# Understanding the relationship of the $L^1$ norm to the total variation distance of probability measures, and the variance bound on it

I am trying to find a bound for variance of an arbitrary distribution $f_Y$ given a bound of a Kullback-Leiber divergence from a zero-mean Gaussian to $f_Y$, as I've explained in this related question. From page 10 of this article, it seems to me that:

$$\frac{1}{2}\left(\int_{-\infty}^{\infty}|p_Z(x)-p_Y(x)|dx\right)^2 \leq D(p_Z\|p_Y)$$

I have two questions:

1) How does this come about? The LHS is somehow related to total variation distance, which is $\sup\left\{|\int_A f_X(x)dx-\int_A f_Y(x)dx|:A \subset \mathbb{R}\right\}$ according to wikipedia article, but I don't see a connection. Can someone elucidate?

2) Section 6 on page 10 of the same article seems to talk about variation bounds, but I can't understand it... Can someone "translate" that to the language that someone with a graduate-level course on probability can understand? (I haven't taken measure theory, unfortunately.)

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Hint for (1): Consider the set $A = \{x: f_X(x) \leq f_Y(x)\}$. – cardinal Oct 2 '11 at 1:30
@cardinal (sorry for the delay) Using your hint I've been able to equate $\mathcal{L}_1$ distance with double of total variation as follows: \begin{align}\int_{-\infty}^\infty |f_Z(x)-f_Y(x)|dx&=\int_A f_Z(x)-f_Y(x) dx+\int_{A^c} f_Y(x)-f_Z(x) dx\\ &=\int_A f_Z(x)dx-\int_A f_Y(x) dx+\int_{A^c} f_Y(x)dx-\int_{A^c}f_Z(x) dx\\&=\int_A f_Z(x)dx-\int_A f_Y(x) dx+1-\int_A f_Y(x)dx-1+\int_Af_Z(x) dx\\ &=2\left(\int_A f_Z(x) dx-\int_A f_Y(x) dx\right)\end{align} and obviously set $A$ yields the supremum in total variation. However, I can't quite connect this to KL divergence... – M.B.M. Oct 12 '11 at 23:48

2) The LHS is essentially the total variation between probability measures $p_Z$ and $p_Y$ (see here). I think "variation bounds" quite literally means bounds on the total variation between the probability measures, as given in the Lemma on p. 11.