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 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.

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 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 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 Lp(μ) and the Minkowski inequality, and also to establish that Lp(μ) is dual to Lq(μ) for 1 ≤ q < ∞.

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

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[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 ∈ 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 ∞.

[edit] Notable special cases

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

\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}.
\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 1Ω, we obtain
\operatorname{E}|X|^r\le\bigl(\operatorname{E}|X|^s\bigr)^{r/s}.
In particular, if the sth absolute moment is finite, then the rth 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 ap = bq. 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, …, 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.

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:

\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