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 \in

[1, \infty)

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 \in

(1, \infty)

and

f \in L p (μ)

and

g \in L q (μ)

, then Hölder's inequality becomes an equality iff

| f | p

and

| g | q

are linearly dependent in

L 1 (μ)

, meaning that there exist real numbers

α, β \geq 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 $L 1 (μ)$ .

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 $p \in$ $[1, \infty)$ .

Hölder's inequality was first found by Leonard James Rogers (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, $1/ \infty$ means zero.
If $p, q \in$ $[1, \infty)$ , 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 $p = \infty$ , then $| | f | | \infty$ stands for the essential supremum of $| f |$ , similarly for $| | g | | \infty$ .
The notation $| | f | | p$ with $1 \leq p \leq \infty$ 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 $f \in L p (μ)$ and $g \in L q (μ)$ , 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 $L 2 (μ)$ , 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 $L 2 (μ)$ . 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 $p, q \in$ $[1, \infty]$ just need to satisfy $1/ p + 1/ q \leq 1$ , 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 $(1,\infty)$ 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}.

Often the following practical form of this is used, for any $(r,s)\in \mathbb {R} _{+}$ :

{\biggl (}\sum _{k=1}^{n}|x_{k}|^{r}\,|y_{k}|^{s}{\biggr )}^{r+s}\leq {\biggl (}\sum _{k=1}^{n}|x_{k}|^{r+s}{\biggr )}^{r}{\biggl (}\sum _{k=1}^{n}|y_{k}|^{r+s}{\biggr )}^{s}.

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} }.

Lebesgue measure

If $S$ is a measurable subset of $R n$ 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 $(\Omega ,{\mathcal {F}},\mathbb {P} ),$ let $\mathbb {E}$ denote the expectation operator. For real- or complex-valued random variables $X$ and $Y$ on $\Omega ,$ Hölder's inequality reads

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

\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.)

Product measure

For two σ-finite measure spaces $(S 1, Σ 1, μ 1)$ and $(S 2, Σ 2, μ 2)$ 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 $S 1$ and $S 2$ , the σ-algebra $Σ$ arises as product σ-algebra of $Σ 1$ and $Σ 2$ , and $μ$ denotes the product measure of $μ 1$ and $μ 2$ . 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 $f = (f 1, ..., f n)$ and $g = (g 1, ..., g n)$ 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 $α, β \geq 0$ , 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 for products.

Proof —

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 $| | f | | p = \infty$ or $| | g | | q = \infty$ , then the right-hand side of Hölder's inequality is infinite. Therefore, we may assume that $| | f | | p$ and $| | g | | q$ are in $(0, \infty)$ .

If $p = \infty$ and $q = 1$ , then $| fg | \leq | | f | | \infty | g |$ almost everywhere and Hölder's inequality follows from the monotonicity of the Lebesgue integral. Similarly for $p = 1$ and $q = \infty$ . Therefore, we may also assume $p, q \in$ $(1, \infty)$ .

\|f\|_{p}=\|g\|_{q}=1.

We now use Young's inequality for products, 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 {\|f\|_{p}^{p}}{p}}+{\frac {\|g\|_{q}^{q}}{q}}={\frac {1}{p}}+{\frac {1}{q}}=1,

which proves the claim.

Under the assumptions $p \in$ $(1, \infty)$ 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 $(0, \infty)$ , 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 (*).

Alternate 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 fg\,d\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 fg\,d\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 $1 \leq p < \infty$ and let $q$ denote the Hölder conjugate. Then, for every $f \in L p (μ)$ ,

\|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 $p = \infty$ and if each set $A$ in the σ-field $Σ$ with $μ (A) = \infty$ contains a subset $B \in Σ$ with $0 < μ (B) < \infty$ (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 $1 \leq p \leq \infty$ ,

\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.

If $1 \leq p < \infty$ , 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 $1 < p < \infty$ separately, we see that $| | g | | q = 1$ and

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

It remains to consider the case $p = \infty$ . For $ε \in$ $(0, 1)$ define

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

Since $f$ is measurable, $A \in Σ$ . By the definition of $| | f | | \infty$ as the essential supremum of $f$ and the assumption $| | f | | \infty > 0$ , we have $μ (A) > 0$ . Using the additional assumption on the σ-field $Σ$ if necessary, there exists a subset $B \in Σ$ of $A$ with $0 < μ (B) < \infty$ . 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}}

\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 $p=\infty$ 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: $0<\mu (B)<\infty .$ (the simplest example is the $\sigma$ -field $\Sigma$ containing just the empty set and $S,$ and the measure $\mu$ with $\mu (S)=\infty .$ ) Then the indicator function $1_{A}$ satisfies $\|1_{A}\|_{\infty }=1,$ but every $g\in L^{1}(\mu )$ 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 $p=\infty ,$ the supremum is in general not attained. As an example, let $S=\mathbb {N} ,\Sigma ={\mathcal {P}}(\mathbb {N} )$ and $\mu$ the counting measure. Define:

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

Then

\|f\|_{\infty }=1.

For

g\in L^{1}(\mu ,\mathbb {N} )

with

0<\|g\|_{1}\leqslant 1,

let

m

denote the smallest natural number with

g(m)\neq 0.

Then

\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 $| | f 1 + f 2 | | p \leq | | f 1 | | p + | | f 2 | | p$ for all $f 1$ and $f 2$ in $L p (μ)$ , see Minkowski inequality.
Hölder's inequality implies that every $f \in L p (μ)$ 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 $r \in$ $(0, \infty]$ and $p 1, \dots, p n \in$ $(0, \infty]$ such that

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

where 1/∞ is interpreted as 0 in this equation. Then, for all measurable real or complex-valued functions $f 1, \dots, f n$ defined on $S$ ,

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

where we interpret any product with a factor of ∞ as ∞ if all factors are positive, but the product is 0 if any factor is 0.

In particular, if $f_{k}\in L^{p_{k}}(\mu )$ for all $k\in \{1,\ldots ,n\}$ then $\prod _{k=1}^{n}f_{k}\in L^{r}(\mu ).$

Note: For $r\in (0,1),$ 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. If $n=1$ then the result is immediate. Let us now pass from $n-1$ to $n.$ Without loss of generality assume that $p_{1}\leq \cdots \leq p_{n}.$

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

Case 2: If $p_{n}<\infty$ then necessarily $r<\infty$ as well, and then

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

are Hölder conjugates in $(1, \infty)$ . 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 $p 1, ..., p n \in$ $(0, \infty]$ and let $θ 1, ..., θ n \in (0, 1)$ denote weights with $θ 1 + ... + θ n = 1$ . Define $p$ as the weighted harmonic mean, that is,

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

Given measurable real- or complex-valued functions $f_{k}$ on $S$ , then the above generalization of Hölder's inequality gives

\left\||f_{1}|^{\theta _{1}}\cdots |f_{n}|^{\theta _{n}}\right\|_{p}\leq \left\||f_{1}|^{\theta _{1}}\right\|_{\frac {p_{1}}{\theta _{1}}}\cdots \left\||f_{n}|^{\theta _{n}}\right\|_{\frac {p_{n}}{\theta _{n}}}=\|f_{1}\|_{p_{1}}^{\theta _{1}}\cdots \|f_{n}\|_{p_{n}}^{\theta _{n}}.

In particular, taking $f_{1}=\cdots =f_{n}=:f$ gives

\|f\|_{p}\leqslant \prod _{k=1}^{n}\|f\|_{p_{k}}^{\theta _{k}}.

Specifying further $θ 1 = θ$ and $θ 2 = 1- θ$ , in the case $n=2,$ we obtain the interpolation result (Littlewood's inequality)

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

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

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

An application of Hölder gives Lyapunov's inequality: If

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

then

\left\||f_{0}|^{\frac {p_{0}(1-\theta )}{p}}\cdot |f_{1}|^{\frac {p_{1}\theta }{p}}\right\|_{p}^{p}\leq \|f_{0}\|_{p_{0}}^{p_{0}(1-\theta )}\|f_{1}\|_{p_{1}}^{p_{1}\theta }

and in particular

\|f\|_{p}^{p}\leqslant \|f\|_{p_{0}}^{p_{0}(1-\theta )}\cdot \|f\|_{p_{1}}^{p_{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 $p \in (1, \infty)$ 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 $g (s) \neq 0$ for $μ$ -almost all $s \in S$ ,

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

If

\|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

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

Note: The expressions:

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

are not norms, they are just compact notations for

\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

{\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:

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

Therefore:

\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 $α \geq 0$ 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, $G \subset F$ a sub-σ-algebra, and $p, q \in$ $(1, \infty)$ 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}}