\leq
    \frac{a(n)}{b(n)} \leq M$ for all $n$.

Let us first examine the case $0 < r < 2$. Remark that the left-hand term of the second equation in (\[eq:7\]) is strictly negative when $1/\delta = 2\sqrt{N}$ and increases to $\infty$ with $1/\delta$. This proves that the solution satisfies $1/\delta
    > 2\sqrt{N}$ and that Proposition \[prop:4\] applies. Furthermore, if we suppose that $\frac{1/\delta}{\sqrt{N}}$ is unbounded when $n \to \infty$, then (up to taking a subsequence) by the first equation $\frac{\frac12+2\gamma}{\delta^2} \sim \log(n)$ whereas, by subtracting the two, $\frac{2 \gamma}{\delta^2} \sim \log(n)$, which is contradictory. So $1/\delta \approx \sqrt{N}$ and we deduce that $N \approx \log(n)$. Then (\[biaisbis\]) yields $$\log{{\left( \frac{b_2^2(n)}{N^{2-r/2} e^{-2BN^{r/2}}} \right)}} \leq
        (4r-12) \log(\delta) + \frac{r}{2} \log(N) - (\log\log(n))^2
        \to -\infty$$ whereas (\[eq:varbcpdebruit\]) gives $$\log{{\left( \frac{\sigma^2(n)}{N^{2-r/2} e^{-2BN^{r/2}}} \right)}} \leq
        \log(\delta) + (\frac56+\frac{r}{2}) \log(N) - (\log\log(n))^2
        \to -\infty.$$ We see that the dominant term is the bound (\[biais\]) on $b_1^2(n)$, hence the result.

When $r = 2$, the same reasoning as above yields $1/\delta >
    2\sqrt{N}$, $1/\delta \approx \sqrt{N}$ and $N \approx
    \log(n)$. Then the right-hand side of (\[biaisbis\]) and (\[eq:varbcpdebruit\]) are of the same order as $N
    e^{-2BN}$, which is the bound (\[biais\]) on $b_1^2(n)$.

Wigner function estimation {#sec.Wigner}
==========================

