I am quite new to Lévy processes and stumbled upon this problem and I cant seem to figure it out. If $(L_t)_{t\geq}$ is a Lévy process and $p>0$. How would one show that $$ L_t \in \mathcal{L}^P \Leftrightarrow L_1 \in \mathcal{L}^P $$ My idea is to use write $$ L_t = L_0 + \sum_{i=1}^t (L_i - L_{i-1}), $$ Where $L_i-L_{i-1} \stackrel{D}{=} L_{i-(i-1)} =L_1$ and thus $L_t \stackrel{D}{=} t L_1$ which would result in the if and only if statement.
Is this the right idea?
It is not correct to conclude that $L_t\stackrel{d}{=}t\cdot L_1$. For a concrete counterexample simply consider Brownian Motion or the homogeneous Poisson process on $(0,\infty)$.
However, your approach is going in the right direction. Assuming $L_0=0$ and writing for $n\in\mathbb N$: $$L_n=\sum_{i=1}^n L_i-L_{i-1}$$ we can conclude that $$\int |L_t|^pd\mathbb P = \int \bigg|\sum_{i=1}^n L_i-L_{i-1}\bigg|^pd\mathbb P\leq \sum_{i=1}^n \int |L_i-L_{i-1}|^pd\mathbb P. $$ Now noting that $L_i-L_{i-1}\stackrel{d}{=}L_1$ we can see that $\mathbb E|L_i-L_{i-1}|^p <\infty$ and hence $\mathbb E|L_t|^p<\infty$.
Now for non-integer $t>0$ we can simply write $$L_t\stackrel{}{=}L_{\lfloor t\rfloor}+(L_{\lfloor t\rfloor+(t-\lfloor t\rfloor)} - L_{\lfloor t\rfloor})$$ and note that the first term is in $L_p$ and the $p$-norm of the second term can be shown to be finite by $$\infty > \mathbb E|L_1|^p = \mathbb E\bigg|L_{\lfloor t\rfloor+(t-\lfloor t\rfloor)} - L_{\lfloor t\rfloor}+ L_{\lfloor t\rfloor +1} -L_{\lfloor t\rfloor+(t-\lfloor t\rfloor)}\bigg|^p$$