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.

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,

$\|fg\|_1 \le \|f\|_p \|g\|_q.$

If, in addition, p, q ∈ (1, ∞) and f ∈ L^p(μ) and g ∈ L^q(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 L^p(μ) and g is in L^q(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 L^p(μ), and also to establish that L^q(μ) is the dual space of L^p(μ) for p ∈ [1, ∞).

Hölder's inequality was first found by 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 p, q ∈ [1, ∞), then || f  ||_p and || g ||_q stand for the (possibly infinite) expressions

$\begin{align} &\left(\int_S |f|^p\,\mathrm{d}\mu\right)^{\frac{1}{p}} \\ &\left(\int_S |g|^q\,\mathrm{d}\mu\right)^{\frac{1}{q}} \end{align}$

• 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 × ∞ 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 || fg ||₁ is finite, then the pointwise products of f  with g and its complex conjugate function are μ-integrable, the estimate

$\biggl|\int_S f\bar g\,\mathrm{d}\mu\biggr|\le\int_S|fg|\,\mathrm{d}\mu =\|fg\|_1$

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

$|\langle f,g\rangle| \le \|f\|_2 \|g\|_2,$

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 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

$\|fg\|_1 \le \|f\|_p \|g\|_q$

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

For the n-dimensional Euclidean space, when the set S is {1, ..., n} with the counting measure, we have

$\sum_{k=1}^n |x_k\,y_k| \le \biggl( \sum_{k=1}^n |x_k|^p \biggr)^{\frac{1}{p}} \biggl( \sum_{k=1}^n |y_k|^q \biggr)^{\frac{1}{q}} \text{ for all }(x_1,\ldots,x_n),(y_1,\ldots,y_n)\in\mathbb{R}^n\text{ or }\mathbb{C}^n.$

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

$\sum_{k=1}^{\infty} |x_k\,y_k| \le \biggl( \sum_{k=1}^{\infty} |x_k|^p \biggr)^{\frac{1}{p}} \left( \sum_{k=1}^{\infty} |y_k|^q \right)^{\frac{1}{q}} \text{ for all }(x_k)_{k\in\mathbb N}, (y_k)_{k\in\mathbb N}\in\mathbb{R}^{\mathbb N}\text{ or }\mathbb{C}^{\mathbb N}.$

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

$\int_S \bigl| f(x)g(x)\bigr| \,\mathrm{d}x \le\biggl(\int_S |f(x)|^p\,\mathrm{d}x\biggr)^{\frac{1}{p}} \biggl(\int_S |g(x)|^q\,\mathrm{d}x\biggr)^{\frac{1}{q}}.$

Probability measure

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

$\mathbb{E}[|XY|] \le \bigl(\mathbb{E}\bigl[|X|^p\bigr]\bigr)^{\frac{1}{p}} \bigl(\mathbb{E}\bigl[|Y|^q\bigr]\bigr)^{\frac{1}{q}}.$

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

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

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

$\int_{S_1}\int_{S_2}|f(x,y)\,g(x,y)|\,\mu_2(\mathrm{d}y)\,\mu_1(\mathrm{d}x) \le\left(\int_{S_1}\int_{S_2}|f(x,y)|^p\,\mu_2(\mathrm{d}y)\,\mu_1(\mathrm{d}x)\right)^{\frac{1}{p}}\left(\int_{S_1}\int_{S_2}|g(x,y)|^q\,\mu_2(\mathrm{d}y)\,\mu_1(\mathrm{d}x)\right)^{\frac{1}{q}}.$

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

$\int_S \sum_{k=1}^n|f_k(x)\,g_k(x)|\,\mu(\mathrm{d}x) \le \left(\int_S\sum_{k=1}^n|f_k(x)|^p\,\mu(\mathrm{d}x)\right)^{\frac{1}{p}}\left(\int_S\sum_{k=1}^n|g_k(x)|^q\,\mu(\mathrm{d}x)\right)^{\frac{1}{q}}.$

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

$\alpha \left (|f_1(x)|^p,\ldots,|f_n(x)|^p \right )= \beta \left (|g_1(x)|^q,\ldots,|g_n(x)|^q \right ),$

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

$a b \le \frac{a^p}p + \frac{b^q}q$

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

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

Integrating both sides gives

$\|fg\|_1 \le \frac{1}p + \frac{1}q = 1,$

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

$\alpha=\|g\|_q^q, \qquad \beta=\|f\|_p^p,$

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 (*).

Alternate (simpler) proof using Jensen's inequality

Recall the Jensen's inequality for the convex function $x^p$ (it is convex because obviously $p\geq1$):

$\int hd\nu\leq\left(\int h^pd\nu \right )^{\frac{1}{p}}$

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 $g^q$, i.e.

$d\nu = \frac{g^q}{\int g^qd\mu}d\mu$

Hence we have, using $\frac{1}{p}+\frac{1}{q}=1$, hence $p(1-q)+q=0$, and letting $h=fg^{1-q}$,

$\int fgd\mu = \left (\int g^qd\mu \right )\int \underbrace{fg^{1-q}}_h\underbrace{\frac{g^q}{\int g^qd\mu}d\mu}_{d\nu} \leq \left (\int g^qd\mu \right ) \left (\int \underbrace{f^pg^{p(1-q)}}_{h^p}\underbrace{\frac{g^q}{\int g^qd\mu}d\mu}_{d\nu} \right )^{\frac{1}{p}} = \left (\int g^qd\mu \right ) \left (\int \frac{f^p}{\int g^qd\mu}d\mu \right )^{\frac{1}{p}}$

Finally, we get

$\int fgd\mu \leq \left(\int f^pd\mu \right )^{\frac{1}{p}} \left(\int g^qd\mu \right )^{\frac{1}{q}}$

This assumes $f,g$ real and non negative, but the extension to complex functions is straightforward (use the modulus of $f,g$). It also assumes that $\|f\|_p,\|g\|_q$ are neither null nor infinity, and that $p,q>1$: all these assumptions can also be lifted as in the proof above.

Extremal equality

Statement

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

$\|f\|_p = \max \left \{ \left| \int_S f g \, \mathrm{d}\mu \right | : g\in L^q(\mu), \|g\|_q \le 1 \right\},$

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

$\|f\|_\infty = \sup \left\{ \left| \int_S f g \,\mathrm{d}\mu \right| : g\in L^1(\mu), \|g\|_1 \le 1 \right \}.$

Proof of the extremal equality

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

$\left |\int_S fg\,\mathrm{d}\mu\right|\le\int_S|fg|\,\mathrm{d}\mu\le\|f\|_p,$

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 || f  ||_p = 0. Therefore, we assume || f  ||_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

$\int_S f g \, \mathrm{d}\mu = \|f\|_p.$

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

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

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

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

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

$\left |\int_S fg\,\mathrm{d}\mu\right| = \int_B\frac{1-\varepsilon}{\mu(B)}\|f\|_\infty\,\mathrm{d}\mu = (1-\varepsilon)\|f\|_\infty.$

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 || 1_A ||_∞ = 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 1_A 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 f (n) = (n − 1)/n for every natural number n. Then || f  ||_∞ = 1. For g ∈ L¹(μ) with 0 < || g ||₁ ≤ 1, let m denote the smallest natural number with g(m) ≠ 0. Then

$\Bigl|\int_S fg\,\mathrm{d}\mu\Bigr| \le \frac{m-1}{m}|g(m)|+\sum_{n=m+1}^\infty|g(n)| = \|g\|_1-\frac{|g(m)|}m<1.$

Applications

• The extremal equality is one of the ways for proving the triangle inequality || f₁ + f₂ ||_p ≤ || f₁ ||_p + || f₂ ||_p for all f₁ and f₂ in L^p(μ), see Minkowski inequality.

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

$\kappa_f(g) = \int_S f g \, \mathrm{d}\mu,\qquad g\in L^q(\mu).$

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

Generalization of Hölder's inequality

Assume that r ∈ (0, ∞) and p₁, …, p_n ∈ (0, ∞] such that

$\sum_{k=1}^n \frac1{p_k}=\frac1r.$

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

$\left\|\prod_{k=1}^n f_k\right\|_r\le \prod_{k=1}^n\|f_k\|_{p_k}.$

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 p₁ ≤ … ≤ p_n.

Case 1: If p_n = ∞, then

$\sum_{k=1}^{n-1}\frac1{p_k}=\frac1r.$

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

$\begin{align} \left \|f_1\cdots f_n \right \|_r &\le \|f_1\cdots f_{n-1}\|_r \|f_n\|_\infty\\ &\le\|f_1\|_{p_1}\cdots\|f_{n-1}\|_{p_{n-1}}\|f_n\|_\infty. \end{align}$

Case 2: If p_n < ∞, then

$p:=\frac{p_n}{p_n-r}, \qquad q:=\frac{p_n}r$

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

$\left \||f_1\cdots f_{n-1}|^r\,|f_n|^r\right \|_1 \le\left \||f_1\cdots f_{n-1}|^r\right \|_p\,\left \||f_n|^r\right \|_q.$

Raising to the power 1/r and rewriting,

$\|f_1\cdots f_n\|_r \le \|f_1\cdots f_{n-1}\|_{pr}\|f_n\|_{qr}.$

Since qr = p_n and

$\sum_{k=1}^{n-1}\frac1{p_k} = \frac1r-\frac1{p_n} = \frac{p_n-r}{rp_n} = \frac1{pr},$

the claimed inequality now follows by using the induction hypothesis.

Interpolation

Let p₁, ..., p_n ∈ (0, ∞] and let θ₁, ..., θ_n ∈ (0, 1) denote weights with θ₁ + ... + θ_n = 1. Define p as the weighted harmonic mean, i.e.,

$\frac1p = \sum_{k=1}^n \frac{\theta_k}{p_k}.$

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

$f_k=|f|^{\theta_k},\quad k\in\{1,\ldots,n\}.$

Then by the above generalization of Hölder's inequality,

$\|f\|_p=\biggl\|\prod_{k=1}^n f_k\biggr\|_p\le \prod_{k=1}^n \|f_k\|_{p_k/\theta_k}=\prod_{k=1}^n \|f\|_{p_k}^{\theta_k}.$

In particular, taking θ₁ = θ and θ₂ = 1 − θ, in the case n = 2, we obtain the interpolation result (Littlewood's inequality)

$\| f\|_{p_\theta}\le \|f\|_{p_1}^\theta \cdot \|f\|_{p_0}^{1-\theta},$

for θ ∈ (0, 1) and

$\frac 1p_\theta= \frac \theta {p_1}+ \frac {1-\theta}{p_0}.$

A similar application of Hölder gives Lyapunov's inequality: If $p=\theta p_0+(1-\theta)p_1$ for $\theta\in(0,1)$, then

$\|f\|_p^p\le\|f\|_{p_0}^{p_0\cdot\theta}\cdot\|f\|_{p_1}^{p_1(1-\theta)}.$

Both Littlewood and Lyapunov imply that if $f\in L^{p_0}\cap L^{p_1}$, then $f\in L^p$ for all $p_0<p<p_1$.

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,

$\|fg\|_1\ge\|f\|_{\frac{1}{p}}\,\|g\|_{-1/(p-1)}.$

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 notations for

$\biggl(\int_S|f|^{\frac{1}{p}}\,\mathrm{d}\mu\biggr)^{\!p}$   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

$\begin{align} \left \||f|^{\frac{1}{p}}\right \|_1 &= \left \||fg|^{\frac{1}{p}}\,|g|^{-\frac{1}{p}}\right \|_1\\ &\le \bigl\||fg|^{\frac{1}{p}}\bigr\|_p\,\bigl\||g|^{-\frac{1}{p}}\bigr\|_q =\|fg\|_1^{\frac{1}{p}}\,\bigl\||g|^{-1/(p-1)}\bigr\|_1^{(p-1)/p}. \end{align}$

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 (Ω, F, ℙ) be a probability space, G ⊂ 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,

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

Remarks:

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

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

• 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

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

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

$\mathbb{E}\bigl[|X|^p1_{\{U=0\}}\bigr] = \mathbb{E}\bigl[1_{\{U=0\}}\underbrace{\mathbb{E}\bigl[|X|^p\big|\,\mathcal{G}\bigr]}_{=\,U^p}\bigr]=0,$

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

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

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

$\frac{\mathbb{E}\bigl[|XY|\big|\,\mathcal{G}\bigr]}{UV}\le1 \qquad\text{a.s. on the set }H:=\{0<U<\infty,\,0<V<\infty\}.$

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

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

Hölder's inequality for increasing seminorms

Let S be a set and let F(S, C) be the space of all complex-valued functions on S. Let N be an increasing seminorm on F(S, C), meaning that, for all real-valued functions f  and g in F(S, C), if f (s) ≥ g(s) ≥ 0 for all s ∈ S, then N(f ) ≥ N(g). The seminorm is also allowed to attain the value ∞. Then, for all p, q ∈ (1, ∞) with 1/p + 1/q = 1, which means that they are conjugate Hölder exponents, and for all complex-valued functions f, g ∈ F(S, C),

$N(|fg|) \le \bigl(N(|f|^p)\bigr)^{\frac{1}{p}} \bigl(N(|g|^q)\bigr)^{\frac{1}{q}}.$

Remark: If (S, Σ, μ) is a measure space and N(f ) is the upper Lebesgue integral of the absolute value | f  |, then the restriction of N to all Σ-measurable functions gives the usual version of Hölder's inequality.

See also

• Cauchy–Schwarz inequality
• Minkowski inequality
• Jensen's inequality

References

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