Consider, on a filtred probability space $ \left (\Omega, \mathcal F, \mathbb F , \mathbb P \right )$ where $ \mathbb F = \left(\mathcal F\_ t \right )\_ {t\geq 0}$ is filtration satisfying the usuual conditions, a non-negative continuous process $X = \left (X\_t \right)\_ {t\geq 0}$ satisfying $\mathbb {E}\left[ \bar X \right ]< \infty$ (where $ \bar X =\sup \_{0\leq t \leq T} X\_t $) and its Snell envelope


$$
\hat{X}\_\theta = \underset{\tau \in \mathcal T\_{\theta,T} } {\text{ess sup}} \ \mathbb {E} \left[ X\_\tau |\mathcal F\_\theta \right]
$$


where $\mathcal T\_{\theta, T} := \{ \tau \quad \mathbb F -\text{stopping time}: \tau \in [0, T] \quad \text{and} \quad \tau \geq \theta \quad \mathbb P -\text{a.s.} \}$ and $T \in \mathbb R\_+$ is a deterministic constant.


I'd like to understand how justify the following inequality:


$$\mathbb E \left [ \sup\_{0\leq t\leq T} \hat X\_t \right] \leq \mathbb E \left [ \sup\_{0\leq t\leq T} \bar X\_t \right] $$


where $\bar X\_t = \mathbb E \left [ \bar X | \mathcal F\_t\right]$ 


**Suplementary question**
Justify the following inequality:
$$\mathbb E \left [ \sup\_{0\leq t\leq T} \hat X\_t^p \right] \leq \mathbb E \left [ \sup\_{0\leq t\leq T} \mathbb E \left [ \bar X ^p| \mathcal F\_t\right] ^p\right] $$



Upper bound concerning Snell envelope
I think it is just a sequence of much stronger inequalities that goes as follows. All the inequalities below are $\Bbb P$-a.s.


1. Let $\tau\in \mathcal T\_{t,T}$ be an arbitrary stopping time, then
$$
 (X\_\tau)(\omega) = (X\_{\tau(\omega)})(\omega)\leq \sup\_{t\leq s\leq T} X\_s(\omega)\tag{1}
$$
since $\Bbb P(\tau\in [0,T]) = 1$.
2. As the latter term in the RHS of $(1)$ is smaller or equal to $\bar X$, we obtain $ X\_\tau\leq \bar X$ for all $\tau\in \mathcal T\_{t,T}$ and thus $\Bbb E[X\_\tau|\mathcal F\_t]\leq\Bbb E[\bar X|\mathcal F\_t]$.
3. As a result, we have that $\operatorname{esssup}\limits\_{\tau\in \mathcal T\_{t,T}}\;\Bbb E[X\_\tau|\mathcal F\_t]\leq \Bbb E[\bar X|\mathcal F\_t]$ which in your notation is $\hat X\_t\leq \bar X\_t$.
4. Applying to $\hat X\_t\leq \bar X\_t$ $\Bbb P$-a.s. first $\sup\_{0\leq t\leq T}$ and then expectation shall make it.


P.S. I have not dealt with esssup over stopping times for a while, so I hope step $3$ is correct - but better if you double-check it.


