In mathematical analysis, Hölder's inequality, named after Otto Hölder, is a fundamental inequality between integrals and an indispensable tool for the study of Lp spaces.
Hölder's inequality (in a slightly different form) was first found by Leonard James Rogers (1888).
As above, let f and g denote measurable real- or complex-valued functions defined on S. If ‖fg‖1 is finite, then the pointwise products of f with g and its complex conjugate function are μ-integrable, the estimate and the similar one for fg hold, and Hölder's inequality can be applied to the right-hand side.
This is also called Cauchy–Schwarz inequality, but requires for its statement that ‖f ‖2 and ‖g‖2 are finite to make sure that the inner product of f and g is well defined.
We may recover the original inequality (for the case p = 2) by using the functions |f | and |g| in place of f and g. If (S, Σ, μ) is a probability space, then p, q ∈ [1, ∞] just need to satisfy 1/p + 1/q ≤ 1, rather than being Hölder conjugates.
A combination of Hölder's inequality and Jensen's inequality implies that for all measurable real- or complex-valued functions f and g on S. For the following cases assume that p and q are in the open interval (1,∞) with 1/p + 1/q = 1.
: For more than two sums, the following generalisation (Lohwater (1982), Chen (2014)) holds, with real positive exponents
Then Tonelli's theorem allows us to rewrite Hölder's inequality using iterated integrals: If f and g are Σ-measurable real- or complex-valued functions on the Cartesian product S, then This can be generalized to more than two σ-finite measure spaces.
Let (S, Σ, μ) denote a σ-finite measure space and suppose that f = (f1, ..., fn) and g = (g1, ..., gn) are Σ-measurable functions on S, taking values in the n-dimensional real- or complex Euclidean space.
By taking the product with the counting measure on {1, ..., n}, we can rewrite the above product measure version of Hölder's inequality in the form If the two integrals on the right-hand side are finite, then equality holds if and only if there exist real numbers α, β ≥ 0, not both of them zero, such that for μ-almost all x in S. This finite-dimensional version generalizes to functions f and g taking values in a normed space which could be for example a sequence space or an inner product space.
, so by Jensen's inequality, where ν is any probability distribution and h any ν-measurable function.
, Finally, we get This assumes that f, g are real and non-negative, but the extension to complex functions is straightforward (use the modulus of f, g).
Alternatively, we can deduce Young's inequality and then resort to the first proof given above.
Then for every f ∈ Lp(μ), where max indicates that there actually is a g maximizing the right-hand side.
contrary to the notation, ‖.‖r is in general not a norm because it doesn't satisfy the triangle inequality.
then Pulling out the essential supremum of |fn| and using the induction hypothesis, we get Case 2: If
as the weighted harmonic mean, that is, Given measurable real- or complex-valued functions
Then for all measurable real- or complex-valued functions f and g on S such that g(s) ≠ 0 for μ-almost all s ∈ S, If then the reverse Hölder inequality is an equality if and only if Note: The expressions:
Application of Hölder's inequality gives Raising to the power p gives us: Therefore: Now we just need to recall our notation.
The Reverse Hölder inequality (above) can be generalized to the case of multiple functions if all but one conjugate is negative.
It was observed by Aczél and Beckenbach[5] that Hölder's inequality can be put in a more symmetric form, at the price of introducing an extra vector (or function): Let
, then: The standard Hölder inequality follows immediately from this symmetric form (and in fact is easily seen to be equivalent to it).
, then: As in the standard Hölder inequalities, there are corresponding statements for infinite sums and integrals.
Then for all real- or complex-valued random variables X and Y on Ω, Remarks: Proof of the conditional Hölder inequality: Define the random variables and note that they are measurable with respect to the sub-σ-algebra.
On the set the right-hand side is infinite and the conditional Hölder inequality holds, too.
Dividing by the right-hand side, it therefore remains to show that This is done by verifying that the inequality holds after integration over an arbitrary Using the measurability of U, V, 1G with respect to the sub-σ-algebra, the rules for conditional expectations, Hölder's inequality and 1/p + 1/q = 1, we see that Let S be a set and let
we have the following implication (the seminorm is also allowed to attain the value ∞): Then: where the numbers
then the restriction of N to all Σ-measurable functions gives the usual version of Hölder's inequality.
Hölder inequality can be used to define statistical dissimilarity measures[7] between probability distributions.
Those Hölder divergences are projective: They do not depend on the normalization factor of densities.