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.
Contents
- Conventions
- Estimates for integrable products
- Generalization for probability measures
- Notable special cases
- Counting measure
- Lebesgue measure
- Probability measure
- Product measure
- Vector valued functions
- Proof of Hlders inequality
- Statement
- Remarks and examples
- Applications
- Generalization of Hlders inequality
- Interpolation
- Reverse Hlder inequality
- Conditional Hlder inequality
- Hlders inequality for increasing seminorms
- References
The numbers p and q above are said to be Hölder conjugates of each other. The special case p = q = 2 gives a form of the Cauchy–Schwarz inequality. Hölder's inequality holds even if || fg ||1 is infinite, the right-hand side also being infinite in that case. Conversely, if f is in Lp(μ) and g is in Lq(μ), then the pointwise product fg is in L1(μ).
Hölder's inequality is used to prove the Minkowski inequality, which is the triangle inequality in the space Lp(μ), and also to establish that Lq(μ) is the dual space of Lp(μ) for p ∈ [1, ∞).
Hölder's inequality was first found by Rogers (1888), and discovered independently by Hölder (1889).
Conventions
The brief statement of Hölder's inequality uses some conventions.
Estimates for integrable products
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. In particular, if f and g are in the Hilbert space L2(μ), then Hölder's inequality for p = q = 2 implies
where the angle brackets refer to the inner product of L2(μ). 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.
Generalization for probability measures
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.
Notable special cases
For the following cases assume that p and q are in the open interval (1,∞) with 1/p + 1/q = 1.
Counting measure
For the n-dimensional Euclidean space, when the set S is {1, ..., n} with the counting measure, we have
If S = N with the counting measure, then we get Hölder's inequality for sequence spaces:
These Hölder inequalities for the counting measure still hold when 1/p + 1/q ≤ 1.
Lebesgue measure
If S is a measurable subset of Rn with the Lebesgue measure, and f and g are measurable real- or complex-valued functions on S, then Hölder inequality is
Probability measure
For the probability space
Let
In particular, if the sth absolute moment is finite, then the r th absolute moment is finite, too. (This also follows from Jensen's inequality.)
These Hölder inequalities for a probability measure still hold when 1/p + 1/q ≤ 1.
Product measure
For two σ-finite measure spaces (S1, Σ1, μ1) and (S2, Σ2, μ2) define the product measure space by
where S is the Cartesian product of S1 and S2, the σ-algebra Σ arises as product σ-algebra of Σ1 and Σ2, and μ denotes the product measure of μ1 and μ2. 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.
Vector-valued functions
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.
Proof of Hölder's inequality
There are several proofs of Hölder's inequality; the main idea in the following is Young's inequality.
If || f ||p = 0, then f is zero μ-almost everywhere, and the product fg is zero μ-almost everywhere, hence the left-hand side of Hölder's inequality is zero. The same is true if || g ||q = 0. Therefore, we may assume || f ||p > 0 and || g ||q > 0 in the following.
If || f ||p = ∞ or || g ||q = ∞, then the right-hand side of Hölder's inequality is infinite. Therefore, we may assume that || f ||p and || g ||q are in (0, ∞).
If p = ∞ and q = 1, then | fg | ≤ || f ||∞ | g | almost everywhere and Hölder's inequality follows from the monotonicity of the Lebesgue integral. Similarly for p = 1 and q = ∞. Therefore, we may also assume p, q ∈ (1, ∞).
Dividing f and g by || f ||p and || g ||q, respectively, we can assume that
We now use Young's inequality, which states that
for all nonnegative a and b, where equality is achieved if and only if ap = bq. Hence
Integrating both sides gives
which proves the claim.
Under the assumptions p ∈ (1, ∞) and || f ||p = || g ||q, equality holds if and only if | f |p = | g |q almost everywhere. More generally, if || f ||p and || g ||q are in (0, ∞), then Hölder's inequality becomes an equality if and only if there exist real numbers α, β > 0, namely
such that
The case || f ||p = 0 corresponds to β = 0 in (*). The case || g ||q = 0 corresponds to α = 0 in (*).
Statement
Assume that 1 ≤ p < ∞ and let q denote the Hölder conjugate. Then, for every f ∈ Lp(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
Remarks and examples
Applications
Generalization of Hölder's inequality
Assume that r ∈ (0, ∞) and p1, …, pn ∈ (0, ∞] such that
Then, for all measurable real- or complex-valued functions f1, …, fn defined on S,
In particular,
Note: For r ∈ (0, 1), contrary to the notation, || . ||r is in general not a norm, because it doesn't satisfy the triangle inequality.
Interpolation
Let p1, ..., pn ∈ (0, ∞] and let θ1, ..., θn ∈ (0, 1) denote weights with θ1 + ... + θn = 1. Define p as the weighted harmonic mean, i.e.,
Given a measurable real- or complex-valued function f on S, define
Then by the above generalization of Hölder's inequality,
In particular, taking θ1 = θ and θ2 = 1 − θ, in the case n = 2, we obtain the interpolation result (Littlewood's inequality)
for
A similar application of Hölder gives Lyapunov's inequality: If
then
Both Littlewood and Lyapunov imply that if
Reverse Hölder inequality
Assume that p ∈ (1, ∞) and that the measure space (S, Σ, μ) satisfies μ(S) > 0. 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:
are not norms, they are just compact notations for
Conditional Hölder inequality
Let (Ω, F, ℙ) be a probability space, G ⊂ F a sub-σ-algebra, and p, q ∈ (1, ∞) Hölder conjugates, meaning that 1/p + 1/q = 1. Then, for all real- or complex-valued random variables X and Y on Ω,
Remarks:
Hölder's inequality for increasing seminorms
Let S be a set and let
Then:
where the numbers
Remark: If (S, Σ, μ) is a measure space and