A bounded likelihood ratio controls KL and log-square moments by Hellinger distance

Put $u=p\mathrm{/}q$ on $\{q>0\}$. The absolute-continuity premise gives $p=0$ $\mu$-almost everywhere on $\{q=0\}$. Define $g(u)=(\sqrt{u}-1{)}^2$ and $a(u)=(u\log u-u+1)\mathrm{/}g(u)$, $b(u)=u(\log u{)}^2\mathrm{/}g(u)$ for $u\mathrm{\ne }1$, with their limiting values at $u=1$ and $u=0$. At zero, set $s=1\mathrm{/}u$; for $s\ge 1$, [F3] gives $0\le \log s\le {\int }_1^s{t}^{-3\mathrm{/}4}dt\le 4s^{1\mathrm{/}4}$. Hence $u\log u=-(\log s)\mathrm{/}s\to 0$ and $u(\log u{)}^2=(\log s{)}^2\mathrm{/}s\le 16s^{-1\mathrm{/}2}\to 0$, so $a(0)=1$ and $b(0)=0$. At one, $\log u\mathrm{/}(u-1)\to 1$ by [F3], $g(u)\mathrm{/}(u-1{)}^2=1\mathrm{/}(\sqrt{u}+1{)}^2\to 1\mathrm{/}4$, and $u(\log u{)}^2\mathrm{/}(u-1{)}^2\to 1$. Also $u\log u-u+1={\int }_1^u\log tdt$, so $[u\log u-u+1]\mathrm{/}(u-1{)}^2\to 1\mathrm{/}2$; therefore $a(1)=2$ and $b(1)=4$. Thus $a,b$ are continuous on $[0,2]$ and bounded there.

For $u\ge 2$, $g(u)=u(1-u^{-1\mathrm{/}2}{)}^2\ge u(1-2^{-1\mathrm{/}2}{)}^2$, while $0\le u\log u-u+1\le u\log u$ and $u(\log u{)}^2$ retain their displayed forms. Hence $a(u)\le C\log u$ and $b(u)\le C(\log u{)}^2$ for a universal $C$. Combining with step 1.1, for every $0\le u\le R$, $u\log u-u+1\le Cg(u)(1+{\log }^{+}R),u(\log u{)}^2\le Cg(u)(1+{\log }^{+}R{)}^2\mathrm{.}$

Integrating the first inequality in step 2.1 against $qd\mu$ gives the KL bound, since $\int q(u\log u-u+1)d\mu =\int p\log (p\mathrm{/}q)d\mu$ by $\int p=\int q=1$ and [F2]. Integrating the second gives the log-square bound. Finally $\int qgd\mu =h^2$ by [F1], so both right sides have the claimed factors of $h^2$.