Let $X := \{X_n\}_{n=0}^\infty$ be a martingales. Let $1 \leq s < \infty$, with $||X||_s < \infty$.
Prove that $Y := \{ |X_n|^s \}_{n=0}^\infty$ is a sub-martingales
I have tried using the definition of martingales with the conditional mean, but i cannot arrive at the solution
It is a direct consequence of Jensen's inequality for conditional expectations: let $X$ be an integrable real valued random variable, $\varphi:\mathbb R\to\mathbb R$ be a convex function such that $\varphi(X)$ is integrable and $\mathcal G$ be a $\sigma$-algebra. Then almost surely, $$ \varphi(\mathbb E[X\vert\mathcal G])\le\mathbb E[\varphi(X)\vert\mathcal G]. $$
So in your case, since $x\mapsto\vert x\vert^s$ is convex, Jensen's inequality for conditional expectation writes for all $n\in\mathbb N$, $$ Y_n=\vert X_n\vert^s=\vert\mathbb E[X_{n+1}\vert X_n]\vert^s\le\mathbb E[\vert X_{n+1}\vert^s\vert X_n]=\mathbb E[Y_{n+1}\vert X_n]. $$
Since $Y_n=\vert X_n\vert^s$, we have $\sigma(Y_n)\subset\sigma(X_n)$ so $$ \mathbb E[Y_{n+1}\vert Y_n]=\mathbb E[\mathbb E[Y_{n+1}\vert X_n]\vert Y_n]\ge\mathbb E[Y_n\vert Y_n]=Y_n, $$ hence $(Y_n)_{n\in\mathbb N}$ is a submartingale. Notice that this would have still held true if $(X_n)_{n\in\mathbb N}$ was only a submartingale.