Can repeated game sustain a discounted average payoff that is not individually rational?

Look at the following payoff matrix, there are two stage Nash equlibria, and we consider a infinitely repeated game with discount factor $$\delta\in(0,1)$$, can players sustain an average payoff of (2,2) under any strategy? (I am confused because the individually rational(i.e. minmax) payoff is (3,3)) If not, why? • How is the mimax payoff (3,3)? The minimum for H is 0. The minimum for L is 3. Which means that if both player play minimax, they both choose L, and the payoff is (5,5). – Acccumulation Mar 15 at 15:49
• @Acccumulation: In the repeated games literature, it's customary to define the set of individually rational payoffs as $\{(u_1,\dots,u_n): u_i\ge\min_{a_{-i}}\max_{a_i}u_i(a_i,a_{-i})\,\forall i\}$. In this case, $(3,3)$ is the minimum of this set. – Herr K. Mar 15 at 16:09

I'm inclined to say $$(2,2)$$ is not sustainable, though I don't have a formal proof at the moment.
To achieve an average payoff of $$(2,2)$$, the players must play $$(H,H)$$ in some stages. But then a player can deviate to playing $$L$$ and avoid any subsequent punishments by ensuring that he can always get at least $$3$$, which is better than $$2$$.