Important inequalities: Young's inequality

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The inequality for means already bounds a product of two real numbers by two equal squares, $\frac{1}{2} (a^{2} + b^{2})$ . What happens when the exponents are a conjugate pair, $p > 1$ and $1/ p + 1/ q = 1$ ?

The reciprocals add to 1

$a$ and $b$ stay positive. Drag $p$ and it stays above $1$ , and $q$ is the number forced by $1/ p + 1/ q = 1$ .

Drag a, b, or p. a and b stay positive, and p stays above 1. q is fixed by 1/p+1/q=1. The upper mark is |ab|. The lower mark is |a|^p/p+|b|^q/q. Set p=2 and a=b. The lower mark is (a²+b²)/2 and the two marks share an x-coordinate.

Set $p = 2$ and $a = b$ . The lower mark is $\frac{1}{2} (a^{2} + b^{2})$ , and the two marks meet. That meeting is equality in the comparison already checked for real numbers.

For complex $a$ and $b$ , and real $p, q > 1$ with $1/ p + 1/ q = 1$ ,

$∣ ab ∣ \leq \frac{∣ a ∣ ^{p}}{p} + \frac{∣ b ∣ ^{q}}{q} .$

The moduli let $a$ and $b$ leave the positive reals. This board keeps $a$ and $b$ positive.

Each factor takes its own exponent

When $n = 2$ , $(0.21)$ is $(0.20)$ . For $n = 2, 3, \dots$ , complex numbers $x_{1}, \dots, x_{n}$ , and real exponents $p_{k} > 1$ whose reciprocals add to $1$ ,

$∣ \prod_{k = 1}^{n} x_{k} ∣ \leq \sum_{k = 1}^{n} \frac{∣ x _{k} ∣ ^{p_{k}}}{p _{k}} .$

Drag a, b, or c. Each stays positive, and the three exponents are frozen at 3, so the reciprocals add to 1. The upper mark is the product. The lower mark is (a³+b³+c³)/3. Set a=b=c and the two marks share an x-coordinate.

The board freezes $n = 3$ and every exponent at $3$ , so each reciprocal is $\frac{1}{3}$ .

The same conjugate split, applied to a sum of products, is Hölder's inequality.