3.2 Product Inequalities
Lemma 3.2.1 (Young’s Inequality). Let $p, q \in (1, \infty)$ such that $1/p + 1/q = 1$, then for any $a, b > 0$,
\[ab \le \frac{a^{p}}{p}+ \frac{b^{q}}{q}\]
and for any $\eps > 0$,
\[ab \le \eps a^{p} + \frac{1}{q}(\eps q)^{-q/p}b^{q}\]
Proof. Since $x \mapsto \exp(x)$ is convex,
\begin{align*}ab&= \exp(\ln(a) + \ln(b)) = \exp\paren{\frac{1}{p}\ln(a^p) + \frac{1}{q}\ln(b^q)}\\&\le \frac{1}{p}\exp(\ln(a^{p})) + \frac{1}{q}\exp(\ln(a^{q})) = \frac{a^{p}}{p}+ \frac{b^{q}}{q}\end{align*}
For any $\eps > 0$, applying the above inequality to $(\eps p)^{1/p}a$ and $b/(\eps p)^{1/p}$ yields
\[ab = (\eps p)^{1/p}a \cdot \frac{b}{(\eps p)^{1/p}}\le \eps a^{p} + \frac{b^{q}}{q}(\eps p)^{-(1/p)q}= \eps a^{p} + \frac{1}{q}(\eps q)^{-q/p}b^{q}\]
$\square$