Hölder’s inequality

Let $(𝑋, Σ, 𝜇)$ be a measure space and $𝑝, 𝑞 \in [1, \infty]$ be Hölder conjugate, i.e. $𝑝 - 1 + 𝑞 - 1 = 1$ . Then for any measurable functions $𝑓, 𝑔 : 𝑋 \to ℂ$ ¹ anal

‖ 𝑓 𝑔 ‖ 1 \leq ‖ 𝑓 ‖ 𝑝 ‖ 𝑔 ‖ 𝑞

where $‖ \cdot ‖ 𝑝$ denotes the [[Lebesgue space| $𝑝$ -norm]].

Proof

Let $𝐴 = ‖ 𝑓 ‖ 𝑝$ and $𝐵 = ‖ 𝑔 ‖ 𝑞$ . If $𝐴 = 0$ or $𝐵 = 0$ the inequality holds trivially, and similarly if $𝐴 = \infty$ or $𝐵 = \infty$ .

Now assume $𝐴, 𝐵 \in (0, \infty)$ , $𝑝, 𝑞 \in (1, \infty)$ and take $𝑥 \in 𝑋$ . Taking Young’s inequality with $𝛼 = | 𝑓 (𝑥) | 𝐴$ and $𝛽 = | 𝑔 (𝑥) | 𝐵$ gives
$𝛼 𝛽 = | 𝑓 (𝑥) 𝑔 (𝑥) | 𝐴 𝐵 \leq | 𝑓 (𝑥) | 𝑝 𝑝 𝐴 𝑝 + | 𝑔 (𝑥) | 𝑞 𝑞 𝐵 𝑞 = 𝛼 𝑝 𝑝 + 𝛽 𝑞 𝑞$
whence
$‖ 𝑓 𝑔 ‖ 1 ‖ 𝑓 ‖ 𝑝 ‖ 𝑔 ‖ 𝑞 = 1 𝐴 𝐵 \int 𝑋 | 𝑓 (𝑥) 𝑔 (𝑥) | 𝑑 𝜇 (𝑥) \leq 1 𝑝 𝐴 𝑝 \int 𝑋 | 𝑓 (𝑥) | 𝑝 𝑑 𝜇 (𝑥) + 1 𝑞 𝐵 𝑞 \int 𝑋 | 𝑔 (𝑥) | 𝑞 𝑑 𝑥 = 𝐴 𝑝 𝑝 𝐴 𝑝 + 𝐵 𝑞 𝑞 𝐵 𝑞 = 1 𝑝 + 1 𝑞 = 1$
wherefore
$‖ 𝑓 𝑔 ‖ 1 \leq ‖ 𝑓 ‖ 𝑝 ‖ 𝑔 ‖ 𝑞$
The only case yet to be handled is $𝑝 = \infty, 𝑞 = 1$ . It follows immediately that $| 𝑓 𝑔 | \leq ‖ 𝑓 ‖ \infty | 𝑔 |$ almost everywhere and thus $‖ 𝑓 𝑔 ‖ 1 \leq ‖ 𝑓 ‖ \infty ‖ 𝑔 ‖ 1$ , as required.

The Cauchy-Schwarz inequality for the [[Lebesgue space| $𝑝$ -norm]] is the case $𝑝 = 𝑞 = 2$ .

tidy | en | SemBr

If in addition, if $𝑝$ and $𝑞$ are finite, $𝑓 \in 𝐿 𝑝 (𝜇)$ , and $𝑔 \in 𝐿 𝑞 (𝜇)$ , then $‖ 𝑓 𝑔 ‖ 1 = ‖ 𝑓 ‖ 𝑝 ‖ 𝑔 ‖ 𝑞$ iff $| 𝑓 | 𝑝, | 𝑔 | 𝑞 \in 𝐿 1 (𝜇)$ are linearly dependent. ↩

Hölder’s inequality

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Hölder’s inequality

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