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,

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

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

${\begin{aligned}&\left(\int _{S}|f|^{p}\,\mathrm {d} \mu \right)^{\frac {1}{p}}\\&\left(\int _{S}|g|^{q}\,\mathrm {d} \mu \right)^{\frac {1}{q}}\end{aligned}}$

• 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

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

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 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

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

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

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

$\sum _{k=1}^{n}|x_{k}\,y_{k}|\leq {\biggl (}\sum _{k=1}^{n}|x_{k}|^{p}{\biggr )}^{\frac {1}{p}}{\biggl (}\sum _{k=1}^{n}|y_{k}|^{q}{\biggr )}^{\frac {1}{q}}{\text{ for all }}(x_{1},\ldots ,x_{n}),(y_{1},\ldots ,y_{n})\in \mathbb {R} ^{n}{\text{ or }}\mathbb {C} ^{n}.$

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

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

These Hölder inequalities for the counting measure still hold when 1/p + 1/q ≤ 1.

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

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

Probability measure

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

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

Let ${\displaystyle 0<r<s}$ and define ${\displaystyle p={\tfrac {s}{r}}.}$ Then ${\displaystyle q={\tfrac {p}{p-1}}}$ is the Hölder conjugate of $p.$ Applying Hölder's inequality to the random variables ${\displaystyle |X|^{r}}$ and ${\displaystyle 1_{\Omega }}$ we obtain

${\displaystyle \mathbb {E} {\bigl [}|X|^{r}{\bigr ]}\leqslant \left(\mathbb {E} {\bigl [}|X|^{s}{\bigr ]}\right)^{\frac {r}{s}}.}$

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.)

These Hölder inequalities for a probability measure still hold when 1/p + 1/q ≤ 1.

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

$\int _{S_{1}}\int _{S_{2}}|f(x,y)\,g(x,y)|\,\mu _{2}(\mathrm {d} y)\,\mu _{1}(\mathrm {d} x)\leq \left(\int _{S_{1}}\int _{S_{2}}|f(x,y)|^{p}\,\mu _{2}(\mathrm {d} y)\,\mu _{1}(\mathrm {d} x)\right)^{\frac {1}{p}}\left(\int _{S_{1}}\int _{S_{2}}|g(x,y)|^{q}\,\mu _{2}(\mathrm {d} y)\,\mu _{1}(\mathrm {d} x)\right)^{\frac {1}{q}}.$

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

$\int _{S}\sum _{k=1}^{n}|f_{k}(x)\,g_{k}(x)|\,\mu (\mathrm {d} x)\leq \left(\int _{S}\sum _{k=1}^{n}|f_{k}(x)|^{p}\,\mu (\mathrm {d} x)\right)^{\frac {1}{p}}\left(\int _{S}\sum _{k=1}^{n}|g_{k}(x)|^{q}\,\mu (\mathrm {d} x)\right)^{\frac {1}{q}}.$

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 normed space which could be for example a sequence space or an inner product 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, 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 $p\geq 1$):

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

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.

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

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

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

Finally, we get

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

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),

$\|f\|_{p}=\max \left\{\left|\int _{S}fg\,\mathrm {d} \mu \right|:g\in L^{q}(\mu ),\|g\|_{q}\leq 1\right\},$

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

$\|f\|_{\infty }=\sup \left\{\left|\int _{S}fg\,\mathrm {d} \mu \right|:g\in L^{1}(\mu ),\|g\|_{1}\leq 1\right\}.$

Proof of the extremal equality

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

$\left|\int _{S}fg\,\mathrm {d} \mu \right|\leq \int _{S}|fg|\,\mathrm {d} \mu \leq \|f\|_{p},$

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 || 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

$\int _{S}fg\,\mathrm {d} \mu =\|f\|_{p}.$

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 ${\displaystyle p=\infty }$ fails whenever there exists a set $A$ of infinite measure in the $\sigma$-field $\Sigma$ with that has no subset $B\in \Sigma$ that satisfies: ${\displaystyle 0<\mu (B)<\infty .}$ (the simplest example is the $\sigma$-field $\Sigma$ containing just the empty set and $S,$ and the measure $\mu$ with ${\displaystyle \mu (S)=\infty .}$) Then the indicator function $1_{A}$ satisfies ${\displaystyle \|1_{A}\|_{\infty }=1,}$ but every ${\displaystyle g\in L^{1}(\mu )}$ has to be $\mu$-almost everywhere constant on $A,$ because it is $\Sigma$-measurable, and this constant has to be zero, because $g$ is $\mu$-integrable. Therefore, the above supremum for the indicator function $1_{A}$ is zero and the extremal equality fails.
• For ${\displaystyle p=\infty ,}$ the supremum is in general not attained. As an example, let ${\displaystyle S=\mathbb {N} ,\Sigma ={\mathcal {P}}(\mathbb {N} )}$ and $\mu$ the counting measure. Define:

${\displaystyle {\begin{cases}f:\mathbb {N} \to \mathbb {R} \\f(n)={\frac {n-1}{n}}\end{cases}}}$

Then ${\displaystyle \|f\|_{\infty }=1.}$ For ${\displaystyle g\in L^{1}(\mu ,\mathbb {N} )}$ with ${\displaystyle 0<\|g\|_{1}\leqslant 1,}$ let $m$ denote the smallest natural number with ${\displaystyle g(m)\neq 0.}$ Then

${\displaystyle \left|\int _{S}fg\,\mathrm {d} \mu \right|\leqslant {\frac {m-1}{m}}|g(m)|+\sum _{n=m+1}^{\infty }|g(n)|=\|g\|_{1}-{\frac {|g(m)|}{m}}<1.}$

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

$\kappa _{f}(g)=\int _{S}fg\,\mathrm {d} \mu ,\qquad g\in L^{q}(\mu ).$

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

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

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

$\left\|\prod _{k=1}^{n}f_{k}\right\|_{r}\leq \prod _{k=1}^{n}\|f_{k}\|_{p_{k}}.$

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 p₁ ≤ … ≤ p_n.

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

${\begin{aligned}\left\|f_{1}\cdots f_{n}\right\|_{r}&\leq \|f_{1}\cdots f_{n-1}\|_{r}\|f_{n}\|_{\infty }\\&\leq \|f_{1}\|_{p_{1}}\cdots \|f_{n-1}\|_{p_{n-1}}\|f_{n}\|_{\infty }.\end{aligned}}$

Case 2: If p_n < ∞, then

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

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

$\left\||f_{1}\cdots f_{n-1}|^{r}\,|f_{n}|^{r}\right\|_{1}\leq \left\||f_{1}\cdots f_{n-1}|^{r}\right\|_{p}\,\left\||f_{n}|^{r}\right\|_{q}.$

Raising to the power 1/r and rewriting,

$\|f_{1}\cdots f_{n}\|_{r}\leq \|f_{1}\cdots f_{n-1}\|_{pr}\|f_{n}\|_{qr}.$

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 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.,

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

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

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

Then by the above generalization of Hölder's inequality,

${\displaystyle \|f\|_{p}=\left\|\prod _{k=1}^{n}f_{k}\right\|_{p}\leqslant \prod _{k=1}^{n}\left\|f_{k}\right\|_{\frac {p_{k}}{\theta _{k}}}=\prod _{k=1}^{n}\|f\|_{p_{k}}^{\theta _{k}}.}$

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

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

for $\theta \in (0,1)$ and

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

A similar application of Hölder gives Lyapunov's inequality: If

${\displaystyle p=\theta p_{0}+(1-\theta )p_{1},\qquad \theta \in (0,1),}$

then

${\displaystyle \|f\|_{p}^{p}\leqslant \|f\|_{p_{0}}^{p_{0}\theta }\cdot \|f\|_{p_{1}}^{p_{1}(1-\theta )}.}$

Both Littlewood and Lyapunov imply that if ${\displaystyle f\in L^{p_{0}}\cap L^{p_{1}},}$ then $f\in L^{p}$ for all ${\displaystyle p_{0}<p<p_{1}.}$

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,

${\displaystyle \|fg\|_{1}\geqslant \|f\|_{\frac {1}{p}}\,\|g\|_{\frac {-1}{p-1}}.}$

If

${\displaystyle \|fg\|_{1}<\infty \quad {\text{and}}\quad \|g\|_{\frac {-1}{p-1}}>0,}$

then the reverse Hölder inequality is an equality if and only if

${\displaystyle \exists \alpha \geqslant 0\quad |f|=\alpha |g|^{\frac {-p}{p-1}}\qquad \mu {\text{-almost everywhere}}.}$

Note: The expressions:

${\displaystyle \|f\|_{\frac {1}{p}}\quad {\text{and}}\quad \|g\|_{\frac {-1}{p-1}},}$

are not norms, they are just compact notations for

${\displaystyle \left(\int _{S}|f|^{\frac {1}{p}}\,\mathrm {d} \mu \right)^{p}\quad {\text{and}}\quad \left(\int _{S}|g|^{\frac {-1}{p-1}}\,\mathrm {d} \mu \right)^{-(p-1)}.}$

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

${\displaystyle {\begin{aligned}\left\||f|^{\frac {1}{p}}\right\|_{1}&=\left\||fg|^{\frac {1}{p}}\,|g|^{-{\frac {1}{p}}}\right\|_{1}\\&\leqslant \left\||fg|^{\frac {1}{p}}\right\|_{p}\left\||g|^{-{\frac {1}{p}}}\right\|_{q}\\&=\|fg\|_{1}^{\frac {1}{p}}\left\||g|^{\frac {-1}{p-1}}\right\|_{1}^{\frac {p-1}{p}}\end{aligned}}}$

Raising to the power p gives us:

${\displaystyle \left\||f|^{\frac {1}{p}}\right\|_{1}^{p}\leqslant \|fg\|_{1}\left\||g|^{\frac {-1}{p-1}}\right\|_{1}^{p-1}.}$

Therefore:

${\displaystyle \left\||f|^{\frac {1}{p}}\right\|_{1}^{p}\left\||g|^{\frac {-1}{p-1}}\right\|_{1}^{-(p-1)}\leqslant \|fg\|_{1}.}$

Now we just need to recall our notation.

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,

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

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

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

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

${\frac {\mathbb {E} {\bigl [}|XY|{\big |}\,{\mathcal {G}}{\bigr ]}}{UV}}\leq 1\qquad {\text{a.s. on the set }}H:=\{0<U<\infty ,\,0<V<\infty \}.$

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

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

Hölder's inequality for increasing seminorms

Let S be a set and let ${\displaystyle F(S,\mathbb {C} )}$ be the space of all complex-valued functions on S. Let N be an increasing seminorm on ${\displaystyle F(S,\mathbb {C} ),}$ meaning that, for all real-valued functions ${\displaystyle f,g\in F(S,\mathbb {C} )}$ we have the following implication (the seminorm is also allowed to attain the value ∞):

${\displaystyle \forall s\in S\quad f(s)\geqslant g(s)\geqslant 0\qquad \Rightarrow \qquad N(f)\geqslant N(g).}$

Then:

${\displaystyle \forall f,g\in F(S,\mathbb {C} )\qquad N(|fg|)\leqslant {\bigl (}N(|f|^{p}){\bigr )}^{\frac {1}{p}}{\bigl (}N(|g|^{q}){\bigr )}^{\frac {1}{q}},}$

where the numbers $p$ and $q$ are Hölder conjugates.^[1]

Remark: If (S, Σ, μ) is a measure space and $N(f)$ is the upper Lebesgue integral of $|f|$ 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
• Young's inequality

Citations

1. For a proof see (Trèves 1967, Lemma 20.1, pp. 205–206).

References

• Grinshpan, A. Z. (2010), "Weighted inequalities and negative binomials", Advances in Applied Mathematics, 45 (4): 564–606, doi:10.1016/j.aam.2010.04.004
• 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. Available at Digi Zeitschriften.
• Kuptsov, L. P. (2001), "Hölder inequality", in Hazewinkel, Michiel, Encyclopedia of Mathematics, Springer, ISBN 978-1-55608-010-4.
• Rogers, L. J. (February 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.
• Trèves, François (1967), Topological Vector Spaces, Distributions and Kernels, Pure and Applied Mathematics. A Series of Monographs and Textbooks, 25, New York, London: Academic Press, MR 0225131, Zbl 0171.10402.

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).
• Tisdell, Chris (2012), Holder's Inequality, Online video on Dr Chris Tisdell's YouTube channel.