}(v)$ and the effective lead $f$ is zero. The working class of the campaign is
$$ \mathcal C \;=\; \{\,c \in X : c'' \text{ concave on } (0,\infty)\,\}, \qquad X = \{\,c(0)=c'(0)=0,\ c',c'',c'''>0 \text{ on } (0,\infty)\,\}, $$so a cost in $\mathcal C$ has $c''' > 0$ and $c'''$ nonincreasing. The proposition adds three explicit hypotheses. Write
$$ a := c''(0^+) > 0,\qquad b := c''(w),\qquad \kappa_0 := c'''(0^+) < \infty,\qquad d := \kappa_0 - c'''(w), $$and require $\Xi = w\,d\,b^2/a^3 \le 1/150.$ Under these it proves, at every finite depth and every positive dial, that all the cubic suffixes are negative, that both players’ per-node masses and locations increase in the dial, and that every ordinary legal closing of either finite-comb family has a strictly positive full closing slope, non-Nash zeros included.
Reach one is without loss
The written proof normalizes the reach to one and the cost to $c(1)=1.$ Nothing is lost: with $\hat c(e) = c(we)/v$ the game is the same game (prizes $1,$ masses and the dial $t=q_1$ unchanged, reach $1$), $\hat c$ is still in $\mathcal C,$ and
$$ \hat c''(e) = \frac{w^2}{v}\,c''(we),\qquad \hat c'''(e) = \frac{w^3}{v}\,c'''(we) \quad\Longrightarrow\quad \hat\Xi = w^{\,3+4-6}\,v^{\,-1-2+3}\,\frac{d\,b^2}{a^3} = \Xi(c,w). $$$\Xi$ is dimensionless — invariant under scaling the cost and under scaling effort with the reach moving along — and the display in the hero is exactly its transform.
§ 2Slope and acceleration of marginal cost
$c'(e)$ is the marginal cost of effort. Its slope $c''(e)$ says how fast marginal cost rises with effort, and it is what the player feels at an interior best response: effort is chosen where marginal cost meets marginal gain, so a unit increase in the marginal gain moves effort by $1/c''(e).$ A large $c''$ is a stiff effort response, a small one a pliant response. The third derivative $c'''(e)$ is the acceleration of marginal cost — the rate at which the slope itself grows. Positive acceleration means marginal cost is convex: each further unit of effort raises marginal cost by more than the last did.
Read this way the class $\mathcal C$ is one sentence: marginal cost steepens, and the steepening is strongest at low effort and fades from there. The four regimes the record uses sit as follows.
| cost | slope $c''$ | acceleration $c'''$ | regime |
|---|---|---|---|
| quadratic $pe^2$ | constant | zero | no steepening |
| cubic $pe^2+qe^3$ | linear | constant | steady steepening |
| class $\mathcal C$ | concave | positive, fading | steepening that tires |
| $\cosh,$ $\exp,$ pure powers above the cubic | convex | growing | steepening that feeds on itself |
The last row is the opposite regime, and it is where the counterexamples to chain order on $X$ live: a knee in the cost, where marginal cost accelerates abruptly just below a player’s atom, is a growing acceleration. The proposition works on the other side of the quadratic-plus-cubic line.
§ 3The condition in words
Three dimensionless ratios carry the whole content of $\Xi.$
| symbol | definition | what it measures on the competitive range $[0,w]$ |
|---|---|---|
| $\rho = w\kappa_0/a$ | initial acceleration times reach, over base slope | the cubic strength: how much a constant acceleration would raise the slope across the range, relative to the slope at zero |
| $\delta = d/\kappa_0 \in [0,1]$ | fade over initial acceleration | the fraction of the acceleration lost by the time effort reaches $w$ |
| $B = b/a$ | slope at the reach over slope at zero | how much stiffer the effort response is at the reach than at zero |
Because the acceleration is nonincreasing, the slope at the reach lies between the base and what constant acceleration would give, $a \le b \le a + \kappa_0 w,$ so $1 \le B \le 1+\rho$ and
$$ \rho\,\delta \;\le\; \Xi \;\le\; \rho\,\delta\,(1+\rho)^2 . $$The second inequality says how the tolerance for fade scales with cubic strength: a nearly quadratic cost ($\rho \ll 1$) may lose up to about $1/(150\rho)$ of its acceleration across the range, a strongly cubic one only about $1/(150\rho(1+\rho)^2)$ — at $\rho = 1,$ one part in six hundred.
The cleanest statement is about the slope itself. For $0 \le e \le w$ the slope sits below the origin tangent $a + \kappa_0 e$ by the integrated fade, and that integral is at most $w\,d$:
$$ 0 \;\le\; a + \kappa_0 e - c''(e) \;=\; \int_0^e \bigl(\kappa_0 - c'''\bigr) \;\le\; w\,d \;=\; \Xi\,a\,\Bigl(\frac ab\Bigr)^{2} \;\le\; \Xi\,a . $$Over the range of efforts the prize can justify, the acceleration of marginal cost is essentially constant: the slope of marginal cost stays within $a/150$ of the straight line it would follow with constant acceleration. On the competitive range the cost is quadratic-plus-cubic, up to a curvature error below two thirds of one percent of its curvature at zero effort.
Quadratic-plus-cubic costs themselves have $d = 0$ and $\Xi = 0$ at every prize; their single crossing was banked earlier by other means. What the proposition buys is a uniform neighbourhood of that cone — any continuous, nonincreasing perturbation of the acceleration, with no cap on depth and no floor on the dial — measured in the one currency the proof needs, the gap at the reach.
§ 4Where it enters the proof
At a response the player’s payoff gap is governed by a kernel built from slopes: the player’s own slope read at rising effort, plus the slopes contributed by the earlier atoms of the history, read at falling distance,
$$ H(\eta) \;=\; c''(z+\eta) + \sum_{\text{old}} \mu_l\, c''(e_l - \eta), \qquad 0 \le \eta \le h,\qquad M = \sum_{\text{old}} \mu_l \le 1 . $$Its slope is a contest between accelerations at two places,
$$ H'(\eta) \;=\; c'''(z+\eta) - \sum_{\text{old}} \mu_l\, c'''(e_l - \eta) \;\ge\; (1-M)\,c'''(w) - M\,d \;\ge\; -M\,d . $$With constant acceleration the right side is $(1-M)c''' \ge 0$: the kernel never tilts down, and its centroid $\psi = G/J$ sits at or below the midpoint $h/2$ of the window — the direction the argument likes. A fading acceleration opens the adverse case: the player responds at a high effort, where the acceleration has worn off, while an earlier atom sits at a low one, where it is still strong; the kernel can tilt down near the end of the window and the centroid drifts past the midpoint. Integration by parts bounds the drift by the fade alone,
$$ hJ - 2G \;=\; \int_0^h \eta\,(h-\eta)\,H'(\eta)\,d\eta \;\ge\; -\frac{M\,d\,h^3}{6}, \qquad (\psi - h/2)_+ \;\le\; \frac{d\,h^2}{24\,a}. $$The constant part of the acceleration cancels through the probability budget; only its decrease is charged. On the other side of the ledger the proof holds a frozen margin $B_3 > h/20$ from the cubic-gap lemma, and the geometric transport, extended to the source, controls the gap ratios by $9(b/a)^2$ uniformly in depth. Putting the three together,
$$ \dot\psi + B_3\,\alpha^3 \;>\; h\,\alpha^3\Bigl(\frac1{20} - \frac{15}{2}\,\Xi\Bigr) \;\ge\; 0 \qquad\text{exactly when}\qquad \Xi \le \frac1{150}, $$and $\Xi \le 1/200$ leaves the quantitative margin $1/80.$ That is the whole role of the number: it keeps the centroid’s drift below a margin the proof already has in hand, at every depth at once.
The drift bound uses only the total variation of $c'''$ on $[0,w],$ so a growing acceleration would pass this one step as well. The other inputs — gap-curvature transport, the cubic-gap margin and rows, the creation-rate envelopes — are banked on $\mathcal C,$ which is why nothing is claimed for convex slopes.
§ 5A prize ceiling for every cost but the polynomial ones
Fix the cost and raise the prize. The reach $w$ grows; the fade $d(w) = \kappa_0 - c'''(w)$ grows, since the acceleration is nonincreasing; and the slope at the reach $b(w)$ grows, since the slope is increasing. All three factors of $\Xi = w\,d\,b^2/a^3$ move the same way, so $\Xi(c,\cdot)$ is nondecreasing in the reach, strictly increasing once the acceleration has faded at all, and unbounded unless the acceleration is constant on the whole half-line. Hence
For each cost the prizes the proposition covers form an interval $(0, v^*(c)].$ The ceiling is infinite exactly for the quadratic-plus-cubic costs. For every other member of $\mathcal C$ the proposition is a small-prize theorem: the competitive range must stop before the efforts at which the acceleration of marginal cost has measurably faded.
The knee cost in Figure 3 makes the point most sharply. Its acceleration is a smoothed step, $c''' = \kappa_0\,[1 - \tanh((e-e_0)/s)]/2,$ full strength below $e_0$ and off above it, so the cost is cubic on the low side of the knee and quadratic on the high side. Below the knee $\Xi$ is exponentially small; as the reach approaches $e_0$ the fade switches on and $\Xi$ shoots through $1/150.$ With $(\kappa_0, e_0, s) = (1, 1, 0.1)$ the crossing is at $w^* = 0.71,$ a fade of $0.32\%$: the condition does not let the players get near the knee. Make the cost more strongly cubic, $(\kappa_0,e_0,s)=(1,3,0.1),$ and the tolerated fade at the crossing falls to $0.02\%,$ the $(1+\rho)^2$ of §3 at work.
| cost ($c''(0^+)=1$ where normalized) | $\Xi(c,w)$ | threshold reach $w^*$ | prize ceiling $v^* = c(w^*)$ |
|---|---|---|---|
| $(1+e)^{\theta}-1-\theta e,$ $\theta = 2.5$ | $(\theta-2)\,w\,(1+w)^{2\theta-7}\bigl[(1+w)^3-(1+w)^\theta\bigr]$ | 0.160 | 0.049 |
| $c'' = 1+\beta(1-e^{-e}),$ $\beta = 0.01$ | $w\beta(1-e^{-w})\bigl(1+\beta(1-e^{-w})\bigr)^2$ | 1.03 | 0.53 |
| same, $\beta = 1$ | 0.077 | 0.0031 | |
| the proposition’s own example $f$ | $\approx w^2/1000$ for small $w$ | 1.45 | 1.56 |
| knee $(\kappa_0,e_0,s) = (1,1,0.1)$ | $\approx 0$ below the knee | 0.71 | 0.31 |
| knee $(1,3,0.1)$ | 2.57 | 6.16 |
§ 6Which costs satisfy it
Membership in $X$ and $\mathcal C$ is tested on the competitive range $(0,w]$ only: that is all the written proof uses, every contact lying below the reach and every kernel argument in $[0,w].$ Whether each of the four banked inputs it cites also reads the cost on $[0,w]$ alone has not been audited, so the rows marked “on the range” are conditional on that audit. The obstructions at the endpoint zero are unaffected by it.
Inside the region, each up to its prize ceiling
| family | why it is in $\mathcal C$ | remark |
|---|---|---|
| $pe^2 + qe^3,$ $p>0,$ $q \ge 0$ | constant acceleration | $\Xi \equiv 0,$ every prize; banked earlier on its own |
| shifted powers $(1+e)^\theta - 1 - \theta e,$ $2<\theta<3$ | $c''' \propto (1+e)^{\theta-3},$ positive and falling | $\Xi \approx (\theta-2)(3-\theta)w^2$ for small $w$ |
| saturating slope $c'' = a + \beta(1-e^{-\lambda e})$ | $c''' = \beta\lambda e^{-\lambda e}$ | the proposition’s example is this plus a cubic term |
| $c'' = a + \kappa\tanh e,$ $\ a + \kappa e/(1+e),$ $\ a + \kappa\log(1+e)$ | accelerations $\kappa\,\mathrm{sech}^2 e,$ $\kappa/(1+e)^2,$ $\kappa/(1+e)$ | thresholds at $\kappa = 0.1$: $w^* = 0.41,\ 0.21,\ 0.29$ |
| knee costs, $c''' = \kappa_0[1-\tanh((e-e_0)/s)]/2$ | a smoothed step down | ceiling just short of the knee |
| quartic $c_0e^2 + de^3 - be^4,$ on the range | in $\mathcal C$ on $(0,w]$ iff $w < d/(4b)$ | $\Xi = 12bw^2\bigl(c_0+3w(d-2bw)\bigr)^2/c_0^3$ |
| quintic $ae^5 - be^4 + de^3 + c_0e^2,$ on the range | in $\mathcal C$ on $(0,w]$ iff $w \le b/(5a)$ and $d > 4bw - 10aw^2$ (Reduce) | $d > 2b^2/(5a)$ covers the whole $(0,b/(5a)]$ |
The quintic without a cubic term, $ae^5 - be^4 + c_0e^2$ with $a,b,c_0>0,$ is in $\mathcal C$ on no range at all: positive acceleration needs $e > 2b/(5a)$ while a nonincreasing one needs $e \le b/(5a).$ A negative quartic term can buy the fade, but only a cubic term can buy the acceleration it fades from.
Outside, and why
| family | obstruction |
|---|---|
| $A(\cosh ke - 1),$ $\ e^{ke} - 1 - ke$ | growing acceleration ($c'''' > 0$) on every interval: the opposite regime |
| $\sinh,$ $\exp,$ the exponential cone, gen-cosh | growing acceleration, and $c'(0) \ne 0$ |
| pure powers $e^\theta,$ $\theta > 3$ | growing acceleration |
| pure powers $e^\theta,$ $2 < \theta \le 3$ | in $\mathcal C,$ but $c''(0^+) = 0$: the first unit of effort is free at the margin; below the cubic also $c'''(0^+) = \infty$ |
| $e^2 + e^\theta,$ $2<\theta<3$ | in $\mathcal C$ with $c''(0^+) = 2,$ but $c'''(0^+) = \infty$: the acceleration explodes at zero |
| polynomials of degree four and up, on all of $(0,\infty)$ | growing acceleration eventually; globally $\mathcal C \cap \text{polynomials} = \{pe^2+qe^3\}$ |
So of the transcendental families the record uses, $\cosh,$ $\sinh$ and $\exp$ are all on the wrong side of the line, and the region is the quadratic-plus-cubic cone together with its small-prize neighbourhood inside $\mathcal C.$
§ 7What the proposition says, and what it does not
It is a sufficient region in cost and prize, with no depth cap and no dial floor, at full symmetry only. On it, both families’ ordinary reachable legal closings have strictly positive full closing slope, non-Nash zeros included, and both players’ per-node masses and locations rise in the dial. It does not assert the full-$\mathcal C$ target prop:sym-concave-curvature-sc, the zero-dial or zero-mass boundary assembly, Nash chain order, or anything at unequal reaches or nonzero lead; the equal-reach extension is registered as prop:equal-reach-small-curvature-drop-sc and is the next brief.
The class tools — $\mathcal C \subset X_{1/2},$ the elasticity band $1 \le ec''/c' \le 2,$ log-concavity of $c$ and $c'$ — are lem:concave-curvature-contact-tools. The proof’s inputs are the gap-curvature transport, the cubic-gap margin and rows, and the creation-rate envelopes of the 2026-10-07/08 sessions, all in module 0’s Proved list.
Numbers on this page: thresholds by root finding on the displayed closed forms (30 digits), polynomial ranges by Reduce, membership by symbolic sign checks; nothing here is a census of equilibria.