I am a worker and I am trying to understand a longer proof. In particular I know that $X$ is a stochastic process $\{X_n \}_{n\in{\mathbb{N}}}$, and I know that $X\in L^p$ (maybe this is unnecessary). I would like to show that for each $p>0$: $$ \sup_{n\in\mathbb{N}} (|X_n|^p) = (\sup_{n\in \mathbb{N}}|X_n|)^p.$$ My attempt:
I know that $p >0$, (note that $|X_n|\ge 0$ so the term $ (|X_n|)^p$ is well defined); I get:$ |X_n| \le \sup_{n\in \mathbb{N}}|X_n| := K$, then, $|X_n|^p \le K^p $ the right side of this inequality does not depend on $n$ hence Iconclude that: \begin{equation} \sup_{n\in\mathbb{N}} (|X_n|^p) \le \sup_{n\in\mathbb{N}} K^p = K^p = (\sup_{n\in \mathbb{N}}|X_n|)^p. \end{equation} Hence: \begin{equation} \sup_{n\in\mathbb{N}} (|X_n|^p) \le (\sup_{n\in \mathbb{N}}|X_n|)^p \;\;\; \forall p>0. \end{equation} On the other hand I can use the inequality just proved and write: \begin{equation} \sup_{n\in\mathbb{N}} (|X_n|^p)= \{[\sup_{n\in\mathbb{N}} (|X_n|^p)]^{\frac{1}{p}}\}^p \ge [\sup_{n\in\mathbb{N}} (|X_n|^{p\cdot\frac{1}{p}} )]^p=[\sup_{n\in\mathbb{N}} (|X_n| )]^p . \end{equation}
Is this correct? I am not self confident. Thank you.
Your answer is in the right direction.
Here is another (fill in the why?s) that is a little more direct solution.
Notice that $f(x)=x^p$ and $g(x)=x^{1/p}$, $p>0$, are increasing functions on $[0,\infty)$.
As $|X_n|\leq\sup_m|X_m|$, one has that $|X_n|^p\leq (\sup_m|X_m|)^p$ (why? )which means that $$\sup_n(|X_n|^p)\leq (\sup_n|X_n|)^p$$
Conversely, since $|X_n|^p\leq(\sup_m|X_m|^p)$, we have |$X_n|\leq (\sup_m|X_m|^p)^{1/p}$ (why?) and so, $\sup_n|X_n|\leq (\sup_m|X_m|^p)^{1/p}$. Consequently (why?) $$(\sup_m|X_n|)^p\leq (\sup_n|X_n|^p)$$