We assume a complete filtered probability space with right continuous filtration.

Let $\mathbb{S}^2$ be the space of adapted and almost surely continuous processes $X = (X_t)_{t \in [0,T]}$ with values in $\mathbb{R}$ and $$ E[ \sup_{t \in [0,T] } | X_t | ^2 ] < \infty. $$ Define the norm $$ || X ||^2 = E \left[ \int_0^T |X_s|^2 ds \right]. $$

Is $\mathbb{S}^2$ a Banach space?

I know it will a Banach space w.r.t the uniform norm.

  • $\begingroup$ No, I don't think so. If $(X_n)_{n \in \mathbb{N}} \subseteq \mathbb{S}^2$, then $X=\lim_n X_n$ exists in $L^2$-sense but there is no reason why $X$ should have continuous sample paths with probability 1. $\endgroup$ – saz Nov 4 '18 at 12:34
  • $\begingroup$ Thank you. I forgot one detail in the definition of $\mathbb{S}^2.$ But, I guess this does not make any difference. $\endgroup$ – White Nov 4 '18 at 12:39
  • $\begingroup$ No, it doesn't make a difference. $\endgroup$ – saz Nov 4 '18 at 12:51

Let $X_t(\omega) =(\frac t T)^{n}$. This sequence converges in the norm to $Y_t(\omega) =1$ for $t=T$ and $0%$ for $t <T$. Since $Y_t$ does not have continuous paths the space is not complete.


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