Hölder's inequality

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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,

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

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.

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.

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).

1 Remarks
2 Notable special cases
3 Proof of Hölder's inequality
4 Generalization
5 Reverse Hölder inequality
6 Conditional Hölder inequality
7 References

[edit] 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 ∞.

[edit] 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

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

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

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

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

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

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

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

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

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

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.)

[edit] Proof of Hölder's inequality

There exist several proofs of Hölder's inequality, we use Young's inequality for the main part.

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

$\|fg\|_1 \le 1,$

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 (*).

[edit] Generalization

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,

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

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.

Proof of the generalization

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

Proof of the reverse Hölder inequality

[edit] 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:

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

[edit] 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), “Hölder inequality”, in Hazewinkel, Michiel, Encyclopaedia of Mathematics, Kluwer Academic Publishers, ISBN 978-1556080104
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, Online e-book in PDF format, Brigham Young University, <http://www.math.byu.edu/~klkuttle/Linearalgebra.pdf>

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Hölder's inequality

From Wikipedia, the free encyclopedia

Contents

[edit] Remarks

[edit] Notable special cases

[edit] Proof of Hölder's inequality

[edit] Generalization

[edit] Reverse Hölder inequality

[edit] Conditional Hölder inequality

[edit] References

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