Suppose $X_n \to X$ in $L^p$. Show that E$|X^p_r| \to E|X^p|$.

Question

Suppose $X_n \to X$ in $L^p$. Show that E$|X^p_r| \to E|X^p|$.

The proof suggested the use of Minkowski's inequality in order to get that: $$ [E|X^p|]^{1/p} \leqslant [E(|X_n - X|^p)]^{1/p} + [E(|X^p_n|)]^{1/p} $$ and letting n $\to \infty$, liminf $E|X^p_r| \geq E|X^p|$.

Using again Minkowski:

$$ [E|X^p_n|]^{1/p} \leqslant [E(|X_n - X|^p)]^{1/p} + [E(|X^p|)]^{1/p} $$ ensuring that limsup $E|X^p_r| \leqslant E|X^p|$

I can't understand two things.

1. Since the Minkowski's inequality state that: $$ ||f+g||_p \leqslant ||f||_p + ||g||_p $$ how in this case has this been used? I thought that, in the first sentence, f = $X^p$ but I can't figure it out where is the "g".

2. Why in the two sentence do I found the liminf and the limsup respectively?

Answer 1

1. Take $f=X-X_n$ and $g=X_n$.

2. If you want to show that $a_n \to a$ and you do not know that the limit exists you show that $\lim \sup a_n \leq a$ and $\lim \inf a_n \geq a$. These two facts imply that $\lim a_n$ exists and equals $a$.

Answer 2

For any norm $||\cdot||$ on a vector space $V$ you have $\bigg|||x||-||y||\bigg|\leq ||x-y||$ . Which in particular means that $x\mapsto ||x||$ is a continuous function.

In particular for $L^{p}(\mu)$ spaces with the $L^{p}(\mu)$ norm $||f||_{p}=\bigg(\int |f|^{p}\,d\mu\bigg)^{\frac{1}{p}}$ you have

Now $\bigg|||X_{n}||_{p}-||X||_{p}\bigg|\leq ||X_{n}-X||_{p}\to 0$

Thus $||X_{n}||_{p}\to ||X||_{p}$

Thus $(E[|X_{n}|^{p}])^{\frac{1}{p}}\to (E[|X|^{p}])^{\frac{1}{p}}$

Which gives you that $E[|X_{n}|^{p}]\to E[|X|^{p}]$

Also by the notion of continuity, whenever you have $x_{n}\to x$ in the norm, i.e. $||x_{n}-x||\to 0$, you must have $||x_{n}||\to ||x||$ due to continuity of $f:(V,||\cdot||)\to\Bbb{R}$ such that $f(x)=||x||$.