RemarkRemark: AI-adaptedProof: Not applicableaudited 2026-08-31

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Lyapunov inequality is equivalent to log-convexity of the reciprocal-exponent norm profile

Fix a measurable $f$ and an interval of exponents on which every $∥ f ∥_{p}$ is finite and positive. Write

$s : = \frac{1}{p}, s_{0} : = \frac{1}{p_{0}}, s_{1} : = \frac{1}{p_{1}} .$

Then Lyapunov interpolation inequality for $L^{p}$ norms says exactly that

$∥ f ∥_{p} \leq ∥ f ∥_{p_{0}}^{θ} ∥ f ∥_{p_{1}}^{1 - θ}$

whenever $s = θ s_{0} + (1 - θ) s_{1}$ . Equivalently,

$∥ f ∥_{1 / s} \leq ∥ f ∥_{1 / s_{0}}^{θ} ∥ f ∥_{1 / s_{1}}^{1 - θ} .$

So the map $s \mapsto ∥ f ∥_{1 / s}$ is log-convex in the sense of Log-convex positive functions on the reciprocal-exponent interval, and conversely that log-convexity is exactly the Lyapunov interpolation statement.

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Sources

John K. Hunter, Measure Theory, Section 6.3 (standard reference, not scraped)