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Theorem (Hölder's inequality). Let (S, ?, ?) be a measure space and let p, q ?[1, ?) with 1/p + 1/q = 1. Then, for all measurablereal- or complex-valued functionsf and g on S,
$\|fg\|_{1}\leq \|f\|_{p}\|g\|_{q}.$
If, in addition, p, q ?(1, ?) and f ? L^{p}(?) and g ? L^{q}(?), then Hölder's inequality becomes an equality iff ||^{p} and |g|^{q} are linearly dependent in L^{1}(?), 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 L^{p}(?) and g is in L^{q}(?), then the pointwise product fg is in L^{1}(?).
Hölder's inequality is used to prove the Minkowski inequality, which is the triangle inequality in the space L^{p}(?), and also to establish that L^{q}(?) is the dual space of L^{p}(?) for p ?[1, ?).
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 ? L^{p}(?) and g ? L^{q}(?), 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
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 spaceL^{2}(?), then Hölder's inequality for p = q = 2 implies
$|\langle f,g\rangle |\leq \|f\|_{2}\|g\|_{2},$
where the angle brackets refer to the inner product of L^{2}(?). 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
$\|fg\|_{1}\leq \|f\|_{p}\|g\|_{q}$
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.
If S is a measurable subset of R^{n} with the Lebesgue measure, and f and g are measurable real- or complex-valued functions on S, then Hölder inequality is
For the probability space$(\Omega ,{\mathcal {F}},\mathbb {P} ),$ let $\mathbb {E}$ denote the expectation operator. For real- or complex-valued random variables$X$ and $Y$ on $\Omega ,$ Hölder's inequality reads
Let $0<r<s$ and define $p={\tfrac {s}{r}}.$ Then $q={\tfrac {p}{p-1}}$ is the Hölder conjugate of $p.$ Applying Hölder's inequality to the random variables $|X|^{r}$ and $1_{\Omega }$ we obtain
where S is the Cartesian product of S_{1} and S_{2}, 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
This can be generalized to more than two measure spaces.
Vector-valued functions
Let (S, ?, ?) denote a measure space and suppose that f = (f_{1}, ..., f_{n}) and g = (g_{1}, ..., g_{n}) 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
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
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
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
where ? is any probability distribution and h any ?-measurable function. Let ? be any measure, and ? the distribution whose density w.r.t. ? is proportional to $g^{q}$, i.e.
$d\nu ={\frac {g^{q}}{\int g^{q}\,d\mu }}d\mu$
Hence we have, using ${\frac {1}{p}}+{\frac {1}{q}}=1$, hence $p(1-q)+q=0$, and letting $h=fg^{1-q}$,
This assumes $f,g$ real and non negative, but the extension to complex functions is straightforward (use the modulus of $f,g$). It also assumes that $\|f\|_{p},\|g\|_{q}$ are neither null nor infinity, and that $p,q>1$: all these assumptions can also be lifted as in the proof above.
Extremal equality
Statement
Assume that 1 p < ? and let q denote the Hölder conjugate. Then, for every f ? L^{p}(?),
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
Since f is measurable, A ? ?. By the definition of ||||_{?} as the essential supremum of f and the assumption ||||_{?} > 0, we have ?(A) > 0. Using the additional assumption on the ? if necessary, there exists a subset B ? ? of A with 0 < ?(B) < ?. Define g on S by
The equality for $p=\infty$ fails whenever there exists a set $A$ of infinite measure in the $\sigma$-field $\Sigma$ with that has no subset $B\in \Sigma$ that satisfies: $0<\mu (B)<\infty .$ (the simplest example is the $\sigma$-field $\Sigma$ containing just the empty set and $S,$ and the measure $\mu$ with $\mu (S)=\infty .$) Then the indicator function$1_{A}$ satisfies $\|1_{A}\|_{\infty }=1,$ but every $g\in L^{1}(\mu )$ has to be $\mu$-almost everywhere constant on $A,$ because it is $\Sigma$-measurable, and this constant has to be zero, because $g$ is $\mu$-integrable. Therefore, the above supremum for the indicator function $1_{A}$ is zero and the extremal equality fails.
For $p=\infty ,$ the supremum is in general not attained. As an example, let $S=\mathbb {N} ,\Sigma ={\mathcal {P}}(\mathbb {N} )$ and $\mu$ the counting measure. Define:
Then $\|f\|_{\infty }=1.$ For $g\in L^{1}(\mu ,\mathbb {N} )$ with $0<\|g\|_{1}\leqslant 1,$ let $m$ denote the smallest natural number with $g(m)\neq 0.$ Then
The extremal equality is one of the ways for proving the triangle inequality ||||_{p}f_{1}||_{p} + ||||_{p} for all f_{1} and f_{2} in L^{p}(?), see Minkowski inequality.
Hölder's inequality implies that every f ? L^{p}(?) defines a bounded (or continuous) linear functional ?_{f} on L^{q}(?) by the formula
The extremal equality (when true) shows that the norm of this functional ?_{f} as element of the continuous dual spaceL^{q}(?)^{*} coincides with the norm of f in L^{p}(?) (see also the article).
Generalization of Hölder's inequality
Assume that r ?(0, ?] and p_{1}, ..., p_{n} ? (0, ?] such that
Note: For r ? (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. For n = 1, the result is obvious. Let us now pass from n - 1 to n. Without loss of generality assume that p_{1}p_{n}.
the claimed inequality now follows by using the induction hypothesis.
Interpolation
Let p_{1}, ..., p_{n} ?(0, ?] and let ?_{1}, ..., ?_{n} ? (0, 1) denote weights with ?_{1} + ... + ?_{n} = 1. Define p as the weighted harmonic mean, i.e.,
Both Littlewood and Lyapunov imply that if $f\in L^{p_{0}}\cap L^{p_{1}},$ then $f\in L^{p}$ for all $p_{0}<p<p_{1}.$
Reverse Hölder inequality
Assume that p ? (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,
Since g is not almost everywhere equal to the zero function, we can have equality if and only if there exists a constant ? >= 0 such that || = ? ||^{-q/p} almost everywhere. Solving for the absolute value of f gives the claim.
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 ?,
and the conditional Hölder inequality holds on this set. On the set
$\{U=\infty ,V>0\}\cup \{U>0,V=\infty \}$
the right-hand side is infinite and the conditional Hölder inequality holds, too. Dividing by the right-hand side, it therefore remains to show that
${\frac {\mathbb {E} {\bigl [}|XY|{\big |}\,{\mathcal {G}}{\bigr ]}}{UV}}\leq 1\qquad {\text{a.s. on the set }}H:=\{0<U<\infty ,\,0<V<\infty \}.$
This is done by verifying that the inequality holds after integration over an arbitrary
$G\in {\mathcal {G}},\quad G\subset H.$
Using the measurability of U, V, 1_{G} with respect to the , the rules for conditional expectations, Hölder's inequality and 1/p + 1/q = 1, we see that
Let S be a set and let $F(S,\mathbb {C} )$ be the space of all complex-valued functions on S. Let N be an increasing seminorm on $F(S,\mathbb {C} ),$ meaning that, for all real-valued functions $f,g\in F(S,\mathbb {C} )$ we have the following implication (the seminorm is also allowed to attain the value ?):
where the numbers $p$ and $q$ are Hölder conjugates.^{[1]}
Remark: If (S, ?, ?) is a measure space and $N(f)$ is the upper Lebesgue integral of $|f|$ then the restriction of N to all functions gives the usual version of Hölder's inequality.
^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
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, MR0225131, Zbl0171.10402.