One question about compensated Poisson process Let $N$ be a Poisson process with parameter $\lambda$, that is, for $a>b\geq0$, there is $$P[N(a,b)=k]=\frac{((a-b)\lambda)^k}{k!}e^{-(a-b)\lambda}.$$ Now denote $N_t=N[0,t)$ and define 
$$
M_t=N_t-\lambda t\qquad 0\leq t<\infty.
$$
we can easily show that $M$ is a Martingale. Now the question is, prove that for any $c>0$, there is
\begin{align*}
\limsup_{t\rightarrow\infty}&P\left[\sup_{s\leq t}M_s\geq c\sqrt{\lambda t}\right]\leq\frac{1}{c\sqrt{2\pi}}\\
   &E\left[\sup_{s\leq u\leq t}\left(\frac{M_u}{u}\right)^2\right]\leq\frac{4t\lambda}{s^2}
\end{align*}
 The main difficulty I meet is how to describe the $sup$ of random variables and how to make use of the fact that $M$ is a martingale to evaluate the probability.
 A: $\newcommand{\ep}{\epsilon}
\newcommand{\ga}{\gamma}
\newcommand{\la}{\lambda}
\newcommand{\Si}{\Sigma}
\newcommand{\R}{\mathbb{R}}
\newcommand{\E}{\operatorname{\mathsf E}} 
\newcommand{\PP}{\operatorname{\mathsf P}}$ 
If $(S_t)_{t\ge0}$ is a nonnegative submartingale, then the following Dood inequalities hold: 
\begin{equation}
 P(\sup_{0\le u\le t} S_u\ge x)\le ES_u/x \tag{1}
\end{equation}
for $x>0$ and 
\begin{equation}
 \|\sup_{0\le u\le t} S_u\|_p\le\frac p{p-1}\,\|S_t\|_p \tag{2}
\end{equation}
for $p>1$, where $t>0$ and $\|X\|_p:=(E|X|^p)^{1/p}$.
Using (1) with $S_u=\max(0,M_u)$, we have 
\begin{multline*}
 P(\sup_{s\le t}M_s\ge c\sqrt{\la t})
 =P(\sup_{s\le t}\max(0,M_s)\ge c\sqrt{\la t}) \\ 
 \le\frac{E\max(0,M_t)}{c\sqrt{\la t}}
 \underset{t\to\infty}\longrightarrow\frac{E\max(0,Z)}{c}=\frac1{c\sqrt{2\pi}}
\end{multline*}
by the central limit theorem, where $Z\sim N(0,1)$. 
Using (2) with $S_u=M_u^2$, for $0<s<t$ we have 
\begin{equation}
 E\sup_{s\le u\le t}\left(\frac{M_u}{u}\right)^2
 \le \frac1{s^2}\,E\sup_{s\le u\le t}M_u^2
  \le \frac1{s^2}\,\Big(\frac 2{2-1}\Big)^2\,EM_t^2
 =\frac{4t\la}{s^2}. 
\end{equation}
