A separating event forces relative entropy

If $D(Q\mathrm{\parallel }P)=+\mathrm{\infty }$, the result holds. Otherwise form distributions on $[m]\times \{0,1\}$ by $\tilde{Q}(i,1)=Q_i{X}_i$, $\tilde{Q}(i,0)=Q_i(1-X_i)$ and the analogous formulas for $\tilde{P}$. Expanding the defining finite sum and summing the common two-outcome kernel gives $D(\tilde{Q}\mathrm{\parallel }\tilde{P})=D(Q\mathrm{\parallel }P)$.

Group the terms of this divergence according to the second coordinate $z$. For each $z$, the group equals $q_z\ln (q_z\mathrm{/}{p}_z)+q_{z}D(\tilde{Q}(\cdot \mid z)\mathrm{\parallel }\tilde{P}(\cdot \mid z))$, where $q_z=\tilde{Q}(z)$ and $p_z=\tilde{P}(z)$; the conditional divergence is nonnegative by applying [F2] to each ratio, and zero-mass cases follow by the stated $+\mathrm{\infty }$ convention. Hence $D(Q\mathrm{\parallel }P)\ge D_{\mathrm{B}\mathrm{e}\mathrm{r}}(Q{X}^{\mathsf{T}}\mathrm{\parallel }P{X}^{\mathsf{T}})$.

For $0<q<p<1$, the binary divergence $K(p,q)=p\ln (p\mathrm{/}q)+(1-p)\ln ((1-p)\mathrm{/}(1-q))$ has ${\mathrm{\partial }}_{p}K=\ln \frac{p(1-q)}{q(1-p)}>0$ and ${\mathrm{\partial }}_{q}K=\frac{q-p}{q(1-q)}<0$. Therefore the hypotheses imply $K(Q{X}^{\mathsf{T}},P{X}^{\mathsf{T}})\ge K(0.94,0.06)=0.88\ln (47\mathrm{/}3)>1$, since $47\mathrm{/}3>9>e^2$. Boundary values follow by limits or give infinite divergence. Combining with step 2.1 proves $D(Q\mathrm{\parallel }P)\ge 1$.