Lemma 3.2.2.label Let $p \in [1, \infty)$, then for each $a, b \ge 0$,
\[|a^{p} - b^{p}| \le p|a - b|(a \vee b)^{p - 1}\]
Proof. Assume without loss of generality that $0 < a < b$, then
\[b^{p} - a^{p} = p\int_{a}^{b} t^{p - 1}dt \le p|a - b|(a \vee b)^{p - 1}\]
$\square$
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