Holder Conjugate
Get Holder Conjugate essential facts below. View Videos or join the Holder Conjugate discussion. Add Holder Conjugate to your PopFlock.com topic list for future reference or share this resource on social media.
Holder Conjugate

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, ?, ?) 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 f ? Lp(?) and g ? Lq(?), then Hölder's inequality becomes an equality iff ||p and |g|q are linearly dependent in L1(?), meaning that there exist real numbers ?, ? >= 0, not both of them zero, such that ?|f |p = ? |g|q ?-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 ||||1 is infinite, the right-hand side also being infinite in that case. Conversely, if f  is in Lp(?) and g is in Lq(?), then the pointwise product fg is in L1(?).

Hölder's inequality is used to prove the Minkowski inequality, which is the triangle inequality in the space Lp(?), and also to establish that Lq(?) is the dual space of Lp(?) 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 ||||p and ||||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 ||||? stands for the essential supremum of ||, similarly for ||||?.
• The notation ||||p with 1 p is a slight abuse, because in general it is only a norm of f  if ||||p is finite and f  is considered as equivalence class of ?-almost everywhere equal functions. If f ? Lp(?) and g ? Lq(?), 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 ||||1 is finite, then the pointwise products of f  with g and its complex conjugate function are ?-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(?), 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(?). This is also called Cauchy-Schwarz inequality, but requires for its statement that ||||2 and ||||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 || and || in place of f  and g.

### Generalization for probability measures

If (S, ?, ?) is a probability space, then p, q ? [1, ?] just need to satisfy 1/p + 1/q , 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

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

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 ?-finite measure spaces (S1, ?1, ?1) and (S2, ?2, ?2) 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 ?-algebra of ?1 and ?2, and ? denotes the product measure of ?1 and ?2. Then Tonelli's theorem allows us to rewrite Hölder's inequality using iterated integrals: If f  and g are 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 measure spaces.

### Vector-valued functions

Let (S, ?, ?) denote a measure space and suppose that f = (f1, ..., fn) and g = (g1, ..., gn) are ?-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 ?, ? >= 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 ?-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.

If ||||p = 0, then f  is zero ?-almost everywhere, and the product fg is zero ?-almost everywhere, hence the left-hand side of Hölder's inequality is zero. The same is true if ||||q = 0. Therefore, we may assume ||||p > 0 and ||||q > 0 in the following.

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

If p = ? and q = 1, then || f||? || 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 ||||p and ||||q, respectively, we can assume that

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

We now use Young's inequality for products, which states that

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

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 ||||p = ||||q, equality holds if and only if ||p = ||q almost everywhere. More generally, if ||||p and ||||q are in (0, ?), then Hölder's inequality becomes an equality if and only if there exist real numbers ?, ? > 0, namely

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

such that

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

The case ||||p = 0 corresponds to ? = 0 in (*). The case ||||q = 0 corresponds to ? = 0 in (*).

## Extremal equality

### Statement

Assume that 1 p < ? and let q denote the Hölder conjugate. Then, for every f ? Lp(?),

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

where max indicates that there actually is a g maximizing the right-hand side. When p = ? and if each set A in the ? with ?(A) = ? contains a subset B ? ? with 0 < ?(B) < ? (which is true in particular when ? is ), then

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

### 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 ||||pf1||p + ||||p for all f1 and f2 in Lp(?), see Minkowski inequality.
• Hölder's inequality implies that every f ? Lp(?) defines a bounded (or continuous) linear functional ?f  on Lq(?) 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 ?f  as element of the continuous dual space Lq(?)* coincides with the norm of f  in Lp(?) (see also the 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 we interpret 1/? as 0 in this equation). Then, for all measurable real- or complex-valued functions f1, ..., fn defined on S,

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

(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,

${\displaystyle f_{k}\in L^{p_{k}}(\mu )\;\;\forall k\in \{1,\ldots ,n\}\implies \prod _{k=1}^{n}f_{k}\in L^{r}(\mu ).}$

Note: For r ? (0, 1), contrary to the notation, ||||r is in general not a norm, because it doesn't satisfy the triangle inequality.

### Interpolation

Let p1, ..., pn ? (0, ?] and let ?1, ..., ?n ? (0, 1) denote weights with ?1 + ... + ?n = 1. Define p as the weighted harmonic mean, i.e.,

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

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 ?1 = ? and ?2 = 1-?, 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 ${\displaystyle \theta \in (0,1)}$ 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, ?, ?) satisfies ?(S) > 0. Then, for all measurable real- or complex-valued functions f  and g on S such that g(s) ? 0 for 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

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

## Conditional Hölder inequality

Let (?, F, P) be a probability space, G ? F a , 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 ?,

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

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 ?.

## 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, ?, ?) is a measure space and ${\displaystyle N(f)}$ is the upper Lebesgue integral of ${\displaystyle |f|}$ then the restriction of N to all functions gives the usual version of Hölder'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) [1994], "Hölder inequality", in Hazewinkel, Michiel (ed.), Encyclopedia of Mathematics, Springer Science+Business Media B.V. / Kluwer Academic Publishers, 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.