# 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.

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,
${\displaystyle \|fg\|_{1}\leq \|f\|_{p}\|g\|_{q}.}$
If, in addition, p, q (1, ∞) and fLp(μ) and gLq(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 Lp(μ) and g is in Lq(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 Lp(μ), and also to establish that Lq(μ) is the dual space of Lp(μ) for p [1, ∞).

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/∞ means zero.
• If p, q [1, ∞), then ||f||p and ||g||q stand for the (possibly infinite) expressions
{\displaystyle {\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 fLp(μ) and gLq(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

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

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

${\displaystyle |\langle f,g\rangle |\leq \|f\|_{2}\|g\|_{2},}$

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

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

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

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

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

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

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

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

### Lebesgue measure

If S is a measurable subset of Rn with the Lebesgue measure, and f and g are measurable real- or complex-valued functions on S, then Hölder inequality is

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

### Probability measure

For the probability space ${\displaystyle (\Omega ,{\mathcal {F}},\mathbb {P} ),}$ let ${\displaystyle \mathbb {E} }$ denote the expectation operator. For real- or complex-valued random variables ${\displaystyle X}$ and ${\displaystyle Y}$ on ${\displaystyle \Omega ,}$ 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 and define ${\displaystyle p={\tfrac {s}{r}}.}$ Then ${\displaystyle q={\tfrac {p}{p-1}}}$ is the Hölder conjugate of ${\displaystyle p.}$ 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 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

${\displaystyle S=S_{1}\times S_{2},\quad \Sigma =\Sigma _{1}\otimes \Sigma _{2},\quad \mu =\mu _{1}\otimes \mu _{2},}$

where S is the Cartesian product of S1 and S2, the 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

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

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, ..., 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

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

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

${\displaystyle \alpha \left(|f_{1}(x)|^{p},\ldots ,|f_{n}(x)|^{p}\right)=\beta \left(|g_{1}(x)|^{q},\ldots ,|g_{n}(x)|^{q}\right),}$

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 = ∞ 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 assume p, q (0, 1) (1,∞). However, to apply Young's inequality for products, we will require p, q (1,∞)

Dividing f and g by ||f||p and ||g||q, respectively, we can assume that

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

We now use Young's inequality for products, which states that whenever ${\displaystyle p,q}$ are in (1,∞) with ${\displaystyle {\frac {1}{p}}+{\frac {1}{q}}=1}$

${\displaystyle ab\leq {\frac {a^{p}}{p}}+{\frac {b^{q}}{q}}}$

for all nonnegative a and b, where equality is achieved if and only if ap = bq. Hence

${\displaystyle |f(s)g(s)|\leq {\frac {|f(s)|^{p}}{p}}+{\frac {|g(s)|^{q}}{q}},\qquad s\in S.}$

Integrating both sides gives

${\displaystyle \|fg\|_{1}\leq {\frac {\|f\|_{p}^{p}}{p}}+{\frac {\|g\|_{q}^{q}}{q}}={\frac {1}{p}}+{\frac {1}{q}}=1,}$

which proves the claim.

Under the assumptions p (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

${\displaystyle \alpha =\|g\|_{q}^{q},\qquad \beta =\|f\|_{p}^{p},}$

such that

${\displaystyle \alpha |f|^{p}=\beta |g|^{q}}$   μ-almost everywhere   (*).

The case ||f||p = 0 corresponds to β = 0 in (*). The case ||g||q = 0 corresponds to α = 0 in (*).

Alternate proof using Jensen's inequality: Recall the Jensen's inequality for the convex function ${\displaystyle x^{p}}$ (it is convex because obviously ${\displaystyle p\geq 1}$):

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

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 ${\displaystyle g^{q}}$, i.e.

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

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

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

Finally, we get

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

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 ${\displaystyle \|f\|_{p},\|g\|_{q}}$ are neither null nor infinity, and that ${\displaystyle p,q>1}$: 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 fLp(μ),

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

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

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

Proof of the extremal equality: By Hölder's inequality, the integrals are well defined and, for 1 ≤ p ≤ ∞,

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

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

${\displaystyle g(x)={\begin{cases}\|f\|_{p}^{1-p}\,|f(x)|^{p}/f(x)&{\text{if }}f(x)\not =0,\\0&{\text{otherwise.}}\end{cases}}}$

By checking the cases p = 1 and 1 < p < ∞ separately, we see that ||g||q = 1 and

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

It remains to consider the case p = ∞. For ε (0, 1) define

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

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

${\displaystyle g(x)={\begin{cases}{\frac {1-\varepsilon }{\mu (B)}}{\frac {\|f\|_{\infty }}{f(x)}}&{\text{if }}x\in B,\\0&{\text{otherwise.}}\end{cases}}}$

Then g is well-defined, measurable and |g(x)| ≤ 1/μ(B) for xB, hence ||g||1 ≤ 1. Furthermore,

${\displaystyle \left|\int _{S}fg\,\mathrm {d} \mu \right|=\int _{B}{\frac {1-\varepsilon }{\mu (B)}}\|f\|_{\infty }\,\mathrm {d} \mu =(1-\varepsilon )\|f\|_{\infty }.}$

### Remarks and examples

• The equality for ${\displaystyle p=\infty }$ fails whenever there exists a set ${\displaystyle A}$ of infinite measure in the ${\displaystyle \sigma }$-field ${\displaystyle \Sigma }$ with that has no subset ${\displaystyle B\in \Sigma }$ that satisfies: ${\displaystyle 0<\mu (B)<\infty .}$ (the simplest example is the ${\displaystyle \sigma }$-field ${\displaystyle \Sigma }$ containing just the empty set and ${\displaystyle S,}$ and the measure ${\displaystyle \mu }$ with ${\displaystyle \mu (S)=\infty .}$) Then the indicator function ${\displaystyle 1_{A}}$ satisfies ${\displaystyle \|1_{A}\|_{\infty }=1,}$ but every ${\displaystyle g\in L^{1}(\mu )}$ has to be ${\displaystyle \mu }$-almost everywhere constant on ${\displaystyle A,}$ because it is ${\displaystyle \Sigma }$-measurable, and this constant has to be zero, because ${\displaystyle g}$ is ${\displaystyle \mu }$-integrable. Therefore, the above supremum for the indicator function ${\displaystyle 1_{A}}$ 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 ${\displaystyle \mu }$ 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 ${\displaystyle m}$ 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 Lp(μ), see Minkowski inequality.
• Hölder's inequality implies that every fLp(μ) defines a bounded (or continuous) linear functional κf on Lq(μ) by the formula
${\displaystyle \kappa _{f}(g)=\int _{S}fg\,\mathrm {d} \mu ,\qquad g\in L^{q}(\mu ).}$
The extremal equality (when true) shows that the norm of this functional κf as element of the continuous dual space Lq(μ)* coincides with the norm of f in Lp(μ) (see also the Lp-space article).

## Generalization of Hölder's inequality

Assume that r (0, ∞] and p1, ..., pn (0, ∞] such that

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

where 1/∞ is interpreted as 0 in this equation. Then for all measurable real or complex-valued functions f1, ..., fn defined on S,

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

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

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

Case 1: If ${\displaystyle p_{n}=\infty }$ then

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

Pulling out the essential supremum of |fn| and using the induction hypothesis, we get

{\displaystyle {\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 ${\displaystyle p_{n}<\infty }$ then necessarily ${\displaystyle r<\infty }$ as well, and then

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

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

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

Raising to the power ${\displaystyle 1/r}$ and rewriting,

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

Since ${\displaystyle qr=p_{n}}$ and

${\displaystyle \sum _{k=1}^{n-1}{\frac {1}{p_{k}}}={\frac {1}{r}}-{\frac {1}{p_{n}}}={\frac {p_{n}-r}{rp_{n}}}={\frac {1}{pr}},}$

the claimed inequality now follows by using the induction hypothesis.

### Interpolation

Let p1, ..., pn (0, ∞] and let θ1, ..., θn ∈ (0, 1) denote weights with θ1 + ... + θn = 1. Define ${\displaystyle p}$ as the weighted harmonic mean, that is,

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

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

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

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

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

Specifying further θ1 = θ and θ2 = 1-θ, in the case ${\displaystyle n=2,}$ we obtain the interpolation result (Littlewood's inequality)

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

for ${\displaystyle \theta \in (0,1)}$ and

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

An application of Hölder gives Lyapunov's inequality: If

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

then

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

and in particular

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

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

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

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

${\displaystyle q:={\frac {p}{p-1}}\in (1,\infty )}$

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, ${\displaystyle \mathbb {P} }$) be a probability space, GF 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,

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

Remarks:

${\displaystyle \mathbb {E} [Z|{\mathcal {G}}]=\sup _{n\in \mathbb {N} }\,\mathbb {E} [\min\{Z,n\}|{\mathcal {G}}]\quad {\text{a.s.}}}$
• 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

${\displaystyle U={\bigl (}\mathbb {E} {\bigl [}|X|^{p}{\big |}\,{\mathcal {G}}{\bigr ]}{\bigr )}^{\frac {1}{p}},\qquad V={\bigl (}\mathbb {E} {\bigl [}|Y|^{q}{\big |}\,{\mathcal {G}}{\bigr ]}{\bigr )}^{\frac {1}{q}}}$

and note that they are measurable with respect to the sub-σ-algebra. Since

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

it follows that |X| = 0 a.s. on the set {U = 0}. Similarly, |Y| = 0 a.s. on the set {V = 0}, hence

${\displaystyle \mathbb {E} {\bigl [}|XY|{\big |}\,{\mathcal {G}}{\bigr ]}=0\qquad {\text{a.s. on }}\{U=0\}\cup \{V=0\}}$

and the conditional Hölder inequality holds on this set. On the set

${\displaystyle \{U=\infty ,V>0\}\cup \{U>0,V=\infty \}}$

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

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

This is done by verifying that the inequality holds after integration over an arbitrary

${\displaystyle G\in {\mathcal {G}},\quad G\subset H.}$

Using the measurability of U, V, 1G with respect to the sub-σ-algebra, the rules for conditional expectations, Hölder's inequality and 1/p + 1/q = 1, we see that

{\displaystyle {\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 ${\displaystyle p}$ and ${\displaystyle q}$ are Hölder conjugates.[1]

Remark: If (S, Σ, μ) is a measure space and ${\displaystyle N(f)}$ is the upper Lebesgue integral of ${\displaystyle |f|}$ then the restriction of N to all Σ-measurable functions gives the usual version of Hölder's inequality.

• Cauchy–Schwarz inequality
• Minkowski inequality
• Jensen's inequality
• Young's inequality for products
• Clarkson's inequalities
• Brascamp–Lieb 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) [1994], "Hölder inequality", Encyclopedia of Mathematics, EMS Press.
• Narici, Lawrence; Beckenstein, Edward (2011). Topological Vector Spaces. Pure and applied mathematics (Second ed.). Boca Raton, FL: CRC Press. ISBN 978-1584888666. OCLC 144216834.
• 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.
• Trèves, François (2006) [1967]. Topological Vector Spaces, Distributions and Kernels. Mineola, N.Y.: Dover Publications. ISBN 978-0-486-45352-1. OCLC 853623322.