Holder Conjugate
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Holder Conjugate

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.

Theorem (Hölder's inequality). Let (S, ?, ?) 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 ? Lp(?) and g ? Lq(?), then Hölder's inequality becomes an equality iff ||p and |g|q are linearly dependent in L1(?), meaning that there exist real numbers ?, ? >= 0, not both of them zero, such that ?|f |p = ? |g|q ?-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 ||||1 is infinite, the right-hand side also being infinite in that case. Conversely, if f  is in Lp(?) and g is in Lq(?), then the pointwise product fg is in L1(?).

Hölder's inequality is used to prove the Minkowski inequality, which is the triangle inequality in the space Lp(?), and also to establish that Lq(?) is the dual space of Lp(?) for p ? [1, ?).

Hölder's inequality was first found by Leonard James Rogers (Rogers (1888)), and discovered independently by Hölder (1889).



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

  • In the definition of Hölder conjugates, 1/ ? means zero.
  • If p, q ? [1, ?), then ||||p and ||||q stand for the (possibly infinite) expressions
  • If p = ?, then ||||? stands for the essential supremum of ||, similarly for ||||?.
  • The notation ||||p with 1 p is a slight abuse, because in general it is only a norm of f  if ||||p is finite and f  is considered as equivalence class of ?-almost everywhere equal functions. If f ? Lp(?) and g ? Lq(?), 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 ||||1 is finite, then the pointwise products of f  with g and its complex conjugate function are ?-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(?), then Hölder's inequality for p = q = 2 implies

where the angle brackets refer to the inner product of L2(?). This is also called Cauchy-Schwarz inequality, but requires for its statement that ||||2 and ||||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 || and || in place of f  and g.

Generalization for probability measures

If (S, ?, ?) is a probability space, then p, q ? [1, ?] just need to satisfy 1/p + 1/q , 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:

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 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

For two ?-finite measure spaces (S1, ?1, ?1) and (S2, ?2, ?2) define the product measure space by

where S is the Cartesian product of S1 and S2, the arises as product ?-algebra of ?1 and ?2, and ? denotes the product measure of ?1 and ?2. Then Tonelli's theorem allows us to rewrite Hölder's inequality using iterated integrals: If f  and g are real- or complex-valued functions on the Cartesian product S, then

This can be generalized to more than two measure spaces.

Vector-valued functions

Let (S, ?, ?) denote a measure space and suppose that f = (f1, ..., fn) and g = (g1, ..., gn) are ?-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 ?, ? >= 0, not both of them zero, such that

for ?-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 ||||p = 0, then f  is zero ?-almost everywhere, and the product fg is zero ?-almost everywhere, hence the left-hand side of Hölder's inequality is zero. The same is true if ||||q = 0. Therefore, we may assume ||||p > 0 and ||||q > 0 in the following.

If ||||p = ? or ||||q = ?, then the right-hand side of Hölder's inequality is infinite. Therefore, we may assume that ||||p and ||||q are in (0, ?).

If p = ? and q = 1, then || f||? || 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 ||||p and ||||q, respectively, we can assume that

We now use Young's inequality for products, 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 ||||p = ||||q, equality holds if and only if ||p = ||q almost everywhere. More generally, if ||||p and ||||q are in (0, ?), then Hölder's inequality becomes an equality if and only if there exist real numbers ?, ? > 0, namely

such that

   ?-almost everywhere   (*).

The case ||||p = 0 corresponds to ? = 0 in (*). The case ||||q = 0 corresponds to ? = 0 in (*).

Extremal equality


Assume that 1 p < ? and let q denote the Hölder conjugate. Then, for every f ? Lp(?),

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

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


  • The extremal equality is one of the ways for proving the triangle inequality ||||pf1||p + ||||p for all f1 and f2 in Lp(?), see Minkowski inequality.
  • Hölder's inequality implies that every f ? Lp(?) defines a bounded (or continuous) linear functional ?f  on Lq(?) by the formula
The extremal equality (when true) shows that the norm of this functional ?f  as element of the continuous dual space Lq(?)* coincides with the norm of f  in Lp(?) (see also the article).

Generalization of Hölder's inequality

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

(where we interpret 1/? 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,

Note: For r ? (0, 1), contrary to the notation, ||||r is in general not a norm, because it doesn't satisfy the triangle inequality.


Let p1, ..., pn ? (0, ?] and let ?1, ..., ?n ? (0, 1) denote weights with ?1 + ... + ?n = 1. Define p as the weighted harmonic mean, i.e.,

Given measurable real- or complex-valued functions on S, then the above generalization of Hölder's inequality gives

In particular, taking gives

Specifying further ?1 = ? and ?2 = 1-?, in the case n = 2, we obtain the interpolation result (Littlewood's inequality)

for and

An application of Hölder gives Lyapunov's inequality: If


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, ?, ?) satisfies ?(S) > 0. Then, for all measurable real- or complex-valued functions f  and g on S such that g(s) ? 0 for all s ? S,


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 (?, F, P) be a probability space, G ? F a , 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 ?,


  • On the right-hand side of the conditional Hölder inequality, 0 times ? as well as ? times 0 means 0. Multiplying a > 0 with ? gives ?.

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 ?):


where the numbers and are Hölder conjugates.[1]

Remark: If (S, ?, ?) is a measure space and is the upper Lebesgue integral of then the restriction of N to all functions gives the usual version of Hölder's inequality.

See also


  1. ^ For a proof see (Trèves 1967, Lemma 20.1, pp. 205-206).


  • 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", in Hazewinkel, Michiel (ed.), Encyclopedia of Mathematics, Springer Science+Business Media B.V. / Kluwer Academic Publishers, ISBN 978-1-55608-010-4.
  • 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.

External links

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