Capacity-achieving output laws bound every row divergence

Fix $x\in \mathcal{X}$. Suppose some output reachable from $x$ has zero $Q^{\ast }$ mass, and put $a:={\sum }_{y:Q^{\ast }(y)=0}W(y\mid x)>0$. For $0<t<1$, let $P_t:=(1-t)P_X^{\ast }+t{\delta }_x$ and $Q_t:=P_{t}W=(1-t)Q^{\ast }+tW(\cdot \mid x)$. By [F2] and [F3], the conditional-entropy part of $I(P_t;W)-I(P_X^{\ast };W)$ is linear in $t$. On coordinates with $Q^{\ast }(y)>0$, differentiability in [F5] bounds the total entropy change in absolute value by $K_{1}t$ for some finite $K_1$ and all small $t$. For zero coordinates with $W(y\mid x)>0$, the entropy contribution is exactly $-tW(y\mid x)\ln t-tW(y\mid x)\ln W(y\mid x)$. All terms other than $-ta\ln t$ have absolute value at most $K_{2}t$ for a finite $K_2$. Since $-\ln t\to +\mathrm{\infty }$ by [F6], the information change is positive for sufficiently small $t$, contradicting maximality in [F1]. Every row therefore has support contained in $\mathrm{supp}{Q}^{\ast }$.

Now fix any $x$ and use the same segment $P_t=(1-t)P_X^{\ast }+t{\delta }_x$. By step 1.1, $Q^{\ast }$ is positive on the support of $W(\cdot \mid x)$, so the right derivative at $t=0$ is finite. Differentiating the entropy expression in [F2] with [F3] and [F5] gives ${\frac{d}{dt}I(P_t;W)\mid }_{0+}=-{\sum }_{y:Q^{\ast }(y)>0}(W(y\mid x)-Q^{\ast }(y))\ln Q^{\ast }(y)-H(W(\cdot \mid x))+{\sum }_u{P}_X^{\ast }(u)H(W(\cdot \mid u))=D(W(\cdot \mid x)\mathrm{\parallel }{Q}^{\ast })-C(W)$, where the $+1$ derivative terms cancel because both rows sum to one, and the last equality is [F2] and [F4].

Since $P_X^{\ast }$ maximizes information and $P_t$ is feasible for $t\ge 0$, the right derivative in step 2.1 is nonpositive. Hence $D(W(\cdot \mid x)\mathrm{\parallel }{Q}^{\ast })\le C(W)$. The input $x$ was arbitrary, proving the bound for all rows.