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.

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,

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

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 Lp(μ) and g is in Lq(μ), then fg is in L1(μ).

For 1 < p, q < ∞ and f ∈ Lp(μ) and g ∈ Lq(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 αβ ≥ 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 Lp(μ), and also to establish that Lq(μ) is the dual space of Lp(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 ∈ Lp(μ) and g ∈ Lq(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

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

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.

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

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

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

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

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

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

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

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

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

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

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

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 1Ω, we obtain

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

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 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{align} \int_{S_1}&\int_{S_2}|f(x,y)\,g(x,y)|\,\mu_2(\mathrm{d}y)\,\mu_1(\mathrm{d}x)\\ &\le\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{align}

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, …, 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

\begin{align} \int_S&\sum_{k=1}^n|f_k(x)\,g_k(x)|\,\mu(\mathrm{d}x)\\ &\le\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{align}

This 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 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

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

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 = 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 αβ > 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 ∈ Lp(μ),

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

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

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

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

Applications

• The extremal equality is one of the ways for proving the triangle inequality ||ƒ1 + ƒ2||p ≤ ||ƒ1||p + ||ƒ2||p  for all ƒ1 and ƒ2 in Lp(μ), see Minkowski inequality.
• Hölder's inequality implies that every ƒ ∈ Lp(μ) defines a bounded (or continuous) linear functional κƒ on Lq(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 κƒ as element of the continuous dual space Lq(μ) coincides with the norm of ƒ in Lp(μ) (see also the Lp-space article).

Generalization of Hölder's inequality

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

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

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

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

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.

Interpolation

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

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

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,

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

As an interpolation result[citation needed] for n = 2,

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

where θ ∈ (0, 1) and

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

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,

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

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   $\biggl(\int_S|g|^{-1/(p-1)}\,\mathrm{d}\mu\biggr)^{-(p-1)}.$

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,

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

Remarks:

$\operatorname{E}[Z|\mathcal{G}]=\sup_{n\in\mathbb{N}}\,\operatorname{E}[\min\{Z,n\}|\mathcal{G}]\quad\text{a.s.}$
• On the right-hand side of the conditional Hölder inequality, 0 times ∞ as well as ∞ times 0 means 0. Multiplying a > 0 with ∞ gives ∞.

References

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