Young's Inequality

Bound a product by a sum of powers using conjugate exponents.

The idea

Theorem (Young's inequality).

Let $p, q > 1$ be conjugate exponents, meaning $\frac{1}{p} + \frac{1}{q} = 1$. Then for all $a, b \ge 0$, $ab \;\le\; \frac{a^{p}}{p} + \frac{b^{q}}{q},$ with equality if and only if $a^{p} = b^{q}$.

The inequality trades a product for a sum. A product couples its two factors, so neither can be estimated alone; a sum splits into pieces that can be handled separately. Young's inequality replaces $ab$ by a sum in which $a$ and $b$ appear apart, at the controlled cost of raising each to a power.

The exponents must be conjugate because the right-hand side is a weighted average of $a^{p}$ and $b^{q}$ with weights $\frac{1}{p}$ and $\frac{1}{q}$, and the weights of an average must total $1$. Within that constraint the choice is a trade-off: a large $p$ softens the penalty on $b$ but punishes a large $a$, so the exponents are picked to favor whichever factor needs the gentler treatment.

The symmetric case $p = q = 2$ gives $ab \le \frac{a^{2}}{2} + \frac{b^{2}}{2}$, the AM-GM bound applied to $a^{2}$ and $b^{2}$.

Ways to work on it

• Walkthrough. Meet conjugate exponents, bound a product by a sum of powers, and see when the bound is tight.
• Proof. See why — the two areas a curve cuts a rectangle into.
• Practice. Evaluate the Young upper bound for a conjugate pair.
• Hardest. Find the parameter that makes a weighted form of the bound tight.

Not sure where to start? Take the ten-question placement test.