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Hölder's inequality

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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 — 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 fLp(μ) and gLq(μ), then Hölder's inequality becomes an equality if and only if |f|p and |g|q are linearly dependent in L1(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.[1] Hölder's inequality holds even if fg1 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 (in a slightly different form) was first found by Leonard James Rogers (1888). Inspired by Rogers' work, Hölder (1889) gave another proof as part of a work developing the concept of convex and concave functions and introducing Jensen's inequality,[2] which was in turn named for work of Johan Jensen building on Hölder's work.[3]



The brief statement of Hölder's inequality uses some conventions.

  • In the definition of Hölder conjugates, 1/∞ means zero.
  • If p, q [1, ∞), then fp and gq stand for the (possibly infinite) expressions
  • If p = ∞, then f stands for the essential supremum of |f|, similarly for g.
  • The notation fp with 1 ≤ p ≤ ∞ is a slight abuse, because in general it is only a norm of f if fp is finite and f is considered as equivalence class of μ-almost everywhere equal functions. If fLp(μ) and gLq(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[edit]

As above, let f and g denote measurable real- or complex-valued functions defined on S. If fg1 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 f2 and g2 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[edit]

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[edit]

For the following cases assume that p and q are in the open interval (1,∞) with 1/p + 1/q = 1.

Counting measure[edit]

For the -dimensional Euclidean space, when the set is with the counting measure, we have

Often the following practical form of this is used, for any :

For more than two sums, the following generalisation (Chen (2015)) holds, with real positive exponents and :

Equality holds iff .

If with the counting measure, then we get Hölder's inequality for sequence spaces:

Lebesgue measure[edit]

If is a measurable subset of with the Lebesgue measure, and and are measurable real- or complex-valued functions on , then Hölder's inequality is

Probability measure[edit]

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 sth absolute moment is finite, then the r th absolute moment is finite, too. (This also follows from Jensen's inequality.)

Product measure[edit]

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[edit]

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[edit]

There are several proofs of Hölder's inequality; the main idea in the following is Young's inequality for products.


If fp = 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 gq = 0. Therefore, we may assume fp > 0 and gq > 0 in the following.

If fp = ∞ or gq = ∞, then the right-hand side of Hölder's inequality is infinite. Therefore, we may assume that fp and gq 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 (1,∞).

Dividing f and g by fp and gq, 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 ap = bq. Hence

Integrating both sides gives

which proves the claim.

Under the assumptions p (1, ∞) and fp = ‖gq, equality holds if and only if |f|p = |g|q almost everywhere. More generally, if fp and gq 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 fp = 0 corresponds to β = 0 in (*). The case gq = 0 corresponds to α = 0 in (*).

Alternative proof using Jensen's inequality:


The function on (0,∞) is convex because , so by Jensen's inequality,

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 that f, g are 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.

We could also bypass use of both Young's and Jensen's inequalities. The proof below also explains why and where the Hölder exponent comes in naturally.


As in the previous proof, it suffices to prove

where and is -measurable (real or complex) function on . To prove this, we must bound by . There is no constant that will make for all . Hence, we seek an inequality of the form

for suitable choices of and .

We wish to obtain on the right-hand side after integrating this inequality. By trial and error, we see that the inequality we wish should have the form

where are non-negative and . Indeed, the integral of the right-hand side is precisely . So, it remains to prove that such an inequality does hold with the right choice of

The inequality we seek would follow from:

which, in turn, is equivalent to

It turns out there is one and only one choice of , subject to , that makes this true: and, necessarily, . (This is where Hölder conjugate exponent is born!) This completes the proof of the inequality at the first paragraph of this proof. Proof of Hölder's inequality follows from this as in the previous proof. Alternatively, we can deduce Young's inequality and then resort to the first proof given above. Young's inequality follows from the inequality (*) above by choosing and multiplying both sides by .

Extremal equality[edit]


Assume that 1 ≤ p < ∞ and let q denote the Hölder conjugate. Then for every fLp(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 fp = 0. Therefore, we assume fp > 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 gq = 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 xB, hence g1 ≤ 1. Furthermore,

Remarks and examples[edit]

  • 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


  • The extremal equality is one of the ways for proving the triangle inequality f1 + f2p ≤ ‖f1p + ‖f2p for all f1 and f2 in Lp(μ), see Minkowski inequality.
  • Hölder's inequality implies that every fLp(μ) defines a bounded (or continuous) linear functional κf on Lq(μ) by the formula
The extremal equality (when true) shows that the norm of this functional κf as element of the continuous dual space Lq(μ)* coincides with the norm of f in Lp(μ) (see also the Lp-space article).

Generalization with more than two functions[edit]


Assume that r (0, ∞] and p1, ..., pn (0, ∞] such that

where 1/∞ is interpreted as 0 in this equation. Then for all measurable real or complex-valued functions f1, ..., fn 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 |fn| 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.


Let p1, ..., pn (0, ∞] and let θ1, ..., θn ∈ (0, 1) denote weights with θ1 + ... + θn = 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 θ1 = θ and θ2 = 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 [4]

Reverse Hölder inequalities[edit]

Two functions[edit]

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 sS,


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 (hidden, click show to reveal.)

Note that p and

are Hölder conjugates. Application of Hölder's inequality gives

Raising to the power p gives us:


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.

Multiple functions[edit]

The Reverse Hölder inequality (above) can be generalized to the case of multiple functions if all but one conjugate is negative. That is,

Let and be such that (hence ). Let be measurable functions for . Then

This follows from the symmetric form of the Hölder inequality (see below).

Symmetric forms of Hölder inequality[edit]

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 be vectors with positive entries and such that for all . If are nonzero real numbers such that , then:

  • if all but one of are positive;
  • if all but one of are negative.

The standard Hölder inequality follows immediately from this symmetric form (and in fact is easily seen to be equivalent to it). The symmetric statement also implies the reverse Hölder inequality (see above).

The result can be extended to multiple vectors:

Let be vectors in with positive entries and such that for all . If are nonzero real numbers such that , then:

  • if all but one of the numbers are positive;
  • if all but one of the numbers are negative.

As in the standard Hölder inequalities, there are corresponding statements for infinite sums and integrals.

Conditional Hölder inequality[edit]

Let (Ω, F, ) be a probability space, GF 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,


  • 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, 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

Hölder's inequality for increasing seminorms[edit]

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 ∞):


where the numbers and are Hölder conjugates.[6]

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.

Distances based on Hölder inequality[edit]

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.

See also[edit]


  1. ^ Roman 2008, p. 303 §12
  2. ^ Maligranda, Lech (1998), "Why Hölder's inequality should be called Rogers' inequality", Mathematical Inequalities & Applications, 1 (1): 69–83, doi:10.7153/mia-01-05, MR 1492911
  3. ^ Guessab, A.; Schmeisser, G. (2013), "Necessary and sufficient conditions for the validity of Jensen's inequality", Archiv der Mathematik, 100 (6): 561–570, doi:10.1007/s00013-013-0522-3, MR 3069109, S2CID 253600514, under the additional assumption that exists, this inequality was already obtained by Hölder in 1889
  4. ^ Wojtaszczyk, P. (1991). Banach Spaces for Analysts. Cambridge Studies in Advanced Mathematics. Cambridge: Cambridge University Press. ISBN 978-0-521-56675-9.
  5. ^ Beckenbach, E. F. (1980). General inequalities 2. International Series of Numerical Mathematics / Internationale Schriftenreihe zur Numerischen Mathematik / Série Internationale d'Analyse Numérique. Vol. 47. Birkhäuser Basel. pp. 145–150. doi:10.1007/978-3-0348-6324-7. ISBN 978-3-7643-1056-1.
  6. ^ For a proof see (Trèves 1967, Lemma 20.1, pp. 205–206).
  7. ^ Nielsen, Frank; Sun, Ke; Marchand-Maillet, Stephane (2017). "On Hölder projective divergences". Entropy. 3 (19): 122. arXiv:1701.03916. Bibcode:2017Entrp..19..122N. doi:10.3390/e19030122.


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