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Young's inequality for products (rational conjugate exponents)

Statement

Let $p, q \in \mathbb{Q}$ with $p > 1$ and $q > 1$ (Order on the rationals) be conjugate exponents, that is

$\frac{1}{p} + \frac{1}{q} = 1, \qquad \text{equivalently} \qquad q = \frac{p}{\,p-1\,}.$

Then for all $a, b \in \mathbb{R}$ with $a \ge 0$ and $b \ge 0$ ,

$ab \;\le\; \frac{a^{p}}{p} + \frac{b^{q}}{q},$

with the rational powers of Rational powers $a^r$ of a positive base (its supplementary clause gives $0^{p} = 0$ , since $p > 0$ ) and with the rationals $p, q$ acting on $\mathbb{R}$ through the canonical embedding (The unique embedding of ℚ into an ordered field).

The conjugate exponent is rational because $p$ is. From $\frac1q = 1 - \frac1p = \frac{p-1}{p}$ one gets $q = p/(p-1)$ , a quotient of rationals with nonzero denominator (Arithmetic on the rationals), hence a rational. This is the observation that keeps Hölder and Minkowski inside the rational world on this page.

Facts & Assumptions

Given: Rationals $p, q > 1$ with $1/p + 1/q = 1$ , and reals $a, b \ge 0$ .

[L1]

Weighted AM-GM with rational weights (Weighted AM-GM inequality with rational weights): for $x_0, x_1 > 0$ and rationals $w_0, w_1 \ge 0$ with $w_0 + w_1 = 1$ , $x_0^{\,w_0} x_1^{\,w_1} \le w_0 x_0 + w_1 x_1$ .

[L2]

Rational power laws (Laws of rational exponents, Rational powers $a^r$ of a positive base): for $u > 0$ and rationals $r, s$ , $u^{r} > 0$ and $\big(u^{r}\big)^{s} = u^{rs}$ , and $u^{1} = u$ ; and $0^{r} = 0$ for rational $r > 0$ .

[L3]

Rational arithmetic (The rationals as equivalence classes of pairs of integers, Arithmetic on the rationals, Order on the rationals): $q = p/(p-1)$ is a quotient of rationals with nonzero denominator, hence rational, and $p \cdot (1/p) = 1$ . Moreover $\mathbb{Q}$ is itself a totally ordered field (The rationals form a totally ordered field), which is what licenses the order arithmetic used on $p$ and $q$ ; being an ordered field it has $1 > 0$ (The multiplicative identity is positive, which is where that fact is proved: The rationals form a totally ordered field states totality, compatibility with addition and closure of the positives under multiplication, and not this), so $p > 1 > 0$ gives $p > 0$ by transitivity and hence $1/p > 0$ , and likewise $1/q > 0$ (Inverses of positives are positive, and reciprocation reverses order, claim 1, applied in $\mathbb{Q}$ ).

[L4]

The embedding $\iota : \mathbb{Q} \to \mathbb{R}$ is an order-preserving field homomorphism, so $\iota(1/p) = \iota(p)^{-1} > 0$ (The unique embedding of ℚ into an ordered field, Inverses of positives are positive, and reciprocation reverses order, Sign rules for products and monotonicity of multiplication).

Proof

technique · direct

1.1

The pair $w_0 := 1/p$ , $w_1 := 1/q$ is a legitimate system of rational weights: both are rational and positive because $p, q > 1 > 0$ , and $w_0 + w_1 = 1$ by hypothesis.

givenL3L4

1.2

Degenerate cases: if $a = 0$ then the left-hand side is $0$ while the right-hand side is $0^{p}/p + b^{q}/q = b^{q}/q \ge 0$ , since $0^{p} = 0$ for $p > 0$ and $b^{q} \ge 0$ ; the case $b = 0$ is symmetric, so the inequality holds whenever $a = 0$ or $b = 0$ .

givenL2L3L4

1.3

For $a > 0$ and $b > 0$ , which is the only case in which this step is used, the left-hand factors simplify: $\big(a^{p}\big)^{1/p} = a^{p \cdot (1/p)} = a^{1} = a$ and, in the same way, $\big(b^{q}\big)^{1/q} = b$ .

L2L3

2.1

Assume now $a > 0$ and $b > 0$ , and put $x_0 := a^{p} > 0$ and $x_1 := b^{q} > 0$ ; applying weighted AM-GM with the weights of step 1.1 gives $\big(a^{p}\big)^{1/p}\big(b^{q}\big)^{1/q} \le \frac{1}{p} a^{p} + \frac{1}{q} b^{q}$ .

step 1.1L1L2

3.1

Substituting, $ab \le \frac{a^{p}}{p} + \frac{b^{q}}{q}$ for all $a, b > 0$ , and together with the degenerate cases this proves the inequality for all $a, b \ge 0$ .

step 2.1step 1.3step 1.2∎

Depends on

Used by

Hölder's inequality for finite sums (rational exponents) Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 67 results over 27 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

J. K. Hunter, An Introduction to Real Analysis (standard reference, not scraped)
Finite inequalities (Cornell University) (standard reference, not scraped)
Young, Hölder, and Minkowski inequalities (Oregon State University) (standard reference, not scraped)
Young's inequality for products (Wikipedia) (standard reference, not scraped)
Hölder's inequality (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 1 (standard reference, not scraped)