A weighted total-variation bound from relative entropy

Define $\varphi (0)=1$ and $\varphi (t)=t\log t-t+1$ for $t>0$. For $0<t\le 1$, the function $h(t)=t-1-\log t$ satisfies $h(1)=0$ and $h^{\mathrm{\prime }}(t)=1-1\mathrm{/}t\le 0$. The mean value theorem on $[t,1]$ gives $h(t)\ge 0$, so $\log t\le t-1$. Now $k(t)=\varphi (t)-(1-t{)}^2\mathrm{/}2$ satisfies $k(1)=0$ and $k^{\mathrm{\prime }}(t)=\log t+1-t\le 0$ by [F4]–[F5]; another application of the mean value theorem gives $k(t)\ge 0$. At $t=0$ the inequality holds because $1\ge 1\mathrm{/}2$.

For $t\ge 1$, put $g(t)=2(1+t)\varphi (t)-(t-1{)}^2$. By [F5], $g(1)=g^{\mathrm{\prime }}(1)=0$ and $g^{\mathrm{\prime }\mathrm{\prime }}(t)=4\log t+2\mathrm{/}t\ge 0$ using [F4] and $\log t\ge 0$ for $t\ge 1$. The mean value theorem applied first to $g^{\mathrm{\prime }}$ and then to $g$ shows $g^{\mathrm{\prime }}(t)\ge 0$ and $g(t)\ge 0$ for $t\ge 1$. Combining with step 1.1 gives $\mathrm{\mid }1-t{\mathrm{\mid }}^2\le 2(1+t)\varphi (t)(t\ge 0)\mathrm{.}$

It remains to consider finite $D(\mu \mathrm{\parallel }\nu )$. Let $r=d\mu \mathrm{/}d\nu$ as in [F1]. For every Borel $f$ with $\mathrm{\mid }f\mathrm{\mid }\le W$, the Radon–Nikodym identity gives $\mid \int fd\mu -\int fd\nu \mid \le \int W\mathrm{\mid }r-1\mathrm{\mid }d\nu \mathrm{.}$ Applying step 2.1 pointwise and [F3] with factors $W\sqrt{1+r}$ and $\sqrt{2\varphi (r)}$ bounds this by $\sqrt{\int W^2(1+r)d\nu }\sqrt{2\int \varphi (r)d\nu }=\sqrt{(\int W^{2}d\nu +\int W^{2}d\mu )2D(\mu \mathrm{\parallel }\nu )}\mathrm{.}$ The moment hypotheses make the first factor finite, and [F1], [F2], and [F6] identify the second integral with the directional relative entropy, using $\int rd\nu =\int 1d\nu =1$. Taking the supremum over $f$ proves the claim.