Young's Inequality

Young’s inequality is the small lever that moves a surprising amount of analysis: it turns a product of two numbers into a sum of their powers, and that single trade is exactly what you need to bootstrap Hölder’s inequality and, through it, Minkowski’s inequality. The proof is one line of convexity once you set it up correctly.

Conjugate exponents

Two real numbers $p, q > 1$ are conjugate exponents when

$$\frac{1}{p} + \frac{1}{q} = 1.$$

Equivalently, $q = \dfrac{p}{p-1}$. The pair $(2, 2)$ is conjugate to itself; $(1, \infty)$ is the limiting case usually treated separately. The defining identity says $1/p$ and $1/q$ are weights that sum to $1$ — that is what lets us read the proof as a convex combination.

The theorem

Theorem (Young’s Inequality). Let $p, q > 1$ be conjugate exponents and let $a, b \geq 0$. Then

$$ab \;\leq\; \frac{a^p}{p} + \frac{b^q}{q}, \tag{1}$$

with equality if and only if $a^p = b^q$.

Proof via convexity of the exponential

If $a = 0$ or $b = 0$ the left side is $0$ and the right side is non-negative, so $(1)$ holds. Assume therefore $a, b > 0$.

The key is the exponential function: it is convex on all of $\mathbb{R}$, so the two-point case of Jensen’s inequality gives, for any reals $x, y$ and weights $1/p, 1/q$ summing to $1$,

$$\exp\!\left(\frac{1}{p}\,x + \frac{1}{q}\,y\right) \;\leq\; \frac{1}{p}\,e^{x} + \frac{1}{q}\,e^{y}. \tag{2}$$

Now choose the inputs so the exponentials become $a^p$ and $b^q$. Using the logarithm, set

$$x \coloneqq p \ln a, \qquad y \coloneqq q \ln b.$$

Then $e^{x} = a^{p}$ and $e^{y} = b^{q}$, while the argument on the left of $(2)$ collapses:

$$\frac{1}{p}\,x + \frac{1}{q}\,y \;=\; \ln a + \ln b \;=\; \ln(ab),$$

so $\exp\!\big(\tfrac{1}{p}x + \tfrac{1}{q}y\big) = ab$. Substituting into $(2)$ yields exactly

$$ab \;\leq\; \frac{a^p}{p} + \frac{b^q}{q}. \quad \square$$

The equality case

The exponential is strictly convex, so $(2)$ is an equality if and only if its two inputs coincide: $x = y$, i.e. $p \ln a = q \ln b$. Exponentiating, this is $a^p = b^q$. Hence equality in $(1)$ holds precisely when $a^p = b^q$.

Summary

• Conjugate exponents $p, q > 1$ satisfy $1/p + 1/q = 1$; the two reciprocals act as weights summing to $1$.
• Young’s inequality: $ab \leq \dfrac{a^p}{p} + \dfrac{b^q}{q}$ for $a, b \geq 0$.
• Proof: write $ab = \exp(\tfrac{1}{p}\cdot p\ln a + \tfrac{1}{q}\cdot q\ln b)$ and apply convexity of $\exp$ — a single use of the two-point Jensen inequality.
• Equality holds iff $a^p = b^q$, because $\exp$ is strictly convex.
• This inequality is the engine behind Hölder’s inequality.