Convergence of a sequence over supremum Given a cadlag-process $X_{t}$ with stationary independent increments (Levy process) for which $E\left[\sup_{s\in[0,t]}\left|X_s\right|\right]<+\infty$ for all $t>0$.
For $n\in \mathbb{N}$ the sequence $K_{n}=\sup_{t\in [n,n+1]}\left|X_t-X_n\right|$ is i.i.d.
Can we conclude, that $\sqrt{n}^{-1}K_{n}\rightarrow 0$ a.s.?
If not, by centering $\tilde{X}_t=X_t-E\left[X_t\right]$ is a martingale. Defining $\tilde{K}_{n}$ analogous via $\tilde{X}_t$, can we conclude
$\sqrt{n}^{-1}\tilde{K}_{n}\rightarrow 0$ a.s.?
Note in both cases $X_t$ and $\tilde{X}_t$ can be written as random walks, for which one can use inequalities e.g. Martingale maximal inequality.
I hope you can help me out. Best regards
 A: Note that for a sequence of nonnegative random variables $\{Z_n\}_{n\geq0}$ the following identity holds
$$\left\{\limsup_{n\to\infty}Z_N=0\right\} 
= \bigcap_{N=1}^\infty\bigcup_{k=1}^\infty \bigcap_{n=k}^\infty \{Z_n\leq 1/N\}
= \bigcap_{N=1}^\infty \left(\limsup_{n\to\infty} \{Z_n> 1/N\} \right)^C.$$
Hence, by the continuity of the probability measure $\mathbb{P}$,
$$\mathbb{P}\left(\limsup_{n\to\infty}Z_n=0\right)
=1 - \lim_{N\to\infty}\mathbb{P}\left(\limsup_{n\to\infty}\{Z_n>1/N\}\right).$$
Since our random variables $Z_n=K_n/\sqrt{n}$ are independent, this is closely related to the finiteness $\sum_{n=1}^\infty \mathbb{P}(Z_n>1/N)$ via the second Borel-Cantelli lemma. If the Lévy process is heavy-tailed, then the random variables $\{K_n\}_{n\geq0}$ will be so too.
Counter-example
Consider the case where $X$ is an $\alpha$-stable process with $1<\alpha<2$. Let $K^+_n =\sup_{t\in[n,n+1]}X_t-X_n$ and $K^-_n=\sup_{t\in[n,n+1]}-(X_t-X_n)$. It is known that $\mathbb{P}(K^\pm_1>x)\sim C_\pm x^{-\alpha}$ as $x\to\infty$ for some positive constants $C_\pm$ (see Doney and Savov's paper https://arxiv.org/abs/1001.4872) except for the spectrally one-sided cases where exactly one of $K_n^+$ or $K_n^-$ is light-tailed, depending on whether $X$ is spectrally negative or positive, respectively (and then, the other one does have said asymptotic behaviour). 
Assume that at least $K_n^+$ is heavy-tailed (that is, $X$ is not spectrally-negative) and otherwise use the same argument for $K_n^-$. Then
$$\mathbb{E}\left(\sup_{s\in[0,t]}|X_s|\right)
=t^{1/\alpha}\mathbb{E}\left(\sup_{s\in[0,1]}|X_s|\right)
\leq t^{1/\alpha}\mathbb{E}\left(K_1^++K_1^-\right)<\infty,$$
however, we also get $$\mathbb{P}\left(K_n/\sqrt{n}>1/N\right)
\geq \mathbb{P}\left(K^+_n>\sqrt{n}/N\right)\sim C_+ N^{\alpha} n^{-\alpha/2}.$$
Since the latter expession is not summable (i.e., $\sum_{n=1}^\infty n^{-\alpha/2}=\infty$), then neither is the expression on the left and so the second Borel-Cantelli lemma implies that
$$\mathbb{P}\left(\limsup_{n\to\infty}\left\{K_n/\sqrt{n}>1/N\right\}\right)=1 \implies\mathbb{P}\left(\limsup_{n\to\infty}K_n/\sqrt{n}=0\right)=0.$$
Sufficient conditions
If you knew that $\mathbb{E}(K_1^p)<\infty$ for some $p>2$, then Markov's inequality yields
$$\sum_{n=1}^\infty \mathbb{P}(K_n/\sqrt{n}>1/N)
\leq \sum_{n=1}^\infty \mathbb{E}(K_1^p)N^pn^{-p/2}<\infty, $$
so the first Borel-Cantelli lemma would instead give
$$\mathbb{P}\left(\limsup_{n\to\infty}\left\{K_n/\sqrt{n}>1/N\right\}\right)=0 \implies\mathbb{P}\left(\limsup_{n\to\infty}K_n/\sqrt{n}=0\right)=1,$$
as required.
