Show That if Equality Holds in This Inequality Then F is Absolutely Continuous on a B

Inequality between integrals in Lp spaces

In mathematical analysis, Hölder's inequality, named after Otto Hölder, is a fundamental inequality between integrals and an indispensable tool for the study of L^p spaces.

Theorem (Hölder's inequality). Let (S, Σ, μ) be a measure space and let p, q ∈ [1, ∞] with 1/p + 1/q = 1. Then for all measurable real- or complex-valued functions f and g on S,

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

If, in addition, p, q ∈ (1, ∞) and f ∈ L^p (μ) and g ∈ L^q (μ), then Hölder's inequality becomes an equality if and only if | f| ^p and |g| ^q are linearly dependent in L ¹(μ), meaning that there exist real numbers α, β ≥ 0, not both of them zero, such that α|f | ^p = β |g| ^q μ-almost everywhere.

The numbers p and q above are said to be Hölder conjugates of each other. The special case p = q = 2 gives a form of the Cauchy–Schwarz inequality. Hölder's inequality holds even if || fg ||₁ is infinite, the right-hand side also being infinite in that case. Conversely, if f is in L^p (μ) and g is in L^q (μ), then the pointwise product fg is in 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 ∈ [1, ∞).

Hölder's inequality 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,^[1] which was in turn named for work of Johan Jensen building on Hölder's work.^[2]

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Conventions [edit]

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 f ∈ L^p (μ) and g ∈ L^q (μ), then the notation is adequate.
• On the right-hand side of Hölder's inequality, 0 × ∞ as well as ∞ × 0 means 0. Multiplying a > 0 with ∞ gives ∞.

Estimates for integrable products [edit]

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 L ²(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 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 [edit]

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 onS.

Notable special cases [edit]

For the following cases assume that p and q are in the open interval (1,∞) with 1/p + 1/q = 1.

Counting measure [edit]

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 ${\displaystyle (r,s)\in \mathbb {R} _{+}}$ :

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

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 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 [edit]

If S is a measurable subset of R ⁿ with the Lebesgue measure, and f and g are measurable real- or complex-valued functions onS, 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 [edit]

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 ${\displaystyle 1<r<s<\infty }$ 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 s ^th absolute moment is finite, then the r  ^th absolute moment is finite, too. (This also follows from Jensen's inequality.)

Product measure [edit]

For two σ-finite measure spaces (S ₁, Σ₁, μ ₁) and (S ₂, Σ₂, μ 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 S ₁ and S ₂ , the σ-algebra Σ arises as product σ-algebra of Σ₁ and Σ₂ , 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: Iff and g are Σ-measurable real- or complex-valued functions on the Cartesian productS, 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 [edit]

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

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

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 [edit]

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 a^p = b^q . 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 (*).

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

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

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 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.

Extremal equality [edit]

Statement [edit]

Assume that 1 ≤ p < ∞ and let q denote the Hölder conjugate. Then for every f ∈ L^p (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 x ∈ B , 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 [edit]

• 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 [edit]

• The extremal equality is one of the ways for proving the triangle inequality || f ₁ + f ₂ || _p ≤ || 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 ∈ L^p (μ) defines a bounded (or continuous) linear functional κ_f on L^q (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 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 [edit]

Assume that r ∈ (0, ∞] and p ₁, ..., p_n ∈ (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 f ₁, ..., f _n 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 | f_n | 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 [edit]

Let p ₁, ..., p _n ∈ (0, ∞] and let θ ₁, ..., θ _n ∈ (0, 1) denote weights with θ ₁ + ... + θ_n = 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 θ ₁ = θ and θ ₂ = 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 ${\displaystyle p_{0}<p<p_{1}.}$

Reverse Hölder inequality [edit]

Assume that p ∈ (1, ∞) and that the measure space (S, Σ, μ) satisfies μ(S) > 0. Then for all measurable real- or complex-valued functions f and g on S such that g(s) ≠ 0 for μ-almost all s ∈ S ,

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

If

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

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

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

Note: The expressions:

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

are not norms, they are just compact notations for

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

Proof of the reverse Hölder inequality: Note that p and

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

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 [edit]

Let (Ω, F, ${\displaystyle \mathbb {P} }$ ) be a probability space, G ⊂ F a sub-σ-algebra, and p, q ∈ (1, ∞) Hölder conjugates, meaning that 1/p + 1/q = 1. Then for all real- or complex-valued random variables X and Y onΩ,

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

Remarks:

• If a non-negative random variable Z has infinite expected value, then its conditional expectation is defined by

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

• 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

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

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, 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

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

Hölder's inequality for increasing seminorms [edit]

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.^[3]

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 [edit]

Hölder inequality can be used to define statistical dissimilarity measures^[4] between probability distributions. Those Hölder divergences are projective: They do not depend on the normalization factor of densities.

See also [edit]

• Cauchy–Schwarz inequality
• Minkowski inequality
• Jensen's inequality
• Young's inequality for products
• Clarkson's inequalities
• Brascamp–Lieb inequality

Citations [edit]

1. ^ Maligranda, Lech (1998), "Why Hölder's inequality should be called Rogers' inequality", Mathematical Inequalities & Applications, 1 (1): 69–83, doi:10.7153/mia-01-05, MR 1492911
2. ^ 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, under the additional assumption that ${\displaystyle \varphi ''}$ exists, this inequality was already obtained by Hölder in 1889
3. ^ For a proof see (Trèves 1967, Lemma 20.1, pp. 205–206).
4. ^ Nielsen, Frank; Sun, Ke; Marchand-Maillet, Stephane (2017). "On Hölder projective divergences". Entropy. 3 (19): 122. arXiv:1701.03916. Bibcode:2017Entrp..19..122N. doi:10.3390/e19030122.

References [edit]

• 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, ISBN0-521-35880-9, 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. ISBN978-1584888666. 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, vol. 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. ISBN978-0-486-45352-1. OCLC 853623322.

External links [edit]

• Chen, Evan (2015), A Brief Introduction to Olympiad Inequalities (PDF) .
• Kuttler, Kenneth (2007), An Introduction to Linear Algebra (PDF), Online e-book in PDF format, Brigham Young University .
• Lohwater, Arthur (1982), Introduction to Inequalities (PDF) .
• Archived at Ghostarchive and the Wayback Machine: Tisdell, Chris (2012), Holder's Inequality, Online video on Dr Chris Tisdell's YouTube channel .

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Source: https://en.wikipedia.org/wiki/H%C3%B6lder%27s_inequality

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