Disjoint supports and singular divergence values

Put $f(t)=\frac{1}{2}\mathrm{\mid }t-1\mathrm{\mid }$ for $t>0$. It is finite-valued and $f(1)=0$. For $x,y>0$ and $s\in [0,1]$, $(1-s)x+sy>0$, so [F5] and [F6] give $f((1-s)x+sy)=\frac{1}{2}\mathrm{\mid }(1-s)(x-1)+s(y-1)\mathrm{\mid }\le (1-s)f(x)+sf(y)\mathrm{.}$ Thus [F7] proves that $f$ is convex, so it is an admissible generator in [F2].

Since $p_a=1>0=q_a$, [F3] gives $D(P\mathrm{\parallel }Q)=+\mathrm{\infty }$. This records the singular direction; $D(Q\mathrm{\parallel }P)$ is also infinite by the same support calculation with $a,b$ exchanged.

For $0<t<1$, $f(t)=\frac{1}{2}(1-t)$, so $f(0)={\lim }_{t\downarrow 0}f(t)=1\mathrm{/}2$. For $t>1$, $f(t)\mathrm{/}t=\frac{1}{2}(1-1\mathrm{/}t)$, so $f_{\mathrm{\infty }}={\lim }_{t\to \mathrm{\infty }}f(t)\mathrm{/}t=1\mathrm{/}2$. The coordinate $a$ contributes $p_a{f}_{\mathrm{\infty }}=1\mathrm{/}2$ by [F2], and coordinate $b$ contributes $q_{b}f(0)=1\mathrm{/}2$. Thus $D_f(P\mathrm{\parallel }Q)=1$.

If $0<\alpha <1$, the only coordinate with $q_x>0$ is $b$, where $p_b=0$, so $S_{\alpha }(P,Q)=0$ and $D_{\alpha }(P\mathrm{\parallel }Q)=+\mathrm{\infty }$ by [F4] and the definition. If $\alpha >1$, coordinate $a$ gives infinite affinity by [F4]. At $\alpha =1$, KL is infinite by step 1.2.