Let $X$ be a random variable and let $p \in (0, \infty)$ such that $\mathbb{E}\left(|X|^p\right) < \infty$. Show that for all $q \in (0, p)$, we have $\mathbb{E}\left(|X|^q\right) \leq \left[\mathbb{E}\left(|X|^p\right)\right]^{\frac{q}{p}}$. You may use the fact that the function $\mathbb{R} \ni x \mapsto |x|^r$ is convex for all $r \geq 1$.
Hello I think we need to use Jensen's inequality, in my book defined as:
Jensen's inequality: Let $X$ be a random variable taking values in the (possibly infinite) interval $(a, b)$ such that $\mathbb{E}(X)$ exists, and let $g:(a, b) \rightarrow \mathbb{R}$ be a convex function such that $\mathbb{E}|g(X)|<\infty$. Then
$$
\mathbb{E}(g(X)) \geq g(E(X)) .
$$
But I don't know how to do this and how to give the formal proof.
The trick is to identify the correct '$X$' and '$g$'. I will switch notation to $\mathbb{E}(g(Y)) \geq g(\mathbb{E}(Y))$ to avoid confusion.
Here, use $Y = |X|^q$, and $g(y) = |y|^{p/q}$. Note that by assumption, $p/q \geq 1$, so $g$ is convex. By Jensen's inequality, $$g(\mathbb{E}(Y)) \leq \mathbb{E}(g(Y)).$$ Writing out $g$, you get $$ |\mathbb{E}(Y)|^{p/q} \leq \mathbb{E}(|Y|^{p/q}). $$ Substituting $Y = |X|^q$, you get $$ |\mathbb{E}(|X|^q)|^{p/q} \leq \mathbb{E}(|X|^p). $$ Since $\mathbb{E}(|X|^q)$ is nonnegative, you can remove the outer absolute value sign and write $$ [\mathbb{E}(|X|^q)]^{p/q} \leq \mathbb{E}(|X|^p). $$ Finally, take $q/p$ power on both sides to get $$ \mathbb{E}(|X|^q) \leq [\mathbb{E}(|X|^p)]^{q/p}. $$