SlowGuess/NeurIPS_2023_OCR huggingface dataset:
1. Except  $b_{i}^{*}$ , there are no other fixed points of  $f_{i}$  with a support that contains the support of  $b_{i}^{*}$ . Formally, there are no fixed points  $b_{i}^{**} \neq b_{i}^{*}$  of  $f_{i}$  with support  $s^{**}$  such that  $s^{*} \subset s^{**}$ .
2. The bids  $b_{i}^{*}$  achieve a higher utility in the original game  $G$ , denoted  $u_{i}^{*}$ , than any other fixed point of Proportional Response Dynamics. Formally, let  $b_{i}^{**}$  be any fixed point other than  $b_{i}^{*}$ , with utility  $u_{i}^{**}$  in the original game  $G$ , then  $u_{i}^{*} > u_{i}^{**}$ .
Proof: Let  $b_i$  be any fixed point of  $f_i$ . By the previous claim it holds that  $\frac{a_{ij}}{p_j} = \frac{u_i}{B_i}$  whenever  $b_{ij} > 0$ . Multiplying by  $p_j$  yields  $a_{ij} = \frac{u_i}{B_i} p_j$ . Summing over  $j$  with  $b_{ij} > 0$  and rearranging yields  $\frac{u_i}{B_i} = \frac{\sum_{j \in s} a_{ij}}{\sum_{j \in s} p_j} = c_s$  as defined in Lemma 7 with support  $s$ . By that lemma, we have that  $c_s^* \geq c_s$  for any set  $s$  distinct from  $s^*$ . Thus, we have that  $u_i^* / B_i = c_s^* \geq c_{s^{**}} = u_i^{**} / B_i$  for  $b_i^{**}$  being a fixed point of  $f_i$  other than  $b_i^*$  with support  $s^{**}$  and utility value  $u_i^{**}$ .
Assume for the sake of contradiction that  $s^* \subset s^{**}$ . If  $j \in s^*$  then  $j \in s^{**}$ . By Claim 1 for every such  $j$  the following inequality holds,
$$
\frac {a _ {i j}}{p _ {j} ^ {*}} = \frac {u _ {i} ^ {*}}{B _ {i}} \geq \frac {u _ {i} ^ {* *}}{B _ {i}} = \frac {a _ {i j}}{p _ {j} ^ {* *}},
$$
implying that  $p_j^{**} \geq p_j^*$ . Subtracting  $\theta_{ij}$  from both sides yields  $b_{ij}^{**} \geq b_{ij}^*$ . Summing over  $j \in s^*$  yields a contradiction:
$$
B _ {i} = \sum_ {j \in s ^ {*}} b _ {i j} ^ {*} \leq \sum_ {j \in s ^ {*}} b _ {i j} ^ {* *} <   \sum_ {j \in s ^ {* *}} b _ {i j} ^ {* *} = B _ {i},
$$
where the first inequality is as explained above, and the last by the strict set containment  $s^* \subset s^{**}$ . Finally, as there are no fixed points with support  $s^{**}$  containing  $s^*$ , by Lemma 7, the inequality stated above is strict, that is  $c_{s^*} > c_{s^{**}}$  and so  $u_i^* > u_i^{**}$ .
Claim 3: If  $b_{i}^{**} \neq b_{i}^{*}$  is a fixed point of  $f_{i}$  then  $b_{i}^{**}$  is not a limit point of any sub-sequence of  $(b_{i}^{t})_{t=0}^{\infty}$ . Proof: The proof considers two cases: (1) When  $u_{i}$  is continuous at  $b^{**}$  (2) when continuity doesn't hold. Let  $(b^{t_k})_{k=1}^{\infty}$  be a converging subsequence of  $(b_{i}^{t})_{t=0}^{\infty}$ .
Case (1): The utility function  $u_{i}$  is continuous at  $b_{i}^{**}$  when for every good  $j$  it holds that  $\theta_{ij} > 0$  or  $b_{ij}^{**} > 0$ . i.e. that there is no good  $j$  with both  $\theta_{ij} = 0$  and  $b_{ij}^{**} = 0$ . This is implied directly from the allocation rule  $x_{ij} = \frac{b_{ij}}{\theta_{ij} + b_{ij}}$  (see the formal definition in Section 2) and the fact that  $u_{i} = \sum_{j} a_{ij} x_{ij}$ . Examine the support of  $b_{i}^{**}$ , by Claim 2 there are no fixed points with support set  $s^{**}$  containing  $s^{*}$ . Therefore  $s^{*} \not\subset s^{**}$  implying that there exists a good  $j \in s^{*} \setminus s^{**}$ . That is, by definition of the supports, there exists  $j$  with  $b_{ij}^{*} > 0$  and  $b_{ij}^{**} = 0$ . Consider such  $j$  and assume for the sake of contradiction that  $b_{i}^{**}$  is indeed a limit point. Then, by definition, for every  $\delta^{**} > 0$  exists a  $T$  s.t. if  $t > T$  then  $\| b_{i}^{t_k} - b_{ij}^{**} \| < \delta^{**}$ . Specifically it means that  $|b_{ij}^{t_k} - b_{ij}^{**}| < \delta^{**}$  whenever  $t > T$ .
By Claim 2,  $u_{i}^{*} > u_{i}^{**}$ . Then, by continuity there exists a  $\delta'$  s.t. if  $\| b_{i}^{**} - b_{i} \| < \delta'$  then  $|u_{i}(b_{i}) - u_{i}^{**}| < u_{i}^{*} - u_{i}^{**}$ . Take  $\delta^{**} < \min \{ \delta', b_{ij}^{*} \}$  and, by the assumption of convergence, there is a  $T$  s.t. for  $t > T$  and we have that  $\| b_{i}^{t_k} - b_{i}^{**} \| < \delta^{**}$ . This implies (I)  $|b_{ij}^{t_k} - 0| < \delta^{**} < b_{ij}^{*}$  as  $b_{ij}^{**} = 0$  and (II)  $|u_{i}^{t_k} - u_{i}^{**}| < u_{i}^{*} - u_{i}^{**} \Rightarrow u_{i}^{t_k} < u_{i}^{*}$ . From these two, we can conclude that
$$
\frac {a _ {i j}}{p _ {j} ^ {t _ {k}}} = \frac {a _ {i j}}{\theta_ {i j} + b _ {i j} ^ {t _ {k}}} > \frac {a _ {i j}}{\theta_ {i j} + b _ {i j} ^ {*}} = \frac {u _ {i} ^ {*}}{B _ {i}} > \frac {u _ {i} ^ {t _ {k}}}{B _ {i}}.
$$
Finally, observe that by rearranging the PRD update rule we get  $b_{ij}^{t+1} = \frac{a_{ij} / p_j^{t_k}}{u_i^{t_k} / B_i} b_{ij}^{t_k}$ , implying that  $b_{ij}^{t_k+1} > b_{ij}^{t_k}$  since  $\frac{a_{ij} / p_j^{t_k}}{u_i^{t_k} / B_i} > 1$  for  $t > T$  and  $b_{ij}^0 > 0$ . This means that for all  $t_k > T$  we have  $b_{ij}^{t_k} > b_{ij}^{T+1}$ . That is,  $b_{ij}^{t_k}$  cannot converge to zero and thus the subsequence cannot converge to  $b_i^{**}$ , a contradiction.
Case (2): When there exists a good  $j$  with  $\theta_{ij} = 0$  and  $b_{ij}^{**} = 0$  we have that  $u_i$  is not continuous at  $b_{i}^{**}$  and the previous idea doesn't work. Instead we will contradict the PRD update rule. Assume of the sake of contradiction that  $b_{i}^{**}$  is a limit point of a subsequence of PRD updates. Therefore for every  $\epsilon$  exists a  $T$  s.t. if  $t_k > T$  then  $|b_{ij}^{t_k} - b_{ij}^{**}| < \epsilon$ . Note that  $b_{ij}^{**} = 0$  in this case and set  $\epsilon < \frac{a_{ij}}{m} B_i$  and so, for  $t_k > T$  it holds that  $\frac{a_{ij}}{b_{ij}^{t_k}} > \frac{m}{B_i}$ . Also note that  $p_j^{t_k} = \theta_{ij}^{t_k} + b_{ij}^{t_k} = b_{ij}^{t_k}$  and that the maximal utility a buyer may have is  $m$  (when it is allocated every good entirely). Then overall we have that  $\frac{a_{ij}}{p_j^{t_k}} > \frac{m}{B_i} > \frac{u_i^{t_k}}{B_i}$ . The PRD update rule is  $b_{ij}^{t_k + 1} = \frac{a_{ij} / p_j^{t_k}}{u_i^{t_k} / B_i} b_i^{t_k}$ . But since the ratio  $\frac{a_{ij} / p_{ij}^{t_k}}{u_i^{t_k} / B_i}$  is greater than 1 it must be that  $b_{ij}^{t_k + 1} > b_{ij}^{t_k}$ . And so every subsequent element of the subsequence is bounded below by  $b_{ij}^{T + 1} > 0$  and as before, we reach a contradiction as the subsequence cannot converge to  $b_i^{**}$ .
Finally we can prove the convergence of the sequence  $(b_i^t)_{t=1}^\infty$ . As the action space  $S_i$  is compact, there exists a converging subsequence  $b_i^{t_k}$  with the limit  $b_i^{**}$ . If  $b_i^{**} = b_i^*$  for any such subsequence, then we are done. Otherwise, assume  $b_i^{**} \neq b_i^*$ . By the previous claim any fixed point of  $f_i$  other than  $b_i^*$  is not a limit point of any subsequence, thus  $b_i^{**}$  is not a fixed point of  $f_i$ . By Lemma 1, any subset of players performing proportional response, strictly increase the potential function unless performed at a fixed point. When discussing a proportional response of a single player, with all others remaining fixed, this implies, by the definition of potential function, that  $\tilde{u}_i$  is increased at each such step. Let  $\epsilon < \tilde{u}_i(f_i(b^{**})) - \tilde{u}_i(b^{**})$ , this quantity is positive since  $b_i^{**}$  is not a fixed point. The function  $\tilde{u}_i \circ f_i$  is a continuous function and  $b_i^{t_k}$  converges to  $b_i^{**}$  therefore there exists a  $T$  such that for all  $t_k > T$  we have that  $|\tilde{u}_i(f_i(b^{**})) - \tilde{u}_i(f_i(b^{t_k}))| < \epsilon$ . Substituting  $\epsilon$  yields  $\tilde{u}_i(f_i(b^{**})) - \tilde{u}_i(f_i(b^{t_k})) < \tilde{u}_i(f_i(b^{**})) - \tilde{u}_i(b^{**})$  which implies  $\tilde{u}_i(b^{**}) < \tilde{u}_i(f_i(b^{t_k})) = \tilde{u}_i(b^{t_k+1}) \leq \tilde{u}_i(b^{t_k+1})$ . That is, the
sequence  $u_{i}(b_{i}^{t_{k}})$  is bounded away from  $\tilde{u}_{i}(b^{**})$  and since  $\tilde{u}_{i}$  is a continuous function, this implies that  $b_{i}^{t_{k}}$  is bounded away from  $b_{i}^{**}$  — a contradiction to convergence.
Appendix C Simultaneous Play by Subsets of Agents
In order to prove Lemma 1, we first need some further definitions and technical lemmas. We use the notation  $D(x\| y)$  to denote the KL divergence between the vectors  $x$  and  $y$ , i.e.,  $D(x\| y) = \sum_{j}x_{j}\ln \left(\frac{x_{j}}{y_{j}}\right)$ . For a subset of the players  $v\subseteq B$ , the subscript  $v$  on vectors denotes the restriction of the vector to the coordinates of the players in  $v$ , that is, for a vector  $b$  we use the notation  $b_{v} = (b_{ij})_{i\in v,j\in [m]}$  to express the restriction to the subset.  $\ell_{\Phi}(b_v;b_v^{\prime})$  denotes the linear approximation of  $\Phi$ ; that is,  $\ell_{\Phi}(b_v;b_v^{\prime}) = \Phi (b_v^{\prime}) + \nabla_{b_v}\Phi (b_v^{\prime})(b_v - b_v^{\prime})$ .
The idea described in the next lemma to present the potential function as a linear approximation term and a divergence term was first described in [5] for a different scenario when all agents act together in a synchronized manner using mirror descent; we extend this idea to our asynchronous setting which requires using different methods and as well as embedding it in a game.
Lemma 9. Fix a subset of the players  $v \subset \mathcal{B}$  and a bid profile  $b_{-v}$  of the other players. Then, for all  $b_v, b_v' \in S_v$  we have that  $\Phi(b_v) = \ell_\Phi(b_v; b_v') - D(p \| p')$ , where  $p = \sum_{i \notin v} b_{ij} + \sum_{i \in v} b_{ij}$  and  $p' = \sum_{i \notin v} b_{ij} + \sum_{i \in v} b_{ij}'$ .
Proof. Calculating the difference  $\Phi(b_v) - \ell_{\Phi}(b_v; b_v')$  yields
$$
\Phi (b _ {v}) - \ell_ {\Phi} (b _ {v}; b _ {v} ^ {\prime}) = \Phi (b _ {v}) - \Phi (b _ {v} ^ {\prime}) - \nabla_ {b _ {v}} \Phi (b _ {v} ^ {\prime}) (b _ {v} - b _ {v} ^ {\prime}).
$$
We rearrange the term  $\Phi (b_v) - \Phi (b_v^{\prime})$  as follows.
$$
\begin{array}{l} \Phi \left(b _ {v}\right) - \Phi \left(b _ {v} ^ {\prime}\right) = \sum_ {i \in v, j \in [ m ]} \left(b _ {i j} - b _ {i j} ^ {\prime}\right) \ln \left(a _ {i j}\right) - \sum_ {j} \left(p _ {j} \ln \left(p _ {j}\right) - p _ {j} ^ {\prime} \ln \left(p _ {j} ^ {\prime}\right)\right) - \sum_ {j} \left(p _ {j} - p _ {j} ^ {\prime}\right) \\ = \sum_ {i \in v, j \in [ m ]} \left(b _ {i j} - b _ {i j} ^ {\prime}\right) \ln \left(a _ {i j}\right) - \sum_ {j} \left(p _ {j} \ln \left(p _ {j}\right) - p _ {j} ^ {\prime} \ln \left(p _ {j} ^ {\prime}\right)\right), \\ \end{array}
$$
where the last equality is since  $\sum_{j}p_{j} = 1$  for any set of prices because the economy is normalized (see the model section in the main text).
The term  $\nabla_{b_v}\Phi (b_v')(b_v - b_v')$  is expanded as follows.
$$
\begin{array}{l} \nabla_ {b _ {v}} \Phi (b _ {v} ^ {\prime}) (b _ {v} - b _ {v} ^ {\prime}) = \sum_ {i \in v, j \in [ m ]} \ln \left(\frac {a _ {i j}}{p _ {j} ^ {\prime}}\right) (b _ {i j} - b _ {i j} ^ {\prime}) \\ = \sum_ {i \in v, j \in [ m ]} \ln \left(a _ {i j}\right) \left(b _ {i j} - b _ {i j} ^ {\prime}\right) - \sum_ {i \in v, j \in [ m ]} \ln \left(p _ {j} ^ {\prime}\right) \left(b _ {i j} - b _ {i j} ^ {\prime}\right). \\ \end{array}
$$
Subtracting the latter from the former cancels out the term  $\sum_{i\in v,j\in [m]}\ln (a_{ij})(b_{ij} - b_{ij}^{\prime})$ , and we are left with the following.
$$
\begin{array}{l} \Phi \left(b _ {v}\right) - \ell_ {\Phi} \left(b _ {v}; b _ {v} ^ {\prime}\right) = \Phi \left(b _ {v}\right) - \Phi \left(b _ {v} ^ {\prime}\right) - \nabla_ {b _ {v}} \Phi \left(b _ {v} ^ {\prime}\right) \left(b _ {v} - b _ {v} ^ {\prime}\right) \\ = \sum_ {i \in v, j \in [ m ]} \ln \left(p _ {j} ^ {\prime}\right) \left(b _ {i j} - b _ {i j} ^ {\prime}\right) - \sum_ {j} \left(p _ {j} \ln \left(p _ {j}\right) - p _ {j} ^ {\prime} \ln \left(p _ {j} ^ {\prime}\right)\right) \\ = - \sum_ {j} p _ {j} \ln (p _ {j}) - \left(p _ {j} ^ {\prime} - \sum_ {i \in v} b _ {i j} ^ {\prime} + \sum_ {i \in v} b _ {i j}\right) \ln \left(p _ {j} ^ {\prime}\right) \\ = - \sum_ {j} p _ {j} \ln (p _ {j}) - \left(\theta_ {v j} + \sum_ {i \in v} b _ {i j}\right) \ln \left(p _ {j} ^ {\prime}\right) \\ = - \sum_ {j} p _ {j} \ln (p _ {j}) - p _ {j} \ln \left(p _ {j} ^ {\prime}\right) \\ = - D (p \| p ^ {\prime}). \\ \end{array}
$$
Lemma 10. For any subset of the players  $v \subset \mathcal{B}$  and any bid profile  $b_{-v}$  of the other players and for every  $b_v, b_v' \in S_v$  it holds that  $D(p \| p') \leq D(b_v \| b_v')$ , with equality only when  $b_v = b_v'$ .
Proof. We begin by proving a simpler case where  $v = \{i\}$  for some player  $i$  and use it to prove the more general statement. Fix  $i$  and  $b_{-i}$ , which implies fixing some  $\theta_i$ . KL divergence is convex in both arguments with equality only if the arguments are equal; formally, for  $\lambda \in (0,1)$  it holds that  $D(\lambda \theta_i + (1 - \lambda) b_i \| \lambda \theta_i + (1 - \lambda) b_i') \leq \lambda D(\theta_i \| \theta_i) + (1 - \lambda) D(b_i \| b_i')$ , which is equivalent to  $D(\lambda \theta_i + (1 - \lambda) b_i \| \lambda \theta_i + (1 - \lambda) b_i') \leq (1 - \lambda) D(b_i \| b_i')$ , with equality only if  $b_i = b_i'$  (since  $D(\theta_i \| \theta_i) = 0$ ). Substituting  $\lambda = \frac{1}{2}$  and noting that  $p_j = \theta_{ij} + b_{ij}$  (and the same for  $p_j'$  and  $b_{ij}'$ ), we obtain the following relation.
$$
\begin{array}{l} D (\frac {1}{2} p \| \frac {1}{2} p ^ {\prime}) = D (\frac {1}{2} \theta_ {i} + \frac {1}{2} b _ {i} \| \frac {1}{2} \theta_ {i} + \frac {1}{2} b _ {i} ^ {\prime}) \\ \leq \frac {1}{2} D \left(b _ {i} \| b _ {i} ^ {\prime}\right). \\ \end{array}
$$
On the other hand, the expression  $D\left(\frac{1}{2} p \parallel \frac{1}{2} p'\right)$  can be evaluated as follows.
$$
\begin{array}{l} D \left(\frac {1}{2} p \| \frac {1}{2} p ^ {\prime}\right) = \sum_ {j} \frac {1}{2} p _ {j} \ln \left(\frac {1 / 2 p _ {j}}{1 / 2 p _ {j} ^ {\prime}}\right) \\ = \frac {1}{2} \sum_ {j} p _ {j} \ln \left(\frac {p _ {j}}{p _ {j} ^ {\prime}}\right) \\ = \frac {1}{2} D (p \| p ^ {\prime}). \\ \end{array}
$$
And therefore, we have  $D(p\| p^{\prime}) \leq D(b_{i}\| b_{i}^{\prime})$ , with equality only if  $b_{i} = b_{i}^{\prime}$ .
Now we can prove the general case, as stated fix  $v$  and  $b_{-v}$  and let  $b_v, b_v' \in S_v$ . We know that for all  $i \in v$  it is true that  $D(p\| p') \leq D(b_i \| b_i')$ , summing those inequalities for all  $i \in v$  yields  $|v|D(p\| p') \leq \sum_{i \in v} D(b_i \| b_i')$ , on the one hand clearly  $D(p\| p') \leq |v|D(p\| p')$  and on the other hand  $\sum_{i \in v} D(b_i \| b_i') = \sum_{i \in v} \sum_j b_{ij} \ln(\frac{b_{ij}}{b_{ij}'}) = D(b_v \| b_v')$  and the result is obtained.
Lemma 11. Let  $v \subseteq \mathcal{B}$ , let  $f_v: S \to S$  be a proportional response update function for members of  $v$  and identity for the others, and let  $b' \in S$  be some bid profile. Then,  $(f_v(b'))_v = \arg \max_{b_v \in S_v} \{\ell_\Phi(b_v; b'_v) - D(b_v \| b'_v)\}$ .
Proof. By adding and removing constants that do not change the maximizer of the expression on the right hand side, we obtain that the maximizer is exactly the proportional response update rule:
$$
\begin{array}{l} \arg \max  _ {b _ {v} \in S _ {v}} \left\{\ell_ {\Phi} \left(b _ {v}; b _ {v} ^ {\prime}\right) - D \left(b _ {v} \| b _ {v} ^ {\prime}\right) \right\} = \arg \max  _ {b _ {v} \in S _ {v}} \left\{\Phi \left(b _ {v} ^ {\prime}\right) + \nabla_ {b _ {v}} \Phi \left(b _ {v} ^ {\prime}\right) \left(b _ {v} - b _ {v} ^ {\prime}\right) - D \left(b _ {v} \| b _ {v} ^ {\prime}\right) \right\} \\ = \arg \max  _ {b _ {v} \in S _ {v}} \left\{\nabla_ {b _ {v}} \Phi \left(b _ {v} ^ {\prime}\right) b _ {v} - D \left(b _ {v} \| b _ {v} ^ {\prime}\right) \right\} \\ = \arg \max _ {b _ {v} \in S _ {v}} \{\nabla_ {b _ {v}} \Phi (b _ {v} ^ {\prime}) b _ {v} - D (b _ {v} \| b _ {v} ^ {\prime}) - \sum_ {i \in v} B _ {i} \ln (\frac {u _ {i} ^ {\prime}}{B _ {i}}) \}. \\ \end{array}
$$
Rearranging the last expression by elements yields the following result,
$$
\begin{array}{l} \nabla_ {b _ {v}} \Phi (b _ {v} ^ {\prime}) b _ {v} - D (b _ {v} \| b _ {v} ^ {\prime}) - \sum_ {i \in v} B _ {i} \ln \left(\frac {u _ {i} ^ {\prime}}{B _ {i}}\right) = \\ = \sum_ {i \in v, j \in [ m ]} b _ {i j} \ln \left(\frac {a _ {i j}}{p _ {j} ^ {\prime}}\right) - \sum_ {i \in v, j \in [ m ]} b _ {i j} \ln \left(\frac {b _ {i j}}{b _ {i j} ^ {\prime}}\right) - \sum_ {i \in v, j \in [ m ]} b _ {i j} \ln \left(\frac {u _ {i} ^ {\prime}}{B _ {i}}\right) \\ = \sum_ {i \in v, j \in [ m ]} b _ {i j} \ln \left(\frac {a _ {i j}}{p _ {j} ^ {\prime}} \frac {b _ {i j} ^ {\prime}}{b _ {i j}} \frac {B _ {i}}{u _ {i} ^ {\prime}}\right) \\ = \sum_ {i \in v, j \in [ m ]} b _ {i j} \ln \left(\frac {\frac {a _ {i j} x _ {i j} ^ {\prime}}{u _ {i} ^ {\prime}} B _ {i}}{b _ {i j}}\right) \\ = -\sum_{i\in v,j\in [m]}b_{ij}\ln \left(\frac{b_{ij}}{\frac{a_{ij}x_{ij}^{\prime}}{u_{i}^{\prime}}B_{i}}\right), \\ \end{array}
$$
which is exactly  $-D(b_v\| (f_v(b'))_v)$ , since  $(f_v(b'))_{ij} = \frac{a_{ij}x_{ij}'}{u_i'} B_i$  for  $i\in v$  by definition. That is, our maximization problem is equivalent to  $\arg \min \{D(b_v\| (f_v(b'))_v)\}$ . Finally, note that KL divergence is minimized when both of its arguments are identical, and  $(f_v(b_v'))_v\in S_v$ , the domain of the minimization.
Proof. (Lemma 1): Let  $v \subseteq \mathcal{B}$  be a subset of players and let  $b \in S$  be some bid profile. By combining the lemmas proved in this section have that
$$
\Phi \left(f _ {v} (b)\right) \geq \ell_ {\Phi} \left(f _ {v} (b); b\right) - D \left(f _ {v} (b) \| b\right) \geq \ell_ {\Phi} (b; b) - D (b \| b) = \Phi (b),
$$
where the first inequality is by Lemmas 9 and 10 with the inequality being strict whenever  $f_v(b) \neq b$ , and the second inequality is by Lemma 11, as  $f_v(b)$  was shown to be the maximizer of this expression over all  $b \in S$ .
An interesting case to note here is when  $v = i$ . In this case, the lemmas above show that if the players' bids are  $b^{t}$  and  $i$  is being activated by the adversary, then the best response bids of  $i$  to  $b_{-i}^{t}$  are the solutions to the optimization problem  $\arg \max_{b_{i} \in S_{i}} \{\ell_{\tilde{u}_{i}}(b_{i}; b_{i}^{t}) - D(p \| p^{t})\}$ . On the other hand, the proportional response to  $b_{-i}^{t}$  is the solution to the optimization problem  $\arg \max_{b_{i} \in S_{i}} \{\ell_{\tilde{u}_{i}}(b_{i}; b_{i}^{t}) - D(b_{i} \| b_{i}^{t})\}$ . This can be seen as a relaxation of the former, as proportional response does not increase  $\tilde{u}_{i}$  (or equivalently the potential) as much as best response does. However, proportional response is somewhat easier to compute.
Appendix D Convergence of Asynchronous Proportional Response Dynamics
Proof. (Theorem 1): Denote the distance between a point  $x$  and a set  $S$  as  $d(x, S) = \inf_{x^* \in S} \| x - x^* \|$ . By Theorem 2 we have that the set of potential maximizing bid profiles is identical to the set of market equilibria. Denote this set by  $ME$  and the maximum value of the potential by  $\Phi^*$ . More specifically, every  $b^* \in ME$  achieves  $\Phi(b^*) = \Phi^*$ , and  $\Phi^*$  is achieved only by elements in  $ME$ .
We start with the following lemma.
Lemma 12. For every  $\epsilon >0$  there exists a  $\delta >0$  such that  $\Phi (b) > \Phi^{*} - \delta$  implies  $d(b,ME) <   \epsilon$
Proof. Assume otherwise that for some  $\epsilon_0$  there exist a sequence  $(b_{t})$  such that  $\Phi (b_{t})\to \Phi^{*}$  but  $\inf_{b^{*}\in ME}\| b_t - b^*\| \geq \epsilon_0$  for all  $t$ . Note that for all  $b^{*}\in ME$  we have that  $\Phi (b^{*}) = \Phi^{*}$  and  $\| b_{t} - b^{*}\| \geq \inf_{b^{\prime}\in ME}\| b_{t} - b^{\prime}\| \geq \epsilon_{0}$  for all  $t$ . Take a condensation point  $b^{**}$  of this sequence and a subsequence  $(t_j)$  that converges to  $b^{**}$ . Thus by our assumption we have  $\Phi (b^{**}) = \lim \Phi (b_{t_j}) = \Phi^*$ . For any  $b^{*}\in ME$  we have  $\| b^{**} - b^{*}\| = \lim \| b_{t_j} - b^*\| \geq \epsilon_0 > 0$ . The former equality must imply  $b^{**}\in ME$ , but the latter implies  $b^{**}\notin ME$ .
Next, for a subset of players  $A \subset \mathcal{B}$  let  $f_{A} : [0,1]^{n} \to [0,1]^{n}$  be the continuous function where  $i \in A$  do a proportional response update and the other players play the identity function. (i.e., do not change their bids, see Section 5 in the main text).
By Lemma 1 from the main text we have that (i) For all  $A$  we have that  $f_{A}(b) = b$  if and only if for all  $i \in A$  it holds that  $f_{i}(b) = b$ ; and (ii)  $\Phi(f_{A}(b)) > \Phi(b)$  unless  $f_{A}(b) = b$ .
Definition. The stable set of  $b^{**}$  is defined to be  $S(b^{**}) = \{i|f_i(b^{**}) = b^{**}\}$ .
A corollary (i) and (ii) above is that if  $A \subseteq S(b^{**})$  then  $f_{A}(b^{**}) = b^{**}$ , but if  $A \setminus S(b^{**}) \neq \emptyset$  then  $\Phi(f_{A}(b^{**})) > \Phi(b^{**})$ .
Lemma 13.: Let  $\Phi(b^{**}) < \Phi^*$ . Then there exists  $\delta > 0$  such that for every  $\|b - b^{**}\| \leq \delta$  and every  $A \setminus S(b^{**}) \neq \emptyset$  we have that  $\Phi(f_A(b)) > \Phi(b^{**})$ .
Proof. Fix a set  $A$  such that  $A \setminus S(b^{**}) \neq \emptyset$  and let  $\alpha = \Phi(f_A(b^{**})) - \Phi(b^{**}) > 0$ . Since  $\Phi(f_A(\cdot))$  is continuous, there exists  $\delta$  so that  $|b - b^{**}| \leq \delta$  implies  $\Phi(f_A(b^{**})) - \Phi(f_A(b)) < \alpha$  and thus  $\Phi(f_A(b)) > \Phi(b^{**})$ . Now take the minimum  $\delta$  for all finitely many  $A$ .
Lemma 14. Let  $\Phi(b^{**}) < \Phi^*$  and let  $F$  be a finite family of continuous functions such that for every  $f \in F$  we have that  $f(b^{**}) = b^{**}$ . Then there exists  $\epsilon > 0$  such that for every  $b$  such that  $\|b - b^{**}\| \leq \epsilon$  and every  $f \in F$  and every  $A \setminus S(b^{**}) \neq \emptyset$  we have that  $\Phi(f_A(f(b))) > \Phi(b^{**})$ .
Proof. Fix  $f \in F$  and let  $\delta$  be as promised by the previous lemma, i.e. for every  $\| z - b^{**} \| \leq \delta$  and every  $A \setminus S(b^{**}) \neq \emptyset$  we have that  $\Phi(f_A(z)) > \Phi(b^{**})$ . Since  $f(b^{**}) = b^{**}$  and  $f$  is continuous there exists  $\epsilon > 0$  so that  $\| b - b^{**} \| \leq \epsilon$  implies  $\| f(b) - f(b^{**}) \| = \| f(b) - b^{**} \| \leq \delta$  and thus  $\Phi(f_A(f(b))) > \Phi(b^{**})$ . Now take the minimum  $\epsilon$  over the finitely many  $f \in F$ .
Definition. a sequence of sets  $A_{t} \subseteq \mathcal{B}$  is called  $T$ -live if for every  $i$  and for every  $t$  there exists some  $t \leq t^{*} \leq t + T$  such that  $i \in S_{t^{*}}$ .
Lemma 15. Fix a sequence  $b = (b_{t})$  where  $b_{t + 1} = f_{A_t}(b_t)$  such that the sequence  $A_{t}$  is  $T$ -live. There are no condensation points of  $(b_{t})$  outside of  $ME$ .
Proof. Assume that exists a condensation point  $b^{*} \notin ME$  and a subsequence that converges to it, then  $\Phi(b^{**}) < \Phi(b^{*})$ . Notice that  $\Phi(b_{t})$  is a non-decreasing sequence and so  $\Phi(b_{t}) \leq \Phi(b^{**})$  for all  $t$ . Let  $F$  be a set of functions achieved by composition of at most  $T$  functions from  $\{f_{A}|A \subset S(b^{**})\}$ . So for every  $f \in F$  we have that  $f(b^{**}) = b^{**}$ , while for every  $B \setminus S(b^{**}) \neq \emptyset$  we have that  $\Phi(f_{B}(b^{**})) > \Phi(b^{**})$ . Let  $\epsilon$  be as promised by the previous lemma, i.e., for every  $\| b - b^{**} \| \leq \epsilon$  and every  $f \in F$  and every  $B$  such that  $B \setminus S(b^{**}) \neq \emptyset$  we have that  $\Phi(f_{B}(f(b))) > \Phi(b^{**})$ . Since the subsequence converges to  $b^{**}$  there exists  $t_{j}$  in the subsequence so that  $\| b_{t_{j}} - b^{**} \| \leq \epsilon$ . Now let  $t > t_{j}$  be the first time that  $A_{t} \setminus S(b^{**}) \neq \emptyset$ . Now  $b_{t+1} = f_{A_{t}}(f(b_{t_{j}}))$ , where  $f$  is the composition of all  $f_{A}$  for the times  $t_{j}$  to  $t$ . We can now apply the previous lemma to get that  $\Phi(b_{t+1}) = \Phi(f_{A_{t}}(f(b_{t_{j}})) > \Phi(b^{**})$ , a contradiction.