Hölder's inequality
In mathematical analysis, Hölder's inequality, named after Otto Hölder, is a fundamental inequality between integrals and an indispensable tool for the study of L^{p} spaces.
- Theorem (Hölder's inequality). Let (S, Σ, μ) be a measure space and let p, q ∈ [1, ∞] with 1/p + 1/q = 1. Then for all measurable real- or complex-valued functions f and g on S,
- If, in addition, p, q ∈ (1, ∞) and f ∈ L^{p}(μ) and g ∈ L^{q}(μ), then Hölder's inequality becomes an equality if and only if |f |^{p} and |g|^{q} are linearly dependent in L^{1}(μ), meaning that there exist real numbers α, β ≥ 0, not both of them zero, such that α|f |^{p} = β |g|^{q} μ-almost everywhere.
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 L^{p}(μ) and g is in L^{q}(μ), then the pointwise product fg is in L^{1}(μ).
Hölder's inequality is used to prove the Minkowski inequality, which is the triangle inequality in the space L^{p}(μ), and also to establish that L^{q}(μ) is the dual space of L^{p}(μ) for p ∈ [1, ∞).
Hölder's inequality was first found by Leonard James Rogers (Rogers (1888)), and discovered independently by Hölder (1889).
Remarks
Conventions
The brief statement of Hölder's inequality uses some conventions.
- In the definition of Hölder conjugates, 1/∞ means zero.
- If p, q ∈ [1, ∞), then ||f ||_{p} and ||g||_{q} stand for the (possibly infinite) expressions
- If p = ∞, then ||f ||_{∞} stands for the essential supremum of |f |, similarly for ||g||_{∞}.
- The notation ||f ||_{p} with 1 ≤ p ≤ ∞ is a slight abuse, because in general it is only a norm of f if ||f ||_{p} is finite and f is considered as equivalence class of μ-almost everywhere equal functions. If f ∈ L^{p}(μ) and g ∈ L^{q}(μ), then the notation is adequate.
- On the right-hand side of Hölder's inequality, 0 × ∞ as well as ∞ × 0 means 0. Multiplying a > 0 with ∞ gives ∞.
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 L^{2}(μ), then Hölder's inequality for p = q = 2 implies
where the angle brackets refer to the inner product of L^{2}(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
Often the following practical form of this is used, for any :
If S = N with the counting measure, then we get Hölder's inequality for sequence spaces:
Lebesgue measure
If S is a measurable subset of R^{n} 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 denote the expectation operator. For real- or complex-valued random variables and on Hölder's inequality reads
Let and define Then is the Hölder conjugate of Applying Hölder's inequality to the random variables and we obtain
In particular, if the s^{th} absolute moment is finite, then the r^{ th} absolute moment is finite, too. (This also follows from Jensen's inequality.)
Product measure
For two σ-finite measure spaces (S_{1}, Σ_{1}, μ_{1}) and (S_{2}, Σ_{2}, μ_{2}) define the product measure space by
where S is the Cartesian product of S_{1} and S_{2}, the σ-algebra Σ arises as product σ-algebra of Σ_{1} and Σ_{2}, and μ denotes the product measure of μ_{1} and μ_{2}. 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 = (f_{1}, ..., f_{n}) and g = (g_{1}, ..., g_{n}) 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 for products.
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 assume p, q ∈ (0, 1) ∪ (1,∞). However, to apply Young's inequality for products, we will require p, q ∈ (1,∞)
Dividing f and g by ||f ||_{p} and ||g||_{q}, respectively, we can assume that
We now use Young's inequality for products, which states that whenever are in (1,∞) with
for all nonnegative a and b, where equality is achieved if and only if a^{p} = b^{q}. 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
- μ-almost everywhere (*).
The case ||f ||_{p} = 0 corresponds to β = 0 in (*). The case ||g||_{q} = 0 corresponds to α = 0 in (*).
Alternate proof using Jensen's inequality: Recall the Jensen's inequality for the convex function (it is convex because obviously ):
where ν is any probability distribution and h any ν-measurable function. Let μ be any measure, and ν the distribution whose density w.r.t. μ is proportional to , i.e.
Hence we have, using , hence , and letting ,
Finally, we get
This assumes f, g real and non negative, but the extension to complex functions is straightforward (use the modulus of f, g). It also assumes that are neither null nor infinity, and that : all these assumptions can also be lifted as in the proof above.
Extremal equality
Statement
Assume that 1 ≤ p < ∞ and let q denote the Hölder conjugate. Then for every f ∈ L^{p}(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 ||_{p} = 0. Therefore, we assume ||f ||_{p} > 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||_{q} = 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} ≤ 1. Furthermore,
Remarks and examples
- The equality for fails whenever there exists a set of infinite measure in the -field with that has no subset that satisfies: (the simplest example is the -field containing just the empty set and and the measure with ) Then the indicator function satisfies but every has to be -almost everywhere constant on because it is -measurable, and this constant has to be zero, because is -integrable. Therefore, the above supremum for the indicator function is zero and the extremal equality fails.
- For the supremum is in general not attained. As an example, let and the counting measure. Define:
- Then For with let denote the smallest natural number with Then
Applications
- The extremal equality is one of the ways for proving the triangle inequality ||f_{1} + f_{2}||_{p} ≤ ||f_{1}||_{p} + ||f_{2}||_{p} for all f_{1} and f_{2} in L^{p}(μ), see Minkowski inequality.
- Hölder's inequality implies that every f ∈ L^{p}(μ) defines a bounded (or continuous) linear functional κ_{f} on L^{q}(μ) by the formula
- The extremal equality (when true) shows that the norm of this functional κ_{f} as element of the continuous dual space L^{q}(μ)^{*} coincides with the norm of f in L^{p}(μ) (see also the L^{p}-space article).
Generalization of Hölder's inequality
Assume that r ∈ (0, ∞] and p_{1}, ..., p_{n} ∈ (0, ∞] such that
where 1/∞ is interpreted as 0 in this equation. Then for all measurable real or complex-valued functions f_{1}, ..., f_{n} defined on S,
where we interpret any product with a factor of ∞ as ∞ if all factors are positive, but the product is 0 if any factor is 0.
In particular, if for all then
Note: For contrary to the notation, ||.||_{r} is in general not a norm because it doesn't satisfy the triangle inequality.
Proof of the generalization: We use Hölder's inequality and mathematical induction. If then the result is immediate. Let us now pass from to Without loss of generality assume that
Case 1: If then
Pulling out the essential supremum of |f_{n}| and using the induction hypothesis, we get
Case 2: If then necessarily as well, and then
are Hölder conjugates in (1, ∞). Application of Hölder's inequality gives
Raising to the power and rewriting,
Since and
the claimed inequality now follows by using the induction hypothesis.
Interpolation
Let p_{1}, ..., p_{n} ∈ (0, ∞] and let θ_{1}, ..., θ_{n} ∈ (0, 1) denote weights with θ_{1} + ... + θ_{n} = 1. Define as the weighted harmonic mean, that is,
Given measurable real- or complex-valued functions on S, then the above generalization of Hölder's inequality gives
In particular, taking gives
Specifying further θ_{1} = θ and θ_{2} = 1-θ, in the case we obtain the interpolation result (Littlewood's inequality)
for and
An application of Hölder gives Lyapunov's inequality: If
then
and in particular
Both Littlewood and Lyapunov imply that if then for all
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
Proof of the reverse Hölder inequality: Note that p and
are Hölder conjugates. Application of Hölder's inequality gives
Raising to the power p gives us:
Therefore:
Now we just need to recall our notation.
Since g is not almost everywhere equal to the zero function, we can have equality if and only if there exists a constant α ≥ 0 such that |fg| = α |g|^{−q/p} almost everywhere. Solving for the absolute value of f gives the claim.
Conditional Hölder inequality
Let (Ω, F, ) 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:
- If a non-negative random variable Z has infinite expected value, then its conditional expectation is defined by
- On the right-hand side of the conditional Hölder inequality, 0 times ∞ as well as ∞ times 0 means 0. Multiplying a > 0 with ∞ gives ∞.
Proof of the conditional Hölder inequality: Define the random variables
and note that they are measurable with respect to the sub-σ-algebra. Since
it follows that |X| = 0 a.s. on the set {U = 0}. Similarly, |Y| = 0 a.s. on the set {V = 0}, hence
and the conditional Hölder inequality holds on this set. 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, 1_{G} with respect to the sub-σ-algebra, the rules for conditional expectations, Hölder's inequality and 1/p + 1/q = 1, we see that
Hölder's inequality for increasing seminorms
Let S be a set and let be the space of all complex-valued functions on S. Let N be an increasing seminorm on meaning that, for all real-valued functions we have the following implication (the seminorm is also allowed to attain the value ∞):
Then:
where the numbers and are Hölder conjugates.[1]
Remark: If (S, Σ, μ) is a measure space and is the upper Lebesgue integral of then the restriction of N to all Σ-measurable functions gives the usual version of Hölder's inequality.
See also
- Cauchy–Schwarz inequality
- Minkowski inequality
- Jensen's inequality
- Young's inequality for products
- Clarkson's inequalities
- Brascamp–Lieb inequality
Citations
- For a proof see (Trèves 1967, Lemma 20.1, pp. 205–206).
References
- Grinshpan, A. Z. (2010), "Weighted inequalities and negative binomials", Advances in Applied Mathematics, 45 (4): 564–606, doi:10.1016/j.aam.2010.04.004
- Hardy, G. H.; Littlewood, J. E.; Pólya, G. (1934), Inequalities, Cambridge University Press, pp. XII+314, ISBN 0-521-35880-9, JFM 60.0169.01, Zbl 0010.10703.
- Hölder, O. (1889), "Ueber einen Mittelwertsatz", Nachrichten von der Königl. Gesellschaft der Wissenschaften und der Georg-Augusts-Universität zu Göttingen, Band (in German), 1889 (2): 38–47, JFM 21.0260.07. Available at Digi Zeitschriften.
- Kuptsov, L. P. (2001) [1994], "Hölder inequality", Encyclopedia of Mathematics, EMS Press.
- Narici, Lawrence; Beckenstein, Edward (2011). Topological Vector Spaces. Pure and applied mathematics (Second ed.). Boca Raton, FL: CRC Press. ISBN 978-1584888666. OCLC 144216834.
- Rogers, L. J. (February 1888), "An extension of a certain theorem in inequalities", Messenger of Mathematics, New Series, XVII (10): 145–150, JFM 20.0254.02, archived from the original on August 21, 2007.
- Trèves, François (1967), Topological Vector Spaces, Distributions and Kernels, Pure and Applied Mathematics. A Series of Monographs and Textbooks, 25, New York, London: Academic Press, MR 0225131, Zbl 0171.10402.
- Trèves, François (2006) [1967]. Topological Vector Spaces, Distributions and Kernels. Mineola, N.Y.: Dover Publications. ISBN 978-0-486-45352-1. OCLC 853623322.
External links
- Kuttler, Kenneth (2007), An Introduction to Linear Algebra (PDF), Online e-book in PDF format, Brigham Young University.
- Lohwater, Arthur (1982), Introduction to Inequalities (PDF).
- Tisdell, Chris (2012), Holder's Inequality, Online video on Dr Chris Tisdell's YouTube channel.