Finite-channel mutual information is concave in the input law

Expanding the finite KL sum in [F1] under $P(x)V(y\mid x)$ and grouping the finite terms gives $I_P(V)=H(PV)-{\sum }_{x}P(x)H(V(\cdot \mid x))\mathrm{.}$ Indeed, the logarithm of the joint-to-product ratio is $\ln V(y\mid x)-\ln (PV)(y)$ on positive joint support; summing the first term gives the negative conditional entropy and summing the second gives the output entropy. Zero-mass coordinates contribute zero by [F1].

For $0<s<t\le 1$, [F3] gives $g^{\mathrm{\prime }}(s)=-\ln s-1>-\ln t-1=g^{\mathrm{\prime }}(t)$, since $\ln$ is strictly increasing by its positive derivative and the mean value theorem. If $0<u<v<w\le 1$, apply [F4] to $g$ on $[u,v]$ and $[v,w]$. The resulting derivative points satisfy ${\xi }_1<{\xi }_2$, so the left secant slope is at least the right secant slope; rearranging is the concavity inequality for $g$ at $v$. Thus $g$ is concave on $(0,1]$.

For $u,w\in [0,1]$ with one endpoint zero, replace that endpoint by a positive sequence tending to zero in the inequality of step 1.2 and use continuity from [F5]. Therefore $g$ is concave on $[0,1]$. Summing the scalar concavity inequalities over coordinates proves that finite entropy $H(q)={\sum }_{y}g(q_y)$ is concave on each probability simplex.

Put $P_{\lambda }=\lambda P_0+(1-\lambda )P_1$. The output law satisfies $P_{\lambda }V=\lambda (P_{0}V)+(1-\lambda )(P_{1}V)$, so step 2.1 gives $H(P_{\lambda }V)\ge \lambda H(P_{0}V)+(1-\lambda )H(P_{1}V)$. The conditional-entropy term in step 1.1 is linear in $P$. Subtracting that linear term yields $I_{P_{\lambda }}(V)\ge \lambda I_{P_0}(V)+(1-\lambda )I_{P_1}(V)$, as claimed.