I'm reading through the paper [Poincaré type and spectral gap inequalities with fractional Laplacians on Hamming cube](https://arxiv.org/abs/1802.04411).


However, I'm having a difficult time understanding the following proof: [Lemma 2.1 page 3](https://i.stack.imgur.com/epHpF.png)).


I understand the general goal of the proof, but the second half in which we establish a lower bound for the integral contains several steps that I find puzzling. Where (and how) do we make use of the fact that $\eta=c\_1^{0.1}$, and why was this specific value chosen? Where does the $\sqrt{3/4}$ come from? Why exactly does $0<\beta\leq 2$ imply that the smallness of $c\_1$ is independent of $\beta$? Any help would be greatly appreciated.



Proof of lower bound on variance
It looks to me like the authors just wanted to put in some ridiculous values which "are clearly sufficient" so that they don't have to work out the details.


The choice $\eta = c\_1^{1/10}$ should ensure that the term $\int\_{|g-1|\leq \eta} |g|^\beta \operatorname{sgn}(g) \, d\mu$ converges to 1 for $c\_1 \rightarrow 0$, uniformly across all $g$ satisfying $\mathbb{E}[g^2] = 1$ and $\mathbb{E}[(g-\mathbb{E}[g])^2 \leq c\_1$ and uniformly across all $\beta$ (this is where you need an upper bound on $\beta$). The term $\sqrt{3/4}$ is then just some value that is sufficiently close to 1 for the remainder of the proof.


To see this convergence, it suffices to show that $\mu(|g-1|\leq \eta)$ goes to 1 using, e.g., Chebyshev's inequality. To this end, we need to use the vague statement "$|1-\mathbb{E}[g]| \ll 1$", which should be made rigorous by establishing a bound on $|\mathbb{E}[g] - 1|$ only depending on $c\_1$. I haven't worked out the details, but I guess something like $|\mathbb{E}[g] - 1| \leq f(c\_1)$, where $f(x) \approx x^{1/2}$ seems reasonable. The choice of $\eta$ is now simply large enough such that \begin{align}
\mu(|g-1| > \eta) \leq \mu(|g-\mathbb{E}[g]| > \eta - f(c\_1)) \leq \mu(|g-\mathbb{E}[g]| > c\_1^{1/3}) \rightarrow 0
\end{align}
for $c\_1 \rightarrow 0$, where of course the $1/3$ exponent is again a rather arbitrary choice by myself, as anything below $1/2$ should work.


