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.

Let (S,Σ,μ) be a measure space and let 1 ≤ p, q ≤ ∞ with 1/p + 1/q = 1. Then, for all measurable real- or complex-valued functions f and g on S,

$\|fg\|_{1}\leq \|f\|_{p}\|g\|_{q}.$

The numbers p and q above are said to be Hölder conjugates of each other. The special case p = q = 2 gives the Cauchy-Schwarz inequality.

This inequality holds even if ||fg||₁ is infinite, the right-hand side also being infinite in that case. In particular, if f is in L^p(μ) and g is in L^q(μ), then fg is in L¹(μ).

For 1 < p, q < ∞ and f ∈ L^p(μ) and g ∈ L^q(μ), 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 α, β ≥ 0, not both of them zero, such that α|f |^p = β|g|^q μ-almost everywhere.

Hölder's inequality is used to prove the triangle inequality in the space L^p(μ) and the Minkowski inequality, and also to establish that L^p(μ) is dual to L^q(μ) for 1 ≤ q < ∞.

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

Remarks

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

• In the definition of Hölder conjugates, 1/∞ means zero.

• If 1 ≤ p, q < ∞, then ||f ||_{p and ||g||q stand for the (possibly infinite) expressions}

${\biggl (}\int _{S}|f|^{p}\,\mathrm {d} \mu {\biggr )}^{1/p}$   and   ${\biggl (}\int _{S}|g|^{q}\,\mathrm {d} \mu {\biggr )}^{1/q}.$

• 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 times ∞ as well as ∞ times 0 means 0. Multiplying a > 0 with ∞ gives ∞.

Notable special cases

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

• In the case of n-dimensional Euclidean space, when the set S is {1, …, n} with the counting measure, we have

$\sum _{k=1}^{n}|x_{k}\,y_{k}|\leq {\biggl (}\sum _{k=1}^{n}|x_{k}|^{p}{\biggr )}^{\!1/p\;}{\biggl (}\sum _{k=1}^{n}|y_{k}|^{q}{\biggr )}^{\!1/q}{\text{ for all }}(x_{1},\ldots ,x_{n}),(y_{1},\ldots .y_{n})\in \mathbb {R} ^{n}{\text{ or }}\mathbb {C} ^{n}.$

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

$\sum \limits _{k=1}^{\infty }|x_{k}\,y_{k}|\leq {\biggl (}\sum _{k=1}^{\infty }|x_{k}|^{p}{\biggr )}^{\!1/p\;}{\biggl (}\sum _{k=1}^{\infty }|y_{k}|^{q}{\biggr )}^{\!1/q}{\text{ for all }}(x_{k})_{k\in \mathbb {N} },(y_{k})_{k\in \mathbb {N} }\in \mathbb {R} ^{\mathbb {N} }{\text{ or }}\mathbb {C} ^{\mathbb {N} }.$

• 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

$\int _{S}{\bigl |}f(x)g(x){\bigr |}\,\mathrm {d} x\leq {\biggl (}\int _{S}|f(x)|^{p}\,\mathrm {d} x{\biggr )}^{\!1/p\;}{\biggl (}\int _{S}|g(x)|^{q}\,\mathrm {d} x{\biggr )}^{\!1/q}.$

• For the probability space $(\Omega ,{\mathcal {F}},\mathbb {P} )$, let $\operatorname {E}$ denote the expectation operator. For real- or complex-valued random variables X and Y on Ω, Hölder's inequality reads

$\operatorname {E} |XY|\leq {\bigl (}\operatorname {E} |X|^{p}{\bigr )}^{1/p}\;{\bigl (}\operatorname {E} |Y|^{q}{\bigr )}^{1/q}.$

Let 0 < r < s and define p = s/r. Then q = p/(p−1) is the Hölder conjugate of p. Applying Hölder's inequality to the random variables |X|^r and 1_Ω, we obtain

$\operatorname {E} |X|^{r}\leq {\bigl (}\operatorname {E} |X|^{s}{\bigr )}^{r/s}.$

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.)

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 Hölders 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}

$\|f\|_{p}=\|g\|_{q}=1.$

We now use Young's inequality, which states that

$ab\leq {\frac {a^{p}}{p}}+{\frac {b^{q}}{q}},$

for all nonnegative a and b, where equality is achieved if and only if a^p = b^q. Hence

$|f(s)g(s)|\leq {\frac {|f(s)|^{p}}{p}}+{\frac {|g(s)|^{q}}{q}},\qquad s\in S.$

Integrating both sides gives

$\|fg\|_{1}\leq {\frac {1}{p}}+{\frac {1}{q}}=1,$

which proves the claim.

Under the assumptions p ∈ (1,∞) and ||f ||_p = ||g||_q = 1, 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 α, β > 0 (namely α = ||g||_q and β = ||f ||_p) such that

$\alpha |f|^{p}=\beta |g|^{q}\,$   μ-almost everywhere   (*)

The case ||f ||_p = 0 corresponds to β = 0 in (*). The case ||g||_q = 0 corresponds to α = 0 in (*).

Extremal equality

Statement

Assume that 1 ≤ p < ∞ and let q denote the Hölder conjugate. Then, for every ƒ ∈ L^p(μ),

$\|f\|_{p}=\max {\Bigl \{}{\Bigl |}\int _{S}fg\,\mathrm {d} \mu {\Bigr |}:g\in L^{q}(\mu ),\,\|g\|_{q}\leq 1{\Bigr \}},$

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

$\|f\|_{\infty }=\sup {\Bigl \{}{\Bigl |}\int _{S}fg\,\mathrm {d} \mu {\Bigr |}:g\in L^{1}(\mu ),\,\|g\|_{1}\leq 1{\Bigr \}}.$

Proof of the extremal equality

By Hölder's inequality, the integrals are well defined and, for 1 ≤ p ≤ ∞,

${\Bigl |}\int _{S}fg\,\mathrm {d} \mu {\Bigl |}\leq \int _{S}|fg|\,\mathrm {d} \mu \leq \|f\|_{p}\,,$

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 ||ƒ ||_p = 0. Therefore, we assume ||ƒ ||_p > 0 in the following.

If 1 ≤ p < ∞, define g on S by

$g(x)={\begin{cases}\|f\|_{p}^{1-p}\,|f(x)|^{p}/f(x)&{\text{if }}f(x)\not =0,\\0&{\text{otherwise.}}\end{cases}}$

By checking the cases p = 1 and 1 < p < ∞ separately, we see that ||g||_q = 1 and

$\int _{S}fg\,\mathrm {d} \mu =\|f\|_{p}\,.$

It remains to consider the case p = ∞. For ε ∈ (0, 1) define

$A={\bigl \{}x\in S:|f(x)|>(1-\varepsilon )\|f\|_{\infty }{\bigr \}}.$

Since ƒ  is measurable, A ∈ Σ. By the definition of ||ƒ ||_∞ as the essential supremum of ƒ and the assumption ||ƒ ||_∞ > 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

$g(x)={\begin{cases}{\frac {1-\varepsilon }{\mu (B)}}{\frac {\|f\|_{\infty }}{f(x)}}&{\text{if }}x\in B,\\0&{\text{otherwise.}}\end{cases}}$

Then g is well-defined, measurable and |g(x)| ≤ 1/μ(B) for x ∈ B, hence ||g||₁ ≤ 1. Furthermore,

${\Bigl |}\int _{S}fg\,\mathrm {d} \mu {\Bigr |}=\int _{B}{\frac {1-\varepsilon }{\mu (B)}}\|f\|_{\infty }\,\mathrm {d} \mu =(1-\varepsilon )\|f\|_{\infty }.$

Remarks and examples

• The equality for p = ∞ fails whenever there exists a set A in the σ-field Σ with μ(A) = ∞ that has no subset B ∈ Σ with 0 < μ(B) < ∞ (the simplest example is the σ-field Σ containing just the empty set and S, and the measure μ with μ(S) = ∞). Then the indicator function 1_{A satisfies ||1A||∞ = 1, but every g ∈ L¹(μ) has to be μ-almost everywhere constant on A, because it is Σ-measurable, and this constant has to be zero, because g is μ-integrable. Therefore, the above supremum for the indicator function 1A is zero and the extremal equality fails.}
• For p = ∞, the supremum is in general not attained. As an example, let S denote the natural numbers (without zero), Σ the power set of S, and μ the counting measure. Define ƒ(n) = (n − 1)/n for every natural number n. Then ||ƒ ||_∞ = 1. For g ∈ L¹(μ) with 0 < ||g||₁ ≤ 1, let m denote the smallest natural number with g(m) ≠ 0. Then

${\Bigl |}\int _{S}fg\,\mathrm {d} \mu {\Bigr |}\leq {\frac {m-1}{m}}|g(m)|+\sum _{n=m+1}^{\infty }|g(n)|=\|g\|_{1}-{\frac {|g(m)|}{m}}<1.$

Applications

• The extremal equality is one of the ways for proving the triangle inequality ||ƒ₁ + ƒ₂||_p ≤ ||ƒ₁||_p + ||ƒ₂||_p  for all ƒ₁ and ƒ₂ in L^p(μ), see Minkowski inequality.

• Hölder's inequality implies that every ƒ ∈ L^p(μ) defines a bounded (or continuous) linear functional κ_ƒ on L^q(μ) by the formula

$\kappa _{f}(g)=\int _{S}fg\,\mathrm {d} \mu ,\qquad g\in L^{q}(\mu ).$

The extremal equality (when true) shows that the norm of this functional κ_ƒ as element of the continuous dual space L^q(μ)^∗ coincides with the norm of ƒ in L^p(μ) (see also the L^p-space article).

Generalization of Hölder's inequality

Assume that r ∈ (0,∞) and p_{1, …, pn ∈ (0,∞] such that}

$\sum _{k=1}^{n}{\frac {1}{p_{k}}}={\frac {1}{r}}.$

Then, for all measurable real- or complex valued functions f₁, …, f_n defined on S,

${\biggl \|}\prod _{k=1}^{n}f_{k}{\biggr \|}_{r}\leq \prod _{k=1}^{n}\|f_{k}\|_{p_{k}}.$

In particular,

$f_{k}\in L^{p_{k}}(\mu )\;\;\forall k\in \{1,\ldots ,n\}\implies \prod _{k=1}^{n}f_{k}\in L^{r}(\mu ).$

Note: For r ∈ (0,1), 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. For n = 1, the result is obvious. Let us now pass from n − 1 to n. Without loss of generality assume that p_{1 ≤ … ≤ pn.}

Case 1: If p_n = ∞, then

$\sum _{k=1}^{n-1}{\frac {1}{p_{k}}}={\frac {1}{r}}.$

Pulling out the essential supremum of |f_n| and using the induction hypothesis, we get

${\begin{aligned}\|f_{1}\cdots f_{n}\|_{r}&\leq \|f_{1}\cdots f_{n-1}\|_{r}\|f_{n}\|_{\infty }\\&\leq \|f_{1}\|_{p_{1}}\cdots \|f_{n-1}\|_{p_{n-1}}\|f_{n}\|_{\infty }.\end{aligned}}$

Case 2: If p_n < ∞, then

$p:={\frac {p_{n}}{p_{n}-r}}$   and   $q:={\frac {p_{n}}{r}}$

are Hölder conjugates in (1,∞). Application of Hölder's inequality gives

${\bigl \|}|f_{1}\cdots f_{n-1}|^{r}\,|f_{n}|^{r}{\bigr \|}_{1}\leq {\bigl \|}|f_{1}\cdots f_{n-1}|^{r}{\bigr \|}_{p}\,{\bigl \|}|f_{n}|^{r}{\bigr \|}_{q}.$

Raising to the power 1/r and rewriting,

$\|f_{1}\cdots f_{n}\|_{r}\leq \|f_{1}\cdots f_{n-1}\|_{pr}\|f_{n}\|_{qr}.$

Since qr = p_n and

$\sum _{k=1}^{n-1}{\frac {1}{p_{k}}}={\frac {1}{r}}-{\frac {1}{p_{n}}}={\frac {p_{n}-r}{rp_{n}}}={\frac {1}{pr}}\,,$

the claimed inequality now follows by using the induction hypothesis.

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,

$\|fg\|_{1}\geq \|f\|_{1/p}\,\|g\|_{-1/(p-1)}.$

If ||fg||₁ < ∞ and ||g||_−1/(p−1) > 0, then the reverse Hölder inequality is an equality if and only if there exists an α ≥ 0 such that

$|f|=\alpha |g|^{-p/(p-1)}\,$    μ-almost everywhere.

Note: ||f ||_1/p and ||g||_−1/(p−1) are not norms, these expressions are just compact notation for

${\biggl (}\int _{S}|f|^{1/p}\,\mathrm {d} \mu {\biggr )}^{\!p}$   and   ${\biggl (}\int _{S}|g|^{-1/(p-1)}\,\mathrm {d} \mu {\biggr )}^{-(p-1)}.$

Proof of the reverse Hölder inequality

Note that p and

$q:={\frac {p}{p-1}}\in (1,\infty )$

are Hölder conjugates. Application of Hölder's inequality gives

${\begin{aligned}{\bigl \|}|f|^{1/p}{\bigr \|}_{1}&={\bigl \|}|fg|^{1/p}\,|g|^{-1/p}{\bigr \|}_{1}\\&\leq {\bigl \|}|fg|^{1/p}{\bigr \|}_{p}\,{\bigl \|}|g|^{-1/p}{\bigr \|}_{q}=\|fg\|_{1}^{1/p}\,{\bigl \|}|g|^{-1/(1-p)}{\bigr \|}_{1}^{(p-1)/p}.\end{aligned}}$

Raising to the power p, rewriting and solving for ||fg||₁ gives the reverse Hölder inequality.

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 $(\Omega ,{\mathcal {F}},\mathbb {P} )$ be a probability space, ${\mathcal {G}}\subset {\mathcal {F}}$ 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,

$\operatorname {E} {\bigl [}|XY|{\big |}\,{\mathcal {G}}{\bigr ]}\leq {\bigl (}\operatorname {E} {\bigl [}|X|^{p}{\big |}\,{\mathcal {G}}{\bigr ]}{\bigr )}^{1/p}\,{\bigl (}\operatorname {E} {\bigl [}|Y|^{q}{\big |}\,{\mathcal {G}}{\bigr ]}{\bigr )}^{1/q}\qquad \mathbb {P} {\text{-almost surely.}}$

Remarks:

• If a non-negative random variable Z has infinite expected value, then its conditional expectation is defined by

$\operatorname {E} [Z|{\mathcal {G}}]=\sup _{n\in \mathbb {N} }\,\operatorname {E} [\min\{Z,n\}|{\mathcal {G}}]\quad {\text{a.s.}}$

• 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

$U={\bigl (}\operatorname {E} {\bigl [}|X|^{p}{\big |}\,{\mathcal {G}}{\bigr ]}{\bigr )}^{1/p},\qquad V={\bigl (}\operatorname {E} {\bigl [}|Y|^{q}{\big |}\,{\mathcal {G}}{\bigr ]}{\bigr )}^{1/q}$

and note that they are measurable with respect to the sub-σ-algebra. Since

$\operatorname {E} {\bigl [}|X|^{p}1_{\{U=0\}}{\bigr ]}=\operatorname {E} {\bigl [}1_{\{U=0\}}\underbrace {\operatorname {E} {\bigl [}|X|^{p}{\big |}\,{\mathcal {G}}{\bigr ]}} _{=\,U^{p}}{\bigr ]}=0,$

it follows that |X| = 0 a.s. on the set {U = 0}. Similarly, |Y| = 0 a.s. on the set {V = 0}, hence

$\operatorname {E} {\bigl [}|XY|{\big |}\,{\mathcal {G}}{\bigr ]}=0\qquad {\text{a.s. on }}\{U=0\}\cup \{V=0\}$

and the conditional Hölder inequality holds on this set. On the set

$\{U=\infty ,V>0\}\cup \{U>0,V=\infty \}$

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

${\frac {\operatorname {E} {\bigl [}|XY|{\big |}\,{\mathcal {G}}{\bigr ]}}{UV}}\leq 1\qquad {\text{a.s. on the set }}H:=\{0<U<\infty ,\,0<V<\infty \}.$

This is done by verifying that the inequality holds after integration over an arbitrary

$G\in {\mathcal {G}},\quad G\subset H.$

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

${\begin{aligned}\operatorname {E} {\biggl [}{\frac {\operatorname {E} {\bigl [}|XY|{\big |}\,{\mathcal {G}}{\bigr ]}}{UV}}1_{G}{\biggr ]}&=\operatorname {E} {\biggl [}\operatorname {E} {\biggl [}{\frac {|XY|}{UV}}1_{G}{\bigg |}\,{\mathcal {G}}{\biggr ]}{\biggr ]}\\&=\operatorname {E} {\biggl [}{\frac {|X|}{U}}1_{G}\cdot {\frac {|Y|}{V}}1_{G}{\biggr ]}\\&\leq {\biggl (}\operatorname {E} {\biggl [}{\frac {|X|^{p}}{U^{p}}}1_{G}{\biggr ]}{\biggr )}^{\!1/p\;}{\biggl (}\operatorname {E} {\biggl [}{\frac {|Y|^{q}}{V^{q}}}1_{G}{\biggr ]}{\biggr )}^{\!1/q}\\&={\biggl (}\operatorname {E} {\biggl [}\underbrace {\frac {\operatorname {E} {\bigl [}|X|^{p}{\big |}\,{\mathcal {G}}{\bigr ]}}{U^{p}}} _{=\,1{\text{ a.s. on }}G}1_{G}{\biggr ]}{\biggr )}^{\!1/p\;}{\biggl (}\operatorname {E} {\biggl [}\underbrace {\frac {\operatorname {E} {\bigl [}|Y|^{q}{\big |}\,{\mathcal {G}}{\bigr ]}}{V^{p}}} _{=\,1{\text{ a.s. on }}G}1_{G}{\biggr ]}{\biggr )}^{\!1/q}\\&=\operatorname {E} {\bigl [}1_{G}{\bigr ]}.\end{aligned}}$

References

• Hardy, G.H.; Littlewood, J.E.; Pólya, G. (1934), Inequalities, Cambridge Univ. Press, ISBN 0521358809
• Hölder, O. (1889), "Ueber einen Mittelwerthsatz", Nachr. Ges. Wiss. Göttingen: 38–47
• Kuptsov, L.P. (2001) [1994], "Hölder inequality", Encyclopedia of Mathematics, EMS Press
• Rogers, L J. (1888), "An extension of a certain theorem in inequalities", Messenger of math, 17: 145–150
• Kuttler, Kenneth (2007), An introduction to linear algebra (PDF), Online e-book in PDF format, Brigham Young University