Proof of H$\ddot{o}$lder's inequality

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H$\ddot{o}$lder's inequality.  Let $X$ and $Y$ are random variables.  Let $1<p<\infty$ and $\frac{1}{p}+\frac{1}{q}=1$. $$|\mathscr{E}(XY)|\leq \mathscr{E}|XY|\leq \left(\mathscr{E}|X|^p\right)^{\frac{1}{p}}\left(\mathscr{E}|Y|^q\right)^{\frac{1}{q}}.$$
See List of Inequalities.

$\bullet$ Proof.

The first inequality is derived by $-|XY|\leq XY\leq|XY|$.  For the second inequality, we need the following lemma.

[Lemma] Let $a$ and $b$ be any positive numbers, and let $p$ and $q$ satisfying $$1<p<\infty\mbox{ and }\frac{1}{p}+\frac{1}{q}=1.$$Then $$\frac{1}{p}a^p+\frac{1}{q}b^q\geq ab$$with the equality if and only if $a^p=b^q$.

Back to the proof.  Define $$a=\frac{|X|}{\left(\mathscr{E}|X|^p\right)^{\frac{1}{p}}}\mbox{ and }b=\frac{|Y|}{\left(\mathscr{E}|Y|^q\right)^{\frac{1}{q}}}.$$According to the lemma above, we have $$\frac{1}{p}a^p+\frac{1}{q}b^q=\frac{1}{p}\frac{|X|^p}{\mathscr{E}|X|^p}+\frac{1}{q}\frac{|Y|^q}{\mathscr{E}|Y|^q}\geq ab=\frac{|X|}{\left(\mathscr{E}|X|^p\right)^{\frac{1}{p}}}\frac{|Y|}{\left(\mathscr{E}|Y|^q\right)^{\frac{1}{q}}}.$$Take expectation for both side, then we have $$1=\frac{1}{p}+\frac{1}{q}\geq\frac{\mathscr{E}|XY|}{\left(\mathscr{E}|X|^p\right)^{\frac{1}{p}}\left(\mathscr{E}|Y|^q\right)^{\frac{1}{q}}}.$$Thus, $$\mathscr{E}|XY|\leq \left(\mathscr{E}|X|^p\right)^{\frac{1}{p}}\left(\mathscr{E}|Y|^q\right)^{\frac{1}{q}}.$$

$\Box$