Finite data processing for f-divergences

For each $y\in \mathcal{Y}$ and $x\in \mathcal{X}$, set $a_{x,y}=p_{x}K(y\mid x)$ and $b_{x,y}=q_{x}K(y\mid x)$. By [F2], $(PK{)}_y={\sum }_x{a}_{x,y}$ and $(QK{)}_y={\sum }_x{b}_{x,y}$. Applying [F4] at each $y$ gives ${\mathrm{\Pi }}_f((PK{)}_y,(QK{)}_y)\le {\sum }_x{\mathrm{\Pi }}_f(p_{x}K(y\mid x),q_{x}K(y\mid x))$. Summing this finite family of inequalities over $y$ is valid for extended real terms because no term is $-\mathrm{\infty }$.

For any $\lambda >0$, the definition of ${\mathrm{\Pi }}_f$ gives ${\mathrm{\Pi }}_f(\lambda a,\lambda b)=\lambda {\mathrm{\Pi }}_f(a,b)$, including $a=0$ or $b=0$; when $\lambda =0$, both sides are interpreted as zero. Thus, for each $x$, the row-sum condition in [F2] yields ${\sum }_y{\mathrm{\Pi }}_f(p_{x}K(y\mid x),q_{x}K(y\mid x))={\sum }_{y}K(y\mid x){\mathrm{\Pi }}_f(p_x,q_x)={\mathrm{\Pi }}_f(p_x,q_x)$. Substituting these identities into step 1.1 and summing over the finite alphabet $\mathcal{X}$ proves $D_f(PK\mathrm{\parallel }QK)\le D_f(P\mathrm{\parallel }Q)$ by [F3].