\section{Proof of Theorem~\ref{thm: OMWU fails}}\label{A1}
We introduce an equivalence of the dynamics of Optimistic MWU which tracing the first pure strategies of the two players in the case where both two players have only two pure strategies.

By the definition of the dynamics sytem of OMWU, when $t$ is even, according to the definition of $A_t$ in Theorem~\ref{thm: OMWU fails}, we have
	\begin{align*}
		\tb{x}_{1,1}^{t+2}&=\frac{\tb{x}_{1,1}^{t+1} e^{2 \eta (A_{t+1} \tb{x}_{2}^{t+1})^1- \eta (A_{t} \tb{x}^{t}_2)^{1}}} { \sum^2_{s=1}\tb{x}_{1,s}^{t+1} e^{2 \eta (A_{t+1} \tb{x}_{2}^{t+1})^s - \eta (A_{t} \tb{x}_2^{t})^s}}\\
		&=\frac{\tb{x}_{1,1}^{t+1} e^{2 \eta  \tb{x}_{2,2}^{t+1}+ \eta  \tb{x}^{t}_{2,2}}} {\tb{x}_{1,1}^{t+1} e^{2 \eta  \tb{x}_{2,2}^{t+1}+ \eta  \tb{x}^{t}_{2,2}}+\tb{x}_{1,2}^{t+1} e^{2 \eta  \tb{x}_{2,1}^{t+1}+ \eta  \tb{x}^{t}_{2,1}}}\\
		&=\frac{\tb{x}_{1,1}^{t+1} e^{2 \eta (1-\tb{x}_{2,1}^{t+1}) + \eta  (1-\tb{x}^{t}_{2,1})}} {\tb{x}_{1,1}^{t+1} e^{2 \eta  (1-\tb{x}_{2,1}^{t+1})+ \eta  (1-\tb{x}^{t}_{2,1})}+(1-\tb{x}_{1,1}^{t+1}) e^{2 \eta  \tb{x}_{2,1}^{t+1}+ \eta  \tb{x}^{t}_{2,1}}}\\
		&=\frac{\tb{x}_{1,1}^{t+1} e^{3\eta-2 \eta \tb{x}_{2,1}^{t+1} - \eta  \tb{x}^{t}_{2,1}}} {\tb{x}_{1,1}^{t+1} e^{3 \eta  -2\eta\tb{x}_{2,1}^{t+1}- \eta  \tb{x}^{t}_{2,1}}+(1-\tb{x}_{1,1}^{t+1}) e^{2 \eta  \tb{x}_{2,1}^{t+1}+ \eta  \tb{x}^{t}_{2,1}}},
	\end{align*}
	where the third equality arises from that $\textbf{x}_1^{t+1}, \ \textbf{x}_1^{t}\in \Delta_2$. 
	By similar computation, it holds that
	\begin{align*}
		\tb{x}_{2,1}^{t+2}=\frac{\tb{x}_{2,1}^{t+1} e^{-3\eta+2 \eta \tb{x}_{1,1}^{t+1} + \eta  \tb{x}^{t}_{1,1}}} {\tb{x}_{2,1}^{t+1} e^{-3 \eta  +2\eta\tb{x}_{1,1}^{t+1}+ \eta  \tb{x}^{t}_{1,1}}+(1-\tb{x}_{2,1}^{t+1}) e^{-2 \eta  \tb{x}_{1,1}^{t+1}- \eta  \tb{x}^{t}_{1,1}}}
	\end{align*}
	
	For the case when $t$ is odd, we have
	\begin{align*}
		&\tb{x}_{1,1}^{t+2}=\frac{\tb{x}_{1,1}^{t+1} e^{-3\eta+2 \eta \tb{x}_{2,1}^{t+1} + \eta  \tb{x}^{t}_{2,1}}} {\tb{x}_{1,1}^{t+1} e^{-3 \eta  +2\eta\tb{x}_{2,1}^{t+1}+ \eta  \tb{x}^{t}_{2,1}}+(1-\tb{x}_{1,1}^{t+1}) e^{-2 \eta  \tb{x}_{2,1}^{t+1}-\eta  \tb{x}^{t}_{2,1}}},\\
		&\tb{x}_{2,1}^{t+2}=\frac{\tb{x}_{2,1}^{t+1} e^{3\eta-2 \eta \tb{x}_{1,1}^{t+1} - \eta  \tb{x}^{t}_{1,1}}} {\tb{x}_{2,1}^{t+1} e^{3 \eta  -2\eta\tb{x}_{1,1}^{t+1}- \eta  \tb{x}^{t}_{1,1}}+(1-\tb{x}_{2,1}^{t+1}) e^{2 \eta  \tb{x}_{1,1}^{t+1}+\eta  \tb{x}^{t}_{1,1}}}.
	\end{align*}
	From the above expression, it can be found that $\tb{x}_{1,1}^{t+2},\ \tb{x}_{2,1}^{t+2}$ can be entirely determined by the vector $\{\tb{x}_{1,1}^{t},\tb{x}_{1,1}^{t+1},\tb{x}_{2,1}^{t},\tb{x}_{2,1}^{t+1}\}$ in the case of when $t$ is odd and when $t$ is even.
	
	Next, we provide two formal functions mapping $(\tb{x}_{1,1}^{t},\tb{x}_{1,1}^{t+1},\tb{x}_{2,1}^{t},\tb{x}_{2,1}^{t+1})$ to $(\tb{x}_{1,1}^{t+1},\tb{x}_{1,1}^{t+2},\tb{x}_{2,1}^{t+1},\tb{x}_{2,1}^{t+2})$. Let 
	\begin{align*}
		\CG_1: &[0,1] \times [0,1] \times [0,1] \times [0,1] \to [0,1] \times [0,1] \times [0,1] \times [0,1]\\
		&(\textbf{z}_1,\textbf{z}_2,\textbf{z}_3,\textbf{z}_4)
		\to \\ &\left(\tb{z}_2,
		\frac{\tb{z}_1 e^{3\eta-2 \eta \tb{z}_4 - \eta  \tb{z}_3}}{\tb{z}_2 e^{3 \eta  -2\eta\tb{z}_4- \eta  \tb{z}_3}+(1-\tb{z}_2) e^{2 \eta  \tb{z}_4+ \eta  \tb{z}_3}}, 
		\right. \\ &\left.
		\tb{z}_4,
		\frac{\tb{z}_4 e^{-3\eta+2 \eta \tb{z}_2 + \eta  \tb{z}_1}}{\tb{z}_4 e^{-3 \eta  +2\eta\tb{z}_2+ \eta  \tb{z}_1}+(1-\tb{z}_4) e^{-2 \eta  \tb{z}_2- \eta  \tb{z}_1}}\right)
	\end{align*} 
	and 
		\begin{align*}
		\CG_2: &[0,1] \times [0,1] \times [0,1] \times [0,1] \to [0,1] \times [0,1] \times [0,1] \times [0,1]\\
		&(\textbf{z}_1,\textbf{z}_2,\textbf{z}_3,\textbf{z}_4)
		\to \\ &\left(\tb{z}_2,
		\frac{\tb{z}_1 e^{-3\eta+2 \eta \tb{z}_4 +\eta  \tb{z}_3}}{\tb{z}_2 e^{-3 \eta  +2\eta\tb{z}_4+ \eta  \tb{z}_3}+(1-\tb{z}_2) e^{-2 \eta  \tb{z}_4+ \eta  \tb{z}_3}}, 
				\right. \\ &\left.
		\tb{z}_4,
		\frac{\tb{z}_4 e^{3\eta-2 \eta \tb{z}_2 - \eta  \tb{z}_1}}{\tb{z}_4 e^{3 \eta  -2\eta\tb{z}_2- \eta  \tb{z}_1}+(1-\tb{z}_4) e^{2 \eta  \tb{z}_2+ \eta  \tb{z}_1}}\right).		
	\end{align*} 
According to their definition, we have
	\begin{align*}
		&\CG_1\left((\tb{x}_{1,1}^t,\tb{x}_{1,1}^{t+1},\tb{x}_{2,1}^t,\tb{x}_{2,1}^{t+1})\right)\\
		=&\left(\tb{x}_{1,1}^{t+1},
		\frac{\tb{x}_{1,1}^{t+1} e^{3\eta-2 \eta \tb{x}_{2,1}^{t+1} - \eta  \tb{x}^{t}_{2,1}}}{\tb{x}_{1,1}^{t+1} e^{3 \eta  -2\eta\tb{x}_{2,1}^{t+1}- \eta  \tb{x}^{t}_{2,1}}+(1-\tb{x}_{1,1}^{t+1}) e^{2 \eta  \tb{x}_{2,1}^{t+1}+ \eta  \tb{x}^{t}_{2,1}}}, 
		\right. \\ &\left.
		\tb{x}_{2,1}^{t+1},
		\frac{\tb{x}_{2,1}^{t+1} e^{-3\eta+2 \eta \tb{x}_{1,1}^{t+1} + \eta  \tb{x}^{t}_{1,1}}}{\tb{x}_{2,1}^{t+1} e^{-3 \eta  +2\eta\tb{x}_{1,1}^{t+1}+ \eta  \tb{x}^{t}_{1,1}}+(1-\tb{x}_{2,1}^{t+1}) e^{-2 \eta  \tb{x}_{1,1}^{t+1}- \eta  \tb{x}^{t}_{1,1}}}\right).
	\end{align*}
	and
			\begin{align*}
		&\CG_2\left((\tb{x}_{1,1}^t,\tb{x}_{1,1}^{t+1},\tb{x}_{2,1}^t,\tb{x}_{2,1}^{t+1})\right)\\
		=&\left(\tb{x}_{1,1}^{t+1},
		\frac{\tb{x}_{1,1}^{t+1} e^{-3\eta+2 \eta \tb{x}_{2,1}^{t+1} +\eta  \tb{x}^{t}_{2,1}}}{\tb{x}_{1,1}^{t+1} e^{-3 \eta  +2\eta\tb{x}_{2,1}^{t+1}+ \eta  \tb{x}^{t}_{2,1}}+(1-\tb{x}_{1,1}^{t+1}) e^{-2 \eta  \tb{x}_{2,1}^{t+1}- \eta  \tb{x}^{t}_{2,1}}}, 
		\right. \\ &\left.
		\tb{x}_{2,1}^{t+1},
		\frac{\tb{x}_{2,1}^{t+1} e^{3\eta-2 \eta \tb{x}_{1,1}^{t+1} - \eta  \tb{x}^{t}_{1,1}}}{\tb{x}_{2,1}^{t+1} e^{3 \eta  -2\eta\tb{x}_{1,1}^{t+1}- \eta  \tb{x}^{t}_{1,1}}+(1-\tb{x}_{2,1}^{t+1}) e^{2 \eta  \tb{x}_{1,1}^{t+1}+ \eta  \tb{x}^{t}_{1,1}}} \right).
	\end{align*}
	
	
	
	Combining our computation for $\tb{x}_{1,1}^{t+2}$ and $\tb{x}_{2,1}^{t+2}$ above, when $t$ is even
		\begin{align*}
		\CG_1\left((\tb{x}_{1,1}^t,\tb{x}_{1,1}^{t+1},\tb{x}_{2,1}^t,\tb{x}_{2,1}^{t+1})\right)=(\tb{x}_{1,1}^{t+1},\tb{x}_{1,1}^{t+2},\tb{x}_{2,1}^{t+1},\tb{x}_{2,1}^{t+2})
	\end{align*}
	and when $t$ is odd, we have
	\begin{align*}
		\CG_2\left((\tb{x}_{1,1}^t,\tb{x}_{1,1}^{t+1},\tb{x}_{2,1}^t,\tb{x}_{2,1}^{t+1})\right)
		=(\tb{x}_{1,1}^{t+1},\tb{x}_{1,1}^{t+2},\tb{x}_{2,1}^{t+1},\tb{x}_{2,1}^{t+2}).
	\end{align*}
	

	By the property of $\CG_1$ and $\CG_2$, it holds that
	\begin{align*}
		(\tb{x}_{1,1}^{2t-1},\tb{x}_{1,1}^{2t},\tb{x}_{2,1}^{2t-1},\tb{x}_{2,1}^{2t})=(\CG_1\circ \CG_2)^t\left((\tb{x}_{1,1}^{-1},\tb{x}_{1,1}^{0},\tb{x}_{2,1}^{-1},\tb{x}_{2,1}^{0})\right)
	\end{align*}
	
	and
	
	\begin{align*}
		(\tb{x}_{1,1}^{2t},\tb{x}_{1,1}^{2t+1},\tb{x}_{2,1}^{2t},\tb{x}_{2,1}^{2t+1})=\CG_2\circ(\CG_1\circ \CG_2)^t\left((\tb{x}_{1,1}^{-1},\tb{x}_{1,1}^{0},\tb{x}_{2,1}^{-1},\tb{x}_{2,1}^{0})\right).
	\end{align*}
The following proposition addresses the behavior of points in the neighborhood of equilibrium for OMWU in periodic game when comparing the convergence in the OMWU in static games.

\OMWUnotconverge*
\begin{proof}
    By computation\footnote{We utilize Matlab for computation.}, the eigenvalues of the Jacobi matrix of $\CG_1\circ \CG_2$ at equilibrium $ \left( 0.5,0.5,0.5,0.5\right) $ are
    \begin{itemize}
        \item $\frac{\eta^2}{2}-\frac{\sqrt{ (\eta^2+\eta+1)(\eta^2-\eta+1) }}{2}+\frac{1}{2}$,
        \item $\frac{\eta^2}{2}-\frac{\sqrt{ (\eta^2+\eta+1)(\eta^2-\eta+1) }}{2}+\frac{1}{2}$,
        \item $\frac{\eta^2}{2}+\frac{\sqrt{ (\eta^2+\eta+1)(\eta^2-\eta+1) }}{2}+\frac{1}{2}$,
        \item $\frac{\eta^2}{2}+\frac{\sqrt{ (\eta^2+\eta+1)(\eta^2-\eta+1) }}{2}+\frac{1}{2}$.
    \end{itemize}
    It can be computed that $\frac{\eta^2}{2}+\frac{\sqrt{ (\eta^2+\eta+1)(\eta^2-\eta+1) }}{2}+\frac{1}{2}>1$ and $\frac{\eta^2}{2}-\frac{\sqrt{ (\eta^2+\eta+1)(\eta^2-\eta+1) }}{2}+\frac{1}{2}<1$. The eigenvectors in the eigenspace corresponding to $\frac{\eta^2}{2}-\frac{\sqrt{ (\eta^2+\eta+1)(\eta^2-\eta+1) }}{2}+\frac{1}{2}$ can be verified to consistently have negative second elements. Thus, the decomposition of any vector in the simplex constrains must contains the eigenvectors corresponding to $\frac{\eta^2}{2}+\frac{\sqrt{ (\eta^2+\eta+1)(\eta^2-\eta+1) }}{2}+\frac{1}{2}$.
    Therefore, by Proposition~\ref{decomposition}, the iterations of these vectors will diverge from the equilibrium.
\end{proof}

Subsequently, we present compelling evidence demonstrating that the KL-divergence tends towards infinity. 
With the initial conditions as stated in Theorem \ref{thm: OMWU fails}, the proof is divided into two stages : 
\begin{itemize}	
		\item \textbf{Stage 1} : we will show that when the current position is far from the boundary of $[0,1]$, the KL-divergence will stably increase.
		Thus both of these variables approach to the boundary.
		Formally, our goal in this stage is to demonstrate the following proposition.
        \KLincreasing*
		\item \textbf{Stage 2} : 
  The convergence of either $x^t_{1}$ or $x^t_{2}$ to the boundary results in the divergence of at least one of them towards infinity, leading to an unbounded KL-divergence. This phenomenon is described by the following proposition.

  when either $x^t_{1}$ or $x^t_{2}$ close to boundary, at least one of them will converge to the boundary, which leads to the KL-divergence goes to infinity. This can be described by the following proposition.

		\OMWUlocallyconverging*
	\end{itemize}
	

 
\subsection{Proof of Stage 1}
In this subsection, we first illustrate that under a stronger conditions than those in Proposition \ref{prop: OMWU fails}, the second elements of the iterative vectors $\tb{x}_1$ and $\tb{x}_2$ increase with every two time iteration.
Following that, we prove that the strong condition can be generalized to the condition in Proposition \ref{prop: OMWU fails}. Before presenting the main lemma in this subsection, we first introduce some necessary transition expressions in the dynamic system.

 
	\begin{lem}\label{lem: fraction}
		If $ 2\mid t $, then 
		\begin{enumerate}
			\item \label{fraction 1} $\frac{\tb{x}^{t+1}_{1,1}}{\tb{x}^{t+1}_{1,2}}=\frac{\tb{x}^{t-1}_{1,1}}{\tb{x}^{t-1}_{1,2}}\cdot e^{-2\eta(2\tb{x}^{t}_{2,2}-\tb{x}^{t-1}_{2,2}-\tb{x}^{t-2}_{2,2})}$;
			\item \label{fraction 2} $\frac{\tb{x}^{t+1}_{2,1}}{\tb{x}^{t+1}_{2,2}}=\frac{\tb{x}^{t-1}_{2,1}}{\tb{x}^{t-1}_{2,2}}\cdot e^{2\eta(2\tb{x}^{t}_{1,2}-\tb{x}^{t-1}_{1,2}-\tb{x}^{t-2}_{1,2})}  $;
			\item \label{fraction 3} $ \frac{\tb{x}^{t+2}_{1,1}}{\tb{x}^{t+2}_{1,2}}=\frac{\tb{x}^{t}_{1,1}}{\tb{x}^{t}_{1,2}}\cdot e^{2\eta(2\tb{x}^{t+1}_{2,2}-\tb{x}^{t}_{2,2}-\tb{x}^{t-1}_{2,2})} $;
			\item \label{fraction 4} $\frac{\tb{x}^{t+2}_{2,1}}{\tb{x}^{t+2}_{2,2}}=\frac{\tb{x}^{t}_{2,1}}{\tb{x}^{t}_{2,2}}\cdot e^{-2\eta(2\tb{x}^{t+1}_{1,2}-\tb{x}^{t}_{1,2}-\tb{x}^{t-1}_{1,2})}  $;
			\item \label{fraction 5} $\frac{\tb{x}^{t+2}_{1,1}}{\tb{x}^{t+2}_{1,2}}=\frac{\tb{x}^{t+1}_{1,1}}{\tb{x}^{t+1}_{1,2}}\cdot e^{-3\eta+2\eta(2\tb{x}^{t+1}_{2,2}+\tb{x}^{t}_{2,2})}$;
			\item \label{fraction 6} $\frac{\tb{x}^{t+1}_{2,1}}{\tb{x}^{t+1}_{2,2}}=\frac{\tb{x}^{t}_{2,1}}{\tb{x}^{t}_{2,2}}\cdot e^{-3\eta+2\eta(2\tb{x}^{t}_{1,2}+\tb{x}^{t-1}_{1,2})}$;
		\end{enumerate}
	\end{lem}

\begin{proof}
		By the definition of dynamic system of OMWU, when $t$ is even, 
		\begin{align}
			\tb{x}_{1,1}^{t+1} = &
			\frac{\tb{x}_{1,1}^t e^{2 \eta (A_t \tb{x}_{2}^t)^1 - \eta (A_{t-1} \tb{x}_2^{t-1})^{1}}} { \sum^2_{s=1}\tb{x}_{1,s}^t e^{2 \eta (A_t \tb{x}_{2}^t)^s - \eta (A_{t-1} \tb{x}_2^{t-1})^s}}\nonumber\\
			=&\frac{\tb{x}_{1,1}^t e^{-2 \eta \tb{x}_{2,2}^t - \eta  \tb{x}^{t-1}_{2,2}}} { \tb{x}_{1,1}^t e^{-2 \eta \tb{x}_{2,2}^t - \eta  \tb{x}^{t-1}_{2,2}}
				+\tb{x}_{1,2}^t e^{-2 \eta \tb{x}_{2,1}^t - \eta  \tb{x}^{t-1}_{2,1}}}\label{x11t+1}
		\end{align}
		and
		\begin{align}
			\tb{x}_{1,2}^{t+1} = &
			\frac{\tb{x}_{1,2}^t e^{2 \eta (A_t \tb{x}_{2}^t)^2 - \eta (A_{t-1} \tb{x}_2^{t-1})^{2}}} { \sum^2_{s=1}\tb{x}_{1,s}^t e^{2 \eta (A_t \tb{x}_{2}^t)^s - \eta (A_{t-1} \tb{x}_2^{t-1})^s}}\nonumber\\
			=&\frac{\tb{x}_{1,2}^t e^{-2 \eta \tb{x}_{2,1}^t - \eta  \tb{x}^{t-1}_{2,1}}} { \tb{x}_{1,1}^t e^{-2 \eta \tb{x}_{2,2}^t - \eta  \tb{x}^{t-1}_{2,2}}
				+\tb{x}_{1,2}^t e^{-2 \eta \tb{x}_{2,1}^t - \eta  \tb{x}^{t-1}_{2,1}}}\nonumber\\
				=&\frac{\tb{x}_{1,2}^t e^{-3\eta+2 \eta \tb{x}_{2,2}^t + \eta  \tb{x}^{t-1}_{2,2}}} { \tb{x}_{1,1}^t e^{-2 \eta \tb{x}_{2,2}^t - \eta  \tb{x}^{t-1}_{2,2}}
					+\tb{x}_{1,2}^t e^{-2 \eta \tb{x}_{2,1}^t - \eta  \tb{x}^{t-1}_{2,1}}}\label{x12t+1}.
		\end{align}
		By the similar computation, we have
		\begin{align}
			\tb{x}_{1,1}^{t} = &\frac{\tb{x}_{1,1}^{t-1} e^{2 \eta \tb{x}_{2,2}^{t-1} + \eta  \tb{x}^{t-2}_{2,2}}} { \tb{x}_{1,1}^{t-1} e^{2 \eta \tb{x}_{2,2}^{t-1} + \eta  \tb{x}^{t-2}_{2,2}}
				+\tb{x}_{1,2}^{t-1} e^{3\eta -2 \eta \tb{x}_{2,2}^{t-1} - \eta  \tb{x}^{t-2}_{2,2}}}\label{x11t},\\
			\tb{x}_{1,2}^{t} = &\frac{\tb{x}_{1,2}^{t-1} e^{3\eta -2 \eta \tb{x}_{2,2}^{t-1} - \eta  \tb{x}^{t-2}_{2,2}}} { \tb{x}_{1,1}^{t-1} e^{2 \eta \tb{x}_{2,2}^{t-1} + \eta  \tb{x}^{t-2}_{2,2}}
				+\tb{x}_{1,2}^{t-1} e^{3\eta -2 \eta \tb{x}_{2,2}^{t-1} - \eta  \tb{x}^{t-2}_{2,2}}}\label{x12t}~,\\
			\tb{x}_{1,1}^{t+2} = &\frac{\tb{x}_{1,1}^{t+1} e^{2 \eta \tb{x}_{2,2}^{t+1} + \eta  \tb{x}^{t}_{2,2}}} { \tb{x}_{1,1}^{t+1} e^{2 \eta \tb{x}_{2,2}^{t+1} + \eta  \tb{x}^{t}_{2,2}}
				+\tb{x}_{1,2}^{t+1} e^{3\eta -2 \eta \tb{x}_{2,2}^{t+1} - \eta  \tb{x}^{t}_{2,2}}}\label{x11t+2}~,\\
			\tb{x}_{1,2}^{t+2} = &\frac{\tb{x}_{1,2}^{t+1} e^{3\eta -2 \eta \tb{x}_{2,2}^{t+1} - \eta  \tb{x}^{t}_{2,2}}} {\tb{x}_{1,1}^{t+1} e^{2 \eta \tb{x}_{2,2}^{t+1} + \eta  \tb{x}^{t}_{2,2}}
				+\tb{x}_{1,2}^{t+1} e^{3\eta -2 \eta \tb{x}_{2,2}^{t+1} - \eta  \tb{x}^{t}_{2,2}}}\label{x12t+2}~,\\
			\tb{x}_{2,1}^{t} = &\frac{\tb{x}_{2,1}^{t-1} e^{-2 \eta \tb{x}_{1,2}^{t-1} - \eta  \tb{x}^{t-2}_{1,2}}} { \tb{x}_{2,1}^{t-1} e^{-2 \eta \tb{x}_{1,2}^{t-1} - \eta  \tb{x}^{t-2}_{1,2}}
				+\tb{x}_{2,2}^{t-1} e^{-3\eta +2 \eta \tb{x}_{1,2}^{t-1} + \eta  \tb{x}^{t-2}_{1,2}}}\label{x21t}~,\\
			\tb{x}_{2,2}^{t} = &\frac{\tb{x}_{2,2}^{t-1} e^{-3\eta +2 \eta \tb{x}_{1,2}^{t-1} + \eta  \tb{x}^{t-2}_{1,2}}} { \tb{x}_{2,1}^{t-1} e^{-2 \eta \tb{x}_{1,2}^{t-1} - \eta  \tb{x}^{t-2}_{1,2}}
				+\tb{x}_{2,2}^{t-1} e^{-3\eta +2 \eta \tb{x}_{1,2}^{t-1} + \eta  \tb{x}^{t-2}_{1,2}}}\label{x22t}~,\\
			\tb{x}_{2,1}^{t+1} = &\frac{\tb{x}_{2,1}^{t} e^{2 \eta \tb{x}_{1,2}^{t} + \eta  \tb{x}^{t-1}_{1,2}}} { \tb{x}_{2,1}^{t} e^{2 \eta \tb{x}_{1,2}^{t} + \eta  \tb{x}^{t-1}_{1,2}}
				+\tb{x}_{2,2}^{t} e^{3\eta-2 \eta \tb{x}_{1,2}^{t} - \eta  \tb{x}^{t-1}_{1,2}}}\label{x21t+1}~,\\
			\tb{x}_{2,2}^{t+1} = &\frac{\tb{x}_{2,2}^{t} e^{3\eta-2 \eta \tb{x}_{1,2}^{t} - \eta  \tb{x}^{t-1}_{1,2}}} { \tb{x}_{2,1}^{t} e^{2 \eta \tb{x}_{1,2}^{t} + \eta  \tb{x}^{t-1}_{1,2}}
				+\tb{x}_{2,2}^{t} e^{3\eta-2 \eta \tb{x}_{1,2}^{t} - \eta  \tb{x}^{t-1}_{1,2}}}\label{x22t+1}~,\\
				\tb{x}_{2,1}^{t+2} = &\frac{\tb{x}_{2,1}^{t+1} e^{-2 \eta \tb{x}_{1,2}^{t+1} - \eta  \tb{x}^{t}_{1,2}}} { \tb{x}_{2,1}^{t+1} e^{-2 \eta \tb{x}_{1,2}^{t+1} - \eta  \tb{x}^{t}_{1,2}}
					+\tb{x}_{2,2}^{t+1} e^{-3\eta+2 \eta \tb{x}_{1,2}^{t+1} + \eta  \tb{x}^{t}_{1,2}}}\label{x21t+2}~,\\
				\tb{x}_{2,2}^{t+2} = &\frac{\tb{x}_{2,2}^{t+1} e^{-3\eta+2 \eta \tb{x}_{1,2}^{t+1} + \eta  \tb{x}^{t}_{1,2}}} { \tb{x}_{2,1}^{t+1} e^{-2 \eta \tb{x}_{1,2}^{t+1} - \eta  \tb{x}^{t}_{1,2}}
					+\tb{x}_{2,2}^{t+1} e^{-3\eta+2 \eta \tb{x}_{1,2}^{t+1} + \eta  \tb{x}^{t}_{1,2}}}\label{x22t+2}.
		\end{align}
		Then we have
		\begin{align}
			&\frac{\tb{x}_{1,1}^{t+1}}{\tb{x}_{1,2}^{t+1}}=\frac{\tb{x}_{1,1}^{t}}{\tb{x}_{1,2}^{t}}\cdot e^{3\eta-2\eta(2 \tb{x}_{2,2}^t +\tb{x}^{t-1}_{2,2})}\label{r: 1 t+1},\\
			&\frac{\tb{x}_{1,1}^{t}}{\tb{x}_{1,2}^{t}}=\frac{\tb{x}_{1,1}^{t-1}}{\tb{x}_{1,2}^{t-1}}\cdot e^{-3\eta+2\eta( 2\tb{x}_{2,2}^{t-1} +  \tb{x}^{t-2}_{2,2})}\label{r: 1 t},\\
			&\frac{\tb{x}_{1,1}^{t+2}}{\tb{x}_{1,2}^{t+2}}=\frac{\tb{x}_{1,1}^{t+1}}{\tb{x}_{1,2}^{t+1}}\cdot e^{-3\eta+2\eta( 2\tb{x}_{2,2}^{t+1} +  \tb{x}^{t}_{2,2})}\label{r: 1 t+2},\\
			&\frac{\tb{x}_{2,1}^{t}}{\tb{x}_{2,2}^{t}}=\frac{\tb{x}_{2,1}^{t-1}}{\tb{x}_{2,2}^{t-1}}\cdot e^{3\eta-2\eta( 2\tb{x}_{1,2}^{t-1} +  \tb{x}^{t-2}_{1,2})}\label{r: 2 t},\\
			&\frac{\tb{x}_{2,1}^{t+1}}{\tb{x}_{2,2}^{t+1}}=\frac{\tb{x}_{2,1}^{t}}{\tb{x}_{2,2}^{t}}\cdot e^{-3\eta+2\eta( 2\tb{x}_{1,2}^{t} +  \tb{x}^{t-1}_{1,2})}\label{r: 2 t+1},\\
			&\frac{\tb{x}_{2,1}^{t+2}}{\tb{x}_{2,2}^{t+2}}=\frac{\tb{x}_{2,1}^{t+1}}{\tb{x}_{2,2}^{t+1}}\cdot e^{3\eta-2\eta( 2\tb{x}_{1,2}^{t+1} +  \tb{x}^{t}_{1,2})}\label{r: 2 t+2}.		
		\end{align}
		Equation~\eqref{r: 1 t+1} follows from the ratio of equation~\eqref{x11t+1} to equation~\eqref{x12t+1}.
		
		Equation~\eqref{r: 1 t} follows from the ratio of equaiton~\eqref{x11t} to equation~\eqref{x12t}.

		Equation~\eqref{r: 1 t+2} follows from the ratio of equaiton~\eqref{x11t+2} to equation~\eqref{x12t+2}, implying item \ref{fraction 5} in the lemma.
		
		Equation~\eqref{r: 2 t}  follows from the ratio of equaiton~\eqref{x21t} to equation~\eqref{x22t}.
		
		Equation~\eqref{r: 2 t+1}  follows from the ratio of equaiton~\eqref{x21t+1} to equation~\eqref{x22t+1}, implying item \ref{fraction 6} in the lemma.
		
		Equation~\eqref{r: 2 t+2}  follows from the ratio of equaiton~\eqref{x21t+2} to equation~\eqref{x22t+2}.
		
		It holds that
		\begin{align*}
			&\frac{\tb{x}_{1,1}^{t+1}}{\tb{x}_{1,2}^{t+1}}=\frac{\tb{x}_{1,1}^{t-1}}{\tb{x}_{1,2}^{t-1}}\cdot e^{-2\eta(2 \tb{x}_{2,2}^t -\tb{x}^{t-1}_{2,2}-\tb{x}^{t-2}_{2,2})},\\
			&\frac{\tb{x}_{1,1}^{t+2}}{\tb{x}_{1,2}^{t+2}}=\frac{\tb{x}_{1,1}^{t}}{\tb{x}_{1,2}^{t}}\cdot e^{2\eta( 2\tb{x}_{2,2}^{t+1} -  \tb{x}^{t}_{2,2}-\tb{x}^{t-1}_{2,2})},\\
			&\frac{\tb{x}_{2,1}^{t+1}}{\tb{x}_{2,2}^{t+1}}=\frac{\tb{x}_{2,1}^{t-1}}{\tb{x}_{2,2}^{t-1}}\cdot e^{2\eta( 2\tb{x}_{1,2}^{t} -  \tb{x}^{t-1}_{1,2}-  \tb{x}^{t-2}_{1,2})},\\
			&\frac{\tb{x}_{2,1}^{t+2}}{\tb{x}_{2,2}^{t+2}}=\frac{\tb{x}_{2,1}^{t}}{\tb{x}_{2,2}^{t}}\cdot e^{-2\eta( 2\tb{x}_{1,2}^{t+1} -  \tb{x}^{t}_{1,2}-  \tb{x}^{t-1}_{1,2})}.
		\end{align*}
		The first equation comes from the combination of equation~\ref{r: 1 t+1} and equation~\ref{r: 1 t}, implying item \ref{fraction 1} in the lemma.
		
		The second equation comes from the combination of equation~\ref{r: 1 t+1} and equation~\ref{r: 1 t+2}, implying item \ref{fraction 3} in the lemma.
		
		The third equation comes from the combination of equation~\ref{r: 2 t} and equation~\ref{r: 2 t+1}, implying item \ref{fraction 2} in the lemma.
		
		The fourth equation comes from the combination of equation~\ref{r: 2 t+1} and equation~\ref{r: 2 t+2}, implying item \ref{fraction 4} in the lemma.
		
	\end{proof}


 
	The following lemma is a simple but useful tool in our proof.
	\begin{lem}\label{lem: ineq}
		Let $ u,v\in [\frac{1}{2},1-\eta^\frac{1}{2}] $, $ 0\le w\le 1 $, if
		\begin{align}
			\frac{1-v}{v}\le \frac{1-u}{u}\cdot e^{w},
		\end{align}
		then we have $ u-v\le w $. If 
		\begin{align}
			\frac{1-v}{v}\ge \frac{1-u}{u}\cdot e^{w},
		\end{align}
		then we have $ u-v\ge \frac{1}{2}w \eta^\frac{1}{2} $.
	\end{lem}
	\begin{proof}
		First, we consider the case when $ \frac{1-v}{v}\le \frac{1-u}{u}\cdot e^{w} $, 
		\begin{align*}
			\frac{1-v}{v}&\le \frac{1-u}{u}\cdot e^{w}\\
			&\le \frac{1-u}{u}\cdot (1+2w),
		\end{align*}
		where the inequality comes from $ e^w\le 1+2w $  when $ 0<w\le 1 $.
		This simplifies to:
		\begin{align*}
			u-v&\le 2v(1-u)w\\
			&\le 2\cdot 1\cdot \frac{1}{2} \cdot w =w,
		\end{align*}
		where the second line arises from $ v\le 1 $ and $ u\ge \frac{1}{2} $.
		
		Then we consider the second part of the lemma,
		\begin{align*}
			\frac{1-v}{v}&\ge \frac{1-u}{u}\cdot e^{w}\\
			&\ge \frac{1-u}{u}\cdot (1+w),
		\end{align*}
		where the inequality comes from $ e^w\ge 1+w $.
		This simplifies to:
		\begin{align*}
			u-v&\ge v(1-u)w\\
			&\ge \frac{1}{2}w\eta^\frac{1}{2},
		\end{align*}
		where the second inequality follows from $ v\ge \frac{1}{2} $ and $ u\le 1-\eta^\frac{1}{2} $.
	\end{proof}
The following lemma shows the effect of initial conditions on $\textbf{x}_{1,2}^1 $ and $\textbf{x}_{2,2}^1$.
 	\begin{lem}\label{lem: initialx1}
	    For the dynamic system in Theorem~\ref{thm: OMWU fails}, when $\textbf{x}_{1,2}^0,\textbf{x}_{2,2}^0>\frac{1}{2}+2p $ for $p\in(0,\frac{1}{4}) $, and $\eta$ is sufficiently small such that $p\ge 16\eta^\frac{1}{2}$, then for any possible $\textbf{x}_{1,1}^{-1},\textbf{x}_{1,2}^{-1}$, we have
     \begin{align*}
     \textbf{x}_{1,2}^1,\textbf{x}_{2,2}^1>\frac{1}{2}+p.
     \end{align*}
	\end{lem}
\begin{proof}
    By equality~\eqref{x11t+1} with $t=0$, it holds that
    \begin{align*}
        \tb{x}_{1,2}^{1} &= 
			\frac{\tb{x}_{1,2}^0 e^{-2 \eta \tb{x}_{2,1}^0 - \eta  \tb{x}^{-1}_{2,1}}}{ \tb{x}_{1,1}^0 e^{-2 \eta \tb{x}_{2,2}^0 - \eta  \tb{x}^{-1}_{2,2}}
				+\tb{x}_{1,2}^0 e^{-2 \eta \tb{x}_{2,1}^0 - \eta  \tb{x}^{-1}_{2,1}} }\\
    &\ge 
			\frac{\tb{x}_{1,2}^0 e^{-3 \eta}}{ \tb{x}_{1,1}^0 
				+\tb{x}_{1,2}^0 }  \\
    &= \tb{x}_{1,2}^0 e^{-3 \eta}\ge (\frac{1}{2}+2p)\cdot (1-3\eta)\ge \frac{1}{2}+p.
    \end{align*}
    The last inequality comes from $p\ge 16\eta^\frac{1}{2}$. Applying the same method, we can also conclude that $\tb{x}_{2,2}^{1} \ge \frac{1}{2}+p$.
\end{proof}
	Next, we give the key lemma for the proof of Proposition~\ref{prop: OMWU fails}. Note that in comparison to  Theorem~\ref{thm: OMWU fails}, the following lemma asks a stronger condition: the initial points $\textbf{x}_{1,2}^0$ and $\textbf{x}_{2,2}^0$ are larger than $\frac{1}{2}$. Later, we will demonstrate how this condition can be relaxed.

%\ping{to relax the condition $x_{1,2}^0=x_{1,2}^1$, the initial point need start from $x_{1,2}^{-1}$ and $x_{1,2}^0$. the initial condition in other place should be updated}



 
	\begin{lem}\label{lem: close to boundary}
		For the same periodic game as Theorem~\ref{thm: OMWU fails} defined by payoff matrices 
		\begin{align}\label{2-periodic game}
			A_t=
			\begin{cases}
				\begin{bmatrix}
					0 & & 1 \\
					1 & & 0
				\end{bmatrix}, & t \textnormal{\ \ is \ odd} \\ 
				\\
				\begin{bmatrix}
					0 & -1 \\
					-1 & 0
				\end{bmatrix}, & t\textnormal{\ \ is \ even}
			\end{cases}
		\end{align}
		Let $0<p<\frac{1}{4}$, and $\eta$ sufficiently small so that $ p\ge 16\eta^\frac{1}{2} $. And the initial points satisfy the condition
  \begin{align*}
      \textbf{x}_{1,2}^{0},\textbf{x}_{2,2}^{0}\ge \frac{1}{2}+2p.
  \end{align*}
  For $t$ is even and $t\ge 4$, if for any $k<t$, $\tb{x}_{1,2}^k, \tb{x}_{2,2}^k \le 1-\sqrt{\eta}$, it holds that
		\begin{enumerate}
			\item\label{result 1} $ \frac{3}{4}p\eta^3\le \tb{x}^{t+1}_{2,2}- \tb{x}^{t-1}_{2,2}\le 12\eta^2 $,	
			\item $ \frac{3}{2}p\eta^\frac{3}{2} \le \tb{x}^{t}_{2,2}-\tb{x}^{t+1}_{2,2}\le 3\eta $,
			\item\label{result 3} $ \frac{3}{4}p\eta^3\le \tb{x}^{t+2}_{1,2}-\tb{x}^{t}_{1,2} \le 12\eta^2 $,
			\item\label{result 4} $  \frac{3}{4}p\eta^3\le \tb{x}^{t+1}_{1,2}-\tb{x}^{t-1}_{1,2} $,
			\item\label{result 5} $  \frac{3}{4}p\eta^3\le \tb{x}^{t+2}_{2,2}-\tb{x}^{t}_{2,2} $,		
			
	        \item $ \frac{3}{2}p\eta^\frac{3}{2} \le \tb{x}^{t+1}_{1,2}-\tb{x}^{t+2}_{1,2}\le 3\eta $.
	\end{enumerate}
	\end{lem}
	\begin{proof}
By Lemma~\ref{lem: initialx1}, we obtain
\begin{align}\label{ineq: x1222}
         \textbf{x}_{1,2}^1,\textbf{x}_{2,2}^1>\frac{1}{2}+p.
\end{align}
		We will do induction on $ t $.\\
  \\
		\textbf{Base case:}\\
		In this part, our goal is to prove the following:
		\begin{enumerate}
			\item \label{base case 1}$ \frac{3}{4}p\eta^3\le \tb{x}^{5}_{2,2}- \tb{x}^{3}_{2,2}\le 12\eta^2 $,
			\item \label{base case 2}$ \frac{3}{2}p\eta^\frac{3}{2} \le \tb{x}^{4}_{2,2}-\tb{x}^{5}_{2,2}\le 3\eta $,
			\item \label{base case 3}$ \frac{3}{4}p\eta^3\le \tb{x}^{6}_{1,2}-\tb{x}^{4}_{1,2} \le 12\eta^2 $,
			
			
			\item \label{base case 4}$  \frac{3}{4}p\eta^3\le \tb{x}^{5}_{1,2}-\tb{x}^{3}_{1,2} $,
			\item \label{base case 5}$  \frac{3}{4}p\eta^3 \le \tb{x}^{6}_{2,2}-\tb{x}^{4}_{2,2} $,			
			
			\item \label{base case 6}$ \frac{3}{2}p\eta^\frac{3}{2} \le \tb{x}^{5}_{1,2}-\tb{x}^{6}_{1,2}\le 3\eta $.
		\end{enumerate}
		We will prove above in accordance with the sequence of the given order.
		
		We will begin by providing an estimate for $ \tb{x}^{2}_{1,2}$.
		From item~\ref{fraction 5} of lemma~\ref{lem: fraction} with $ t=0 $, we obtain
		\begin{align*}
			\frac{\tb{x}^{2}_{1,1}}{\tb{x}^{2}_{1,2}}=\frac{\tb{x}^{1}_{1,1}}{\tb{x}^{1}_{1,2}}\cdot e^{-3\eta+2\eta(2\tb{x}^{1}_{2,2}+\tb{x}^{0}_{2,2})}.
		\end{align*}
		From $ \tb{x}^{0}_{2,2},\tb{x}^{1}_{2,2}\in [\frac{1}{2}+p,1] $, it holds that
		\begin{align*}
			\frac{\tb{x}^{1}_{1,1}}{\tb{x}^{1}_{1,2}}\cdot e^{6p\eta}
			\le \frac{\tb{x}^{2}_{1,1}}{\tb{x}^{2}_{1,2}}
			\le\frac{\tb{x}^{1}_{1,1}}{\tb{x}^{1}_{1,2}}\cdot e^{3\eta}.
		\end{align*}
		Because $ \tb{x}^{1}_{1,1}+\tb{x}^{1}_{1,2}=1 $, $ \tb{x}^{2}_{1,1}+ \tb{x}^{2}_{1,2}=1$, combining with lemma~\ref{lem: ineq},
		\begin{align*}
			3p\eta^\frac{3}{2}\le \tb{x}^{1}_{1,2}-\tb{x}^{2}_{1,2}\le 3\eta,
		\end{align*}
		which leads to
		\begin{align}\label{ineq: x122}
			\tb{x}^{2}_{1,2}\ge \tb{x}^{1}_{1,2}-3\eta\ge \frac{1}{2}+p-3\eta \ge \frac{1}{2}+\frac{1}{2}p,
		\end{align}
		where the frist inequality follows from $\textbf{x}_{1,2}^1\in [\frac{1}{2}+p,1]$, and the last inequality arises from $p\ge 16\eta^\frac{1}{2}$.
		
		  Then we provide the estimate for $\textbf{x}_{2,2}^1-\textbf{x}_{2,2}^3$, by item \ref{fraction 2} in Lemma~\ref{lem: fraction}, 
		\begin{align*}
			\frac{\textbf{x}_{2,1}^3}{\textbf{x}_{2,2}^3}=\frac{\textbf{x}_{2,1}^1}{\textbf{x}_{2,2}^1}\cdot e^{2\eta(2\textbf{x}_{1,2}^2-\textbf{x}_{1,2}^1-\textbf{x}_{1,2}^0)}.
		\end{align*}
		From inequality~\eqref{ineq: x122}, we have $\tb{x}^{2}_{1,2}\ge \frac{1}{2}$, thus together with $\textbf{x}_{1,2}^1$, $\textbf{x}_{1,2}^0\in[\frac{1}{2}+p,1]$,
		\begin{align*}
			-1\le 2\textbf{x}_{1,2}^2-\textbf{x}_{1,2}^1-\textbf{x}_{1,2}^0\le 1.
		\end{align*}
		Combining with Lemma~\ref{lem: ineq}, we have
		\begin{align}\label{ineq: x2321}
		-2\eta\le	\textbf{x}_{2,2}^1-\textbf{x}_{2,2}^3\le 2\eta.
		\end{align}
		
		
		Then we show the estimate for $\textbf{x}_{2,2}^3-\textbf{x}_{2,2}^2$, by item \ref{fraction 6} in Lemma~\ref{lem: fraction}, we have
		\begin{align*}
			\frac{\textbf{x}_{2,1}^3}{\textbf{x}_{2,2}^3}=\frac{\textbf{x}_{2,1}^2}{\textbf{x}_{2,2}^2}\cdot e^{-3\eta +2\eta(2\textbf{x}_{1,2}^2+\textbf{x}_{1,2}^1)}.
		\end{align*}
		From inequality~\eqref{ineq: x122}, we have
		\begin{align*}
		\frac{\textbf{x}_{2,1}^2}{\textbf{x}_{2,2}^1}\cdot e^{4p\eta}\le	\frac{\textbf{x}_{2,1}^3}{\textbf{x}_{2,2}^3}\le \frac{\textbf{x}_{2,1}^2}{\textbf{x}_{2,2}^1}\cdot e^{3\eta}.
		\end{align*}
	According to Lemma~\ref{lem: ineq}, we obtain
	\begin{align}\label{ineq: x2322}
		2p\eta^\frac{3}{2}\le \textbf{x}_{2,2}^3-\textbf{x}_{2,2}^2\le 3\eta.
	\end{align}

	
		
		Now we provide the estimate for $\textbf{x}_{1,2}^4-\textbf{x}_{1,2}^2$. From item \ref{fraction 3} in Lemma~\ref{lem: fraction}, 
		\begin{align*}
			\frac{\textbf{x}_{1,1}^4}{\textbf{x}_{1,2}^4}=\frac{\textbf{x}_{1,1}^2}{\textbf{x}_{1,2}^2}\cdot e^{2\eta(2\textbf{x}_{2,2}^3-\textbf{x}_{2,2}^2-\textbf{x}_{2,2}^1)}.
		\end{align*}
			From inequalities~\eqref{ineq: x2321} and \eqref{ineq: x2322}, it holds that
		\begin{align*}
			 \frac{\textbf{x}_{1,1}^2}{\textbf{x}_{1,2}^2}\cdot e^{-2\eta^2}\le \frac{\textbf{x}_{1,1}^4}{\textbf{x}_{1,2}^4}\le \frac{\textbf{x}_{1,1}^2}{\textbf{x}_{1,2}^2}\cdot e^{8\eta^2}.
		\end{align*}
		By Lemma~\ref{lem: ineq}, we have
		\begin{align}\label{ineq: x1412}
			-8\eta^2\le \textbf{x}_{1,2}^4-\textbf{x}_{1,2}^2\le 2\eta^2.
		\end{align}
		
		Next we provide the estimate of $\textbf{x}_{1,2}^3-\textbf{x}_{1,2}^4$, by item \ref{fraction 5} in Lemma~\ref{lem: fraction},
		\begin{align*}
			\frac{\textbf{x}_{1,1}^4}{\textbf{x}_{1,2}^4}
			=\frac{\textbf{x}_{1,1}^3}{\textbf{x}_{1,2}^3}\cdot e^{-3\eta+2\eta(2\textbf{x}_{2,2}^3+\textbf{x}_{2,2}^2)}.
		\end{align*}
		According to the estimate of $\textbf{x}_{2,2}^2$ and $\textbf{x}_{2,2}^3$ in the inequalities~\eqref{ineq: x2321} and \eqref{ineq: x2322}, it holds that
		\begin{align*}
			\frac{\textbf{x}_{1,1}^3}{\textbf{x}_{1,2}^3}\cdot e^{3p\eta}\le 
			\frac{\textbf{x}_{1,1}^4}{\textbf{x}_{1,2}^4}
			\le\frac{\textbf{x}_{1,1}^3}{\textbf{x}_{1,2}^3}\cdot e^{3\eta}.
		\end{align*}
		Using Lemma~\ref{lem: ineq}, we have
		\begin{align}\label{ineq: x1314}
			\frac{3}{2}p\eta^\frac{3}{2} \le \textbf{x}_{1,2}^3-\textbf{x}_{1,2}^4\le 3\eta.
		\end{align}
		
		
		\textit{proof of item \ref{base case 1}.}\\
		According to item \ref{fraction 2} lemma~\ref{lem: fraction} with $ t=4 $, we have
		\begin{align*}
			\frac{\tb{x}^{5}_{2,1}}{\tb{x}^{5}_{2,2}}&=\frac{\tb{x}^{3}_{2,1}}{\tb{x}^{3}_{2,2}}\cdot e^{2\eta(2\tb{x}^{4}_{1,2}-\tb{x}^{3}_{1,2}-\tb{x}^{2}_{1,2})}.
		\end{align*}
		Combining with the inequalities~(\ref{ineq: x1412}) and \eqref{ineq: x1314}, it holds that
		\begin{align*}
			\frac{\tb{x}^{5}_{2,1}}{\tb{x}^{3}_{2,2}}\cdot e^{\frac{3}{2}p\eta^\frac{5}{2}}\le \frac{\tb{x}^{3}_{2,1}}{\tb{x}^{3}_{2,2}}\le \frac{\tb{x}^{5}_{2,1}}{\tb{x}^{5}_{2,2}}\cdot e^{12\eta^2}.
		\end{align*}
		By lemma~\ref{lem: ineq}, we have
		\begin{align}\label{ineq: y3y1}
			\frac{3}{4}p\eta^3 \le \tb{x}^{5}_{2,2}-\tb{x}^{3}_{2,2}\le 12\eta^2.		
		\end{align}
		
		
		\textit{proof of item \ref{base case 2}.}\\
		From item \ref{fraction 6} in lemma~\ref{lem: fraction} with $ t=4 $,
		\begin{align*}
			\frac{\tb{x}^{5}_{2,1}}{\tb{x}^{5}_{2,2}}=\frac{\tb{x}^{4}_{2,1}}{\tb{x}^{4}_{2,2}}\cdot e^{-3\eta+2\eta(2\tb{x}^{4}_{1,2}+\tb{x}^{3}_{1,2})}.
		\end{align*}
		Using inequalities~(\ref{ineq: x1412}) and \eqref{ineq: x1314},
		\begin{align*}
			\frac{\tb{x}^{4}_{2,1}}{\tb{x}^{4}_{2,2}}\cdot e^{3p\eta}\le \frac{\tb{x}^{5}_{2,1}}{\tb{x}^{5}_{2,2}}\le \frac{\tb{x}^{4}_{2,1}}{\tb{x}^{4}_{2,2}}\cdot e^{3\eta}.
		\end{align*}
		Then we can use lemma~\ref{lem: ineq}, and obtain
		\begin{align}\label{ineq: y2y3}
			\frac{3}{2}p\eta^\frac{3}{2}\le \tb{x}^{4}_{2,2}-\tb{x}^{5}_{2,2}\le 3\eta.
		\end{align}
		
		
		
		\textit{proof of item \ref{base case 3}.}\\
		From item \ref{fraction 3} in lemma~\ref{lem: fraction} with $ t=4 $,
		\begin{align*}
			\frac{\tb{x}^{6}_{1,1}}{\tb{x}^{6}_{1,2}}=\frac{\tb{x}^{4}_{1,1}}{\tb{x}^{4}_{1,2}}\cdot e^{2\eta(2\tb{x}^{5}_{2,2}-\tb{x}^{4}_{2,2}-\tb{x}^{3}_{2,2})} .
		\end{align*}
		According to inequalities (\ref{ineq: y3y1}) and (\ref{ineq: y2y3}), we have
		\begin{align*}
			-3\eta \le -(\tb{x}^{4}_{2,2}-\tb{x}^{5}_{2,2})+(\tb{x}^{5}_{2,2}-\tb{x}^{3}_{2,2})
			\le -\frac{3}{2}p\eta^\frac{3}{2}+12\eta^2,
		\end{align*}
		which is equivalent to
		\begin{align*}
			-3\eta \le -(\tb{x}^{4}_{2,2}-\tb{x}^{5}_{2,2})+(\tb{x}^{5}_{2,2}-\tb{x}^{3}_{2,2})\le -\frac{3}{4}p\eta^\frac{3}{2},
		\end{align*} 
		where the right-side inequality follows from $ p\ge 16\eta^\frac{1}{2} $. Then
		\begin{align*}
			\frac{\tb{x}^{6}_{1,1}}{\tb{x}^{6}_{1,2}}\cdot e^{2\eta\cdot \frac{3}{4}p\eta^\frac{3}{2}} \le \frac{\tb{x}^{4}_{1,1}}{\tb{x}^{4}_{1,2}}\le \frac{\tb{x}^{6}_{1,1}}{\tb{x}^{6}_{1,2}}\cdot e^{2\eta\cdot 3\eta}, \\
			\frac{\tb{x}^{6}_{1,1}}{\tb{x}^{6}_{1,2}}\cdot e^{ \frac{3}{2}p\eta^\frac{5}{2}} \le \frac{\tb{x}^{4}_{1,1}}{\tb{x}^{4}_{1,2}}\le \frac{\tb{x}^{6}_{1,1}}{\tb{x}^{6}_{1,2}}\cdot e^{6\eta^2}.
		\end{align*}
		By lemma~\ref{lem: ineq}, it can be concluded that
		\begin{align*}
			\frac{3}{4}p\eta^3 \le \tb{x}^{6}_{1,2}-\tb{x}^{4}_{1,2}\le 6\eta^2\le 12\eta^2.
		\end{align*}
		\textit{proof of item \ref{base case 4}.}\\
		By (\ref{ineq: y3y1}), (\ref{ineq: y2y3}), \eqref{ineq: x2321} and \eqref{ineq: x2322}, we have
		\begin{align*}
			\tb{x}^{4}_{2,2}-\frac{3}{2}p\eta^\frac{3}{2}\ge \tb{x}^{5}_{2,2}\ge \tb{x}^{3}_{2,2}\ge \textbf{x}_{2,2}^2.
		\end{align*}
		From item \ref{fraction 1} in lemma~\ref{lem: fraction},
		\begin{align*}
			\frac{\tb{x}^{5}_{1,1}}{\tb{x}^{5}_{1,2}}=\frac{\tb{x}^{3}_{1,1}}{\tb{x}^{3}_{1,2}}\cdot e^{-2\eta(2\tb{x}^{4}_{2,2}-\tb{x}^{3}_{2,2}-\tb{x}^{2}_{2,2})}.
		\end{align*}
		Therefore, we obtain
        \begin{align*}
            \frac{\tb{x}^{5}_{1,1}}{\tb{x}^{5}_{1,2}}\cdot e^{6p\eta^\frac{5}{2}}\le\frac{\tb{x}^{3}_{1,1}}{\tb{x}^{3}_{1,2}}.
        \end{align*}
  By Lemma~\ref{lem: ineq},
		\begin{align}\label{ineq: x3x1}
			\tb{x}^{5}_{1,2}- \tb{x}^{3}_{1,2}\ge 3p\eta^3.
		\end{align} 
		\textit{proof of item \ref{base case 5}.}\\
		By (\ref{ineq: x1314}) and (\ref{ineq: x3x1}),
		\begin{align*}
			\tb{x}^{5}_{1,2}\ge \tb{x}^{3}_{1,2}\ge \tb{x}^{4}_{1,2}+\frac{3}{2}p\eta^\frac{3}{2}.
		\end{align*}
		From item \ref{fraction 4} in lemma \ref{lem: fraction} with $ t=4 $,
		\begin{align*}
			\frac{\tb{x}^{6}_{2,1}}{\tb{x}^{6}_{2,2}}=\frac{\tb{x}^{4}_{2,1}}{\tb{x}^{4}_{2,2}}\cdot e^{-2\eta(2\tb{x}^{5}_{1,2}-\tb{x}^{4}_{1,2}-\tb{x}^{3}_{1,2})}.
		\end{align*}
		Then it holds that
        \begin{align*}
            \frac{\tb{x}^{6}_{2,1}}{\tb{x}^{6}_{2,2}}\cdot e^{3p\eta^\frac{5}{2}} \le \frac{\tb{x}^{4}_{2,1}}{\tb{x}^{4}_{2,2}}.
        \end{align*}
        Combining with Lemma~\ref{lem: ineq}, it holds that
		\begin{align*}
			\tb{x}^{6}_{2,2}-\tb{x}^{4}_{2,2}\ge \frac{3}{2}p\eta^3.
		\end{align*}
		\textit{proof of item \ref{base case 6}.}\\
		By (\ref{ineq: y3y1}), (\ref{ineq: y2y3}) and \eqref{ineq: x2321}, 
		\begin{align*}
			\tb{x}^{4}_{2,2}\ge \tb{x}^{5}_{2,2}\ge \tb{x}^{3}_{2,2}\ge \frac{1}{2}+\frac{1}{2}p.
		\end{align*}
		According to item \ref{fraction 5} of lemma \ref{lem: fraction},
		\begin{align*}
			\frac{\tb{x}^{6}_{1,1}}{\tb{x}^{6}_{1,2}}=\frac{\tb{x}^{5}_{1,1}}{\tb{x}^{5}_{1,2}}\cdot e^{-3\eta+2\eta(2\tb{x}^{5}_{2,2}+\tb{x}^{4}_{2,2})},
		\end{align*}
		which leads to
		\begin{align*}
			\frac{\tb{x}^{5}_{1,1}}{\tb{x}^{5}_{1,2}}\cdot e^{3p\eta}\le \frac{\tb{x}^{6}_{1,1}}{\tb{x}^{6}_{1,2}}\le \frac{\tb{x}^{5}_{1,1}}{\tb{x}^{5}_{1,2}}\cdot e^{3\eta}.
		\end{align*}
		Combining with lemma~\ref{lem: ineq}, we have
		\begin{align*}
			\frac{3}{2}p\eta^\frac{3}{2}\le \tb{x}^{5}_{1,2}-\tb{x}^{6}_{1,2}\le 3\eta.	
		\end{align*}


\textbf{Induction step:}


		Now assume that the result in the lemma holds for $ t=k $, 
		\begin{enumerate}
			\item \label{info 1}$ \frac{3}{4}p\eta^3\le \tb{x}^{k+1}_{2,2}- \tb{x}^{k-1}_{2,2}\le 12\eta^2 $,	
			\item \label{info 2}$ \frac{3}{2}p\eta^\frac{3}{2} \le \tb{x}^{k}_{2,2}-\tb{x}^{k+1}_{2,2}\le 3\eta $,
			\item \label{info 3}$ \frac{3}{4}p\eta^3\le \tb{x}^{k+2}_{1,2}-\tb{x}^{k}_{1,2} \le 12\eta^2 $,
			
			
			\item \label{info 4}$  \frac{3}{4}p\eta^3\le \tb{x}^{k+1}_{1,2}-\tb{x}^{k-1}_{1,2} $,
			\item \label{info 5}$  \frac{3}{4}p\eta^3\le \tb{x}^{k+2}_{2,2}-\tb{x}^{k}_{2,2} $,		
			
			\item \label{info 6}$ \frac{3}{2}p\eta^\frac{3}{2} \le \tb{x}^{k+1}_{1,2}-\tb{x}^{k+2}_{1,2}\le 3\eta $.
		\end{enumerate}	
		Then we prove the case of $ t=k+2 $, just as follows,
		\begin{enumerate}
			\item \label{induction 1} $ \frac{3}{4}p\eta^3\le \tb{x}^{k+3}_{2,2}- \tb{x}^{k+1}_{2,2}\le 12\eta^2 $,	
			\item \label{induction 2}$ \frac{3}{2}p\eta^\frac{3}{2} \le \tb{x}^{k+2}_{2,2}-\tb{x}^{k+3}_{2,2}\le 3\eta $,
			\item \label{induction 3}$ \frac{3}{4}p\eta^3\le \tb{x}^{k+4}_{1,2}-\tb{x}^{k+2}_{1,2} \le 12\eta^2 $,
			
			
			\item \label{induction 4}$  \frac{3}{4}p\eta^3\le \tb{x}^{k+3}_{1,2}-\tb{x}^{k+1}_{1,2} $,
			\item \label{induction 5}$  \frac{3}{4}p\eta^3\le \tb{x}^{k+4}_{2,2}-\tb{x}^{k+2}_{2,2} $,			
			
			\item \label{induction 6}$ \frac{3}{2}p\eta^\frac{3}{2} \le \tb{x}^{k+3}_{1,2}-\tb{x}^{k+4}_{1,2}\le 3\eta $.
		\end{enumerate}
		\textit{proof of item \ref{induction 1}.}\\
		According to item \ref{fraction 2} in lemma~\ref{lem: fraction} with $ t\rightarrow k+2 $, we have
		\begin{align*}
			\frac{\tb{x}^{k+3}_{2,1}}{\tb{x}^{k+3}_{2,2}}&=\frac{\tb{x}^{k+1}_{2,1}}{\tb{x}^{k+1}_{2,2}}\cdot e^{2\eta(2\tb{x}^{k+2}_{1,2}-\tb{x}^{k+1}_{1,2}-\tb{x}^{k}_{1,2})}.
		\end{align*}
		Combining with the item~\ref{info 3} and \ref{info 6} in induction assumption, it holds that
		\begin{align*}
			\frac{\tb{x}^{k+3}_{2,1}}{\tb{x}^{k+3}_{2,2}}\cdot e^{\frac{3}{2}p\eta^\frac{5}{2}}\le \frac{\tb{x}^{k+1}_{2,1}}{\tb{x}^{k+1}_{2,2}}\le \frac{\tb{x}^{k+3}_{2,1}}{\tb{x}^{k+3}_{2,2}}\cdot e^{6\eta^2}.
		\end{align*}
		By lemma~\ref{lem: ineq}, we have
		\begin{align}\label{ineq: ty3y1}
			\frac{3}{4}p\eta^3 \le \tb{x}^{k+3}_{2,2}-\tb{x}^{k+1}_{2,2}\le 6\eta^2 \le 12\eta^2.		
		\end{align}
		
		
		\textit{proof of item \ref{induction 2}.}\\
		From item \ref{fraction 6} in lemma~\ref{lem: fraction} with $ t\rightarrow k+2 $,
		\begin{align*}
			\frac{\tb{x}^{k+3}_{2,1}}{\tb{x}^{k+3}_{2,2}}=\frac{\tb{x}^{k+2}_{2,1}}{\tb{x}^{k+2}_{2,2}}\cdot e^{-3\eta+2\eta(2\tb{x}^{k+2}_{1,2}+\tb{x}^{k+1}_{1,2})}.
		\end{align*}
		By induction assumption, we have $ \tb{x}^{k+2}_{1,2}\ge \tb{x}^{k}_{1,2}\ge \tb{x}^{k-2}_{1,2}\ge \cdots \ge \tb{x}^{0}_{1,2}\ge \frac{1}{2}+p $,
		Using item~\ref{info 6} in the induction assumption, we have $ \tb{x}^{k+1}_{1,2}\ge \tb{x}^{k+2}_{1,2}\ge \frac{1}{2}+p $, then 
		\begin{align*}
			\frac{\tb{x}^{k+2}_{2,1}}{\tb{x}^{k+2}_{2,2}}\cdot e^{6p\eta}\le \frac{\tb{x}^{k+3}_{2,1}}{\tb{x}^{k+3}_{2,2}}\le \frac{\tb{x}^{k+2}_{2,1}}{\tb{x}^{k+2}_{2,2}}\cdot e^{3\eta}.
		\end{align*}
		Then we can use lemma~\ref{lem: ineq}, and obtain
		\begin{align}\label{ineq: ty2y3}
			\frac{3}{2}p\eta^\frac{3}{2}\le \tb{x}^{k+2}_{2,2}-\tb{x}^{k+3}_{2,2}\le 3\eta.
		\end{align}
		
		
		
		\textit{proof of item \ref{induction 3}.}\\
		From item \ref{fraction 3} in lemma~\ref{lem: fraction} with $ t\rightarrow k+2 $,
		\begin{align*}
			\frac{\tb{x}^{k+4}_{1,1}}{\tb{x}^{k+4}_{1,2}}=\frac{\tb{x}^{k+2}_{1,1}}{\tb{x}^{k+2}_{1,2}}\cdot e^{2\eta(2\tb{x}^{k+3}_{2,2}-\tb{x}^{k+2}_{2,2}-\tb{x}^{k+1}_{2,2})} .	
		\end{align*}
		and
		\begin{align*}
			2\tb{x}^{k+3}_{2,2}-\tb{x}^{k+2}_{2,2}-\tb{x}^{k+1}_{2,2}=-(\tb{x}^{k+2}_{2,2}-\tb{x}^{k+3}_{2,2})+(\tb{x}^{k+3}_{2,2}-\tb{x}^{k+1}_{2,2}).
		\end{align*}
		According to inequalities (\ref{ineq: ty3y1}) and (\ref{ineq: ty2y3}), we have
		\begin{align*}
			-3\eta \le -(\tb{x}^{k+2}_{2,2}-\tb{x}^{k+3}_{2,2})+(\tb{x}^{k+3}_{2,2}-\tb{x}^{k+1}_{2,2})\le -\frac{3}{2}p\eta^\frac{3}{2}+12\eta^2,
		\end{align*}
		which is equivalent to
		\begin{align*}
			-3\eta \le -(\tb{x}^{k+2}_{2,2}-\tb{x}^{k+3}_{2,2})+(\tb{x}^{k+3}_{2,2}-\tb{x}^{k+1}_{2,2})\le -\frac{3}{4}p\eta^\frac{3}{2}.
		\end{align*} 
		The right side inequality follows from $ p\ge 16\eta^\frac{1}{2} $. Then
		\begin{align*}
			\frac{\tb{x}^{k+4}_{1,1}}{\tb{x}^{k+4}_{1,2}}\cdot e^{2\eta\cdot \frac{3}{4}p\eta^\frac{3}{2}} \le \frac{\tb{x}^{k+2}_{1,1}}{\tb{x}^{k+2}_{1,2}}\le \frac{\tb{x}^{k+4}_{1,1}}{\tb{x}^{k+4}_{1,2}}\cdot e^{2\eta\cdot 3\eta}, \\
			\frac{\tb{x}^{k+4}_{1,1}}{\tb{x}^{k+4}_{1,2}}\cdot e^{ \frac{3}{2}p\eta^\frac{5}{2}} \le \frac{\tb{x}^{k+2}_{1,1}}{\tb{x}^{k+2}_{1,2}}\le \frac{\tb{x}^{k+4}_{1,1}}{\tb{x}^{k+4}_{1,2}}\cdot e^{6\eta^2}.
		\end{align*}
		It can be concluded that
		\begin{align*}
			\frac{3}{4}p\eta^3 \le \tb{x}^{k+4}_{1,2}-\tb{x}^{k+2}_{1,2}\le 6\eta^2\le 12\eta^2.
		\end{align*}
		
		
		\textit{proof of item \ref{induction 4}.}\\
		According to item \ref{info 2} and item \ref{info 5} in induction assumption, we have $ \tb{x}^{k+2}_{2,2}\ge \tb{x}^{k}_{2,2}\ge \tb{x}^{k+1}_{2,2}+\frac{3}{2}p\eta^\frac{3}{2} $.
		From item \ref{fraction 1} in lemma~\ref{lem: fraction},
		\begin{align*}
			\frac{\tb{x}^{k+3}_{1,1}}{\tb{x}^{k+3}_{1,2}}=\frac{\tb{x}^{k+1}_{1,1}}{\tb{x}^{k+1}_{1,2}}\cdot e^{-2\eta(2\tb{x}^{k+2}_{2,2}-\tb{x}^{k+1}_{2,2}-\tb{x}^{k}_{2,2})}
		\end{align*}
		Therefore, we obtain
        \begin{align*}
            \frac{\tb{x}^{k+3}_{1,1}}{\tb{x}^{k+3}_{1,2}}\cdot e^{3p\eta^\frac{5}{2}}
            \le \frac{\tb{x}^{k+1}_{1,1}}{\tb{x}^{k+1}_{1,2}}.
        \end{align*}
        Using Lemma~\ref{lem: ineq}, 
		\begin{align}\label{ineq: tx3x1}
			\tb{x}^{k+3}_{1,2}- \tb{x}^{k+1}_{1,2}\ge \frac{3}{2}p\eta^3.
		\end{align} 
		
		
		\textit{proof of item \ref{induction 5}.}\\
		By item \ref{info 6} in induction assumption and (\ref{ineq: tx3x1}),
		\begin{align*}
			\tb{x}^{k+3}_{1,2}\ge \tb{x}^{k+1}_{1,2}\ge \tb{x}^{k+2}_{1,2}+\frac{3}{2}p\eta^\frac{3}{2} .
		\end{align*}
		From item \ref{fraction 4} in lemma \ref{lem: fraction},
		\begin{align*}
			\frac{\tb{x}^{k+4}_{2,1}}{\tb{x}^{k+4}_{2,2}}=\frac{\tb{x}^{k+2}_{2,1}}{\tb{x}^{k+2}_{2,2}}\cdot e^{-2\eta(2\tb{x}^{k+3}_{1,2}-\tb{x}^{k+2}_{1,2}-\tb{x}^{k+1}_{1,2})}.
		\end{align*}
        Therefore, we obtain
        \begin{align*}
            \frac{\tb{x}^{k+4}_{2,1}}{\tb{x}^{k+4}_{2,2}}\cdot e^{3p\eta^\frac{5}{2}}
            \le \frac{\tb{x}^{k+2}_{2,1}}{\tb{x}^{k+2}_{2,2}}.
        \end{align*}
		Then it holds that
		\begin{align*}
			\tb{x}^{k+4}_{2,2}-\tb{x}^{k+2}_{2,2}\ge\frac{3}{2}p\eta^3.
		\end{align*}
		
		
		\textit{proof of item \ref{induction 6}.}\\
		By (\ref{ineq: ty3y1}), (\ref{ineq: ty2y3}) and induction assumption, 
		\begin{align*}
			\tb{x}^{k+2}_{2,2}\ge \tb{x}^{k+3}_{2,2}\ge \tb{x}^{k+1}_{2,2}\ge \frac{1}{2}+p.
		\end{align*}
		According to item \ref{fraction 5} in lemma \ref{lem: fraction},
		\begin{align*}
			\frac{\tb{x}^{k+4}_{1,1}}{\tb{x}^{k+4}_{1,2}}=\frac{\tb{x}^{k+3}_{1,1}}{\tb{x}^{k+3}_{1,2}}\cdot e^{-3\eta+2\eta(2\tb{x}^{k+3}_{2,2}+\tb{x}^{k+2}_{2,2})},
		\end{align*}
		which leads to
		\begin{align*}
			\frac{\tb{x}^{k+3}_{1,1}}{\tb{x}^{k+3}_{1,2}}\cdot e^{6p\eta}\le \frac{\tb{x}^{k+4}_{1,1}}{\tb{x}^{k+4}_{1,2}}\le \frac{\tb{x}^{k+3}_{1,1}}{\tb{x}^{k+3}_{1,2}}\cdot e^{3\eta}.
		\end{align*}
		Combining with lemma~\ref{lem: ineq},
		\begin{align*}
			\frac{3}{2}p\eta^\frac{3}{2}\le 3p\eta^\frac{3}{2}\le \tb{x}^{k+3}_{1,2}-\tb{x}^{k+4}_{1,2}\le 3\eta.		
		\end{align*}
	\end{proof}
It is natural that the closer the points are to the boundary, the larger the KL-divergence. The following lemma quantifies the KL-divergence of the points to the equilibrium and the distance between the points and the boundary in $\ell_1$ norm.
\begin{lem}\label{lem: kl and boundary}
    For two point $\textbf{x}=(\textbf{x}_1,\textbf{x}_2)\in \Delta_2 $, $\textbf{x}'=(\textbf{x}_1',\textbf{x}_2')\in \Delta_2$, and $c\ge 0$ such that given $p\in (0,\frac{1}{2})$
    \begin{align*}
        \textbf{x}_2-c\ge\textbf{x}'_2\ge \frac{1}{2}+p,
    \end{align*}
    then we have
    \begin{align*}
        \KL((\frac{1}{2},\frac{1}{2}), (\textbf{x}_1,\textbf{x}_2))-\KL((\frac{1}{2},\frac{1}{2}), (\textbf{x}_1',\textbf{x}_2'))\ge pc.
    \end{align*}
\end{lem}
\begin{proof}
    By the definition of KL-divergence, it holds that
    \begin{align*}
        \KL\left((\frac{1}{2},\frac{1}{2}), (\textbf{x}_1,\textbf{x}_2)\right)=
        \frac{1}{2}\ln(\frac{1}{2\textbf{x}_1})+        \frac{1}{2}\ln(\frac{1}{2\textbf{x}_2}).
    \end{align*}
    Thus,
    \begin{align*}
        &\KL\left((\frac{1}{2},\frac{1}{2}), (\textbf{x}_1,\textbf{x}_2)\right)-\KL\left((\frac{1}{2},\frac{1}{2}), (\textbf{x}_1',\textbf{x}_2')\right)\\&=
        \frac{1}{2}\ln\left(\frac{\textbf{x}_1'\textbf{x}_2'}{\textbf{x}_1\textbf{x}_2}\right)= \frac{1}{2}\ln \left(\frac{(1-\textbf{x}_2')\textbf{x}_2'}{(1-\textbf{x}_2)\textbf{x}_2}\right)\\
        &\ge \frac{1}{2}\ln\left(\frac{(1-\textbf{x}_2+c)(\textbf{x}_2-c)}{(1-\textbf{x}_2)\textbf{x}_2}\right)\\
        &=\frac{1}{2}\ln\left(1+\frac{2\textbf{x}_2-1-c}{(1-\textbf{x}_2)\textbf{x}_2}\right)\\
        &\ge \frac{1}{2}\ln\left(1+\frac{2\textbf{x}_2-1-(\textbf{x}_2-\frac{1}{2})}{(1-\textbf{x}_2)\textbf{x}_2}\right)\\
        &\ge \frac{1}{2}\ln(1+4pc)\ge pc.
    \end{align*}
    The second equality arises from $\textbf{x}=(\textbf{x}_1,\textbf{x}_2)\in \Delta_2$ and $\textbf{x}'=(\textbf{x}_1',\textbf{x}_2')\in \Delta_2$. The third line follows from the decreasing monotonicity with respect to $\textbf{x}_2'$. The fifth line and the last line are a result of $\textbf{x}_2-c\ge \frac{1}{2}+p$.   
\end{proof}

When we apply Lemma~\ref{lem: kl and boundary} to Lemma~\ref{lem: close to boundary}, the following proposition is derived.
\begin{prop}\label{prop: KL increasing}
    For the periodic game defined in \ref{2-periodic game_m}, let $p\in (0,\frac{1}{4})$ and sufficient small $\eta$ such that $p\ge 16\eta^\frac{1}{2}$, when $t\ge 3$, if 
    \begin{align*}
\textbf{x}_{1,2}^0,\textbf{x}_{2,2}^0\ge \frac{1}{2}+2p,
    \end{align*}
    and for any $k<t$, $\textbf{x}_{1,2}^k,\textbf{x}_{2,2}^k\le 1-\sqrt{\eta}$, then it holds that
    \begin{align*}
        \KL\left((\textbf{x}^*_1,\textbf{x}_2^*), (\textbf{x}_{1}^{t+2},\textbf{x}_{2}^{t+2})\right)-
         \KL\left((\textbf{x}^*_1,\textbf{x}_2^*), (\textbf{x}_{i,1}^{t},\textbf{x}_{i,2}^{t})\right)
        \ge \frac{3}{4}p^2\eta^3.
    \end{align*}
\end{prop}
\begin{proof}
    Under the conditions in the lemma, according to item \ref{result 1}, \ref{result 3}, \ref{result 4}, and \ref{result 5} in Lemma~\ref{lem: close to boundary}, we have for any $t\ge 3$, and $i=1,2$
    \begin{align*}
        \textbf{x}_{i,2}^{t+2}-\textbf{x}_{i,2}^{t}\ge \frac{3}{8}p\eta^3.
    \end{align*}
    Futhermore, when $t$ is even,
    \begin{align*}
        \textbf{x}_{i,2}^{t+2}>\textbf{x}_{i,2}^{t}>\cdots> \textbf{x}_{i,2}^{4}>\frac{1}{2}+\frac{1}{2}p,
    \end{align*}
    where the last inequality comes from the proof in Lemma~\ref{lem: close to boundary} and the condition $p\ge 16\eta^\frac{1}{2}$.
    And when $t$ is odd,
    \begin{align*}
        \textbf{x}_{i,2}^{t+2}>\textbf{x}_{i,2}^{t}>\cdots> \textbf{x}_{i,2}^{3}>\frac{1}{2}+\frac{1}{2}p.
    \end{align*}
    Then using Lemma~\ref{lem: kl and boundary}, we have that
    \begin{align*}
        \KL\left((\textbf{x}^*_1,\textbf{x}_2^*), (\textbf{x}_{1}^{t+2},\textbf{x}_{2}^{t+2})\right)-
         \KL\left((\textbf{x}^*_1,\textbf{x}_2^*), (\textbf{x}_{1}^{t},\textbf{x}_{2}^{t})\right)
        \ge \frac{3}{8}p^2\eta^3.
    \end{align*}
\end{proof}

Proposition~\ref{prop: KL increasing} auctually states that when $\textbf{x}_{1,2}^{0},\textbf{x}_{1,2}^{0}> \frac{1}{2}$, and the iterative points never enter the neighborhood of boundary, it holds that KL-divergence will always increase with an constant depending on $\eta$ and the distance of $(\textbf{x}_{1}^{0},\textbf{x}_{2}^{0})$ to the equilibrium. Note that we have no constraint on the value of $(\textbf{x}_{1}^{-1},\textbf{x}_{2}^{-1})$.

In fact, the condition $\textbf{x}_{1,2}^{0},\textbf{x}_{1,2}^{0}> \frac{1}{2}$ is not necessary. 
All possible cases of $(\textbf{x}_1^0,\textbf{x}_2^0)$ are following
    \begin{enumerate}\label{cases: intial condition}
    \item $\textbf{x}_{1,2}^{0}>\frac{1}{2},\textbf{x}_{2,2}^{0}>\frac{1}{2}$,
        \item $\textbf{x}_{1,2}^{0}<\frac{1}{2},\textbf{x}_{2,2}^{0}>\frac{1}{2}$,
        \item $\textbf{x}_{1,2}^{0}<\frac{1}{2},\textbf{x}_{2,2}^{0}<\frac{1}{2}$,
        \item $\textbf{x}_{1,2}^{0}>\frac{1}{2},\textbf{x}_{2,2}^{0}<\frac{1}{2}$,
    \end{enumerate}
    Proposition~\ref{prop: KL increasing} states the result that under the initial condition in the first case. The key lemma in the proof of Proposition~\ref{prop: KL increasing} is Lemma~\ref{lem: close to boundary}. Then combining the relation between KL-divergence and the distance to the boundary in $\ell_1$ norm, the result in the proposition is concluded.
    
  In Figure \ref{Figure: OMWUwithInitial22}\footnote{The step size of 200 is employed here to enhance the clarity in observing the increment of strategies with a probability greater than $\frac{1}{2}$ after every two iterations.}, we present the trajectories of mixed strategies of both players under the initial condition $\textbf{x}_{1}^0=\textbf{x}_{2}^0=(0.45,0.55)$. It illustrates the findings from Lemma~\ref{lem: close to boundary}.  Similar result are observed in the other cases as in Lemma~\ref{lem: close to boundary}, as shown in Figure \ref{Figure: OMWUwithInitial11}, \ref{Figure: OMWUwithInitial12}, and \ref{Figure: OMWUwithInitial21}.
\begin{figure}[h]
    \centering
    \subfigure[]{
        \includegraphics[width=1.5in]{OMWUwithInitial11.png}
        \label{Figure: OMWUwithInitial11}
    }
    \subfigure[]{
	\includegraphics[width=1.5in]{OMWUwithInitial12.png}
        \label{Figure: OMWUwithInitial12}
    }
    \subfigure[]{
	\includegraphics[width=1.5in]{OMWUwithInitial21.png}
        \label{Figure: OMWUwithInitial21}
    }
    \subfigure[]{
	\includegraphics[width=1.5in]{OMWUwithInitial22.png}
        \label{Figure: OMWUwithInitial22}
    }
    \caption{Trajectories of strategies for both players when using OMWU in the periodic game defined in (\ref{2-periodic game_m}).}
    \label{em2}
\end{figure}
    
    Then we can extend Proposition~\ref{prop: KL increasing} to encompass general initial conditions:
    \begin{prop}\label{prop: before4.2}
    For the periodic game defined in \ref{2-periodic game_m}, let $p=\frac{1}{2}\min (\lvert \textbf{x}_{1,1}^0-\textbf{x}_{1,1}^*\rvert ,\lvert \textbf{x}_{2,1}^0-\textbf{x}_{2,1}^* \rvert)$, and sufficient small $\eta$ such that $p\ge 16\eta^\frac{1}{2}$, when $t\ge 3$, if for any $k<t$, $\textbf{x}_{1,2}^k,\textbf{x}_{2,2}^k\le 1-\sqrt{\eta}$, then it holds that
    \begin{align*}
        \KL\left((\textbf{x}^*_1,\textbf{x}_2^*), (\textbf{x}_{1}^{t+2},\textbf{x}_{2}^{t+2})\right)-
         \KL\left((\textbf{x}^*_1,\textbf{x}_2^*), (\textbf{x}_{1}^{t},\textbf{x}_{2}^{t})\right)
        \ge \frac{3}{4}p^2\eta^3.
    \end{align*}
\end{prop}
Here, both $\textbf{x}_{1,1}^* $ and $ \textbf{x}_{2,1}^*$ are equal to $\frac{1}{2}$ in game (\ref{2-periodic game_m}). Hence, parameter $p$ defined in Proposition \ref{prop: before4.2} satisfies the initial condition in Proposition \ref{prop: KL increasing}. 
The conclusions of Proposition~\ref{prop: OMWU fails} can be directly derived from Proposition~\ref{prop: before4.2}.

 
\subsection{Proof of Stage 2}

%Assume $n$ is an even number in the follows. 
%Denote $\CG_1$ as the dynamical system in the odd number step, and $\CG_2$  as the dynamical system in the even number step.
	
Lemma \ref{lem: close to boundary} states that the iterative vector   $(\tb{x}^{n}_{1,1},\tb{x}^{n+1}_{1,1},\tb{x}^{n}_{2,1},\tb{x}^{n+1}_{2,1})^\top$ will be close to one of $(0,0,\cdot,\cdot)$, $(1,1,\cdot,\cdot)$, $(\cdot,\cdot,0,0)$, $(\cdot,\cdot,1,1)$ after a sufficient number of steps.
	Without loss of generality, let's assume that the iterative vector is close to $(0,0,\cdot,\cdot)$. The other cases are symmetric to it.
	
The following lemma demonstrates that the composition of $\CG_1\circ \CG_2$ has continuous fixed points on the boundary.


\begin{lem}\label{lem: fix point}
     For every $a \in [0,1]$, $(0,0,a,\frac{ae^{-3\eta}}{ae^{-3\eta} + (1-a)})$ is a fixed point of $\CG_1 \circ \CG_2$. 
\end{lem}
\begin{proof}
    Recall the definition of $\CG_1 $ and $ \CG_2$, for any $a\in(0,1)$, we have
    \begin{align*}
		&\CG_2\left((0,0,a,\frac{ae^{-3\eta}}{ae^{-3\eta} + (1-a)})\right)\\
		=&\left(0,0, 
		\frac{ae^{-3\eta}}{ae^{-3\eta} + (1-a)},
		\frac{\frac{ae^{-3\eta}}{ae^{-3\eta} + (1-a)} e^{3\eta}}{\frac{ae^{-3\eta}}{ae^{-3\eta} + (1-a)} e^{3 \eta }+(1-\frac{ae^{-3\eta}}{ae^{-3\eta} + (1-a)})}\right)\\
        =&\left(0,0,\frac{ae^{-3\eta}}{ae^{-3\eta} + (1-a)},a\right),
	\end{align*} 
	and 
	\begin{align*}
		\CG_1\left((0,0,\frac{ae^{-3\eta}}{ae^{-3\eta} + (1-a)},a)\right)
		=\left(0,0, 
		a,\frac{ae^{-3\eta}}{ae^{-3\eta} + (1-a)}\right).
	\end{align*} 
    Thus, it holds that
    \begin{align*}
        \CG_1 \circ \CG_2\left((0,0,a,\frac{ae^{-3\eta}}{ae^{-3\eta} + (1-a)})\right)=\left(0,0,a,\frac{ae^{-3\eta}}{ae^{-3\eta} + (1-a)}\right).
    \end{align*}
\end{proof}
The following lemma provides the description for the eigenvalues and eigenvectors of Jacobi matrix at the fixed points.
\begin{lem}\label{lem: eigenvalue}
    For $a\in(0,1)$, all the eigenvalues of the Jacobian matrix of $\CG_1 \circ \CG_2$ at points 
    $(0,0,a,\frac{ae^{-3\eta}}{ae^{-3\eta} + (1-a)})$ are no larger than $1$. Moreover, the central eigenspace corresponds to eigenvalue $1$ is generated by
    vectors $(0,0,*_1,*_2)$.
\end{lem}
\begin{proof}
    It can be computed that\footnote{We employed Matlab for computation.}
    \begin{align*}
         0,0,1,e^{-\frac{2\eta a(1-a)(e^{3\eta}-1) }{a+(1-a)e^{3\eta}}}
    \end{align*}
    are the eigenvalues of Jacobian matrix of $\CG_1 \circ \CG_2$ at points 
    $(0,0,a,\frac{ae^{-3\eta}}{ae^{-3\eta} + (1-a)})$.

    Here, since $a\in(0,1)$, it holds that $e^{-\frac{2\eta a(1-a)(e^{3\eta}-1) }{a+(1-a)e^{3\eta}}}< 1$. And the eigenvector corresponding to $ 1 $ is $\tb{w}_1= (0,0,e^{-3\eta}(a+(1-a)e^{3\eta})^2,1)^\top $, with its first two elements being zero.
\end{proof}


	\begin{lem}\label{lem: first two converge to zero}
		Consider the composition dynamical system $\CG_1 \circ \CG_2$ which maps $(\tb{x}^{n-1}_{1,1},\tb{x}^{n}_{1,1},\tb{x}^{n-1}_{2,1},\tb{x}^{n}_{2,1})^\top$ to $(\tb{x}^{n+1}_{1,1},\tb{x}^{n+2}_{1,1},\tb{x}^{n+1}_{2,1},\tb{x}^{n+2}_{2,1})^\top$. 
		Then, for any points $ \tb{v} $ close to $(0,0,a,\frac{ae^{-3\eta}}{ae^{-3\eta} + (1-a)})^\top$ for arbitrary $a\in(0,1)$, the first two elements of $ \tb{v} $ will finally converge to zero.
	\end{lem}
	\begin{proof}
		By Lemma~\ref{lem: fix point} for any $ a\in (0,1) $, $(0,0,a,\frac{ae^{-3\eta}}{ae^{-3\eta} + (1-a)})$ is a fixed point of the dynamical system $\CG_1 \circ \CG_2$. Denote the Jacobian matrix of $\CG_1 \circ \CG_2$ at $(0,0,a,\frac{ae^{-3\eta}}{ae^{-3\eta} + (1-a)})$ as $ J $. 
		
		According to Lemma~\ref{lem: eigenvalue}, the neighborhood $\CW$ of the boundary of simplex can be decomposed as follows,
		\begin{align*}
			\CW=\CW_1+\CW_0,
		\end{align*}
		where $ \CW_1  $ is the eigenspace corresponding to eigenvalue $1$ and is spanned by $ \tb{w}_1=(0,0,e^{-3\eta}(a+(1-a)e^{3\eta})^2,1)^\top $, as mentioned in Lemma~\ref{lem: eigenvalue}.
		Additionally, $ \CW_0 $ is the eigenspace corresponding to eigenvalues with modules smaller than $ 1 $.
		Naturally, for any point $ \tb{v} $ close to $(0,0,a,\frac{ae^{-3\eta}}{ae^{-3\eta} + (1-a)})^\top$, $ \tb{v} $ can be decomposed as $ (\tb{v}_1+\tb{v}_0) $, where $ \tb{v}_1\in \CW_1 $ and $ \tb{v}_0\in \CW_0 $. By Proposition~\ref{decomposition}, with $ n $ enough large, $ (\CG_1 \circ \CG_2)^n(\tb{v}) $ will converge to the space $ (0,0,a,\frac{ae^{-3\eta}}{ae^{-3\eta} + (1-a)})^\top+\CW_1 $. Consequently, it can be concluded that the first two elements of the iterative vector converge to zero.
	\end{proof}
 
	With the above preparation, we are ready to prove Proposition~\ref{prop: converge to boundary}.
	\begin{proof}[Proof of Proposition~\ref{prop: converge to boundary}]
        We only discuss about the case when $(\tb{x}^{n-1}_{1,1},\tb{x}^{n}_{1,1},\tb{x}^{n-1}_{2,1},\tb{x}^{n}_{2,1})^\top$ is close to $(0,0,\cdot,\cdot)$, while the other case are similar.

        It can be computed by the update rule of dynamic of (OMWU), when $(\tb{x}^{n-1}_{1,1},\tb{x}^{n}_{1,1},\tb{x}^{n-1}_{2,1},\tb{x}^{n}_{2,1})^\top$ is close to $(0,0,\cdot,\cdot)$, it is auctually close to $(0,0,a,\frac{ae^{-3\eta}}{ae^{-3\eta} + (1-a)})$ for some $a\in(0,1)$.
        
	    According to Lemma~\ref{lem: first two converge to zero}, when $(\tb{x}^{n-1}_{1,1},\tb{x}^{n}_{1,1},\tb{x}^{n-1}_{2,1},\tb{x}^{n}_{2,1})^\top$ lies in the neighborhood of $(0,0,a,\frac{ae^{-3\eta}}{ae^{-3\eta} + (1-a)})$, it holds that $\tb{x}^{n-1}_{1,1}$ and $\tb{x}^{n}_{1,1}$ will converge to $0$. This results in the fact that KL-divergence tends to infinity.
	\end{proof}