Convergence in$ L_p$ Let $p \geq 1.$ Suppose that $X_n \overset{L_p}{\rightarrow}$ X. I know this implies that $\mathbb{E}|X_n|^p \rightarrow \mathbb{E}|X|^p.$ But does this also imply $EX_n^p \rightarrow \mathbb{E}X^p?$
For $p = 1,$ $|\mathbb{E}(X_n) - \mathbb{E}(X)| \leq \mathbb{E}|X_n - X|$ and hence $L_1$ convergence does imply convergence of means. But what happens when $p \geq2.$
 A: You know that $\mathbb E(|X_n-X|^p)\to0$ and you are wondering whether $\mathbb E(X_n^p)$ converges to $\mathbb E(X^p)$ or not, where $p\gt1$ is either an integer or a real number and then every $X_n$ is almost surely nonnegative. Well... note that $|\mathbb E(X_n^p)-\mathbb E(X^p)|=|\mathbb E(X_n^p-X^p)|\leqslant\mathbb E(|X_n^p-X^p|)$ hence it suffices to show that $\mathbb E(|X_n-X|^p)\to0$ implies $\mathbb E(|X_n^p-X^p|)\to0$.
To do that, note that for every $(x,y)$, $|x^p-y^p|\leqslant p|x-y|\,|x|^{p-1}+p|x-y|\,|y|^{p-1}$, hence
$$
\mathbb E(|X_n^p-X^p|)\leqslant p\mathbb E(|X_n-X|\,|X_n|^{p-1})+p\mathbb E(|X_n-X|\,|X|^{p-1}).
$$
Applying Minkowski inequality to both expectations in the RHS, with $\frac1p+\frac1q=1$, yields
$$
\mathbb E(|X_n^p-X^p|)\leqslant p\mathbb E(|X_n-X|^p)^{1/p}\,(\mathbb E(|X_n|^p)^{1/q}+\mathbb E(|X|^{p})^{1/q}).
$$
Furthermore, $|X_n|^p\leqslant|X|^p+|X_n^p-X^p|$ and $(x+y)^{1/q}\leqslant x^{1/q}+yx^{-1/p}$ for every positive $x$ and $y$ hence $\mathbb E(|X_n|^p)^{1/q}\leqslant\mathbb E(|X|^p)^{1/q}+\mathbb E(|X_n^p-X^p|)\,\mathbb E(|X|^p)^{-1/p}$. Finally,
$$
\mathbb E(|X_n^p-X^p|)\leqslant p\mathbb E(|X_n-X|^p)^{1/p}\,(2\mathbb E(|X|^p)^{1/q}+\mathbb E(|X_n^p-X^p|)\,\mathbb E(|X|^p)^{-1/p}),
$$
that is,
$\mathbb E(|X_n^p-X^p|)\leqslant u_p(\mathbb E(|X_n-X|^p)^{1/p})$ with
$$
u_p(x)=\frac{2p\mathbb E(|X|^p)\,x}{\mathbb E(|X|^p)^{1/p}-px}.
$$
Since $\mathbb E(|X_n-X|^p)^{1/p}\to0$ when $n\to\infty$ and $u_p(x)\to0$ when $x\to0$, this proves that $\mathbb E(|X_n^p-X^p|)\to0$ when $n\to\infty$.
