Suppose $\mu$ is a probability measure on $S$ and $f:S\rar\R_0^+$ is such that $\int f\,d\mu=1$. Then $$ \int|f-1|\,d\mu\leq\sqrt{-2\Ent(f)}~. $$ This is called Pinsker's inequality. Hint: Use the elementary inequality: $3(x-1)^2\leq(4+2x)(x\log x-x+1)$, which holds for all $x\geq0$. Thus if $\Ent(P^nf)$ converges to $0$, then $P^nf$ converges in $L_1(\mu)$ to the constant function $1$.
By Cauchy-Schwarz we infer from the elementary inequality: $3(x-1)^2\leq(4+2x)(x\log x-x+1)$ that \begin{eqnarray*} \sqrt3\int|f-1|\,d\mu &\leq&\int\sqrt{4+2f}\sqrt{f\log f-f+1}\,d\mu\\ &\leq&\sqrt{\int 4+2f\,d\mu}\sqrt{\int f\log f-f+1\,d\mu} =\sqrt6\sqrt{-\Ent(f)}~. \end{eqnarray*}