Proof. By definition, \(z_1'=P_{d_x}\!\big(z_1\odot\mathrm{Exp}(\eta_{x}A((1+m)z_2-mz_4))\big)\). By definition of \(\mathop{\mathrm{\mathscr{D}}}_{{{\Phi}_m}}\), the normalizer is nonzero, and the exponential factor is strictly positive in every coordinate, so \(z_{1,i}'=0\iff z_{1,i}=0\). The same argument holds for \(z_2'\). If \(Z \in \mathop{\mathrm{relint}}\smash{\widehat{\Delta}}\) then all coordinates of \(z_1, z_2, z_3,z_4\) are strictly positive. Combined with the strict positivity of the exponential function yields the claim. ◻

Proof of Theorem 1. Let \(Z=[z_1,z_2,z_3,z_4]\in\smash{\widehat{\Delta}}\). By the definition of \(\Phi_m\), \(\Phi_m(Z)=Z\) is equivalent to \[\begin{align} \begin{cases} z_1 &= P_{d_x}\!\left( z_1\odot \mathrm{Exp}\!\left(\eta_{x}A((1+m)z_2-mz_4)\right) \right),\\ z_2 &= P_{d_y}\!\left( z_2\odot \mathrm{Exp}\!\left(\eta_{y}B((1+m)z_1-mz_3)\right) \right),\\ z_3&=z_1, \qquad z_4=z_2. \end{cases} \end{align}\] This is equivalent to \[\begin{align} \begin{cases} z_3&=z_1, \qquad z_4=z_2,\\ z_1 &= P_{d_x}\!\left( z_1\odot \mathrm{Exp}(\eta_{x}A z_2) \right),\\ z_2 &= P_{d_y}\!\left( z_2\odot \mathrm{Exp}(\eta_{y}B z_1) \right). \end{cases} \end{align}\] Define \[C_x := \sum_{\ell=1}^{d_x} z_{1,\ell}\exp\!\left(\eta_{x}(Az_2)_\ell\right).\] Since \(z_1\in\Delta_{d_x}\), we have \(\mathop{\mathrm{supp}}(z_1) = S_x\neq\varnothing\) and \(C_x>0\). By Proposition 1, the zero pattern of the first block is preserved by the update. Hence, the coordinates outside \(S_x\) impose no additional condition. Therefore, \[\begin{align} z_1 &= P_{d_x}\!\left( z_1\odot \mathrm{Exp}(\eta_{x}A z_2) \right) \\ &\iff \forall i\in S_x,\quad z_{1,i} = z_{1,i} \frac{\exp(\eta_{x}(Az_2)_i)}{C_x} \\ &\iff \forall i\in S_x,\quad \exp(\eta_{x}(Az_2)_i)=C_x \\ &\iff \forall i,j\in S_x,\quad (Az_2)_i=(Az_2)_j \, . \end{align}\] For the last step, we used that \(\eta_{x}>0\). Combined with the observation that the argument for \(z_2\) is identical, we note that \(Z\) is a fixed point of \({ \left.\kern-\nulldelimiterspace \Phi_m \vphantom{|} \right|_{\smash{\widehat{\Delta}}} }\) if and only if \[z_3=z_1\in\Delta_{d_x},\qquad z_4=z_2\in\Delta_{d_y},\] and the payoff vectors \(Az_2\) and \(Bz_1\) are constant on the supports of \(z_1\) and \(z_2\), respectively. ◻

Proof. We prove the identity for the \(x\)-player; the proof for the \(y\)-player is analogous.

By iterating the \(m\)-optEW update, for every coordinate \(i \in \{1,\dots, d_x\}\), \[[x_{t+1}]_i = \frac{ [x_1]_i \exp\!\left( \eta_x \left[ A \sum_{s=1}^t\big((1+m)y_s-my_{s-1}\big) \right]_i \right) }{ \sum_{k=1}^{d_x} [x_1]_k \exp\!\left( \eta_x \left[ A \sum_{s=1}^t\big((1+m)y_s-my_{s-1}\big) \right]_k \right) } .\] The sum simplifies to \[\sum_{s=1}^t((1+m)y_{s}-my_{s-1}) = m(y_{t} - y_{0})+\overline{y}_{1:t} =: Y_t^{(m)}\, .\] Since \(Z_0\in\mathop{\mathrm{relint}}\smash{\widehat{\Delta}}\), all iterates remain strictly positive, so the coordinate-wise logarithm of the ratio is well defined. Taking the logarithm of the quotient gives \[\log\frac{[x_{t+1}]_i}{[x_t]_i} = \eta_x[A(Y_t^{(m)}-Y_{t-1}^{(m)})]_i + \eta_x\Big( \psi^x(Z_{1:t-1})-\psi^x(Z_{1:t})\Big).\] Finally, \[Y_t^{(m)}-Y_{t-1}^{(m)} = (1+m)y_t-my_{t-1}.\] This proves the first identity. The second follows from an analogous argument. ◻

Proof. We prove case (I), the case (II) is analogous. Set \[g:=(Ay_\infty)_{\hat{\imath}}-(Ay_\infty)_j>0, \qquad \beta:=\frac{[x_{\infty}]_{j}}{2}>0.\] For any \(Z=[x,y,x',y']\), define the continuous local payoff-difference map \[\Gamma_x^{(m)}(Z) := [A((1+m)y-my')]_{\hat{\imath}} - [A((1+m)y-my')]_j .\] Since \(Z_\infty=[x_\infty,y_\infty,x_\infty,y_\infty]\), we have \((1+m)y_\infty-my_\infty=y_\infty.\) Therefore \[\Gamma_x^{(m)}(Z_\infty) = (Ay_\infty)_{\hat{\imath}}-(Ay_\infty)_j = g.\] By continuity, there exists \(\delta_1>0\) such that \[\|Z-Z_\infty\|<\delta_1 \quad\Longrightarrow\quad \Gamma_x^{(m)}(Z)>\frac{g}{2}.\] Moreover, since \(j\in\mathop{\mathrm{supp}}(x_\infty)\), we have \([x_\infty]_j>0\). Thus there exists \(\delta_2>0\) such that \[\|Z-Z_\infty\|<\delta_2 \quad\Longrightarrow\quad x_j>\beta .\] Let \[\delta:=\min\{\delta_1,\delta_2\}, \qquad \epsilon:=\frac{\eta_x g}{2}, \qquad \tau:=\frac{\delta}{\beta}.\] We first show the exponential growth statement. Since the normalization factors in the update operation cancel, we obtain \[R_{t+1} = R_t \exp\!\left( \eta_x\Gamma_x^{(m)}(Z_t) \right).\] If \(\|Z_t-Z_\infty\|<\delta\), then \(\|Z_t-Z_\infty\|<\delta_1\), and hence \[\Gamma_x^{(m)}(Z_t)>\frac{g}{2}.\] Therefore \[R_{t+1} = R_t \exp\!\left( \eta_x\Gamma_x^{(m)}(Z_t) \right) \geqslant R_t\exp\!\left(\frac{\eta_x g}{2}\right) = e^\epsilon R_t .\] This proves the exponential growth statement in case (I).

It remains to prove the bounded-ratio statement. Let \(Z=[x,y,x',y']\in\smash{\widehat{\Delta}}\) satisfy \[\|Z-Z_\infty\|<\delta .\] Since \(\hat{\imath}\notin\mathop{\mathrm{supp}}(x_\infty)\), we have \([x_\infty]_{\hat{\imath}}=0\). Thus \[x_{\hat{\imath}} = |x_{\hat{\imath}}-[x_\infty]_{\hat{\imath}}| \leqslant \|Z-Z_\infty\| < \delta.\] On the other hand, since \(\delta\leqslant\delta_2\), we have \[x_j>\beta.\] Consequently, \[R(Z) = \frac{x_{\hat{\imath}}}{x_j} < \frac{\delta}{\beta} = \tau.\] This proves the bounded-ratio statement in case (I). ◻

Proof. Again, we only show the first identity; the second identity follows the analogous argument. Using the definition of the KL-divergence, Proposition 4 and that \(x_{t+1}\in\Delta_{d_x}\), we obtain \[\begin{align} \frac{1}{\eta_x}\mathop{\mathrm{KL}}(x_{t+1},x_{t}) &= \frac{1}{\eta_x} \left\langle x_{t+1}, \mathrm{Log}({x_{t+1}}\odiv{x_{t}}) \right\rangle \\ &= x_{t+1}^{\top}A(2y_{t}-y_{t-1}) + \psi^x(Z_{1:t-1})-\psi^x(Z_{1:t}) . \end{align}\] Rearranging yields \[\psi^x(Z_{1:t-1})-\psi^x(Z_{1:t}) = -x_{t+1}^{\top}A(2y_{t}-y_{t-1}) + \frac{1}{\eta_x}\mathop{\mathrm{KL}}(x_{t+1},x_{t}) .\] Analogously, \[\psi^y(Z_{1:t-1})-\psi^y(Z_{1:t}) = (2x_{t}-x_{t-1})^\top A y_{t+1} + \frac{1}{\eta_y}\mathop{\mathrm{KL}}(y_{t+1},y_{t}) .\] Adding the two identities gives the third claim. ◻

Proof. Let \(Z^\star = [x^\star, y^\star, x^\star, y^\star] \in \mathop{\mathrm{\mathscr{Z}}}^{\star}\). Applying the three-point identity (Proposition 3), Proposition 4 and Proposition 5 yields \[\begin{align} \frac{1}{\eta_{x}}\mathop{\mathrm{KL}}(x^\star, x_{t})- \frac{1}{\eta_{x}}\mathop{\mathrm{KL}}(x^\star, x_{t+1}) &= \frac{1}{\eta_x}\mathop{\mathrm{KL}}(x_{t+1},x_{t}) + \frac{1}{\eta_x} \left\langle x^\star-x_{t+1}, \mathrm{Log}({x_{t+1}}\odiv{x_{t}}) \right\rangle\\ &\overset{(1)}{=} \frac{1}{\eta_x}\mathop{\mathrm{KL}}(x_{t+1},x_{t}) + \left\langle x^\star-x_{t+1}, A(2y_{t}-y_{t-1}) \right\rangle \\ &= \psi^x(Z_{1:t-1})-\psi^x(Z_{1:t}) + \left\langle x^\star, A(2y_{t}-y_{t-1}) \right\rangle . \end{align}\] For \((1)\) we used that \(\left\langle x^\star - x_{t+1}, c \mathbf{1} \right\rangle = 0\) for any \(c\) (here \(c = \eta_{x}(\psi^x(Z_{1:t-1}) - \psi^x(Z_{1:t}))\)). Analogously, \[\frac{1}{\eta_{y}}\mathop{\mathrm{KL}}(y^\star, y_{t})- \frac{1}{\eta_{y}}\mathop{\mathrm{KL}}(y^\star, y_{t+1}) = \psi^y(Z_{1:t-1})-\psi^y(Z_{1:t}) - \left\langle y^\star, A^\top(2x_{t}-x_{t-1}) \right\rangle .\] Therefore, \(\frac{1}{\eta_{x}}\mathop{\mathrm{KL}}(x^\star, x_{t})- \frac{1}{\eta_{x}}\mathop{\mathrm{KL}}(x^\star, x_{t+1}) + \frac{1}{\eta_{y}}\mathop{\mathrm{KL}}(y^\star, y_{t})- \frac{1}{\eta_{y}}\mathop{\mathrm{KL}}(y^\star, y_{t+1})\) is equal to \[\begin{align} &\psi(Z_{1:t-1})-\psi(Z_{1:t}) + \left\langle x^\star, A(2y_{t}-y_{t-1}) \right\rangle - \left\langle y^\star, A^\top(2x_{t}-x_{t-1}) \right\rangle\\ &\quad= \psi(Z_{1:t-1})-\psi(Z_{1:t}) + \left\langle x^\star, A(y_{t}-y_{t-1}) \right\rangle - \left\langle y^\star, A^\top(x_{t}-x_{t-1}) \right\rangle \\ &\phantom{xxxxxxxxxxxxxxxx}+ \left\langle x^\star, Ay_{t} \right\rangle - \left\langle y^\star, A^\top x_{t} \right\rangle . \end{align}\] Summing over \(t=1,\ldots,T\) yields \[\begin{align} &\frac{1}{\eta_{x}}\left(\mathop{\mathrm{KL}}(x^\star, x_{1})- \mathop{\mathrm{KL}}(x^\star, x_{T+1}) \right)+ \frac{1}{\eta_{y}}\left(\mathop{\mathrm{KL}}(y^\star, y_{1})- \mathop{\mathrm{KL}}(y^\star, y_{T+1}) \right)\\ &\qquad \qquad= \psi(Z_{1:0})-\psi(Z_{1:T}) + \left\langle x^\star, A(y_{T}-y_{0}) \right\rangle - \left\langle y^\star, A^\top( x_{T}-x_{0}) \right\rangle \\ &\qquad \qquad\qquad \qquad\quad + \sum_{t=1}^T \left( \left\langle x^\star,Ay_{t} \right\rangle - \left\langle y^\star,A^\top x_{t} \right\rangle \right)\\ &\qquad \qquad\geqslant \psi(Z_{1:0})-\psi(Z_{1:T}) - \left( \left\langle x^\star, A y_{0} \right\rangle - \left\langle y^\star, A^\top x_{0} \right\rangle \right) \, , \end{align}\] where the last inequality is due to \(Z^\star \in \mathop{\mathrm{\mathscr{Z}}}^{\star}\). Rearranging \[\psi(Z_{1:T}) \geqslant \psi(Z_{1:0}) - \underbrace{ \left( \frac{1}{\eta_{x}} \mathop{\mathrm{KL}}(x^\star, x_{1}) + \frac{1}{\eta_{y}}\mathop{\mathrm{KL}}(y^\star, y_{1}) \right)}_{=: k_1} - \left( \left\langle x^\star, A y_{0} \right\rangle - \left\langle y^\star, A^\top x_{0} \right\rangle \right)\, .\] Since, by assumption, \(Z_0 \in \mathop{\mathrm{relint}}\smash{\widehat{\Delta}}\) the term \(k_1\) is bounded by a constant. Thus \((\psi(Z_{1:t}))_{t\in \mathop{\mathrm{\mathbb{N}}}}\) is bounded from below by a constant independent of \(T\).

Next, we prove the descent inequality, that is, we show that \((V(Z_{1:t}))_{t\in \mathop{\mathrm{\mathbb{N}}}}\) is non-increasing. By definition, \(V(Z_{1:t})-V(Z_{1:t+1})\) is equal to \[\begin{align} \psi(Z_{1:t-1})-&\psi(Z_{1:t}) - \mathop{\mathrm{gap}}(Z_t) + \mathop{\mathrm{gap}}(Z_{t+1})\\ + \frac{1}{2\eta_x}\mathop{\mathrm{KL}}(x_{t},x_{t-1}) + &\frac{1}{2\eta_y}\mathop{\mathrm{KL}}(y_{t},y_{t-1}) - \frac{1}{2\eta_x}\mathop{\mathrm{KL}}(x_{t+1},x_{t}) - \frac{1}{2\eta_y}\mathop{\mathrm{KL}}(y_{t+1},y_{t})\,. \end{align}\] Using the entropy identity from Proposition 5 and the definition of the gap function, yields \[\begin{align} V(Z_{1:t})-&V(Z_{1:t+1})\\ &= -(x_{t+1})^\top A(2y_{t}-y_{t-1}) + (2x_{t}-x_{t-1})^\top Ay_{t+1} - x_{t}^\top Ay_{t-1}\\ &\qquad+ x_{t-1}^\top Ay_{t} + (x_{t+1})^\top Ay_{t} - x_{t}^\top Ay_{t+1} \\ &\qquad+ \frac{1}{2\eta_x}\mathop{\mathrm{KL}}(x_{t},x_{t-1}) + \frac{1}{2\eta_y}\mathop{\mathrm{KL}}(y_{t},y_{t-1}) + \frac{1}{2\eta_x}\mathop{\mathrm{KL}}(x_{t+1},x_{t}) + \frac{1}{2\eta_y}\mathop{\mathrm{KL}}(y_{t+1},y_{t})\, , \intertext{After regrouping the bilinear terms, we obtain} &= -(x_{t+1}-x_{t})^\top A(y_{t}-y_{t-1}) + (x_{t}-x_{t-1})^\top A(y_{t+1}-y_{t}) \\ &+ \frac{1}{2\eta_x}\mathop{\mathrm{KL}}(x_{t},x_{t-1}) + \frac{1}{2\eta_y}\mathop{\mathrm{KL}}(y_{t},y_{t-1}) + \frac{1}{2\eta_x}\mathop{\mathrm{KL}}(x_{t+1},x_{t}) + \frac{1}{2\eta_y}\mathop{\mathrm{KL}}(y_{t+1},y_{t})\, . \end{align}\] By Fenchel’s inequality and the definition of the operator norm, \[\begin{align} -(x_{t+1}-x_{t})^\top A(y_{t}-y_{t-1}) &\geqslant - \frac{3\eta_y}{2} \|A^\top(x_{t+1}-x_{t})\|_\infty^2 - \frac{1}{6\eta_y} \|y_{t}-y_{t-1}\|_1^2 \\ &\geqslant - \frac{3\eta_y\|A^\top\|_{{1 \rightarrow \infty}}^2}{2} \|x_{t+1}-x_{t}\|_1^2 - \frac{1}{6\eta_y} \|y_{t}-y_{t-1}\|_1^2 \, , \end{align}\] and analogously, \[\begin{align} (x_{t}-x_{t-1})^\top A(y_{t+1}-y_{t}) &\geqslant - \frac{3\eta_x\|A\|_{{1 \rightarrow \infty}}^2}{2} \|y_{t+1}-y_{t}\|_1^2 - \frac{1}{6\eta_x} \|x_{t}-x_{t-1}\|_1^2 . \end{align}\] Therefore, by Pinsker’s inequality (cf. Proposition 2), \[\begin{align} V(Z_{1:t})-V(Z_{1:t+1}) \geqslant& \left( \frac{1}{4\eta_x} - \frac{3\eta_y\|A^\top\|_{{1 \rightarrow \infty}}^2}{2} \right) \|x_{t+1}-x_{t}\|_1^2 \\ &+ \left( \frac{1}{4\eta_y} - \frac{3\eta_x\|A\|_{{1 \rightarrow \infty}}^2}{2} \right) \|y_{t+1}-y_{t}\|_1^2 \\ &+ \frac{1}{12\eta_x}\|x_{t}-x_{t-1}\|_1^2 + \frac{1}{12\eta_y}\|y_{t}-y_{t-1}\|_1^2 . \end{align}\] By the step size condition, (note that \(\|A^\top\|_{{1 \rightarrow \infty}} = \|A\|_{{1 \rightarrow \infty}}\)) \[\frac{1}{4\eta_x} - \frac{3\eta_y\|A^\top\|_{{1 \rightarrow \infty}}^2}{2} \geqslant 0, \qquad \frac{1}{4\eta_y} - \frac{3\eta_x\|A\|_{{1 \rightarrow \infty}}^2}{2} \geqslant 0.\] Hence \(V(Z_{1:t+1})\leqslant V(Z_{1:t})\) for all \(t\).

To show part 3, we note that \(V(Z_{1:1})\) is a constant. Further, since \(\psi(Z_{1:t})\) is bounded from below and the KL-divergence is non-negative, while \(\mathop{\mathrm{gap}}(Z_t)\) is bounded on the compact set \(\smash{\widehat{\Delta}}\), the sequence \((V(Z_{1:t}))_{t\in \mathop{\mathrm{\mathbb{N}}}}\) is bounded from below. Summing over \(t\) gives \[\sum_{t=1}^\infty \left( \frac{1}{12\eta_{x}} \|x_{t}-x_{t-1}\|_1^2 + \frac{1}{12\eta_{y}} \|y_{t}-y_{t-1}\|_1^2 \right) <\infty \, .\] Consequently, \[\|x_{t+1}-x_{t}\|_1\to 0, \qquad \|y_{t+1}-y_{t}\|_1\to 0.\] Because \[Z_t=(x_{t},y_{t},x_{t-1},y_{t-1}),\] we conclude that \[\|Z_t-Z_{t-1}\|\to 0 \, .\] This completes the proof. ◻

Proof of Theorem 2. By Proposition 1, the orbit remains in \(\smash{\widehat{\Delta}}\). Since \(\smash{\widehat{\Delta}}\) is compact, the sequence \((Z_t)_{t\in\mathop{\mathrm{\mathbb{N}}}}\) has at least one accumulation point.

We first show that every accumulation point of \((Z_t)_{t\in\mathop{\mathrm{\mathbb{N}}}}\) is a fixed point of \(\Phi_1\). Let \(\overline{Z}\) be an accumulation point. Then there exists a subsequence \((Z_{t_k})_{k\in\mathop{\mathrm{\mathbb{N}}}}\) such that \(Z_{t_k}\to \overline{Z} .\) By Lemma 2, Part 3, we have \(\left\| Z_{t+1}-Z_t \right \|\to 0 .\) Hence \[\left\| Z_{t_k+1}-Z_{t_k} \right \|\to 0,\] and therefore \[Z_{t_k+1}\to \overline{Z} .\] By continuity of \(\Phi_1\) on \(\smash{\widehat{\Delta}}\), \[\overline{Z} = \lim Z_{t_k+1} = \lim \Phi_1(Z_{t_k}) = \Phi_1(\overline{Z})\, .\] Thus, every accumulation point of the orbit belongs to the fixed-point set \(\widetilde{\mathop{\mathrm{\mathscr{Z}}}}({ \left.\kern-\nulldelimiterspace \Phi_1 \vphantom{|} \right|_{\smash{\widehat{\Delta}}} })\).

It remains to show that the distance to \(\widetilde{\mathop{\mathrm{\mathscr{Z}}}}({ \left.\kern-\nulldelimiterspace \Phi_1 \vphantom{|} \right|_{\smash{\widehat{\Delta}}} })\) converges to zero. Suppose, for contradiction, that \(d(Z_t,\widetilde{\mathop{\mathrm{\mathscr{Z}}}}({ \left.\kern-\nulldelimiterspace \Phi_1 \vphantom{|} \right|_{\smash{\widehat{\Delta}}} }))\) does not converge to \(0\). Then there exist \(\epsilon>0\) and a subsequence \((Z_{t_k})_{k\in\mathop{\mathrm{\mathbb{N}}}}\) such that \[d(Z_{t_k},\widetilde{\mathop{\mathrm{\mathscr{Z}}}}({ \left.\kern-\nulldelimiterspace \Phi_1 \vphantom{|} \right|_{\smash{\widehat{\Delta}}} }))\geqslant\epsilon \qquad \text{for all } k\in\mathop{\mathrm{\mathbb{N}}}.\] By compactness of \(\smash{\widehat{\Delta}}\), after passing to a further subsequence if necessary, we may assume that \[Z_{t_k}\to \overline{Z}\] for some \(\overline{Z}\in\smash{\widehat{\Delta}}\). By the first part of the proof, \(\overline{Z}\) is a fixed point, i.e. \(\overline{Z}\in\widetilde{\mathop{\mathrm{\mathscr{Z}}}}({ \left.\kern-\nulldelimiterspace \Phi_1 \vphantom{|} \right|_{\smash{\widehat{\Delta}}} })\). Since the distance function \(Z\mapsto d(Z,\widetilde{\mathop{\mathrm{\mathscr{Z}}}}({ \left.\kern-\nulldelimiterspace \Phi_1 \vphantom{|} \right|_{\smash{\widehat{\Delta}}} }))\) is continuous, we obtain \[d(Z_{t_k},\widetilde{\mathop{\mathrm{\mathscr{Z}}}}({ \left.\kern-\nulldelimiterspace \Phi_1 \vphantom{|} \right|_{\smash{\widehat{\Delta}}} }))\to d(\overline{Z},\widetilde{\mathop{\mathrm{\mathscr{Z}}}}({ \left.\kern-\nulldelimiterspace \Phi_1 \vphantom{|} \right|_{\smash{\widehat{\Delta}}} }))=0,\] contradicting \(d(Z_{t_k},\widetilde{\mathop{\mathrm{\mathscr{Z}}}}({ \left.\kern-\nulldelimiterspace \Phi_1 \vphantom{|} \right|_{\smash{\widehat{\Delta}}} }))\geqslant\epsilon\) for all \(k\). Therefore \[d(Z_t,\widetilde{\mathop{\mathrm{\mathscr{Z}}}}({ \left.\kern-\nulldelimiterspace \Phi_1 \vphantom{|} \right|_{\smash{\widehat{\Delta}}} }))\to 0.\] Since this holds for any \(Z_0 \in \mathop{\mathrm{relint}}\smash{\widehat{\Delta}}\), \(\widetilde{\mathop{\mathrm{\mathscr{Z}}}}({ \left.\kern-\nulldelimiterspace \Phi_1 \vphantom{|} \right|_{\smash{\widehat{\Delta}}} })\) is globally attracting with respect to \(\mathop{\mathrm{relint}}\smash{\widehat{\Delta}}\). ◻

Proof. \(\Omega_{\omega}\) is a decreasing intersection of nonempty compact sets in a compact space, hence nonempty and compact.

Suppose, for contradiction, that \(\Omega_{\omega}\) is disconnected, say \(\Omega_{\omega}=K_1\sqcup K_2\) (disjoint union) with \(K_1,K_2\) nonempty compact and \(2\delta:=d(K_1,K_2)>0\). Define the \(1\)-Lipschitz function \(g(t):=d(U_t,K_1)\), so that \[|g(t+1)-g(t)|\leqslant\|U_{t+1}-U_t\|\xrightarrow[t\to\infty]{}0 .\] Because \(K_1\subseteq\Omega_{\omega}\), we have \(g(t)<\delta/2\) for infinitely many \(t\). Since \(K_2\subseteq\Omega_{\omega}\) and \(d(K_2,K_1)\geqslant 2\delta\), we have \(g(t)>3\delta/2\) for infinitely many \(t\). A real sequence with vanishing increments that lies below \(\delta/2\) and above \(3\delta/2\) infinitely often must take values in the band \([\delta/2,3\delta/2]\) at arbitrarily large times; in particular, there are times \(s_k\to\infty\) with \(|g(s_k)-\delta|\to 0\). By compactness, pass to a convergent subsequence \(U_{s_k}\to U^\star\); then \(U^\star\in\Omega_{\omega}\) and \(d(U^\star,K_1)=\lim_k g(s_k)=\delta\). But \(U^\star\in\Omega_{\omega}=K_1\sqcup K_2\) forces \(d(U^\star,K_1)=0\) (if \(U^\star\in K_1\)) or \(d(U^\star,K_1)\geqslant 2\delta\) (if \(U^\star\in K_2\)), contradicting \(d(U^\star,K_1)=\delta\in(0,2\delta)\). Hence \(\Omega_{\omega}\) is connected. ◻

Proof. This is the standard fact that every countable metric space is totally disconnected; see, e.g., [31]. ◻

Proof. By Lemma 2, Part 3, \(\|Z_{t+1}-Z_t\|\to 0\), and \(\smash{\widehat{\Delta}}\) is compact, so by Proposition 6 the \(\omega\)-limit set \(\Omega_{\omega}\) is nonempty, compact, and connected.

Note that every point of \(\Omega_{\omega}\) is a fixed point of \(\Phi_1\). If \(Z_{t_k}\to Z'\) then, using continuity of \(\Phi_1\) and \(\|Z_{t_k+1}-Z_{t_k}\|\to 0\), \[Z'=\lim_k Z_{t_k+1}=\lim_k \Phi_1(Z_{t_k})=\Phi_1(Z') .\] Hence \(\Omega_{\omega}\subseteq\widetilde{\mathop{\mathrm{\mathscr{Z}}}}({ \left.\kern-\nulldelimiterspace \Phi_1 \vphantom{|} \right|_{\smash{\widehat{\Delta}}} })\). Since \(\widetilde{\mathop{\mathrm{\mathscr{Z}}}}({ \left.\kern-\nulldelimiterspace \Phi_1 \vphantom{|} \right|_{\smash{\widehat{\Delta}}} })\) is countable, \(\Omega_{\omega}\) is a connected subset of a countable set, so by Proposition 7, \(\Omega_{\omega}\) is a single point, \(\Omega_{\omega}=\{Z_\infty\}\) with \(Z_\infty\in\widetilde{\mathop{\mathrm{\mathscr{Z}}}}({ \left.\kern-\nulldelimiterspace \Phi_1 \vphantom{|} \right|_{\smash{\widehat{\Delta}}} })\).

Finally, we observe that a sequence in a compact space whose \(\omega\)-limit set is the single point \(Z_\infty\) converges to \(Z_\infty\). If not, there would exist \(\varepsilon>0\) and a subsequence with \(\|Z_{t_k}-Z_\infty\|\geqslant\varepsilon\), which by compactness has a further subsequence converging to some \(Z''\in\Omega_{\omega}\) with \(Z''\neq Z_\infty\), contradicting \(\Omega_{\omega}=\{Z_\infty\}\). Thus \(Z_t\to Z_\infty\). The stated form and support conditions are Theorem 1. ◻

Proof of Theorem 3. By Corollary 7, \(Z_t\to Z_\infty\) with \(Z_\infty=(x_\infty,y_\infty,x_\infty,y_\infty)\in\widetilde{\mathop{\mathrm{\mathscr{Z}}}}({ \left.\kern-\nulldelimiterspace \Phi_1 \vphantom{|} \right|_{\smash{\widehat{\Delta}}} })\), and \(Ay_\infty\) is constant on \(S_x=\mathop{\mathrm{supp}}(x_\infty)\) while \(A^\top x_\infty\) is constant on \(S_y=\mathop{\mathrm{supp}}(y_\infty)\). It remains to show \((x_\infty,y_\infty)\) is a Nash equilibrium.

9.2.2.1 Reduction to a profitable deviation.

Let \(v_x:=(Ay_\infty)_i\) for any \(i\in S_x\); then \(x_\infty^\top Ay_\infty=v_x\). Since \(\max_{x\in\Delta_{d_x}}x^\top Ay_\infty=\max_i (Ay_\infty)_i\), the strategy \(x_\infty\) is a best response to \(y_\infty\) iff \((Ay_\infty)_i\leqslant v_x\) for all \(i\). Symmetrically, with \(v_y:=(A^\top x_\infty)_l\) for \(l\in S_y\) (note \(B=-A^\top\), so the \(y\)-player maximizes \(y^\top Bx=-y^\top A^\top x\), equivalently minimizes \(y^\top A^\top x\)), \(y_\infty\) is a best response to \(x_\infty\) iff \((A^\top x_\infty)_k\geqslant v_y\) for all \(k\). Consequently, if \((x_\infty,y_\infty)\) is not a Nash equilibrium, then at least one of the following holds:

• there exist \(\hat{\imath}\) and \(j\in S_x\) with \((Ay_\infty)_{\hat{\imath}}>(Ay_\infty)_j=v_x\); since \((Ay_\infty)_i=v_x\) for \(i\in S_x\), necessarily \(\hat{\imath}\notin S_x\);

• there exist \(\hat{\jmath}\) and \(l\in S_y\) with \((A^\top x_\infty)_{\hat{\jmath}}<(A^\top x_\infty)_l=v_y\); necessarily \(\hat{\jmath}\notin S_y\).

We only show case (I) since case (II) follows exactly the same argument.

Define \(R_t:=x_{t,\hat{\imath}}/x_{t,j}\). This ratio is well defined and strictly positive for every \(t\) by Proposition 1 and since \(Z_0 \in \mathop{\mathrm{relint}}\smash{\widehat{\Delta}}\) by assumption. Since \(\hat{\imath}\notin S_x\), \(j\in S_x\), and \((Ay_\infty)_{\hat{\imath}}>(Ay_\infty)_j\), Lemma 1 provides \(\delta,\epsilon,\tau>0\) such that, for all \(t\in\mathop{\mathrm{\mathbb{N}}}\), \[\label{eq:F4growth} \|Z_t-Z_\infty\|<\delta \quad\Longrightarrow\quad R_{t+1}\geqslant e^{\epsilon}R_t,\tag{7}\] and, for all \(Z=[x,y,x',y']\), \[\label{eq:F4bound} \|Z-Z_\infty\|<\delta \quad\Longrightarrow\quad \frac{x_{\hat{\imath}}}{x_{j}}<\tau .\tag{8}\] By Corollary 7, \(Z_t\to Z_\infty\). Thus, there exists \(T_\star\in\mathop{\mathrm{\mathbb{N}}}\) with \[\label{eq:tail95in95ball} \|Z_t-Z_\infty\|<\delta\qquad\text{for all }t\geqslant T_\star .\tag{9}\] Convergence of the whole sequence ensures that every iterate with \(t\geqslant T_\star\) is in \(B_\delta(Z_\infty)\). Hence 7 applies at every \(t\geqslant T_\star\), and by induction \[\label{eq:Rt95lower} R_t\;\geqslant\;e^{\epsilon(t-T_\star)}\,R_{T_\star} \qquad(t\geqslant T_\star).\tag{10}\] Since \(R_{T_\star}>0\) and \(\epsilon>0\), the right-hand side of 10 diverges, so \(R_t\to+\infty\). On the other hand, 9 and 8 give the uniform bound \[\label{eq:Rt95upper} R_t=\frac{x_{t,\hat{\imath}}}{x_{t,j}}<\tau \qquad(t\geqslant T_\star).\tag{11}\] Inequalities 10 and 11 are incompatible. This excludes Case (I).

We conclude that neither deviation can occur, so \(x_\infty\) is a best response to \(y_\infty\) and \(y_\infty\) is a best response to \(x_\infty\). With the support-equalization of Theorem 1, \((x_\infty,y_\infty)\) is a Nash equilibrium, i.e. \(Z_\infty\in\mathop{\mathrm{\mathscr{Z}}}^{\star}\). By Corollary 7, \(Z_t\to Z_\infty\in\mathop{\mathrm{\mathscr{Z}}}^{\star}\). ◻

Proof. Using the block structure of \(\Phi_m = [(\Phi_m)_1,(\Phi_m)_2,(\Phi_m)_3,(\Phi_m)_4]\), where \((\Phi_m)_k:\mathop{\mathrm{\mathbb{R}}}^{2(d_x + d_y)} \rightarrow \mathop{\mathrm{\mathbb{R}}}^{d_x}\) for \(k = 1,3\) and \((\Phi_m)_k:\mathop{\mathrm{\mathbb{R}}}^{2(d_x + d_y)} \rightarrow \mathop{\mathrm{\mathbb{R}}}^{d_y}\) for \(k= 2,4\), we compute the Jacobian block-wise \(J_{1,1} = \frac{\partial {(\Phi_m})_1}{\partial z_1}\), \(J_{1,2} = \frac{\partial {(\Phi_m})_1}{\partial z_2}\) and \(J_{1,4} = \frac{\partial {(\Phi_m})_1}{\partial z_4}\). The blocks \(J_{2,1},J_{2,2}\) and \(J_{2,3}\) follow analogously. Since for any \(i, j\) and for \(u \in \mathbb{R}^d \setminus \{ x \in \mathop{\mathrm{\mathbb{R}}}^d : \mathbf{1}^\top x = 0\}\) , we have \[\frac{\partial (P_d)_i}{\partial u_j} = \left\{ \begin{align} & - \frac{u_i}{\left\langle \mathbf{1},u \right\rangle^2} & \text{if } i \neq j \\ & \frac{1}{ \left\langle \mathbf{1},u \right\rangle} - \frac{u_i}{ \left\langle \mathbf{1},u \right\rangle^2} & \text{if } i = j \end{align} \right. \,,\] the Jacobian for \(P_d\) is \[\begin{align} \label{id:Jacob} J_{P_d}(u) = \frac{1}{ \left\langle \mathbf{1},u \right\rangle} \big( I_d - P_d(u) \mathbf{1}^\top \big) \end{align}\tag{12}\] Now let \(({\Phi^{\mathrm{u}}_m})_1(z_1, z_2, z_3, z_4) = z_1 \odot \mathrm{Exp}( \eta_{x}A((1+m)z_2 - mz_4))\) denote the first component of the unnormalized exponential weight updates. Then, \[J_{1, 1} = \frac{\partial {(\Phi^{\mathrm{u}}_m})_1}{\partial z_1} = J_{P_{d_x}}( ({\Phi^{\mathrm{u}}_m})_1(z_1, z_2, z_3, z_4))\mathop{\mathrm{diag}}\left( \mathrm{Exp}( \eta_{x}A((1+m)z_2 - mz_4)) \right)\, .\] As \(z_2 = z_4\), we have \(\mathrm{Exp}( \eta_{x}A((1+m)z_2 - mz_4)) = \mathrm{Exp}(\eta_{x}A z_2)\). Thus, by 12 \[\begin{align} J_{1, 1} & = \frac{1}{\left\langle z_1 , \mathrm{Exp}(\eta_{x}A z_2) \right\rangle} \big( \mathop{\mathrm{Id}}_{d_x} - P_{d_x}\Big(z_1 \odot \mathrm{Exp}(\eta_{x}A z_2)\Big) \mathbf{1}^\top \big) \mathop{\mathrm{diag}}(\mathrm{Exp}(\eta_{x}A z_2)) \\ &\overset{(\star)}{ = } \frac{1}{\left\langle z_1, \mathrm{Exp}(\eta_{x}A z_2) \right\rangle} \big( \mathop{\mathrm{Id}}_{d_x} - z_1 \mathbf{1}^\top \big) \mathop{\mathrm{diag}}(\mathrm{Exp}(\eta_{x}A z_2)) \\ & = \frac{1}{\left\langle z_1 ,\mathrm{Exp}(\eta_{x}A z_2) \right\rangle} \big( \mathop{\mathrm{diag}}(\mathrm{Exp}(\eta_{x}A z_2)) - z_1 \mathrm{Exp}(\eta_{x}A z_2)^\top \big) \\ & = \mathop{\mathrm{diag}}(\mathrm{Exp}(\eta_{x}(A z_2 - v^x\mathbf{1}))) - z_1 \mathrm{Exp}(\eta_{x}(A z_2 - v^x\mathbf{1}))^\top \,. \end{align}\] We used the fixed-point equation in the final line \[\frac{1}{\left\langle z_1 , \mathrm{Exp}(\eta_{x}A z_2) \right\rangle} = e^{-\eta_{x}v^x} \,,\] and consequently also \(P_{d_x}(z_1 \odot \mathrm{Exp}(\eta_{x}A z_2)) = z_1\) in \((\star)\). The computation of the block \(J_{1,2}\) follows similar arguments. We have, \[\begin{align} J_{1,2} & = J_{P_{d_x}}(z_1 \odot \mathrm{Exp}(\eta_{x}A z_2)) \mathop{\mathrm{diag}}(z_1) \mathop{\mathrm{diag}}(\mathrm{Exp}( \eta_{x}A z_2)) (1+m)\eta_{x}A \\ & = \frac{1+m}{\left\langle z_1 , \mathrm{Exp}(\eta_{x}A z_2) \right\rangle} \big( \mathop{\mathrm{Id}}_{d_x} - P_{d_x}(z_1 \odot \mathrm{Exp}(\eta_{x}A z_2)) \mathbf{1}^\top \big) \mathop{\mathrm{diag}}(z_1) \mathop{\mathrm{diag}}(\mathrm{Exp}( \eta_{x}A z_2 )) \eta_{x}A \\ & = \frac{1+m}{\left\langle z_1 , \mathrm{Exp}(\eta_{x}A z_2) \right\rangle} \big( \mathop{\mathrm{Id}}_{d_x} - z_1 \mathbf{1}^\top \big) \mathop{\mathrm{diag}}\big(z_1 \odot \mathrm{Exp}(\eta_{x}A z_2 )\big) \eta_{x}A \\ & = (1+m)\big( \mathop{\mathrm{Id}}_{d_x} - z_1 \mathbf{1}^\top \big) \mathop{\mathrm{diag}}\big(P_{d_x}(z_1 \odot \mathrm{Exp}(\eta_{x}A z_2 ))\big) \eta_{x}A \\ & = (1+m) \big( \mathop{\mathrm{Id}}_{d_x} - z_1 \mathbf{1}^\top \big) \mathop{\mathrm{diag}}\big(z_1) \eta_{x}A \\ & = (1+m) \big( \mathop{\mathrm{diag}}\big(z_1) - z_1 z_1^\top \big) \eta_{x}A \,. \end{align}\] The proof is completed by noting that \(J_{1,4}\) follows by the same arguments. ◻

Proof of Lemma 4. The complex \(\lambda \in \mathbb{C}^\star\) is an eigenvalue of \(J_{\Phi_m}(Z)|_L\), if and only if there is some \(\mathbf{h} = [h_1, h_2, h_3, h_4] \in L_\mathbb{C}\) such that \(J_{\Phi_m}(Z) \mathbf{h} = \lambda \mathbf{h}\).

Assuming \(\lambda \neq 0\), the last two block rows imply \(h_3 = \lambda^{-1} h_1\) and \(h_4 = \lambda^{-1} h_2\). Substituting these into the first two rows yields the reduced system: \[\begin{align} (\mathop{\mathrm{diag}}(\mathbf{w}_1) - z_1 \mathbf{w}_1^\top) h_1 + \eta_{x}H(z_1) A (1 + m - m\lambda^{-1}) h_2 &= \lambda h_1 \tag{14} \\ (\mathop{\mathrm{diag}}(\mathbf{w}_2) - z_2 \mathbf{w}_2^\top) h_2 + \eta_{y}H(z_2) B (1 + m - m\lambda^{-1}) h_1 &= \lambda h_2 \tag{15} \end{align}\] We then proceed by a case disjunction:

10.2.0.2 Case 1: \(h_{1, k} \neq 0\) for some \(k \notin S_x\)

If there exists \(k \notin S_x\) such that \(h_{1, k} \neq 0\), then for such \(k\), we have \(z_{1,k} = 0\), so the \(k\)-th row of \(H(z_1)\) is zero, and the \(k\)-th row of \(z_1 \mathbf{w}_1^\top\) is zero. Equation 14 for the \(k\)-th component becomes: \[\mathbf{w}_{1, k} \, h_{1,k} + 0 = \lambda\, h_{1,k} \,,\] therefore \(\lambda = \mathbf{w}_{1, k}\), since \(h_{1, k} \neq 0\).

Conversely, for any \(k \notin S_x\), consider the left-eigenvector \(\mathbf{v} = (e_k, 0 , 0, 0)\) then \[\mathbf{v}^\top J_{\Phi_m}(Z) = (e_k^\top, 0 , 0, 0)J_{\Phi_m}(Z) = (\mathbf{w}_{1,k}e_k^\top, 0 , 0 , 0) = \mathbf{w}_{1,k} \mathbf{v}^\top \,.\] So \(\mathbf{w}_{1, k}\) is a non-zero eigenvalue of \(J_{\Phi_m}(Z)\). It remains to show that the corresponding right eigenvector lies in \(L_{\mathop{\mathrm{\mathbb{C}}}}\). If \([h_1, h_2, \lambda^{-1} h_1, \lambda^{-1}h_2]\) is a (right-)eigenvector with eigenvalue \(\mathbf{w}_{1, k} > 0\), then necessarily, due to Equation 14 , \[\lambda \mathbf{1}^\top h_1 = \mathbf{1}^\top (\mathop{\mathrm{diag}}(\mathbf{w}_1) - z_1 \mathbf{w}_1^\top) h_1 + \eta_{x}\mathbf{1}^\top H(z_1) A (1 + m - m\lambda^{-1}) h_2 \, .\] Now \[\mathbf{1}^\top (\mathop{\mathrm{diag}}(\mathbf{w}_1) - z_1 \mathbf{w}_1^\top) = \mathbf{1}^\top \mathop{\mathrm{diag}}(\mathbf{w}_1) - \mathbf{1}^\top z_1 \mathbf{w}_1^\top =\mathbf{w}_1^\top - \mathbf{w}_1^\top = 0 \, ,\] and similarly \[\mathbf{1}^\top H(z_1) = 0 \,.\] So \(\mathbf{1}^\top h_1 = 0\). Symmetrically, we also have \(\mathbf{1}^\top h_2 = 0\), so \([h_1, h_2, \lambda^{-1}h_1, \lambda^{-1}h_2] \in L_\mathbb{C}\).

10.2.0.3 Case 2: \(h_{2, \ell} \neq 0\) for some \(\ell \notin S_y\)

Similarly, if there exists \(\ell \notin S_y\) such that \(h_{2, \ell} \neq 0\), then necessarily \(\lambda = \mathbf{w}_{2, \ell}\), and conversely these are indeed eigenvalues of \(J_{\Phi_m}(Z)|_{L}\).

10.2.0.4 Case 3

If neither Case 1 nor Case 2 holds, then \(h_1\) is supported on \(S_x\) and \(h_2\) on \(S_y\). On the support \(S_x\) (resp. \(S_y\)), all coordinates of the vector \(\mathbf{w}_1\) (resp. \(\mathbf{w}_2\)) are \(1\). The term \((\mathop{\mathrm{diag}}(\mathbf{w}_1) - z_1 \mathbf{w}_1^\top) h_1\) becomes \((I - z_1 \mathbf{1}^\top) h_1\). Since \(\mathbf{1}^\top h_1 = 0\), this simplifies to \(h_1\). Equations 14 and 15 become: \[\begin{align} h_1 + (1 + m - m\lambda^{-1}) \eta_{x}H(z_1) A h_2 &= \lambda h_1 \quad \text{i.e.}\quad (\lambda - 1) h_1 = (1 + m - m\lambda^{-1}) \eta_{x}H(z_1) A h_2 \\ h_2 + (1 + m - m\lambda^{-1}) \eta_{y}H(z_2) B h_1 &= \lambda h_2 \quad \text{i.e.} \quad (\lambda - 1) h_2 = (1 + m - m\lambda^{-1}) \eta_{y}H(z_2) B h_1 \,. \end{align}\] To rewrite these in the form 13 , divide by \(1+m-m\lambda^{-1} = ((1+m)\lambda-m)/\lambda\), which is nonzero if \(\lambda \neq m/(m+1)\). If instead \(\lambda = m/(m+1)\), then the equations simplify to \((\lambda-1)h_1 = 0\), \((\lambda-1)h_2 = 0\). Since \(\lambda=m/(m+1)\neq 1\), this forces \(h_1=h_2=0\). Hence \(m/(m+1)\) never arises as an eigenvalue through Case 3, and can occur only via Cases 1 or 2, i.e.as an element of \(W\).

Conversely, we check that any pair \((\lambda, \mathbf{h}) \in \mathbb{C}^\star \times L_\mathbb{C}\setminus\{0\}\) satisfying these conditions is indeed an eigenvalue-eigenvector pair for \(J_{\Phi_m}(Z)|_L\). Assume first that \(\lambda \neq \frac{m}{m+1}\) and that 13 holds. Define \[\mathbf{h} := \bigl[h_1,h_2,\lambda^{-1}h_1,\lambda^{-1}h_2\bigr] \in L_{\mathbb{C}}\, .\] Since \([h_1,h_2]\neq [0,0]\), we have \(\mathbf{h}\neq 0\). Moreover, \(h_1,h_2\) are supported on \(S_x,S_y\), respectively, and \(\mathbf{w}_1=\mathbf{1}\) on \(S_x\), \(\mathbf{w}_2=\mathbf{1}\) on \(S_y\). Hence, using \(\mathbf{1}^\top h_1=\mathbf{1}^\top h_2=0\), \[\bigl(\mathop{\mathrm{diag}}(\mathbf{w}_1)-z_1\mathbf{w}_1^\top\bigr)h_1=h_1\, , \qquad \bigl(\mathop{\mathrm{diag}}(\mathbf{w}_2)-z_2\mathbf{w}_2^\top\bigr)h_2=h_2 \,.\] Also, \[1+m-m\lambda^{-1} = \frac{(1+m)\lambda-m}{\lambda}\,.\] Therefore, by 13 , \[\bigl(1+m-m\lambda^{-1}\bigr)\eta_x H(z_1)Ah_2 = (\lambda-1)h_1\,,\] and similarly \[\bigl(1+m-m\lambda^{-1}\bigr)\eta_y H(z_2)Bh_1 = (\lambda-1)h_2\,.\] Substituting these identities into the first two block rows of \(J_{\Phi_m}(Z)\mathbf{h}\), and using \(h_3=\lambda^{-1}h_1\), \(h_4=\lambda^{-1}h_2\) for the last two block rows, gives \[J_{\Phi_m}(Z)\mathbf{h}=\lambda \mathbf{h}\, .\] Thus \(\lambda\in \mathrm{eig}\left( J_{\Phi_m}(Z)|_L\right)\).

It remains to consider the case \(\lambda\in W\). This is exactly the off-support case treated above: the corresponding off-support coordinate gives a nonzero eigenvector of \(J_{\Phi_m}(Z)\), and the summation argument shows that this eigenvector lies in \(L_{\mathop{\mathrm{\mathbb{C}}}}\). Hence \[\lambda\in \mathrm{eig}\left( J_{\Phi_m}(Z)|_L\right)\,.\] ◻

Proof of Lemma 5. We prove the two implications separately and isolate the case \(\lambda=1\).

\((i) \Rightarrow (ii)\), first assume \(\lambda\neq 1\). Let \([h_1, h_2, \lambda^{-1}h_1, \lambda^{-1} h_2] \in L_\mathbb{C}\setminus\{0\}\) satisfying the condition of the statement. Then since \(\mathbf{1}^\top h_1 = 0\), and since \(h_{1, i} = 0\) for all \(i \notin S_x\), we have \[\sum_{i \in S_x} h_{1, i} = \sum_{i =1}^{d_x} h_{1, i} = 0 \,.\] so \(\Pi_{S_x} h_1 \in L_{x, \mathbb{C}}\). Moreover, the eigenvalue equations imply that \(h_1 \neq 0\), so \(\Pi_{S_x} h_1 \neq 0\). Finally, since \(h_1\) is supported on \(S_x\), we have \(\Pi_{S_x}^{\top} \Pi_{S_x} h_1 = h_1\) so \[\begin{gather} \eta_{x}\eta_{y}\Pi_{S_x} (H(z_1) A H(z_2) B) \Pi_{S_x}^{\top} \Pi_{S_x} h_1 = \eta_{x}\eta_{y}\Pi_{S_x} (H(z_1) A H(z_2) B) h_1 \\ = \eta_{x}\Pi_{S_x} H(z_1) A \bigg( \frac{\lambda(\lambda - 1)}{(1+m)\lambda - m} h_2 \bigg) = \bigg(\frac{\lambda(\lambda - 1)}{(1+m)\lambda - m} \bigg)^2 \Pi_{S_x}h_1 \,. \end{gather}\] Therefore \(\Pi_{S_x} h_1\in L_{x,\mathop{\mathrm{\mathbb{C}}}}\) is an eigenvector of \(\eta_{x}\eta_{y}{ \left.\kern-\nulldelimiterspace M_x(Z) \vphantom{|} \right|_{L_x} }\) with eigenvalue \(Q_m(\lambda)\).

\((i) \Rightarrow (ii)\), now consider \(\lambda=1\). Then \(\mu=0\) and \(Q_m(1)=0\). If \(h_1\neq0\), then \(\Pi_{S_x}h_1\in L_{x,\mathop{\mathrm{\mathbb{C}}}}\setminus\{0\}\) and \[H(z_2)Bh_1=0,\] so \[\Pi_{S_x}H(z_1)AH(z_2)B\Pi_{S_x}^{\top}\Pi_{S_x}h_1=0.\] Thus \(0=Q_m(1)\) is an eigenvalue of \(\Pi_{S_x}H(z_1)AH(z_2)B\Pi_{S_x}^{\top}|_{L_x}\).

Suppose instead that \(h_1=0\). Since \([h_1,h_2]\neq [0,0]\), we have \(h_2\neq0\). Moreover \(h_2\) is supported on \(S_y\) and belongs to \(L_{y}\), and the first equation gives \[H(z_1)Ah_2=0.\] Define \[T:=\Pi_{S_x}H(z_1)A\Pi_{S_y}^{\top}:L_{y,\mathop{\mathrm{\mathbb{C}}}}\to L_{x,\mathop{\mathrm{\mathbb{C}}}}.\] Then \(\ker T\neq\{0\}\), so \[\mathop{\mathrm{rank}}T<\dim L_{y}=n_y-1.\] Now write \[\Pi_{S_x}H(z_1)AH(z_2)B\Pi_{S_x}^{\top}\big|_{L_{x}} = T\circ R\] where \[R:=\Pi_{S_y}H(z_2)B\Pi_{S_x}^{\top}:L_{x,\mathop{\mathrm{\mathbb{C}}}}\to L_{y,\mathop{\mathrm{\mathbb{C}}}}.\] Hence \[\mathop{\mathrm{rank}}(T\circ R)\leqslant\mathop{\mathrm{rank}}T < n_y-1.\] Using the assumption \(n_x\geqslant n_y\), we get \[\mathop{\mathrm{rank}}(T\circ R)< n_y-1\leqslant n_x-1=\dim L_{x}.\] Therefore \[\Pi_{S_x}H(z_1)AH(z_2)B\Pi_{S_x}^{\top}\big|_{L_{x}}\] is singular, and so \(0=Q_m(1)\) is an eigenvalue.

\((ii) \Rightarrow (i)\), first assume \(\lambda\neq 1\). Let \[\mu = \frac{\lambda(\lambda -1)}{(1+m) \lambda - m} \neq 0 \,,\] then the assumption is that there exist \(v_1 \in L_{x, \mathbb{C}} \setminus \{0\}\) such that \[\eta_{x}\eta_{y}\Pi_{S_x} (H(z_1) A H(z_2) B) \Pi_{S_x}^{\top} v_1 = \mu^2 v_1 \, .\] Define the vectors \[h_1 = \Pi_{S_x}^\top v_1 \quad \text{and} \quad h_2 = \frac{\eta_{y}}{\mu} H(z_2) B \Pi_{S_x}^\top v_1\] then \[\eta_{y}H(z_2) B h_1 = \mu h_2\] and \[\eta_{x}\Pi_{S_x} H(z_1) A h_2 = \frac{\eta_{x}\eta_{y}}{\mu} \Pi_{S_x} H(z_1) AH(z_2) B \Pi_{S_x}^\top v_1 = \frac{1}{\mu} \mu^2 v_1 = \mu v_1 \, ,\] so \[\eta_{x}H(z_1) A h_2 = \eta_{x}\Pi_{S_x}^\top \Pi_{S_x} H(z_1) A h_2 = \mu \Pi_{S_x}^\top v_1 = \mu h_1 \,.\]

\((ii) \Rightarrow (i)\), now consider \(\lambda=1\). Since \(Q_m(1)=0\), assume that there exists \(v_1\in L_{x,\mathbb{C}}\setminus\{0\}\) such that \[\Pi_{S_x}H(z_1)AH(z_2)B\Pi_{S_x}^{\top}v_1=0.\] Set \[\widehat h_1:=\Pi_{S_x}^{\top}v_1, \qquad r:=H(z_2)B\widehat h_1 .\] Then \(r\) is supported on \(S_y\) and satisfies \(\mathbf{1}^\top r=0\), hence \(r\in L_{y,\mathbb{C}}\). Moreover, \[\Pi_{S_x}H(z_1)Ar = \Pi_{S_x}H(z_1)AH(z_2)B\Pi_{S_x}^{\top}v_1 = 0.\] Since \(H(z_1)Ar\) is supported on \(S_x\), this implies \[H(z_1)Ar=0.\]

If \(r=0\), choose \[h_1=\widehat h_1, \qquad h_2=0.\] Then \[H(z_2)Bh_1=0, \qquad H(z_1)Ah_2=0.\] If \(r\neq0\), choose \[h_1=0, \qquad h_2=r.\] Then \[H(z_2)Bh_1=0, \qquad H(z_1)Ah_2=H(z_1)Ar=0.\] In both cases, \([h_1,h_2]\neq[0,0]\), and the vectors have the required support and tangent-space properties. Hence 16 holds for \(\lambda=1\). ◻

Proof of Theorem 5. From Lemma 5, for every \(\lambda \in \mathbb{C}\setminus\{0,\tfrac{m}{m+1}\}\), \[\lambda \in Q_m^{-1}\Big(\widetilde{\mathrm{eig}}\left( \eta_{x}\eta_{y}{ \left.\kern-\nulldelimiterspace M_x(Z) \vphantom{|} \right|_{L_x} }\right)\Big) \iff \eqref{eq:linSys}.\] By Lemma 4, every \(\lambda \in \mathrm{eig}\left( J_{\Phi_m}(Z)|_L\right)\setminus\{0,\tfrac{m}{m+1}\}\) either satisfies 13 or lies in \(W\). It remains to treat \(\lambda = m/(m+1)\). As the pole of \(Q_m\), it is never an element of \(Q_m^{-1}(\mu)\) for any finite \(\mu\), so it never belongs to the right-hand set; and by the Case 3 analysis above, it is an eigenvalue of \(J_{\Phi_m}(Z)|_L\) if and only if \(m/(m+1)\in W\). Thus adjoining or removing \(\lambda=m/(m+1)\) changes neither side, and \[\widetilde{\mathrm{eig}}\left( J_{\Phi_m}(Z)|_{L}\right)\setminus\{0\} = \left(W \cup Q_m^{-1}\Big(\widetilde{\mathrm{eig}}\left( \eta_{x}\eta_{y}{ \left.\kern-\nulldelimiterspace M_x(Z) \vphantom{|} \right|_{L_x} }\right)\Big)\right)\setminus\{0\},\] as claimed. ◻

Proof. We first show that we can assume without loss of generality that \(x\) is the origin. Define \(g(u)=f(x+u)-x,\) for \(u\in \mathop{\mathrm{\mathscr{O}}}-x\). Then \(g\) is \(\mathop{\mathrm{\mathscr{C}}}^1\), \(Dg(0)=Df(x)\), and \(g\bigl((B_\delta(0)\cap V)\bigr)\subset V\). Thus, it suffices to prove the claim for \(x=0\).

The set \(\mathop{\mathrm{\mathscr{O}}}\cap V\) is open in \(V\), so \({ \left.\kern-\nulldelimiterspace f \vphantom{|} \right|_{\mathop{\mathrm{\mathscr{O}}}\cap V} }\) is differentiable as a map defined on \(V\). For any \(v\in V\) and any \(t\) small enough, we have \(tv\in \mathop{\mathrm{\mathscr{O}}}\cap V\), hence \[Df(0)v = \lim_{t\to 0}\frac{f(tv)-f(0)}{t} = \lim_{t\to 0}\frac{f|_{\mathop{\mathrm{\mathscr{O}}}\cap V}(tv)-f|_{\mathop{\mathrm{\mathscr{O}}}\cap V}(0)}{t} = D(f|_{\mathop{\mathrm{\mathscr{O}}}\cap V})(0)v.\] Moreover, due to the local invariance assumption, for \(t\) small enough, both \(f(tv)\) and \(f(0)\) lie in \(V\), so \[\frac{f(tv)-f(0)}{t}\in V.\] Passing to the limit gives \(Df(0)v\in V\). ◻

Proof. We assume again without loss of generality that \(x = 0\) (otherwise consider \(G(u) = F(x + u) - x\)). We also denote \(B_\delta = B_\delta(0)\).

Since \(F(V\cap B_\delta)\subset V,\) the restriction \(f:={ \left.\kern-\nulldelimiterspace F \vphantom{|} \right|_{V\cap B_\delta} } : V\cap B_\delta \to V\) is well defined and by Lemma 6, \[Df(0)={ \left.\kern-\nulldelimiterspace DF(0) \vphantom{|} \right|_{V} }.\] Let \(k=\dim V\), and let \(T_c:\mathbb{R}^k\to V\) be a linear isomorphism. Define \[\widetilde{F} := T_c^{-1}\circ f\circ T_c\] on the open set \(T_c^{-1}(V\cap B_\delta)\subset \mathbb{R}^k\). Then \(\widetilde{F}\) is of class \(\mathop{\mathrm{\mathscr{C}}}^1\), satisfies \(\widetilde{F}(0)=0\), and \[D\widetilde{F}(0) =T_c^{-1}\circ Df(0)\circ T_c =T_c^{-1}\circ { \left.\kern-\nulldelimiterspace DF(0) \vphantom{|} \right|_{V} }\circ T_c.\] Hence \(D\widetilde{F}(0)\) is similar to \({ \left.\kern-\nulldelimiterspace DF(0) \vphantom{|} \right|_{V} }\), so they have the same spectrum and therefore the same spectral radius.

Since \(T_c\) is a homeomorphism, Lyapunov stability, attraction, and instability are preserved under this change of coordinates (see, e.g., Section 3 [32]). Therefore, by Proposition 8, the fixed point \(0\) is asymptotically stable relative to \(V\) when \(\rho\left({ \left.\kern-\nulldelimiterspace DF(0) \vphantom{|} \right|_{V} }\right)<1\), and unstable relative to \(V\) when \(\rho\left({ \left.\kern-\nulldelimiterspace DF(0) \vphantom{|} \right|_{V} }\right)>1\). ◻

Proof. Choose \(\delta>0\) small enough that, for every \(h\in {L^{\parallel}_{Z}}\cap B_\delta(0)\), all on-support coordinates of \(Z+h\) are positive. Since \(h\in{L^{\parallel}_{Z}}\), the off-support coordinates of \(Z+h\) are zero, and the four block sums are still equal to \(1\). Thus \(Z+h\in \mathop{\mathrm{\mathscr{D}}}_{\Phi_m}\) for \(\delta\) small enough.

Let \(U:=Z+h\). By Proposition 1, the first two blocks of \(\Phi_m(U)\) have no mass outside \(S_x\) and \(S_y\), respectively. The last two blocks of \(\Phi_m(U)\) are the first two blocks of \(U\), and hence are also supported on \(S_x\) and \(S_y\). Moreover, the first two blocks are normalized by \(P_{d_x}\) and \(P_{d_y}\), while the last two blocks already have coordinate sum \(1\). Hence, all four blocks of \(\Phi_m(U)\) have the same support constraints and affine-sum constraints as \(Z+{L^{\parallel}_{Z}}\). Therefore \[\Phi_m\bigl(Z+({L^{\parallel}_{Z}}\cap B_\delta(0))\bigr)\subseteq Z+{L^{\parallel}_{Z}}.\]

The final assertion follows from Lemma 6 applied with \(x=Z\), \(V={L^{\parallel}_{Z}}\), and \(f=\Phi_m\). ◻

Proof. By Theorem 5, the nonzero spectrum of \({J_{\Phi_m}}(Z)|_L\) is \(W\cup Q_m^{-1}\big(\mathrm{eig}\left( \eta_{x}\eta_{y}M_x(Z)|_{L_x}\right)\big)\) if \(n_x \geqslant n_y\) and by Remark 3, we have the analogous identity with \(M_y(Z)|_{L_y}\) if \(n_y \geqslant n_x\). Since the proof in the second case follows the same argument, we assume without loss of generality that \(n_x \geqslant n_y\). In this case, \(\rho\left({J_{\Phi_m}}(Z)|_L\right)>1\) forces (a) some \(\mathbf{w}\in W\) with \(\mathbf{w}>1\), or (b) some \(\mu\in\mathrm{eig}\left( \eta_{x}\eta_{y}M_x(Z)|_{L_x}\right)\) with a preimage \(\lambda\in Q_m^{-1}(\mu)\), \(|\lambda|>1\). We treat the two cases separately.

(a) Off-support instability. Suppose first that \(\mathbf{w}_{1,i}>1\) for some \(i\notin S_x\). Then \[(Ay)_i>v_x=(Ay)_j \qquad \text{for every } j\in S_x .\] Fix such a \(j\in S_x\). We apply Lemma 1, case (I), to the fixed point \[Z=[x,y,x,y].\] Now perturb \(Z\) by giving the off-support coordinate \(i\) a small positive mass. For \(\varepsilon>0\), let \(x^\varepsilon:=(1-\varepsilon)x+\varepsilon e_i\) and define \(Z^\varepsilon_0:=(x^\varepsilon,y,x^\varepsilon,y).\) Note that this perturbation ensures that \([x_t]_i > 0\). This allows us to apply Lemma 1. Hence there exist constants \(\delta>0\), \(\epsilon_0>0\), and \(\tau>0\) such that any orbit satisfying \(\|Z_t-Z\|<\delta\) also satisfies the exponential growth, while every point \(Z'\in\smash{\widehat{\Delta}}\) with \(\|Z'-Z\|<\delta\) satisfies the ratio bound.

Then \(Z^\varepsilon_0\in\smash{\widehat{\Delta}}\), \(\|Z^\varepsilon_0-Z\|=O(\varepsilon),\) and \(R_0=\frac{x^\varepsilon_i}{x^\varepsilon_j}>0 .\) By Proposition 1, coordinates \(i\) and \(j\) remain positive along the orbit, so \(R_t\) is well defined for all \(t\). Assume for contradiction that the orbit initialized at \(Z^\varepsilon_0\) remains in \(B_\delta(Z)\) for all \(t\geqslant 0\). Then Lemma 1 gives \[R_t\geqslant e^{\epsilon_0 t}R_0 \, .\] Since \(R_0>0\), this implies \(R_t\to+\infty\). On the other hand, since the orbit remains in \(B_\delta(Z)\), the bounded-ratio part of the same lemma gives \[R_t<\tau \qquad \text{for all } t\geqslant 0,\] a contradiction. Therefore, the orbit initialized at \(Z^\varepsilon_0\) must leave \(B_\delta(Z)\).

Since \(Z^\varepsilon_0\to Z\) as \(\varepsilon\downarrow0\), we have constructed arbitrarily small perturbations of \(Z\) whose orbits leave the fixed neighbourhood \(B_\delta(Z)\). Hence \(Z\) is not stable.

The case \(\mathbf{w}_{2,k}>1\) for some \(k\notin S_y\) is identical, using Lemma 1, case (II).

(b) Support (in-face) instability. Here the unstable eigenvector lies in \(({L^{\parallel}_{Z}})_{\mathop{\mathrm{\mathbb{C}}}}\) (it is supported on \(S_x,S_y\); Lemma 4, Case 3), and \(\rho\left({J_{\Phi_m}}(Z)|_{{L^{\parallel}_{Z}}}\right)>1\). By Lemma 7, \({{\Phi}_m}\) maps a (ambient) neighbourhood of \(Z\) in \(Z+{L^{\parallel}_{Z}}\) into \(Z+{L^{\parallel}_{Z}}\) and \({ \left.\kern-\nulldelimiterspace {J_{\Phi_m}}(Z) \vphantom{|} \right|_{{L^{\parallel}_{Z}}} }\) is the corresponding differential, so Theorem 9 applies with \(V={L^{\parallel}_{Z}}\). Hence, \(Z\) is unstable relative to \(Z+{L^{\parallel}_{Z}}\). Because \(Z\in\mathop{\mathrm{relint}}{\mathop{\mathrm{\mathscr{F}}}_Z}\), a relative neighbourhood of \(Z\) in \(Z+{L^{\parallel}_{Z}}\) is contained in \({\mathop{\mathrm{\mathscr{F}}}_Z}\subseteq\smash{\widehat{\Delta}}\); instability relative to \(Z+{L^{\parallel}_{Z}}\) therefore gives instability relative to \(\smash{\widehat{\Delta}}\). ◻

Proof. Let \(Z=[x,y,x,y]\) and \(H=[h_1,h_2,h_3,h_4]\in L\). We use the block structure of the operator \(\Phi_m = [(\Phi_m)_1,(\Phi_m)_2,(\Phi_m)_3,(\Phi_m)_4]\), where \((\Phi_m)_k:\mathop{\mathrm{\mathbb{R}}}^{2(d_x + d_y)} \rightarrow \mathop{\mathrm{\mathbb{R}}}^{d_x}\) for \(k = 1,3\) and \((\Phi_m)_k:\mathop{\mathrm{\mathbb{R}}}^{2(d_x + d_y)} \rightarrow \mathop{\mathrm{\mathbb{R}}}^{d_y}\) for \(k= 2,4\). Since \[\bigl(x\odot \mathrm{Exp}(\eta_{x}Ay)\bigr)^\top \mathbf{1}>0 \qquad\text{and}\qquad \bigl(y\odot \mathrm{Exp}(\eta_{y}Bx)\bigr)^\top \mathbf{1}>0,\] continuity implies that there exists \(\delta>0\) such that for every \(H\in L\cap B_\delta(0)\) the first two coordinates of \({\Phi_m}(Z+H)\) are well-defined. By definition of the projections \(P_{d_x}\) and \(P_{d_y}\), we then have \[({\Phi_m})_1(Z+H)^\top \mathbf{1} = 1 \qquad\text{and}\qquad ({\Phi_m})_2(Z+H)^\top \mathbf{1} = 1.\] Moreover, the last two coordinates of \({\Phi_m}\) are just copies of the first two coordinates of the input, so \[({\Phi_m})_3(Z+H) = x+h_1 \qquad\text{and}\qquad ({\Phi_m})_4(Z+H) = y+h_2.\] Since \(H\in L\), we have \(h_1^\top \mathbf{1} = h_2^\top \mathbf{1}=0\), hence \[({\Phi_m})_3(Z+H)^\top \mathbf{1} = (x+h_1)^\top \mathbf{1} = 1 \qquad\text{and}\qquad ({\Phi_m})_4(Z+H)^\top \mathbf{1} = (y+h_2)^\top \mathbf{1} = 1.\] Therefore all four blocks of \({\Phi_m}(Z+H)\) satisfy the affine constraints defining \(Z+L\), so \({\Phi_m}(Z+H)\in Z+L\) for every \(H\in L\cap B_\delta\). ◻

Proof of Theorem 4. By Lemma 8, there exists \(\delta>0\) such that \[{\Phi_m}\bigl(Z + (L\cap B_\delta(0))\bigr)\subset Z+L.\] Applying Lemma 6 with \(x=Z\), \(V=L\), and \(f={\Phi_m}\), we obtain \[D{\Phi_m}(Z)L\subset L \qquad\text{and}\qquad D\bigl({\Phi_m}|_{(Z+L)\cap \mathop{\mathrm{\mathscr{D}}}_{\Phi_m}}\bigr)(Z)=D{\Phi_m}(Z)|_L.\] This proves the first claim.

Assume first that \(\rho\left(J_{\Phi_m}(Z)|_L\right)<1\). Applying Theorem 9 with \(V=L\), \(x=Z\), and \(F={\Phi_m}\) shows that \(Z\) is asymptotically stable relative to \(Z+L\). Since \(\smash{\widehat{\Delta}}\subset Z+L\), this implies that \(Z\) is asymptotically stable for \({\Phi_m}|_{\smash{\widehat{\Delta}}}\).

Assume now that \(\rho\left(J_{\Phi_m}(Z)|_L\right)>1\). Applying Theorem 10 gives the result. ◻

Proof. For a mixed equilibrium, \(\dim (L_x) \geqslant 1\), hence \(\mathrm{eig}\left( { \left.\kern-\nulldelimiterspace M_x(Z) \vphantom{|} \right|_{L_x} }\right) \neq \varnothing\). Furthermore, since \({ \left.\kern-\nulldelimiterspace M_x(Z) \vphantom{|} \right|_{L_x} }\) is non-singular, \(0\notin \mathrm{eig}\left( { \left.\kern-\nulldelimiterspace M_x(Z) \vphantom{|} \right|_{L_x} }\right)\). Hence, there exists a \(\mu \in \mathrm{eig}\left( { \left.\kern-\nulldelimiterspace M_x(Z) \vphantom{|} \right|_{L_x} }\right)\) with \(\mu \neq 0\). Solving \(Q_m(\lambda) = \mu\) reduces to finding the roots of the polynomials \[\begin{align} \begin{cases} p_+(\lambda) &= \lambda^2 - (1+(m+1) s) \lambda + ms\, ,\\ p_-(\lambda) &= \lambda^2 - (1-(m+1) s) \lambda - m s\, , \end{cases} \end{align}\] where \(s^2 = \mu\). We will show that not all roots can simultaneously lie in the closed unit disk \(\mathop{\mathrm{\overline{\mathbb{D}}}}\). To simplify notation, let \(a_+ := (1+(m+1) s)\) and \(b_+ := ms\), similarly \(a_- := (1-(m+1) s)\) and \(b_- := -m s\). Assume, for contradiction, that all roots lie in \(\mathop{\mathrm{\overline{\mathbb{D}}}}\). By the Schur–Cohn criterion for quadratic polynomials (cf. Lemma 9), \(\left| a_+ - \overline{a}_+ b_+ \right| \leqslant 1 - \left| b_+ \right|^2\) and \(\left| a_- - \overline{a}_- b_- \right| \leqslant 1 - \left| b_- \right|^2\). Hence \[\begin{align} \begin{cases} &\left| 1 + s - m(m+1) \left| s \right|^2 \right| \leqslant 1 - m^2\left| s \right|^2\, ,\\[1ex] &\left| 1 - s - m(m+1) \left| s \right|^2 \right| \leqslant 1 - m^2\left| s \right|^2\, . \end{cases} \end{align}\] Squaring and adding gives \[\begin{align} \label{ineq:roots3727} \left| 1 + s - m(m+1) \left| s \right|^2 \right|^2 + \left| 1 - s - m(m+1) \left| s \right|^2 \right|^2 \leqslant 2(1 - m^2\left| s \right|^2)^2\, . \end{align}\tag{17}\] Note that \(1- m(m+1) \left| s \right|^2 \in \mathop{\mathrm{\mathbb{R}}}\). Hence, we use that for \(r \in \mathop{\mathrm{\mathbb{R}}}\) and \(c \in \mathbb{C}\), \(\left| r+c \right|^2 + \left| r-c \right|^2 = 2r^2 + r(c+\overline{c}) - r(c+\overline{c}) + 2\left| c \right|^2\). Thus, 17 is equivalent to \[\begin{align} &2(1 - m(m+1) \left| s \right|^2)^2 + 2\left| s \right|^2 \leqslant 2(1 - m^2\left| s \right|^2)^2\\ \Leftrightarrow \qquad&\left| s \right|^2\left((1-2m) + m^2(2m+1) \left| s \right|^2 \right) \leqslant 0\, . \end{align}\] However, since \(\left| s \right|^2 > 0\) and \(m \in (0,1/2]\), we have \((1-2m) \geqslant 0\) and \(m^2(2m+1) \left| s \right|^2 > 0\). Thus \[\begin{align} \left| s \right|^2\left((1-2m) + m^2(2m+1) \left| s \right|^2 \right) > 0\, , \end{align}\] which gives a contradiction. Thus, there exists a \(\lambda \in Q_m^{-1}(\mu)\), such that \(\lambda \not \in \mathop{\mathrm{\overline{\mathbb{D}}}}\) and therefore \(\left| \lambda \right| > 1\). Hence \(\rho\left(J_{\Phi_m}(Z)|_L\right)>1\), and Theorem 4 gives the claim. ◻

Proof. Denote the two roots by \(\lambda_+,\lambda_-.\) Then \[a=\lambda_++\lambda_-, \qquad b=\lambda_+\lambda_- .\] We first prove the necessary direction. Assume that both roots lie in the closed unit disk. Then \[\left| b \right| = \left| \lambda_+\lambda_- \right| \leqslant 1 .\] Moreover, \[\begin{align} a-\overline{a} b &= \lambda_+ + \lambda_- - (\overline{\lambda}_+ + \overline{\lambda}_-)\lambda_+\lambda_- \\ &= \lambda_+(1-\left| \lambda_- \right|^2) + \lambda_-(1-\left| \lambda_+ \right|^2) . \end{align}\] Thus, using the triangle inequality, \[\begin{align} \left| a-\overline{a} b \right| &\leqslant \left| \lambda_+ \right|(1-\left| \lambda_- \right|^2) + \left| \lambda_- \right|(1-\left| \lambda_+ \right|^2). \end{align}\] Writing \(r=\left| \lambda_+ \right|\) and \(s=\left| \lambda_- \right|\), we have \[\begin{align} 1-r^2s^2 - r(1-s^2) - s(1-r^2) = (1-r)(1-s)(1+r+s+rs) \geqslant 0 . \end{align}\] Hence \[\left| a-\overline{a} b \right| \leqslant 1-r^2s^2 = 1-\left| b \right|^2 .\] For the sufficient direction, assume \[\left| b \right|<1 \qquad\text{and}\qquad \left| a-\overline{a} b \right|<1-\left| b \right|^2 .\] Suppose, for contradiction, that at least one root does not lie in the open unit disk. Without loss of generality, let \[r=\left| \lambda_+ \right|\geqslant 1, \qquad s=\left| \lambda_- \right|.\] Since \(\left| b \right|=rs<1\), we have \(s<1\). Using the same identity as above and the reverse triangle inequality, \[\begin{align} \left| a-\overline{a} b \right| &= \left| \lambda_+(1-s^2) + \lambda_-(1-r^2) \right| \\ &\geqslant r(1-s^2)-s(r^2-1) \\ &= (r+s)(1-rs). \end{align}\] Since \(r\geqslant 1\) and \(s<1\), \[r+s-(1+rs) = (r-1)(1-s) \geqslant 0 .\] Therefore \[\left| a-\overline{a} b \right| \geqslant (r+s)(1-rs) \geqslant (1+rs)(1-rs) = 1-r^2s^2 = 1-\left| b \right|^2 ,\] which contradicts the assumed strict inequality. Hence, both roots must lie in the open unit disk. ◻

Proof. This follows from the continuity of a polynomial’s roots with respect to its coefficients. ◻

Proof. First note that \(Q_m(\lambda)\) must be in \(\mathop{\mathrm{\mathbb{R}}}\) by assumption. Since \(\left| \lambda \right| = 1\) we write \(\lambda = e^{i \theta}\) and for easier notation define \(c_m := 2m +1\). By Euler’s identity, \(e^{i \theta} -1 = e^{i \frac{\theta}{2}}(2 i \sin(\theta/2))\) and \((m+1)e^{i \theta/2} - m e^{- i \theta/2} = \cos(\theta/2) + i c_m \sin(\theta/2)\). Hence, using trigonometric identities, we find that \(Q_m(e^{i\theta})\) is equal to \[\begin{align} &- \frac{4 \sin^2\left( \frac{\theta}{2} \right) e^{2 i \theta}}{ \left(\cos\left( \frac{\theta}{2}\right)+ i c_m\sin\left( \frac{\theta}{2}\right)\right)^2}\nonumber \\ &\overset{(1)}{=} - 4 \sin^2\left( \frac{\theta}{2} \right) \cos^2 \left( \frac{ \theta}{2}\right)\left( \frac{\left( 1+ i \tan \left( \frac{\theta}{2}\right)\right)^2}{ 1+ i c_m \tan \left( \frac{\theta}{2}\right)}\right)^2\nonumber\\ &= - 4 \sin^2\left( \frac{\theta}{2} \right) \cos^2 \left( \frac{ \theta}{2}\right)\left( \underbrace{ \frac{1 - \tan^2 \left(\frac{\theta}{2} \right) + 2 c_m \tan^2 \left(\frac{\theta}{2} \right)}{1+c_m^2 \tan^2 \left(\frac{\theta}{2} \right)} }_{:= \xi}+ i \underbrace{\frac{ \tan \left(\frac{\theta}{2} \right) \left(c_m \tan^2 \left(\frac{\theta}{2} \right) + 2 - c_m \right)}{1+c_m^2 \tan^2 \left(\frac{\theta}{2} \right)}}_{=: \rho} \right)^2\, . \end{align}\] Since \(\tan(\pi/2)\) is undefined, the equality in \((1)\) is invalid if \(\theta = \pi\), that is \(\lambda = -1\). However, we note that \(Q_m(-1) = \frac{4}{(2m +1)^2} > 0\), hence, this case is excluded by the assumption that \(Q_m(\lambda)\) is negative. Since \(Q_m(e^{i \theta}) = Q_m(\lambda)\) must be real, \(\rho \xi\) must be zero. Since \[1 - \tan^2 \left(\frac{\theta}{2} \right) + 2 c_m \tan^2 \left(\frac{\theta}{2} \right) = 1 + (4m +1 )\tan^2 \left(\frac{\theta}{2} \right) > 0\, ,\] \(\rho\) must be zero. This implies that one of the following conditions must hold: \[\begin{align} (1) \qquad \theta = 0 \qquad \text{ or, } \qquad (2) \qquad \tan^2 \left(\frac{\theta}{2} \right) = \frac{c_m -2}{ c_m}\,. \end{align}\] In Case \((1)\), \(Q_m(e^{i 0}) = 0\), hence \(\sigma = 0\), and in Case (2), we simplify \(Q_m(e^{i \theta}) = -\frac{2m-1}{m^2(2m+1)}\) which implies that \(\sigma = \sigma^\star(m)\).

It remains to show that for \(m > \frac{1}{2}\), \(\sigma=\sigma^\star(m)\) is attained by a root on the unit circle. Choose \[\theta := 2\arctan\left(\sqrt{\frac{c_m-2}{c_m}}\right), \qquad \lambda := e^{i\theta}.\] Due to the assumption \(m > \frac{1}{2}\), \(\frac{c_m-2}{c_m}=\frac{2m-1}{2m+1}>0\). Using \(\sigma \in \mathop{\mathrm{\mathbb{R}}}\), hence \(\rho = 0\), by the same computations as before, \[Q_m(\lambda) = -\frac{4(c_m-2)}{c_m(c_m-1)^2} = -\frac{2m-1}{m^2(2m+1)} = -\sigma^\star(m)\,\] and \(\left| \lambda \right| = 1\). ◻

Proof. Let \(s=\sqrt{\sigma}\). If \(\sigma=0\), then \[P_{0,m}(\lambda)=\lambda^2(1-\lambda)^2,\] so the roots are \(0,0,1,1\).

Assume now that \(\sigma>0\). We factor \[P_{\sigma,m}(\lambda) = \prod_{\delta\in\{-1,1\}} \left( \lambda(1-\lambda)+\delta i s((m+1)\lambda-m) \right).\] Equivalently, \[P_{\sigma,m}(\lambda)=\prod_{\delta\in\{-1,1\}}p_\delta(\lambda),\] where \[p_\delta(\lambda) = \lambda^2-a_\delta\lambda+b_\delta, \qquad a_\delta=1+\delta i(m+1)s, \qquad b_\delta=\delta i m s .\] For each \(\delta\), \[\left| b_\delta \right|^2=m^2\sigma,\] and \[a_\delta-\overline{a}_\delta b_\delta = 1-m(m+1)\sigma+\delta i\sqrt{\sigma}.\] Hence \[\left| a_\delta-\overline{a}_\delta b_\delta \right|^2 = (1-m(m+1)\sigma)^2+\sigma,\] and therefore \[\begin{align} \label{eq:SC-difference} (1-\left| b_\delta \right|^2)^2 -\left| a_\delta-\overline{a}_\delta b_\delta \right|^2 = \sigma\Big((2m-1)-m^2(2m+1)\sigma\Big). \end{align}\tag{18}\]

Suppose first that \(m>1/2\) and \[0<\sigma<\sigma^\star(m) = \frac{2m-1}{m^2(2m+1)}.\] Then the right-hand side of 18 is positive. Moreover, \[m^2\sigma<m^2\sigma^\star(m) = \frac{2m-1}{2m+1}<1.\] Thus, for both \(\delta=\pm1\), \[\left| b_\delta \right|<1 \qquad\text{and}\qquad \left| a_\delta-\overline{a}_\delta b_\delta \right|<1-\left| b_\delta \right|^2.\] By Lemma 9, both roots of each quadratic factor \(p_\delta\) lie in the open unit disk. Hence every root of \(P_{\sigma,m}\) satisfies \(\left| \lambda \right|<1\).

Next let \(m>1/2\) and \(\sigma=\sigma^\star(m)\). By the unit-circle calculation in Lemma 11, choosing \[\theta = 2\arctan\sqrt{\frac{2m-1}{2m+1}}, \qquad \lambda=e^{i\theta},\] gives \[Q_m(\lambda) = -\frac{2m-1}{m^2(2m+1)} = -\sigma^\star(m).\] Equivalently, \(P_{\sigma^\star(m),m}(\lambda)=0\), and by construction \(\left| \lambda \right|=1\). Thus at least one root lies on the unit circle.

Now let \(m>1/2\) and \(\sigma>\sigma^\star(m)\). Suppose, for contradiction, that all roots of \(P_{\sigma,m}\) lie in the closed unit disk. Then the roots of each quadratic factor \(p_\delta\) lie in the closed unit disk. By the necessary part of Lemma 9, \[\left| a_\delta-\overline{a}_\delta b_\delta \right| \leqslant 1-\left| b_\delta \right|^2.\] Squaring gives \[(1-\left| b_\delta \right|^2)^2 -\left| a_\delta-\overline{a}_\delta b_\delta \right|^2 \geqslant 0.\] This contradicts 18 , whose right-hand side is negative when \(\sigma>\sigma^\star(m)\). Hence, at least one root satisfies \(\left| \lambda \right|>1\).

Finally, suppose that \(m\leqslant 1/2\). If \(\sigma=0\), then \(\lambda=1\) is a double root, so the desired conclusion holds. If \(\sigma>0\) and all roots lay in the open unit disk, then Lemma 9 applied to each \(p_\delta\) would imply \[(1-\left| b_\delta \right|^2)^2 -\left| a_\delta-\overline{a}_\delta b_\delta \right|^2 >0.\] But by 18 , \[(1-\left| b_\delta \right|^2)^2 -\left| a_\delta-\overline{a}_\delta b_\delta \right|^2 = \sigma\Big((2m-1)-m^2(2m+1)\sigma\Big)<0\] for every \(m\leqslant 1/2\) and every \(\sigma>0\). This contradiction shows that not all roots lie in the open unit disk. Therefore, at least one root satisfies \(\left| \lambda \right|\geqslant 1\). ◻

Proof. Write \(Z^\star = [x^\star,y^\star,x^\star,y^\star]\). Since the equilibrium is fully mixed, \[{ \left.\kern-\nulldelimiterspace M_x(Z^\star) \vphantom{|} \right|_{L_x} } = - H(x^\star) A H(y^\star) A^\top .\] Write \(H_x:=H(x^\star)\) and \(H_y:=H(y^\star)\). Suppose, for contradiction, that \({ \left.\kern-\nulldelimiterspace M_x(Z^\star) \vphantom{|} \right|_{L_x} }\) is singular. Then there exists \(0\neq h\in L_x\) such that \[M_x(Z^\star)h=0 \quad \Leftrightarrow \quad H_x A H_y A^\top h=0 \,.\] Thus, \(A H_y A^\top h \in \ker H_x\). But since \(\ker H_x=\mathop{\mathrm{span}}(\mathbf{1}_{d_x})\), there exists \(c\in\mathop{\mathrm{\mathbb{R}}}\) such that \[A H_y A^\top h = c\mathbf{1}_{d_x}.\] Taking the inner product with \(h\), and using \(h\in L_x\), gives \[0 = c\, h^\top \mathbf{1}_{d_x} = h^\top A H_y A^\top h = (A^\top h)^\top H_y (A^\top h).\] Since \(H_y\) is positive semidefinite with kernel \(\mathop{\mathrm{span}}(\mathbf{1}_{d_y})\), it follows that \(A^\top h \in \mathop{\mathrm{span}}(\mathbf{1}_{d_y}).\) Thus, for some \(c'\in\mathop{\mathrm{\mathbb{R}}}\), \[A^\top h = c' \mathbf{1}_{d_y}.\] Because \(h\in L_x\), we have \(\mathbf{1}_{d_x}^\top h=0\). Since \(x^\star\in\mathop{\mathrm{relint}}\Delta_{d_x}\), for all sufficiently small \(\epsilon\neq 0\), \[x(\epsilon) := x^\star+\epsilon h\] still belongs to \(\mathop{\mathrm{relint}}\Delta_{d_x}\), and \(x(\epsilon)\neq x^\star\).

We show that \((x(\epsilon),y^\star)\) is again a Nash equilibrium: Since \(y^\star\) is fully mixed at the zero-sum equilibrium, \(A y^\star = v\mathbf{1}_{d_x}\) for some \(v\in\mathop{\mathrm{\mathbb{R}}}\). Similarly, since \(x^\star\) is fully mixed at equilibrium, \(A^\top x^\star = v\mathbf{1}_{d_y}.\) Using \(A^\top h=c' \mathbf{1}_{d_y}\), we obtain \[A^\top x(\epsilon) = A^\top x^\star+\epsilon A^\top h = (v+\epsilon c')\mathbf{1}_{d_y}\,.\] Hence, \(x(\epsilon)\) is a best response to \(y^\star\) and vice versa. Consequently \((x(\epsilon),y^\star)\) is a fully mixed Nash equilibrium distinct from \([x^\star,y^\star]\), contradicting uniqueness. Therefore, no such nonzero \(h\in L_x\) exists, and \({ \left.\kern-\nulldelimiterspace M_x(Z^\star) \vphantom{|} \right|_{L_x} }\) is non-singular. ◻

Proof of Theorem 6. We show that the modulus of the eigenvalues of the Jacobian is strictly less than or strictly greater than one. This implies stability under the assumed step size conditions. Let \(Z^\star\) denote the unique Nash equilibrium and \(M_x(Z^\star)\) the matrix corresponding to it. We first note that \(M_x(Z^\star)\) has a real, non-positive spectrum. This is due to \(H(z_1^\star)\) and \(A H(z_2^\star)A^\top\) being symmetric positive semidefinite, and the non-zero spectrum of \(H(z_1^\star)A H(z_2^\star)A^\top\) coincides with the non-zero spectrum of the symmetric positive semidefinite matrix \(H(z_1^\star)^{\frac{1}{2}}A H(z_2^\star)A^\top H(z_1^\star)^{\frac{1}{2}}\). Thus, the spectrum of \(M_x(Z^\star) = - H(z_1^\star)A H(z_2^\star)A^\top\) is real and non-positive. Furthermore, since \(Z^\star\) is by assumption full support, \({ \left.\kern-\nulldelimiterspace M_x(Z^\star) \vphantom{|} \right|_{L_x} }\) is non-singular (cf. Lemma 13). Hence, \(\mathrm{eig}\left( { \left.\kern-\nulldelimiterspace M_x(Z^\star) \vphantom{|} \right|_{L_x} }\right) \subset (-\infty, 0)\). Since the entries in \(M_x(Z^\star)\) are finite, \(\delta := \max(\left| \lambda \right| : \lambda \in \mathrm{eig}\left( { \left.\kern-\nulldelimiterspace M_x(Z^\star) \vphantom{|} \right|_{L_x} }\right))\), is finite and \(\mathrm{eig}\left( \eta_{x}\eta_{y}{ \left.\kern-\nulldelimiterspace M_x(Z^\star) \vphantom{|} \right|_{L_x} }\right) \subset [-\eta_{x}\eta_{y}\delta, 0)\). Hence, by Theorem 5, the spectrum of \(J_{\Phi_m}(Z^\star)|_{L}\) are either the values in \(W\) or the solutions to \[Q_m(\lambda) = -\sigma \qquad \sigma \in (0,\eta_{x}\eta_{y}\delta]\,.\] Due to the assumption that the Nash equilibrium has full support, all values in \(W\) are strictly less than \(1\). Hence, we can ignore them for this proof. Under the assumption that \(m \in ( 1/2,1]\), the claim follows by noting that \(Q_m(\lambda) = - \sigma\) if and only if \(P_{\sigma, m}(\lambda) = 0\). Combining this with Lemma 12 implies the claim. The case \(m \in (0, 1/2]\), follows from Lemma 13 and Corollary 1. ◻

Proof of Theorem 7. Since the Nash equilibrium is, by assumption, fully supported, \(W = \varnothing\). In the \(2\)-dimensional case, \(L_x\) is the one-dimensional space of vectors proportional to \((-1, 1)\). Moreover, \[H(x^\star) = p(1-p) \begin{bmatrix} \phantom{-}1 & -1\\ -1 & \phantom{-}1 \end{bmatrix}, \qquad H(y^\star) = q(1-q) \begin{bmatrix} \phantom{-}1 & -1\\ -1 & \phantom{-}1 \end{bmatrix}\] Let \(r=(1,-1)^\top\). Then \(H(x^\star)=p(1-p)rr^\top\) and \(H(y^\star)=q(1-q)rr^\top\). Hence

\[M_x(Z^\star) = H(x^\star) A H(y^\star) B = p(1-p)q(1-q)\, r r^\top A r r^\top B .\] Since \[r^\top A r=\Delta_A, \qquad r^\top B r=\Delta_B,\] we obtain \[M_x(Z^\star) = p(1-p)q(1-q)\Delta_A \begin{bmatrix} e-g &\phantom{-} f-h\\ g-e &\phantom{-} h-f \end{bmatrix}.\] Restricting to \(L_x= \mathop{\mathrm{span}}\{(-1,1)^\top\}\), \[\mathrm{eig}\left( M_x(Z^\star)|_{L_x}\right) = \left\{ p(1-p)q(1-q)\Delta_A\Delta_B \right\}.\] We write this eigenvalue as \[\mu = p(1-p)q(1-q)\Delta_A\Delta_B.\] Thus, the local criterion becomes \[Q_m(\lambda)=\eta_{x}\eta_{y}\,\mu.\] If \(\Delta_A\Delta_B>0\), then \(\mu>0\), and by Corollary 4, the mixed equilibrium is locally unstable for every \(\eta_{x}\eta_{y}>0\), and every \(m\).

If \(\Delta_A\Delta_B<0\) and \(m \leqslant\frac{1}{2}\), instability follows from Corollary 1. Note that in this case \(\eta_{x}\eta_{y}> E\) for all strictly positive step sizes.

Thus, assume \(\Delta_A\Delta_B<0\) and \(m > \frac{1}{2}\), then we have local stability if all solutions to \[Q_m(\lambda)=-\eta_{x}\eta_{y}|\mu| \, ,\] are of modulus strictly less than \(1\). Applying Lemma 12 and Theorem 5 gives that the full-support mixed equilibrium is locally asymptotically stable when \[\begin{align} \eta_{x}\eta_{y}\left| \mu \right| \in \left(0,\sigma^\star(m) \right)\,. \end{align}\] When \(\eta_{x}\eta_{y}\left| \mu \right| > \sigma^\star(m)\), by Lemma 12 (3) it is unstable.

The degenerate case \(\Delta_A\Delta_B=0\) gives \(\mu=0\) and therefore a unit eigenvalue; in this case, the linear criterion is inconclusive. ◻

Proof of Theorem 8. We denote the \(i^{\mathrm{th}}\) row of the matrices \(A,B\) by \(A_{i,:}\) and \(B_{i,:}\). Define \[\begin{align} M_A= \begin{bmatrix} A_{1,:} - A_{2,:}\\ \vdots\\ A_{1,:} - A_{d_x,:}\\ \mathbf{1}^\top \\ \end{bmatrix}, \qquad M_B= \begin{bmatrix} B_{1,:} - B_{2,:}\\ \vdots\\ B_{1,:} - B_{d_y,:}\\ \mathbf{1}^\top \\ \end{bmatrix}\, . \end{align}\] Furthermore, we denote by \(M_A^{(i)}\) (respectively \(M_B^{(i)}\)) the matrix \(M_A\) where the \(i^{\mathrm{th}}\) column is substituted by \(e_{d_x}\) (respectively \(e_{d_y}\)). We first show the following:

1. \(\det(M_A) \neq 0\) and \(\det(M_B)\neq 0\);

2. Under the assumption that the Nash equilibrium is fully mixed and unique the Nash equilibrium is the solutions to the linear equations: \(M_A y = e_{d_x}\) and \(M_B x = e_{d_y}\). Hence \[y^\star_i = \frac{\det(M_A^{(i)})}{\det(M_A)} \qquad \text{ and } \qquad x^\star_i = \frac{\det(M_B^{(i)})}{\det(M_B)}\, .\]

Due to Theorem 1, \([Ay^\star]_1= [Ay^\star]_i\) for \(i \in \{2, \dots, d_x\}\) and \(y^\star \in \Delta_{d_y}\) Equivalently, \[[Ay^\star]_1-[Ay^\star]_i=0 \text{ for } i \in \{2, \dots, d_x\} \text{, and } \sum_{i=1}^{d_y} y_i^\star=1.\] By the definition of \(M_A\), these equations are exactly \[M_Ay^\star=e_{d_x}.\] Now, assume for contradiction that \(M_A\) is singular. Then \(M_A z = 0\) has a non-zero solution and, due to the last row in \(M_A\), \(\left\langle z,\mathbf{1} \right\rangle = 0\). We define \(\hat{y}(\alpha) = y^\star + \alpha z\). Then \(M_A\hat{y}(\alpha) = M_A y^\star + \alpha M_A z = e_{d_x}\) and for a sufficiently small \(\alpha>0\), \(\hat{y}(\alpha) \in \Delta_{d_y}\) (recall \(y^\star \in \mathop{\mathrm{relint}}\Delta_{d_y}\) by assumption). Thus, \([A\hat{y}(\alpha)]_1 = [A\hat{y}(\alpha)]_i\) for all \(i \in \{2, \dots, d_x\}\), that is, the payoff is constant. Conversely, \(B x^\star\) is constant by definition of \(x^\star\). Thus, \([x^\star, \hat{y}(\alpha)]\) is a Nash equilibrium since \(x^\star \in \Delta_{d_x}\) is a best response to \(\hat{y}(\alpha)\) and vice versa. This contradicts the uniqueness of the Nash equilibrium.

Applying Cramer’s rule gives \(x^\star_i\) and \(y^\star_i\) as closed-form solutions in the entries of the game matrices, and consequently, the linear operator \({ \left.\kern-\nulldelimiterspace M_x(Z^\star) \vphantom{|} \right|_{L_x} }\) can be expressed in closed-form. So far, we do not require any assumptions on the dimensions besides \(d_x = d_y\).

The claim follows from Theorem 5 combined with the observation that the characteristic polynomial of \({ \left.\kern-\nulldelimiterspace M_x(Z^\star) \vphantom{|} \right|_{L_x} }\) has degree at most \(4\) and thus roots are expressible by radicals. ◻