Capacity of a finite relay channel with output feedback to source and relay

Apply [F2]. Its outer bound has a finite time-sharing variable $Q$ and bounds $I(X,X_r;Y\mathrm{\mid }Q)$ and $I(X;Y,Y_r\mathrm{\mid }{X}_r,Q)$. Apply [F5] first with input $(X,X_r)$ and no side information, then with input $X$ and side information $X_r$. Averaging over $Q$ cannot increase either conditional mutual information, so both are bounded by the corresponding quantities under the averaged joint input law $P_{X,X_r}$. This gives the displayed maximum as an outer bound.

For achievability, view $(Y_r,Y)$ as the relay's observation and $Y$ as the destination output. This is a finite relay channel whose relay encoder can causally use the observation pair because it has full feedback in [F1]. Apply the block-Markov construction of [F3] with the relay observation ${\tilde{Y}}_r=(Y_r,Y)$; its decoding term becomes $I(X;Y,Y_r\mathrm{\mid }{X}_r)$, while the destination decoding term is $I(X,X_r;Y)$. Every joint input law factors as $P_{X_r}{P}_{X\mathrm{\mid }{X}_r}$, so the construction achieves every rate below the displayed minimum.

The outer bound in step 1.1 and achievable rates in step 1.2 coincide for every joint input law. The joint-input probability simplex is compact. Each mutual information is continuous there because it is a finite sum of terms $-p\ln p$, with $p\ln p$ extended continuously by $0$ at $p=0$; the minimum of the two continuous functions is continuous. It therefore attains a maximum, and taking that maximum and closure proves the capacity formula.