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

1.1L2L3

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.

1.2L2L3

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

2.1step 1.2L1L2

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

3.1step 1.2step 2.1L2∎

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

Holder's inequality for finite sums and conjugate real exponents

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources