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 Lp 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 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. 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 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,[1] which was in turn named for work of Johan Jensen building on Hölder's work.[2]



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 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 productsEdit

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 measuresEdit

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 casesEdit

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

Counting measureEdit

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  :


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


Equality holds iff  .

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


Lebesgue measureEdit

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 measureEdit

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 measureEdit

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 functionsEdit

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 inequalityEdit

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 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

    μ-almost everywhere   (*).

The case ||f||p = 0 corresponds to β = 0 in (*). The case ||g||q = 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.

Extremal equalityEdit


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 ||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 xB, hence ||g||1 ≤ 1. Furthermore,


Remarks and examplesEdit

  • 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 + f2||p ≤ ||f1||p + ||f2||p 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 of Hölder's inequalityEdit

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




and in particular


Both Littlewood and Lyapunov imply that if   then   for all  

Reverse Hölder inequalityEdit

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.

(Reverse) Hölder inequality with multiple negative normsEdit

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

As was proven by Wing-Sum Cheung in 2000[3]

Conditional Hölder inequalityEdit

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 seminormsEdit

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.[4]

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 inequalityEdit

Hölder inequality can be used to define statistical dissimilarity measures[5] between probability distributions. Those Hölder divergences are projective: They do not depend on the normalization factor of densities.

See alsoEdit


  1. ^ 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
  2. ^ 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, under the additional assumption that   exists, this inequality was already obtained by Hölder in 1889
  3. ^ Cheung, Wing-Sum (2001). "Generalizations of Hölder's inequality". International Journal of Mathematics and Mathematical Sciences. 26 (1): 7–10. doi:10.1155/S0161171201005658. ISSN 0161-1712.
  4. ^ For a proof see (Trèves 1967, Lemma 20.1, pp. 205–206).
  5. ^ 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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