Better lower bound on min-entropy In “Randomness Condensers for Efficiently Samplable,
Seed-Dependent Sources” by Dodis, Ristenpart, and Vadhan (PDF), I have seen the statement that:


  any tuple of distributions $(X,Z)$ is $\varepsilon$-close
  to $(X′, Z)$ such that $$H_\infty(X′\mid Z) ≥ H_2(X\mid Z) − \log (1/\varepsilon)$$ offering a bound on min-entropy which may be better than $H_\infty(X′\mid Z) ≥ H_2(X\mid Z)/2$. 


I can not see why this is the case. Could anyone please shed some light? Does it involve the Leftover Hash Lemma!?
The proof (courtesy of M. Skorski) relies on the so-called "smoothing" technique [Cac97,RW,Sko15]. 

For some $0&lt;\epsilon&lt;1$, let $B$ be the subset of heaviest points in $X$ whose mass add up to $\epsilon$ -- i.e.:
  $$\sum_{x\in B} p_X(x)=\epsilon, \text{ and } \min_{x\in B}p_X(x)\geq\max_{x\in\bar{B}}p_X(x).$$
Let $X'$ denote the conditional distribution of $X$ on $\bar{B}$ -- i.e.:
  $$
  p_{X'}(x):=
    \begin{cases}
      0 &amp; x\in B,\\
      \frac{p_{X}(x)}{(1-\epsilon)} &amp; x\in\bar{B}.
    \end{cases}
  $$
Note that the statistical distance between $X$ and $X'$, $\Delta(X,X')=\epsilon$.$^1$ Let $m:=\max_{x\in\bar{B}}p_X(x)$. Now, let's consider the collision entropy of $X$:
  $$H_2(X):=-\log\sum_{x\in X} p_X^2(x)\leq-\log\sum_{x\in B} p_X^2(x)\leq^2-\log (m\epsilon).$$ 
It follows that $-\log(m)\geq H_2(X)-\log(1/\epsilon)$, and the results follows as $H_\infty(X')=-\log(m/(1-\epsilon))$.

A similar argument can be made for a joint distribution too.

References:
[Cac97]: C. Cachin. Smooth Entropy and Renyi Entropy, EUROCRYPT'97.
[RW]: R. Renner and S. Wolf. Smooth Renyi Entropy and its Applications.
[Sko15]: M. Skorski. How to Smooth Entropy? SOFSEM'16

Footnotes:
1. Proof: 
  \begin{align}
    \Delta(X,X')
      &amp;=\frac{1}{2}\cdot\sum_{x\in X}\left|p_X(x)-p_{X'}(x)\right|\\
      &amp;=\frac{1}{2}\cdot\sum_{x\in B}\left|p_X(x)\right|+\sum_{x\in \bar{B}}\left|p_X(x)-p_{X'}(x)\right|\\
      &amp;=\frac{1}{2}\cdot\left(\epsilon+ \sum_{x\in \bar{B}}p_X(x)\left|1-\frac{1}{1-\epsilon}\right|\right)\\
      &amp;=\frac{1}{2}\cdot\left(\epsilon+ (1-\epsilon)\left|1-\frac{1}{1-\epsilon}\right|\right)=\epsilon
  \end{align}
2. Proof: $\sum_{x\in B} p_X^2(x)\geq \sum_{x\in B}m\cdot p_X(x)= m\cdot\epsilon$