# Banach space :space of all adapted processes continuous equipped wih specific norm is complete

Let $\mathbb{B}$ be space of all adapted processes continuous equipped with the norm $\lVert Y\rVert_{\mathbb{B}}^2=E\left[\sup_{t\in [0,T]} |Y_{t}|^{2}\right] < \infty$,
$(B,\lVert\cdot\rVert_{\mathbb{B}})$ is complete ( is Banach space ).

Indeed, suppose $(X^{n})_{n\in \mathbb{N}}$ is a Cauchy sequence in $(\mathbb B,\lVert\cdot\rVert_{\mathbb{B}})$. Then we can find a subsequence $(n_{k})_{k\in \mathbb{N}}$ such that \begin{align} \sum_{n=1}^{\infty} \lVert X^{n_{k+1}}-X^{n_k}\rVert_{\mathbb B} &<\infty \mbox{ by the triangular inequality, }\\ E\left[ \sum_{n=1}^{\infty}\sup_{t\geq 0}|X_{t}^{n_{k+1}}-X_{t}^{n_k}|^{2}]^{\frac{1}{2}}\right] &\leq \sum_{n=1}^{\infty}E\left[\sup_{t \geq 0}|X^{n_{k+1}}-X^{n_k}|^{2}\right]^{\frac{1}{2}} \\ &=\sum_{n=1}^{\infty}\lVert X^{n_{k+1}}-X^{n_k}\rVert_{\mathbb{B}}<\infty\\ \mbox{Then } E[ \sum_{n=1}^{\infty}\sup_{t \geq 0}|X_{t}^{n_{k+1}}-X_{t}^{n_k}|^{2}]^{\frac{1}{2}} &<\infty \end{align} and so for almost every $\omega \in \Omega$ $$\sum_{k=1}^{\infty}\sup_{t \geq 0}|X_{t}^{n_{k+1}}-X_{t}^{n_{k}}|<\infty$$ then $(\mathbb{B},|.|_{\mathbb{\infty}})$ with $\lVert\cdot\rVert_{\infty}=\sup_{0\leq t \leq T }|X_{t}|$ then $(B,\lVert\cdot\rVert_{\infty})$ is complete then there exists a process $X\in \mathbb{B}$ such that $(X_{t}^{n_{k}}(\omega))_{k\in \mathbb{N}}$

$$(X_{t}^{n_{k}}(\omega))_{k\in \mathbb{N}} \xrightarrow[n \longrightarrow \infty]{C.U} X(\omega) \hspace{4mm} \forall t \geq 0$$ now: \begin{align} |X^{n}-X|_{\mathbb{B}}^{2}&=E[\sup_{t \geq 0}|X_{t}^{n}-X_{t}|^{2}]\\ \text{by Fatou's lemma} & \leq \lim_{k \to \infty} \inf E\left[ \sup_{t \leq 0}|X_{t}^{n}-X_{t}^{n_{k}}|^{2}\right]^{2}\\ &=\lim_{k \to \infty} \inf|X^{n}-X^{n_{k}}|_{\mathbb{B}}^{2}\xrightarrow[n \longrightarrow \infty]{} 0 \end{align} since $(X^{n})_{n\in \mathbb{N}}$ is Cauchy sequence $(\lim_{n\to \infty}\sup |X^{n}-X^{n_{k}}|=0)$.

My question: Am i right? If not, please correct. Please respond. I'll be grateful for any help offered!