First, we prove (5.152). Both the upper bound and lower bound estimates can be proved in the similar way:

	$\displaystyle\Tri(\beta,\alpha,y)$	$\displaystyle=\frac{e^{y}}{\Gamma(\fa-\fb)}\int_{0}^{\infty}e^{yt}t^{\fa-\fb-1}(1+t)^{\fb-1}dt$
		$\displaystyle\leq\frac{e^{y}}{\Gamma(\fa-\fb)}\cdot\int_{0}^{\infty}e^{yt}t^{\fa-\fb-1}dt$
		$\displaystyle=\frac{e^{y}\cdot(-y)^{\beta-\alpha}}{\Gamma(\fa-\fb)}\cdot\int_{0}^{\infty}e^{-u}u^{\fa-\fb-1}du$
(5.154)			$\displaystyle=e^{y}\cdot(-y)^{\beta-\alpha}.$

Similarly,

	$\displaystyle\Tri(\fb,\fa,y)$	$\displaystyle\geq\frac{e^{y}}{\Gamma(\fa-\fb)}\int_{0}^{1}e^{yt}t^{\fa-\fb-1}(1+t)^{\fb-1}dt$
		$\displaystyle\geq C_{n}\cdot e^{y}\int_{0}^{1}e^{yt}t^{\fa-\fb-1}dt$
(5.155)			$\displaystyle\geq C_{n}\cdot e^{y}\cdot(-y)^{\fb-\fa}.$

Next, we prove the upper bound estimate for $\Ku$. Notice in the proof of Lemma 5.9 we do not need the condition $Q\leq 1$ for the upper bound on $\Ku$. So we have

(5.156)

$$\Ku(\beta,\alpha,y)\leq C_{n}\cdot\frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)}\cdot(-y)^{\frac{1-2\alpha}{4}}\cdot e^{y+G(u_{0})}.$$

To prove (5.153), we need an upper bound estimate for $e^{y+G(u_{0})}$. This follows from elementary computations. In fact,

$$e^{y+G(u_{0})}=e^{y-u_{0}^{2}+2\sqrt{-y}u_{0}}\cdot(u_{0})^{2Q+\gamma_{n}}\\ \leq C_{n}\cdot e^{y-u_{0}^{2}+2\sqrt{-y}u_{0}}\cdot(-y)^{Q+\frac{\gamma_{n}}{2}}.$$

Notice that $u_{0}$ satisfies $G^{\prime}(u_{0})=0$, i.e.,

(5.157)

$$u_{0}^{2}-\sqrt{-y}\cdot u_{0}-\frac{2Q+\gamma_{n}}{2}=0,$$

so we have

(5.158)

$$e^{y+G(u_{0})}\leq C_{n}\cdot e^{y+\sqrt{-y}u_{0}}\cdot(-y)^{Q+\frac{\gamma_{n}}{2}}.$$

By (5.115), it is straightforward that

(5.159)

$\displaystyle y+\sqrt{-y}u_{0}=\frac{y}{2}\Big(1-\sqrt{1+\frac{4Q+2\gamma_{n}}{-y}}\Big)=\frac{2Q+\gamma_{n}}{1+\sqrt{1+\frac{4Q+2\gamma_{n}}{-y}}}\in[C_{n}^{-1},C_{n}],$

for some dimensional constant $C_{n}>0$. Therefore,

(5.160)

$\displaystyle e^{y+G(u_{0})}\leq C_{n}(-y)^{Q+\frac{1}{4}+\frac{1}{2n}},$

and hence

(5.161)

$\displaystyle\Ku(\fb,\fa,y)\leq C_{n}(-y)^{Q+\frac{1}{n}}=C_{n}(-y)^{-\beta}.$

This completes the proof. ∎