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 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. 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 real numbers 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 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 1 ≤ p < ∞.

Hölder's inequality was first found by L. J. 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 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 ∞.

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 products of f with g and its complex conjugate function, respectively, are μ-integrable, the estimates}

${\biggl |}\int _{S}f{\bar {g}}\,\mathrm {d} \mu {\biggr |}\leq \int _{S}|fg|\,\mathrm {d} \mu =\|fg\|_{1}$

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

$|\langle f,g\rangle |\leq \|f\|_{2}\|g\|_{2}\,,$

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 measures

If (S, Σ, μ) is a probability space, then 1 ≤ p, q ≤ ∞ 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

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

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

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:

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

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

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

Probability measure

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

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

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

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

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.)

Product measure

For two σ-finite measure spaces (S₁, Σ₁, μ₁) and (S₂, Σ₂, μ₂) define the product measure space by

$S=S_{1}\times S_{2},\quad \Sigma =\Sigma _{1}\otimes \Sigma _{2},\quad \mu =\mu _{1}\otimes \mu _{2},$

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

${\begin{aligned}\int _{S_{1}}&\int _{S_{2}}|f(x,y)\,g(x,y)|\,\mu _{2}(\mathrm {d} y)\,\mu _{1}(\mathrm {d} x)\\&\leq {\biggl (}\int _{S_{1}}\int _{S_{2}}|f(x,y)|^{p}\,\mu _{2}(\mathrm {d} y)\,\mu _{1}(\mathrm {d} x){\biggr )}^{\!1/p\;}{\biggl (}\int _{S_{1}}\int _{S_{2}}|g(x,y)|^{q}\,\mu _{2}(\mathrm {d} y)\,\mu _{1}(\mathrm {d} x){\biggr )}^{\!1/q}.\end{aligned}}$

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 = (f₁, …, f_n) and g = (g₁, …, 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

${\begin{aligned}\int _{S}&\sum _{k=1}^{n}|f_{k}(x)\,g_{k}(x)|\,\mu (\mathrm {d} x)\\&\leq {\biggl (}\int _{S}\sum _{k=1}^{n}|f_{k}(x)|^{p}\,\mu (\mathrm {d} x){\biggr )}^{\!1/p\;}{\biggl (}\int _{S}\sum _{k=1}^{n}|g_{k}(x)|^{q}\,\mu (\mathrm {d} x){\biggr )}^{\!1/q}.\end{aligned}}$

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

$\alpha (|f_{1}(x)|^{p},\ldots ,|f_{n}(x)|^{p})=\beta (|g_{1}(x)|^{q},\ldots ,|g_{n}(x)|^{q})$    for μ-almost all x in S.

This finite-dimensional version generalizes to functions f and g taking values in a sequence space.

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 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 ||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 real numbers α, β > 0, namely

$\alpha =\|g\|_{q}^{q}$   and   $\beta =\|f\|_{p}^{p}$

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.

Interpolation

Let p_{1, …, pn ∈ (0, ∞] and let θ1, …, θn ∈ (0, 1) denote weights with θ1+ … + θn = 1. Define p as the weighted harmonic mean, i.e.,}

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

Given a measurable real- or complex-valued functions f on S, define

$f_{k}=|f|^{\theta _{k}},\quad k\in \{1,\ldots ,n\}.$

Then by the above generalization of Hölder's inequality,

$\|f\|_{p}={\biggl \|}\prod _{k=1}^{n}f_{k}{\biggr \|}_{p}\leq \prod _{k=1}^{n}\|f_{k}\|_{p_{k}/\theta _{k}}=\prod _{k=1}^{n}\|f\|_{p_{k}}^{\theta _{k}}.$

In particular, taking θ₁ = θ and θ₂ = 1 - θ, in the case n =2, we obtain the interpolation result

$\|f\|_{p_{\theta }}\leq \|f\|_{p_{1}}^{\theta }\cdot \|f\|_{p_{0}}^{1-\theta },$

for θ ∈ (0, 1) and

${\frac {1}{p}}_{\theta }={\frac {\theta }{p_{1}}}+{\frac {1-\theta }{p_{0}}}.$

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

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

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 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{{citation}}: CS1 maint: unrecognized language (link). Available at Digi Zeitschriften.
• 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 Mathematics, New Series, XVII (10): 145–150, JFM 20.0254.02, archived from the original on August 21, 2007 {{citation}}: Unknown parameter |month= ignored (help).

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).