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Contents
  1. C73-15: Payoff characterization for a fixed mirror-descent rule
  2. 1) FORMAL RESTATEMENT
  3. 2) QUICK LITERATURE/CONTEXT CHECK
  4. 3) ATTACK PLAN
  5. 4) WORK
  6. 5) VERIFICATION
  7. 6) FINAL
  8. Reproducibility and scope.
  9. References

Research attempt and adversarial verification

9 October 2026 Original-target status: UNRESOLVED

C73-15: Payoff characterization for a fixed mirror-descent rule

Input: C73-15.tex.  Tools: web browsing [YES]; Python computation [YES].

1) FORMAL RESTATEMENT

Fix finite nonempty action sets, proper lower-semicontinuous strictly convex regularizers $h_i$ on their simplexes, differentiable and finite on the relative interiors, and a sequence $\eta_t>0$. Let $X=\prod_i\Delta(S_i)$ and $v_i(x)_a=u_i(a,x_{-i})$. For an interior starting point define \[ D_{h_i}(p,x)=h_i(p)-h_i(x)-\langle\nabla h_i(x),p-x\rangle, \quad x_i^{t+1}={\operatorname*{arg\,max}}_{p\in\Delta(S_i)} \{\eta_t\langle v_i(x^t),p\rangle-D_{h_i}(p,x_i^t)\}. \] As in the input, assume uniqueness of every update and that all iterates remain where the required gradients are defined. The target is a condition on $u$ for \[ \forall x^1\in\operatorname{relint}X,\qquad \operatorname{dist}(x^t,{\operatorname{NE}}(G))\longrightarrow0. \] The rule, including every step size, is fixed before testing a game. The input does not select one particular regularizer or schedule, so a theorem for a special choice is a partial classification, not a solution for the whole family.

The main specialization below is negative entropy $h_i(p)=\sum_a p_a\log p_a$, with $0\log0=0$ and natural logarithms. It gives \[ x_{i,a}^{t+1}= \frac{x_{i,a}^{t}\exp(\eta_t v_{i,a}(x^t))} {\sum_bx_{i,b}^{t}\exp(\eta_t v_{i,b}(x^t))}. \tag{MD} \] The formula is the unique optimizer: the first-order conditions of the strictly concave entropy-regularized objective give the displayed positive probabilities, and normalization determines the multiplier. Hence the interior-domain condition is satisfied for every finite step.

2) QUICK LITERATURE/CONTEXT CHECK

Bichler et al. [potentialness] distinguish potential structure from general learning convergence and explicitly discuss the absence of a comprehensive classification. Their empirical potentialness measure is not a necessary-and-sufficient theorem for the particular discrete rule above. The following schedule-sensitive statements and counterexamples are proved directly.

3) ATTACK PLAN

Proof track. Obtain exact log-ratio recursions for entropy; classify the finite-total-step case using initializations near every vertex; identify positive subclasses when total learning time is infinite.

Construction track. Try a one-player game with summable steps, a zero-sum mixed-equilibrium game, and an identical-payoff game with a large constant step. These test three different failed generalizations.

Tools and purpose. Entropy first-order conditions make updates explicit; log odds remove normalization; uniform payoff bounds control finite total movement; closedness of the Nash set permits vertex tests; strict convexity supplies an energy inequality; invariant diagonal profiles reduce a two-player update to one variable; exact logarithmic identities certify a two-cycle.

4) WORK

Theorem (Complete classification for summable entropic steps): For the entropy rule (MD), suppose $S=\sum_{t\geq1}\eta_t<\infty$. Then the universal convergence property holds if and only if the game is strategically constant: \[ u_i(a_i,s_{-i})=u_i(b_i,s_{-i}) \quad\text{for all }i,a_i,b_i,s_{-i}. \tag{SC} \] Equivalently, ${\operatorname{NE}}(G)=X$.
Proof: If (SC) holds, all pure actions of a player have the same payoff against every mixed opponents' profile. Every mixed profile is therefore Nash, proving sufficiency.

For necessity choose $M$ with $|u_i(s)|\leq M$ for every player and pure profile. Multilinearity gives $|v_{i,a}(x)|\leq M$ on $X$. Iterating the ratio form of (MD), \[ \frac{x_{i,a}^{t}}{x_{i,b}^{t}} =\frac{x_{i,a}^{1}}{x_{i,b}^{1}} \exp\left(\sum_{s<t}\eta_s[v_{i,a}(x^s)-v_{i,b}(x^s)]\right). \] Fix an arbitrary pure profile $a^*$. Take interior starting points with $x_{i,a_i^*}^{1}\geq1-\delta$. Summing the ratios to $a_i^*$ gives, for every $t$, \[ \frac{1-x_{i,a_i^*}^{t}}{x_{i,a_i^*}^{t}} \leq e^{2MS}\frac{\delta}{1-\delta}. \] Thus every iterate is within a quantity tending to zero with $\delta$ of the vertex $a^*$. If the assumed universal Nash-set convergence holds, compactness provides a subsequential Nash limit with the same closeness bound. Letting $\delta\downarrow0$ and using that the Nash set is closed shows $a^*\in{\operatorname{NE}}(G)$. Every pure profile is therefore Nash. For fixed $i,s_{-i}$, the profiles $(a_i,s_{-i})$ and $(b_i,s_{-i})$ give both opposite best-response inequalities, forcing equality of the two payoffs. This proves (SC). ∎

The necessity argument uses the universal interior-initialization quantifier. It does not assert that one preselected interior starting point must fail in every nonconstant game.

Proposition (Two positive tests when total learning time is infinite): Under (MD), if $\sum_t\eta_t=\infty$, every one-player finite game converges to its optimal-action simplex. Also, every finite game in which each player has a strictly dominant pure action converges to the corresponding pure Nash equilibrium.
Proof: For one player, let $c_a$ denote its fixed payoff and $S_t=\sum_{s<t}\eta_s$. The exact formula is \[ x_a^t=\frac{x_a^1e^{c_aS_t}}{\sum_bx_b^1e^{c_bS_t}}. \] If $c_a<\max_b c_b$, its ratio to a maximizing action tends to zero. Ratios between maximizing actions remain their initial ratios, proving the asserted convergence.

If $a_i^*$ strictly dominates every other action, finiteness gives $v_{i,a_i^*}(x)-v_{i,b}(x)\geq\delta_i>0$ for every $x$ and $b\ne a_i^*$. The same ratio recursion bounds $x_{i,b}^t/x_{i,a_i^*}^t$ by its initial value times $e^{-\delta_iS_t}$. All the undesired masses tend to zero. The dominant profile is Nash by definition. ∎

Theorem (Zero-sum does not guarantee last-iterate convergence): For matching pennies \[ A=\begin{pmatrix}1&-1\\-1&1\end{pmatrix},\qquad B=-A, \] the simultaneous entropy rule fails to approach the Nash set from every interior initialization other than the equilibrium, for every positive step sequence.
Proof: Write $p,q$ for the first-action probabilities and $z=\log(p/(1-p))$, $w=\log(q/(1-q))$. The unique Nash equilibrium is $p=q=1/2$: a nonindifferent opponent would force a pure best reply, and no pure pair is mutually optimal; indifference requires exactly these probabilities. The ratio recursion gives \[ z^{t+1}=z^t+2\eta_t\tanh(w^t/2),\qquad w^{t+1}=w^t-2\eta_t\tanh(z^t/2). \] Set $E(z,w)=\log\cosh(z/2)+\log\cosh(w/2)$. Its Hessian is diagonal with strictly positive entries $\tfrac14\operatorname{sech}^2(z/2)$ and $\tfrac14\operatorname{sech}^2(w/2)$. The displacement above is orthogonal to $\nabla E$. Strict convexity therefore gives $E(z^{t+1},w^{t+1})>E(z^t,w^t)$ unless $(z^t,w^t)=(0,0)$. Starting elsewhere, energy stays at least $E(z^1,w^1)>0$. Convergence to the unique Nash profile would make $(z^t,w^t)\to(0,0)$ and the energy tend to zero, which is impossible. This argument uses no lower bound on $\eta_t$ and no assumption on its total sum. ∎
Theorem (An exact two-cycle in an exact potential game): Let both payoff matrices be $A=B=\left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right)$, take $\eta_t=4\log3$, and initialize $p^1=q^1=3/4$. Then the entropy iterates alternate between $(3/4,3/4)$ and $(1/4,1/4)$. Neither profile is Nash.
Proof: The common payoff itself is an exact potential. Symmetry preserves $p=q$. For the common log odds $z$, the payoff difference between the first and second actions is $1-2p=-\tanh(z/2)$, so \[ z^{t+1}=z^t-4\log3\,\tanh(z^t/2). \] At $z=\log3$, $\tanh(z/2)=1/2$, and the next value is $-\log3$. Oddness gives the reverse step. At $(3/4,3/4)$ the expected payoff is $3/8$ and switching to the second pure action gives $3/4$; at $(1/4,1/4)$ switching to the first gives the same improvement. The periodic orbit is disjoint from the closed Nash set, so its distance from that set is positive. ∎
This is a counterexample to arbitrary-step potential convergence, not to the summable-step classification, whose assumptions it does not satisfy.

5) VERIFICATION

For a one-player two-action game with rewards $(1,0)$, initialization $(1/2,1/2)$, and $\eta_t=2^{-t}$, the exact limit is $(e/(1+e),1/(1+e))$, not an optimum. This tiny case checks the finite-total-step obstruction without interactions. A one-action player contributes no condition to (SC), and a wholly constant game trivially passes every rule.

The script verifies the potential two-cycle with a floating-point round-trip error below $10^{-12}$; its proof is the exact identity involving $\log3$ above. It also checks 100 strict matching-pennies energy increases from log odds $(0.7,-0.4)$ with step $0.1$. Minimal pseudocode is 

\begin{lstlisting} for t:     for each player and action:         new_log_weight = log(old_weight) + eta[t] * payoff_vector[action]     normalize each player's new weights simultaneously \end{lstlisting}
Simultaneity matters: using one player's new probabilities in the other player's update would be a different algorithm. The proof never treats negative entropy as differentiable on the boundary. None of the failures concerns convergence of time-averaged play to a correlated equilibrium.

\Needspace{8\baselineskip}

6) FINAL

LABEL: UNRESOLVED.

Strongest checked partial result. A complete payoff-array classification for entropy with summable positive steps; positive infinite-total-step subclasses; an all-positive-schedules matching-pennies obstruction; and an exact potential-game two-cycle for a specified constant step.

Exact first gap. For arbitrary fixed regularizers and nonsummable schedules, no necessary-and-sufficient payoff-array criterion is proved.

Next three moves.

  1. For nonsummable entropy steps, identify a payoff condition that controls last-iterate rather than averaged regret and survives all interior initializations.
  2. For a specified potential array, determine a sharp step-size stability bound; the exact cycle supplies a concrete large-step test.
  3. Extend the finite-total-step vertex argument to regularizers for which a uniform, initialization-sensitive movement bound can actually be proved.
Counterexample perspective. Two actions and two players already suffice to break both zero-sum and potential-based universal claims. They do not establish that an arbitrary specified rule has no useful convergent class.

Reproducibility and scope.
This is one of eleven companion reports. The bundle includes verify.py and its executed results. Finite experiments supplement the proofs; they are not proofs of asymptotic claims. Literature coverage is targeted, and no novelty or formal proof-assistant certification is claimed.

References

  • [potentialness] Martin Bichler, Davide Legacci, Panayotis Mertikopoulos, Matthias Oberlechner, and Bary Pradelski. Characterizing the Convergence of Game Dynamics via Potentialness. arXiv:2503.16285, version 1 (2025). https://arxiv.org/html/2503.16285v1.