= Uniform payoff bound against Tit for tat
{title2=$\sum_{t<T}u_t\le RT+(T_{\rm tempt}-R)$}
In a symmetric two-action stage game write $R,P,T_{\rm tempt},S$ for mutual cooperation, mutual defection, sole defection and sole cooperation <payoffs>. Suppose $T_{\rm tempt}\ge R$, $P\le R$ and $T_{\rm tempt}+S\le2R$. Against an initially cooperative <Tit for tat>, every action <sequence> has limiting upper average <payoff> at most $R$.
For a deviator's actions $a_t$, put $a_{-1}=C$, $h(C)=0$ and $h(D)=T_{\rm tempt}-R$. The opponent at time $t$ plays $a_{t-1}$. Check the four transitions: $C$ to $C$ and $C$ to $D$ satisfy
$$
u(a_t,a_{t-1})\le R+h(a_t)-h(a_{t-1})
$$
with equality; $D$ to $C$ uses $S\le2R-T_{\rm tempt}$, and $D$ to $D$ uses $P\le R$. A <telescoping sum> gives the displayed finite-horizon bound. Divide by the horizon and let it increase. Mutual <Tit for tat> yields $R$ every round, so it is a <Nash equilibrium> under the <long-run average payoff> criterion. The argument is pathwise and therefore covers arbitrary deviations representable by <finite-state machines>, as well as action <sequences> outside that class.
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