exercise 1.21 of chapter 1 of Revuz and Yor's This is the exercise 1.21 of chapter 1 of Revuz and Yor's:
Let $X=B^+$ or $|B|$ where $B$ is the standard linear BM, $p$ be a real number $>1$ and $q$ its conjugate number ($q^{-1}+p^{-1}=1$).


*

*Prove that the r.v. $J_p=\sup_{t\geq0}(X_t-t^{p/2})$ is a.s. strictly positive and finite and has the same law as $\sup_{t\geq0}(X_t/(1+t^{p/2}))^q$.

*Using time-inversion, show that
$$\sup_{t\geq0}(X_t/(1+t^{p/2}))\overset{(d)}{=}\sup_{u\leq1}\left(\frac{1}{1+u^{p/2}}\right)\left(\frac{X_u}{u^{1-p/2}}\right)$$
and conclude that $E[J_p]<\infty$.

*Prove that there exists a constant $C_p(X)$ such that for any positive r.v. $L$
$$E[X_L]\leq C_p(X)||L^{1/2}||_p.$$

*Let $L_\mu$ be a random time such that
$$X_{L_\mu}-\mu L_{\mu}^{p/2}=\sup_{t\geq0}(X_t-\mu t^{p/2}).$$
Prove that $L_\mu$ is a.s. unique and that the constant $C_p(X)=p^{1/p}(qE[J_p])^{1/q}$ is the best possible.

*Prove that
$$E[X_{L_1}|J_p]=qJ_p \text{, }E\left[L_1^{p/2}|J_p\right]=\frac{q}{p}J_p.$$
By the hints and the reference cited in the notes of the book, I almost solved this question. My only question is : In the proof of 4, I think I need $EL_\mu^{p/2}<+\infty$ (or $EX_{L_\mu}<+\infty$) firstly, but can't figure it out. Any help, thanks!
 A: Indeed, one needs to verify that $E[L^{p/2}_\mu]<+\infty.$ Consider the inequality 
$$
E[L^{p/2}_\mu]\leq 1+\sum^\infty_{n=0}2^{(n+1)p/2}P(2^n<L_\mu\leq 2^{n+1}).
$$
To bound the probability $P(2^n<L_\mu\leq 2^{n+1})$ few crude estimates are enough. The event $\{2^n<L_\mu\leq 2^{n+1}\}$ implies that the supremum of the function $t\mapsto X_t-\mu t^{p/2}$ is attained on $(2^n,2^{n+1}].$ Since this supremum is positive, it follows that for some $t\in (2^n,2^{n+1}]$
$$
X_t>\mu t^{p/2}.
$$
Further it follows that $X_t>\mu 2^{np/2}$ and, finally, 
$$
\sup_{0\leq t\leq 2^{n+1}}|B_t|>\mu 2^{np/2}.
$$
Consequently,
$$
P(2^n<L_\mu\leq 2^{n+1})\leq P\bigg(\sup_{0\leq t\leq 2^{n+1}}|B_t|>\mu 2^{np/2}\bigg)\leq 4P\bigg(B_1>\frac{\mu}{\sqrt{2}}2^{(p-1)\frac{n}{2}}\bigg).
$$
Using tail estimates for the normal distribution it is easy to verify that the following series  converges:
$$
\sum^\infty_{n=0}2^{\frac{pn}{2}}P\bigg(B_1>\frac{\mu}{\sqrt{2}}2^{(p-1)\frac{n}{2}}\bigg)<\infty.
$$
These estimates show that $L_\mu$ actually has all moments.
