Sorry for the defintional overload but I promise there's an interesting question at the end! Please skip to the last section if you're familiar with one-shot information theory.
Defintions
The max-relative entropy between two states is defined as
$$D_{\max }(\rho \| \sigma):=\log \min \{\lambda: \rho \leq \lambda \sigma\},$$
where $\rho\leq \sigma$ should be read as $\sigma - \rho$ is positive semidefinite. If we consider probability distributions i.e. diagonal $\rho$, $\sigma$, we have
$$D_{\max}(p_X\|q_X) = \sup_{x\in\mathcal{X}}\log\frac{p(x)}{q(x)}$$
There is also a smoothed version of this quantity and this is given by taking the infimum of $D_{\max}(\rho\|\sigma)$ over all states $\bar{\rho}$ which are within an $\varepsilon$ ball of $\rho$ according to trace distance. We have
$$D^{\varepsilon}_{\max}(\rho\|\sigma) = \inf\limits_{\bar{\rho}\in\mathcal{B}^{\varepsilon}(\rho)}D_{\max}(\bar{\rho}\|\sigma)$$
Now consider the case where $\rho = \rho_{AB}$ (some bipartite state) and $\sigma_B$ is some state. A quantity known as the max-information that $B$ has about $A$ is given by
$$I_{\max}(A:B)_\rho = \inf_{\sigma_B}D_{\max}(\rho_{AB}||\rho_A\otimes\sigma_B)$$
The smoothed max-information is
$$I^{\varepsilon}_{\max}(A;B)_{\rho_{AB}} = \inf\limits_{\bar{\rho}_{AB}\in\mathcal{B}^{\varepsilon}(\rho_{AB})} I_{\max}(A;B)_{\bar{\rho}}$$
And finally, the partially smoothed max-information defined in this paper says your ball is restricted to those nearby states that share the same marginal on $A$. This is denoted with a dot above the restricted register as below.
$$I^{\varepsilon}_{\max}(\dot{A} ; B)_{\rho_{AB}} = \inf\limits_{\bar{\rho}_{AB}\in\mathcal{B}^{\varepsilon}(\rho_{AB})\atop{\bar{\rho}_A = \rho_A}} I_{\max}(A;B)_{\bar{\rho}}$$
AEP
It was shown ((27) of the same paper) that the partially smoothed max-information satisfies as asymptotic equipartition property. That is, for i.i.d. copies of the state $\rho_{AB}$, we have
$$\lim_{\varepsilon\rightarrow 0}\lim_{n\rightarrow\infty}\frac{1}{n}I^{\varepsilon}_{\max}(\dot{A}^n ; B^n)_{\rho_{AB}^{\otimes n}} = I(A:B)_{\rho}$$
Classical case
All the above defintions also apply to the case where $\rho$ and $\sigma$ are chosen to be diagonal i.e. classical probability distributions. For some classical channel $W_{Y|X}$ and input distribution $p_X$, we obtain $p_{XY} = p_X W_{Y|X}$. Let $q_Y$ be some other distribution.
$$I^{\varepsilon}_{\max}(\dot{X} ; Y)_{p_{XY}} = \inf_{\|\bar{p} - p\|_1 \leq \varepsilon \atop{\bar{p}_X = p_X}}I_{\max}(X;Y)_{\bar{p}_{XY}} = \inf_{\|\bar{p} - p\|_1 \leq \varepsilon \atop{\bar{p}_X = p_X}}\inf_{q_Y} D_{\max}(\bar{p}_{XY}\|p_X\times q_Y)$$
The main observation is that this quantity is independent of $p_X$ for any choice of $p_X$ that shares the same support. The support of $p_X$ entirely determines the one-shot quantity. To see why, note that once we have the optimal $\bar{p}$, the right hand side can be written as follows for some $\tilde{W}_{Y|X}$
$$\inf_{q_Y}\sup_{x\in\mathcal{X}, y\in\mathcal{Y}}\log \frac{\bar{p}_{XY}(x,y)}{p_{X}(x)q_{Y}(y)} = \inf_{q_Y}\sup_{x\in\mathcal{X}, y\in\mathcal{Y}}\log \frac{p_{X}(x)\tilde{W}_{Y|X}(y|x)}{p_{X}(x)q_{Y}(y)}$$
because we had imposed the restriction $\bar{p}_X = p_X$. Now the $p_X(x)$ just cancels out provided it is nonzero.
Question
How does the above independence of $I^{\varepsilon}_{\max}(\dot{X};Y)_{p_{XY}}$ on $p_X$ square with the AEP. In particular, if I consider $p_{X}^{\otimes n}$ and $W_{Y|X}^{\otimes n}$, we have $p_{XY}^{\otimes n} = p_{X}^{\otimes n}W_{Y|X}^{\otimes n}$. Then, one would have
$$\lim_{\varepsilon\rightarrow 0}\lim_{n\rightarrow\infty}\frac{1}{n}I^{\varepsilon}_{\max}(\dot{X}^n;Y^n)_{p_{XY}^{\otimes n}} = I(X:Y)_{p_{XY}}$$
The left hand side is independent of the choice of $p_{X}$ but the right hand side definitely depends on the choice of $p_X$. The right side, when maximized over all $p_X$ gives us the capacity of the channel.
