This lower bound estimate can be obtained by analyzing the regularity scale $\mathfrak{s}(\bm{x})$. Denote by $w\equiv\frac{L_{T}(\bm{x})}{T}$ and recall that the two end points $T_{-},T_{+}$ satisfy

(4.279)

$\displaystyle\begin{cases}L_{T}(T_{-})=T^{\frac{n-2}{n}}\\ L_{T}(T_{+})=T^{\frac{n-2}{n}},\end{cases}$

then we have $w\in[T^{-\frac{2}{n}},1]$. So it follows that

(4.280)

$$\rho_{\delta,\nu,\mu}^{(k+\alpha)}=F(w)\cdot\mathfrak{r}(\bm{x})^{\nu+k+\alpha}\cdot T^{\frac{\nu+k+\alpha}{n}+\mu},$$

where $F(w)\equiv e^{\delta\cdot T(1-w^{\frac{n}{2}})}\cdot w^{\frac{\nu+k+\alpha}{2}}$. By the definition of $\mathfrak{r}(\bm{x})$, immediately we have

(4.281)

$\displaystyle T^{-1}\leq\mathfrak{r}(\bm{x})\leq 1$

for all $\bm{x}\in\mathcal{M}_{T}$, so it follows that

(4.282)

$\displaystyle\rho_{\delta,\nu,\mu}^{(k+\alpha)}\geq\begin{cases}F(w)\cdot T^{(\frac{1}{n}-1)(\nu+k+\alpha)+\mu},&\nu+k+\alpha\geq 0,\\ F(w)\cdot T^{\frac{\nu+k+\alpha}{n}+\mu},&\mu+\nu+k+\alpha<0,\end{cases}$

Now it suffices to compute the lower bound of $F(w)$. To this end, there are two cases to analyze depending on the sign of $\nu+k+\alpha$. First, let $\nu+k+\alpha\leq 0$, then obviously $F(w)\geq F(1)=1$ and hence

(4.283)

$$\rho_{\delta,\nu,\mu}^{(k+\alpha)}(\bm{x})\geq T^{(\frac{1}{n}-1)(\nu+k+\alpha)+\mu}.$$

Next, we consider the case $\nu+k+\alpha>0$. Simple calculus shows that $F(w)$ achieves its minimum in $[T^{-\frac{2}{n}},1]$ either at $w=1$ or at $w=T^{-\frac{2}{n}}$. Notice that $F(T^{-\frac{2}{n}})\gg F(1)$ as $T\gg 1$. This tells us that

(4.284)

$$\rho_{\delta,\nu,\mu}^{(k+\alpha)}(\bm{x})\geq T^{\frac{\nu+k+\alpha}{n}+\mu}.$$

The proof is done.

∎