A finite information-capacity supremum need not be attained

Write $q_x=W(1\mid x)\in (0,1)$ and $r={\sum }_x{p}_x{q}_x$. There is some $x_0$ with $p_{x_0}>0$, so $r\ge p_{x_0}{q}_{x_0}>0$ and $1-r\ge p_{x_0}(1-q_{x_0})>0$. The output entropy is $h_2(r)\le h_2(1\mathrm{/}2)=\ln 2$: concavity and symmetry in [F2] give $h_2(1\mathrm{/}2)\ge \frac{1}{2}{h}_2(r)+\frac{1}{2}{h}_2(1-r)=h_2(r)$. Every row has entropy $h_2(q_x)>0$, and at least one term has positive weight, so ${\sum }_x{p}_x{h}_2(q_x)>0$. By [F1], $I(p;W)=h_2(r)-{\sum }_x{p}_x{h}_2(q_x)<\ln 2$ for every input law.

For each $k$, choose the law assigning probability $1\mathrm{/}2$ to $2k$ and $1\mathrm{/}2$ to $2k-1$. Its output is uniform because the two row probabilities sum to one, while its conditional entropy is $h_2({ϵ}_k)$ by symmetry. Hence its mutual information is $\ln 2-h_2({ϵ}_k)$. Since ${ϵ}_k\downarrow 0$ and binary entropy is continuous with $h_2(0)=0$ by [F2]–[F3], these values tend to $\ln 2$.

Step 1.1 bounds every input law strictly below $\ln 2$, while step 1.2 gives a sequence of feasible values tending to $\ln 2$. Therefore the supremum is finite and equals $\ln 2$, but no input law attains it.