# 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.

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.[1] Hölder's inequality holds even if fg1 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 (in a slightly different form) was first found by Leonard James Rogers (1888). Inspired by Rogers' work, Hölder (1889) gave another proof as part of a work developing the concept of convex and concave functions and introducing Jensen's inequality,[2] which was in turn named for work of Johan Jensen building on Hölder's work.[3]

## 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 fp and gq 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 fp with 1 ≤ p ≤ ∞ is a slight abuse, because in general it is only a norm of f if fp 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 fg1 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 f2 and g2 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 ${\displaystyle n}$-dimensional Euclidean space, when the set ${\displaystyle S}$ is ${\displaystyle \{1,\dots ,n\}}$ 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 ${\displaystyle (r,s)\in \mathbb {R} _{+}}$:

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

For more than two sums, the following generalisation (Chen (2015)) holds, with real positive exponents ${\displaystyle \lambda _{i}}$ and ${\displaystyle \lambda _{a}+\lambda _{b}+\cdots +\lambda _{z}=1}$:

${\displaystyle \sum _{k=1}^{n}|a_{k}|^{\lambda _{a}}\,|b_{k}|^{\lambda _{b}}\cdots |z_{k}|^{\lambda _{z}}\leq {\biggl (}\sum _{k=1}^{n}|a_{k}|{\biggr )}^{\lambda _{a}}{\biggl (}\sum _{k=1}^{n}|b_{k}|{\biggr )}^{\lambda _{b}}\cdots {\biggl (}\sum _{k=1}^{n}|z_{k}|{\biggr )}^{\lambda _{z}}.}$

Equality holds iff ${\displaystyle |a_{1}|:|a_{2}|:\cdots :|a_{n}|=|b_{1}|:|b_{2}|:\cdots :|b_{n}|=\cdots =|z_{1}|:|z_{2}|:\cdots :|z_{n}|}$.

If ${\displaystyle S=\mathbb {N} }$ 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 ${\displaystyle S}$ is a measurable subset of ${\displaystyle \mathbb {R} ^{n}}$ with the Lebesgue measure, and ${\displaystyle f}$ and ${\displaystyle g}$ are measurable real- or complex-valued functions on ${\displaystyle S}$, then Hölder's 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 1 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 fp = 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 gq = 0. Therefore, we may assume fp > 0 and gq > 0 in the following.

If fp = ∞ or gq = ∞, then the right-hand side of Hölder's inequality is infinite. Therefore, we may assume that fp and gq 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 (1,∞).

Dividing f and g by fp and gq, 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 fp = ‖gq, equality holds if and only if |f|p = |g|q almost everywhere. More generally, if fp and gq 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 fp = 0 corresponds to β = 0 in (*). The case gq = 0 corresponds to α = 0 in (*).

Alternative proof using Jensen's inequality:

Proof

The function ${\displaystyle x\mapsto x^{p}}$ on (0,∞) is convex because ${\displaystyle p\geq 1}$, so by Jensen's inequality,

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

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.

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

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 {\begin{aligned}\int fg\,\mathrm {d} \mu =&\left(\int g^{q}\,\mathrm {d} \mu \right)\int \underbrace {fg^{1-q}} _{h}\underbrace {{\frac {g^{q}}{\int g^{q}\,\mathrm {d} \mu }}\mathrm {d} \mu } _{\mathrm {d} \nu }\\\leq &\left(\int g^{q}\mathrm {d} \mu \right)\left(\int \underbrace {f^{p}g^{p(1-q)}} _{h^{p}}\underbrace {{\frac {g^{q}}{\int g^{q}\,\mathrm {d} \mu }}\,\mathrm {d} \mu } _{\mathrm {d} \nu }\right)^{\frac {1}{p}}\\=&\left(\int g^{q}\,\mathrm {d} \mu \right)\left(\int {\frac {f^{p}}{\int g^{q}\,\mathrm {d} \mu }}\,\mathrm {d} \mu \right)^{\frac {1}{p}}.\end{aligned}}}

Finally, we get

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

This assumes that f, g are 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.

We could also bypass use of both Young's and Jensen's inequalities. The proof below also explains why and where the Hölder exponent comes in naturally.

Proof

As in the previous proof, it suffices to prove

${\displaystyle \int _{X}|h|\mathrm {d} \nu \leq \left(\int _{X}|h|^{p}\mathrm {d} \nu \right)^{\frac {1}{p}}}$

where ${\displaystyle \nu (X)=1}$ and ${\displaystyle h}$ is ${\displaystyle \nu }$-measurable (real or complex) function on ${\displaystyle X}$. To prove this, we must bound ${\displaystyle |h|}$ by ${\displaystyle |h|^{p}}$. There is no constant ${\displaystyle C}$ that will make ${\displaystyle |h(x)|~\leq ~C|h(x)|^{p}}$ for all ${\displaystyle x>0}$. Hence, we seek an inequality of the form

${\displaystyle |h(x)|~\leq ~a'|h(x)|^{p}+b',\quad for\quad all\quad x>0}$

for suitable choices of ${\displaystyle a'}$ and ${\displaystyle b'}$.

We wish to obtain ${\displaystyle A:=\|f\|_{p}}$ on the right-hand side after integrating this inequality. By trial and error, we see that the inequality we wish should have the form

${\displaystyle |h(x)|~\leq ~aA^{1-p}|h(x)|^{p}+bA,\quad for\quad all\quad x>0,}$

where ${\displaystyle a,b}$ are non-negative and ${\displaystyle a+b=1}$. Indeed, the integral of the right-hand side is precisely ${\displaystyle A}$. So, it remains to prove that such an inequality does hold with the right choice of ${\displaystyle a,b.}$

The inequality we seek would follow from:

${\displaystyle {\tfrac {y}{A}}~\leq ~a({\tfrac {y}{a}})^{p}+b,\quad for\quad all\quad y>0,}$

which, in turn, is equivalent to

${\displaystyle (*)\quad z~\leq ~az^{p}+b,\quad for\quad all\quad z>0.}$

It turns out there is one and only one choice of ${\displaystyle a,b}$, subject to ${\displaystyle a+b=1}$, that makes this true: ${\displaystyle a={\tfrac {1}{p}}}$ and, necessarily, ${\displaystyle b=1-{\tfrac {1}{p}}}$. (This is where Hölder conjugate exponent is born!) This completes the proof of the inequality at the first paragraph of this proof. Proof of Hölder's inequality follows from this as in the previous proof. Alternatively, we can deduce Young's inequality and then resort to the first proof given above. Young's inequality follows from the inequality (*) above by choosing ${\displaystyle z={\tfrac {a}{b^{q-1}}}}$ and multiplying both sides by ${\displaystyle b^{q}}$.

## 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:

Proof

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 fp = 0. Therefore, we assume fp > 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 gq = 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 g1 ≤ 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 + f2p ≤ ‖f1p + ‖f2p 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 with more than two functions

### Statement

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:

Proof

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 — For ${\displaystyle \theta \in (0,1)}$ and ${\displaystyle {\frac {1}{p_{\theta }}}={\frac {\theta }{p_{1}}}+{\frac {1-\theta }{p_{0}}}}$,

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

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}[4]

## Reverse Hölder inequalities

### Two functions

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:

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

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 (hidden, click show to reveal.)

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.

### Multiple functions

The Reverse Hölder inequality (above) can be generalized to the case of multiple functions if all but one conjugate is negative. That is,

Let ${\displaystyle p_{1},...,p_{m-1}<0}$ and ${\displaystyle p_{m}\in \mathbb {R} }$ be such that ${\displaystyle \sum _{k=1}^{m}{\frac {1}{p_{k}}}=1}$ (hence ${\displaystyle 0). Let ${\displaystyle f_{k}}$ be measurable functions for ${\displaystyle k=1,...,m}$. Then
${\displaystyle \left\|\prod _{k=1}^{n}f_{k}\right\|_{1}\geq \prod _{k=1}^{n}\left\|f_{k}\right\|_{p_{k}}.}$

This follows from the symmetric form of the Hölder inequality (see below).

## Symmetric forms of Hölder inequality

It was observed by Aczél and Beckenbach[5] that Hölder's inequality can be put in a more symmetric form, at the price of introducing an extra vector (or function):

Let ${\displaystyle f=(f(1),\dots ,f(m)),g=(g(1),\dots ,g(m)),h=(h(1),\dots ,h(m))}$ be vectors with positive entries and such that ${\displaystyle f(i)g(i)h(i)=1}$ for all ${\displaystyle i}$. If ${\displaystyle p,q,r}$ are nonzero real numbers such that ${\displaystyle {\frac {1}{p}}+{\frac {1}{q}}+{\frac {1}{r}}=0}$, then:

• ${\displaystyle \|f\|_{p}\|g\|_{q}\|h\|_{r}\geq 1}$ if all but one of ${\displaystyle p,q,r}$ are positive;
• ${\displaystyle \|f\|_{p}\|g\|_{q}\|h\|_{r}\leq 1}$ if all but one of ${\displaystyle p,q,r}$ are negative.

The standard Hölder inequality follows immediately from this symmetric form (and in fact is easily seen to be equivalent to it). The symmetric statement also implies the reverse Hölder inequality (see above).

The result can be extended to multiple vectors:

Let ${\displaystyle f_{1},\dots ,f_{n}}$ be ${\displaystyle n}$ vectors in ${\displaystyle \mathbb {R} ^{m}}$ with positive entries and such that ${\displaystyle f_{1}(i)\dots f_{n}(i)=1}$ for all ${\displaystyle i}$. If ${\displaystyle p_{1},\dots ,p_{n}}$ are nonzero real numbers such that ${\displaystyle {\frac {1}{p_{1}}}+\dots +{\frac {1}{p_{n}}}=0}$, then:

• ${\displaystyle \|f_{1}\|_{p_{1}}\dots \|f_{n}\|_{p_{n}}\geq 1}$ if all but one of the numbers ${\displaystyle p_{i}}$ are positive;
• ${\displaystyle \|f_{1}\|_{p_{1}}\dots \|f_{n}\|_{p_{n}}\leq 1}$ if all but one of the numbers ${\displaystyle p_{i}}$ are negative.

As in the standard Hölder inequalities, there are corresponding statements for infinite sums and integrals.

## 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:

Proof

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.[6]

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.

## Distances based on Hölder inequality

Hölder inequality can be used to define statistical dissimilarity measures[7] between probability distributions. Those Hölder divergences are projective: They do not depend on the normalization factor of densities.

3. ^ Guessab, A.; Schmeisser, G. (2013), "Necessary and sufficient conditions for the validity of Jensen's inequality", Archiv der Mathematik, 100 (6): 561–570, doi:10.1007/s00013-013-0522-3, MR 3069109, S2CID 253600514, under the additional assumption that ${\displaystyle \varphi ''}$ exists, this inequality was already obtained by Hölder in 1889