Hölder's inequality
Statement

Assume that 1 ≤ p < ∞ and let q denote the Hölder conjugate. Then for every f ∈ L (μ),

where max indicates that there actually is a g maximizing the right-hand side. When p = ∞ and if each set A in the σ-field Σ with μ (A) = ∞ contains a subset B ∈ Σ with 0 < μ (B) < ∞ (which is true in particular when μ is σ-finite), then

Proof of the extremal equality:

By Hölder's inequality, the integrals are well defined and, for 1 ≤ p ≤ ∞,

hence the left-hand side is always bounded above by the right-hand side.

Conversely, for 1 ≤ p ≤ ∞, observe first that the statement is obvious when ‖ f ‖ = 0. Therefore, we assume ‖ f ‖ > 0 in the following.

If 1 ≤ p < ∞, define g on S by

By checking the cases p = 1 and 1 < p < ∞ separately, we see that ‖ g ‖ = 1 and

It remains to consider the case p = ∞. For ε ∈ (0, 1) define

Since f is measurable, A ∈ Σ. By the definition of ‖ f ‖ as the essential supremum of f and the assumption ‖ f ‖ > 0, we have μ (A) > 0. Using the additional assumption on the σ-field Σ if necessary, there exists a subset B ∈ Σ of A with 0 < μ (B) < ∞. Define g on S by

Then g is well-defined, measurable and | g (x) | ≤ 1/ μ (B) for x ∈ B, hence ‖ g ‖ ≤ 1. Furthermore,
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