If $X_n\to X$ in $L^p$, is $E(X_n^p)\to E(X^p)$ true?

Question

Let $p$ denote a positive integer. If $X_n\to X$ in $L^p$, is $E(X_n^p)\to E(X^p)$ true?

If each $X_n$ and $X$ are nonnegative, then it follows directly from Minkowski's inequality.

Then how about the general case? I separated each $X_n$ into its positive and negative parts as $X_n=X_n^+-X_n^-$, but it might not be the case that $X_n^+\to X^+$ and $X_n^-\to X^-$ in $L^p$.

Answer 1

For a simple, self-contained, proof, one can start from the binomial identity $$X_n^p-X^p=\sum_{k=1}^p{p\choose k}(X_n-X)^kX^{p-k}$$ which yields $$|E(X_n^p)-E(X^p)|\leqslant\sum_{k=1}^p{p\choose k}E(|X_n-X|^k|X|^{p-k})$$ Now, for each $1\leqslant k\leqslant p$, by convexity, $$E(|X_n-X|^k|X|^{p-k})\leqslant E(|X_n-X|^p)^{k/p}E(|X|^p)^{1-k/p}$$ hence each term in the sum on the RHS goes to $0$ as soon as $X_n\to X$ in $L^p$.

The method yields an effective upper bound of $|E(X_n^p)-E(X^p)|$ in terms of $\|X_n-X\|_p$ and $\|X\|_p$, which we leave to the reader.

Answer 2

We have $$|\mathbb{E}[X_n^p - X^p]| \leq p \mathbb{E}[(|X_n| + |X|)^{p-1} |X_n - X|]$$ since for any $x,y \in \mathbb{R}$, $|x^p - y^p| \leq p (|x| + |y|)^p |x-y|$ (for example, apply the mean value theorem to $t \mapsto t^p$).

So by Holder's inequality, \begin{align*} |\mathbb{E}[X_n^p - X^p]| \leq& p \|X_n - X\|_p \mathbb{E}[(|X_n| + |X|)^p]^{\frac{p-1}{p}} \\ \leq & 2^{p-1} p \|X_n - X\|_p (\|X_n\|_p^{p} + \|X\|_p^p)^{\frac{p-1}{p}} \to 0 \end{align*} since $\|X_n\|_p$ is a bounded sequence and $\|X_n - X\|_p \to 0$.