Clarification of a step in the proof of $E|X_n-X|^r \to 0 \implies E|X_n|^r \to E|X|^r$

Question

I do not understand how the $2$nd step comes from the $1$st in the following :

Does the following inequality hold? $$E|X+Y|^r \leq E|X|^r+E|Y|^r$$ where $0<r \leq 1$. (If it holds then it is clear, but I do not think that it holds in general).

Basically I want the proof of CASE $1$ from the accepted answer of this question : absolute value to the power of some positive number, is it a norm? Unfortunately, the author of the proof gave "it is clear from the inequality" to prove it, not caring about poor idiots like me.

Answer 1

It suffices to prove $$|x+y|^p\leq|x|^p+|y|^p$$ for all $x,y \in \Bbb{R}$ and $p \in (0,1]$.

This can be easily proved by calculating derivatives, but I present you a proof using convexity and concavity, which are techniques allowed in high school contest-math.

The case $p=1$ is just the triangle inequality. If $xy = 0$, then the inequality is also trivial, so assume $p \in (0,1)$.

We first prove this for $x,y > 0$. By the concavity of $x \mapsto x^p$ on $(0,\infty)$,

\begin{align} \frac{y}{x+y}0^p + \frac{x}{x+y}(x+y)^p &\le x^p \\ \frac{x}{x+y}0^p + \frac{y}{x+y}(x+y)^p &\le y^p. \end{align}

Add them up so that $x^p + y^p \ge (x+y)^p$.

Hence, (using the result for $p=1$,) for all $x,y \in \Bbb{R}$, $$|x+y|^p \le(|x| + |y|)^p \le |x|^p + |y|^p.$$

Use the linearity of expectation to finish the rest.

$$\forall p \in (0,1], |X+Y|^p \le |X|^p + |Y|^p \implies E|X+Y|^p \leq E|X|^p + E|Y|^p$$ Now substitue $Y = X_n - X$ and conclude.