For a Levy process $(X_t)$, it is intuitively that, for any $\varepsilon>0$, $$ P(\sup_{0\le s\le t}|X_s|>\varepsilon)\to0\;\;\mbox{as}\;\;t\to0. $$ Is there any useful inequality to claim this or some related references?
If we apply the Markov's inequality, then $$ P(\sup_{0\le s\le t}|X_s|>\varepsilon)\le\frac{E(\sup_{0\le s\le t}|X_s|)}{\varepsilon}. $$ But, how to see that $E(\sup_{0\le s\le t}|X_s|)\to0$ as $t\to0$?
Thanks.
Lévy processes are, in general, not integrable and therefore Markov's inequality is not good enough.
Note that
$$\mathbb{P} \left( \sup_{0 \leq s \leq t} |X_s| > \varepsilon \right) \xrightarrow[]{t \to 0} 0\tag{1}$$
is equivalent to saying that $\sup_{0 \leq s \leq t} |X_s|$ converges in probability to $0$ as $t \to 0$. By the definition of Lévy processes, we know that $(X_t)_{t \geq 0}$ has (almost surely) right-continuous sample paths and $X_0 = 0$; this implies in particular that
$$\lim_{t \to 0} \sup_{s \leq t} |X_s| = 0 \qquad \text{a.s.},$$
i.e. $\sup_{s \leq t} |X_s|$ converges almost surely to $0$. Since pointwise convergence (almost surely) implies convergence in probability, this proves $(1)$.