Let $0<p<1$ and $X_1,\ldots,X_n$ be random variables with finite absolute moments of order $p$. Suppose that the random variables $X_1,\ldots,X_n$ converge in mean of order $p$ to a random variable $X$ as $n\to\infty$ (i.e. $\operatorname E|X_n-X|^p\to0$ as $n\to\infty$).
Does it follow that $\operatorname E|X_n|^p\to\operatorname E|X|^p$ as $n\to\infty$?
It is indeed the case when $p\ge1$ since $\|X\|_p=(\operatorname E|X|^p)^{1/p}$ is a norm when $p\ge 1$ and we can use the reverse triangle inequality $|\|X_n\|-\|X\||\le\|X_n-X\|$ to obtain the result, but $\|X\|_p$ does not satisfy the triangle inequality when $0<p<1$.
Any help is much appreciated!
First remark that $(u+v)^p \leq u^p + v^p$ for all $u,v$ non-negative. Indeed, the function $u \mapsto (u+v)^p - u^p$ is non-increasing by concavity of $x \mapsto x^p$ and its values at $u=0$ is $v^p$.
This implies $|X_n|^p \leq (|X_n-X|+|X|)^p \leq |X_n-X|^p + |X|^p$ and similarly $|X|^p \leq |X-X_n|^p + |X_n|^p$.
Taking expectations, we deduce $$ \bigl|E(|X_n|^p) - E(|X|^p)\bigr| \leq E(|X_n-X|^p) $$ for all $n$, and the convergence follows.