Nonnegativity of finite relative entropy with a positive reference

If $t=1$, then $\ln t=t-1=0$. Otherwise let $a=\min \{1,t\}$ and $b=\max \{1,t\}$. By [F2] and [F3], $\ln$ is continuous on $[a,b]$; by [F5], it is differentiable on $(a,b)$ with derivative $1\mathrm{/}u$. Applying [F6] gives $c\in (a,b)$ with $\ln t-\ln 1=(t-1)\mathrm{/}c$, so $\ln t=(t-1)\mathrm{/}c$ by [F4]. If $t>1$, then $c>1$; if $0<t<1$, then $0<c<1$ and $t-1<0$. In either case $(t-1)\mathrm{/}c<t-1$. Thus $\ln t\le t-1$, with equality only at $t=1$.

Let $S=\{i:p_i>0\}$. By [F1], [A1], and step 1.1 applied to $q_i\mathrm{/}{p}_i$ for each $i\in S$, $D(p\mathrm{\parallel }q)=-{\sum }_{i\in S}{p}_i\ln (q_i\mathrm{/}{p}_i)\ge {\sum }_{i\in S}(p_i-q_i)=1-{\sum }_{i\in S}{q}_i\ge 0$. If $D(p\mathrm{\parallel }q)=0$, both inequalities are equalities. Equality in the last requires $q_i=0$ outside $S$, and [A1] then forces $S=[m]$. Equality in the first requires each nonnegative gap from step 1.1, multiplied by $p_i>0$, to vanish; hence $q_i\mathrm{/}{p}_i=1$ for every $i$. Conversely $p=q$ gives $D(p\mathrm{\parallel }q)=0$ by [F1].