I was going through this paper in which the weak $L_\alpha$ norm $$\Vert X\Vert_{\alpha,w}^\alpha = \sup_{\lambda>0} \lambda^\alpha \mathbb P\bigg[\sup_{0\leq t\leq T} |X_t|>\lambda\bigg]$$ was defined.
The space $$\Lambda^\alpha :=\bigg\lbrace X \text{ cadlag and adapted } : \Vert X\Vert_{\alpha,w}<\infty\bigg\rbrace $$ was introduced.
Now it was claimed that $\Lambda^\alpha$ is a complete linear metric space for all $0<\alpha<2$.
It's easy to show that this space is indeed linear, but i have some trouble verifying that it is complete. Here is my attempt to far
Let $(X^{(n)})_{n\in\mathbb N}$ be a Cauchy sequence (w.r.t. $\Vert\cdot\Vert_{\alpha,w}$) in $\Lambda^\alpha$, meaning for all $\epsilon>0$ and sufficiently large $M$ we have $$\epsilon> \Vert X^{(n)}-X^{(m)}\Vert_{\alpha,w}^\alpha = \sup_{\lambda>0}\lambda^\alpha \mathbb P\bigg[\sup_{0\leq t\leq T} |X_t^{(n)}-X_t^{(m)}|>\lambda\bigg]\geq \sup_{\lambda>0} \lambda^\alpha \mathbb P[|X_t^{(n)}-X_t^{(m)}|>\lambda]$$ for $n,m>M$ and each $t\in[0,T]$. This implies that the finite-dimensional projections of $X^{(n)}$ converge weakly. Now i would have to verify some tightness criterion of the sequence $X^{(n)}$ to show that it converges weakly in $D([0,T])$ and show that its limit $X$ satisfies $\Vert X\Vert_{\alpha,w}<\infty$.
For the tightness criterion, one might use the characterization using the modulus of continuity $w'$, meaning one has to show that $$\lim_{\delta\to 0}\limsup_{n\to\infty} \mathbb P[w'(X^{(n)},0,T,\delta)>\epsilon]=0.$$ Now by the triangle inequality we get $$\mathbb P[w'(X^{(n)},0,T,\delta)>\epsilon]\leq \mathbb P[w'(X^{(n)}-X^{(m)},0,T,\delta)>\epsilon]+\mathbb P[w'(X^{(m)},0,T,\delta)>\epsilon] \hspace{1cm} (*)$$ which i think is sufficient to show that $\lim_{\delta\to 0}\limsup_{n\to\infty} (*)=0$.
Is this proof correct? Is there an easier way to show the completeness?
This is essentially correct, but the "I think" at the end can be made more precise. Indeed, first, one has $$ \omega'\left(X^{(n)}-X^{(m)},0,T,\delta)>\varepsilon\right)\leqslant 2\sup_{0\leqslant t\leqslant T}\left\lvert X_t^{(n)}-X_t^{(m)}\right\rvert. $$ For a fixed $\eta$, choose $N$ such that for each $m,n\geqslant N$, $\lVert X^{(n)}-X^{(m)}\rVert^\alpha_{\alpha,w}\leq \eta$. Then for $m,n\geqslant N$, $$ \mathbb P[w'(X^{(n)},0,T,\delta)>\varepsilon]\leqslant\varepsilon^{-\alpha}\eta+\mathbb P[w'(X^{(m)},0,T,\delta)>\varepsilon]. $$ Taking $\limsup_{n\to\infty}$, keeping $m\geqslant N$ fixed, and using tightness of $X^{(m)}$ gives that $$ \lim_{\delta\to 0}\limsup_{n\to\infty}\mathbb P[w'(X^{(n)},0,T,\delta)>\varepsilon]\leqslant\varepsilon^{-\alpha}\eta. $$ Don't forget to show that the limit $X$ is adapted.