TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02

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Holder's inequality for finite sums and conjugate real exponents

Statement

Let $p, q > 1$ with $1 / p + 1 / q = 1$ . For real families $(a_{i})_{i < n}$ and $(b_{i})_{i < n}$ , $\sum_{i < n} ∣ a_{i} b_{i} ∣ \leq {(\sum_{i < n} ∣ a_{i} ∣^{p})}^{1 / p} {(\sum_{i < n} ∣ b_{i} ∣^{q})}^{1 / q} .$

Facts & Assumptions

Given: A natural $n$ , conjugate exponents $p, q > 1$ , and real families $a_{i}, b_{i}$ for $i < n$ .

[L1]

Young's inequality says $u v \leq u^{p} / p + v^{q} / q$ for $u, v \geq 0$ (Young's inequality for conjugate real exponents).

[L2]

Finite sums obey termwise addition and scalar multiplication, including the empty-sum convention, and $∣ a b ∣ = ∣ a ∣ ∣ b ∣$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products, Basic properties of the absolute value).

[L3]

For $r > 0$ , the zero-base convention gives $0^{r} = 0$ , while positive-base real powers are positive and obey the real-power laws (Real powers for positive bases, with the zero-base positive-exponent convention, The exponent, product, quotient, and iterated-power laws for positive real bases and real exponents, The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ ).

Proof

technique · direct

1.1

Put $A = (\sum_{i < n} ∣ a_{i} ∣^{p})^{1 / p}$ and $B = (\sum_{i < n} ∣ b_{i} ∣^{q})^{1 / q}$ . If $A = 0$ or $B = 0$ , the real-power laws and zero convention show that the corresponding nonnegative power sum is zero; finite-sum order then makes every corresponding term zero, so the claim follows.

L2L3

1.2

Suppose $A, B > 0$ , and set $u_{i} = ∣ a_{i} ∣ / A$ , $v_{i} = ∣ b_{i} ∣ / B$ . Then $\sum u_{i}^{p} = \sum v_{i}^{q} = 1$ .

L2L3

2.1

Apply [L1] to $u_{i}, v_{i}$ and sum over $i < n$ to obtain $\sum u_{i} v_{i} \leq 1 / p + 1 / q = 1$ .

step 1.2L1L2

3.1

Multiplying by $A B$ and using $∣ a_{i} b_{i} ∣ = ∣ a_{i} ∣ ∣ b_{i} ∣$ gives the asserted inequality.

step 1.2step 2.1L2∎

Depends on

Used by

Dependency tree · two levels

28 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. Lebl, Basic Analysis, Logarithm and Exponential (standard reference, not scraped)
MIT OpenCourseWare 18.100B Real Analysis, Spring 2025 full lecture notes (standard reference, not scraped)