Can we show that $\sup_{t\in[0,\:T]}X_t\in\mathcal L^2(\text P)\Leftrightarrow\sup_{t\in[0,\:T]}\left\|X_t\right\|_{\mathcal L^2(\text P)}<\infty$?

Question

Let

• $T>0$
• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
• $(X_t)_{t\in[0,\:T]}$ be a real-valued stochastic process on $(\Omega,\mathcal A,\operatorname P)$

Please consider the following properties:

1. $\sup_{t\in[0,\:T]}X_t\in\mathcal L^2(\text P)$
2. $\sup_{t\in[0,\:T]}\left\|X_t\right\|_{\mathcal L^2(\text P)}<\infty$

It's easy to see that (1.) implies (2.).

Can we show the other direction too?

It's worth noting that if $X$ is almost surely continuous, then $$\sup_{t\in[0,\:T]}X_t\in\mathcal L^2(\text P)<\infty\Leftrightarrow\operatorname E\left[\sup_{t\in[0,\:T]}\left|X_t\right|^2\right]=\operatorname E\left[\left|X_T\right|^2\right]<\infty\Leftrightarrow X_T\in\mathcal L^2(\operatorname P)\;.\tag 1$$ So, at least in that case (2.) implies (1.) too.

Answer 1

Not only this is false in general (as already answered by John Dawkins), but also this is false for a.s. continuous processes (contrary to your claim). Indeed, let for $T=1$, $\xi$ be a random variable such that $\mathsf{P}(\xi = n) = \frac{1}{n(n+1)}, n\ge 1$, and $f_n$ be a positive continuous function supported by $[1/(n+1), 1/n]$ with $\max f_n = n$. Define $X_t = f_\xi(t)$. Then $$\mathsf E[X_t] \le \frac1t \mathsf P(\xi = \lfloor1/t\rfloor )\le 1,\ t\in[0,1]$$ but $\mathsf E[\sup_{[0,1]} X_t] = \mathsf{E}[\xi] = +\infty$.

Answer 2

No. Let $B_s$, $s\ge 0$, be a standard Brownian motion, and consider $X_t:={B_{t/(1-t)}\over \sqrt{t/(1-t)}}$, for $0< t\le 1=:T$. (And $X_0:=0$.) Then $E[X^2_t]\le 1$ for all $t\in[0,1]$ but $\sup_{0\le t\le 1}X_t=+\infty$ by the law of the iterated logarithm.