If, in addition, p, q ∈(1, ∞) and f ∈ Lp(μ) and g ∈ Lq(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|qμ-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 ‖fg‖1 is infinite, the right-hand side also being infinite in that case. Conversely, if f is in Lp(μ) and g is in Lq(μ), then the pointwise productfg is in L1(μ).
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
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 ∈ Lp(μ) and g ∈ Lq(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 ∞.
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
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 spaceL2(μ), then Hölder's inequality for p = q = 2 implies
where the angle brackets refer to the inner product of L2(μ). This is also called Cauchy–Schwarz inequality, but requires for its statement that ‖f‖2 and ‖g‖2 are finite to make sure that the inner product of f and g is well defined. We may recover the original inequality (for the case p = 2) by using the functions |f| and |g| in place of f and g.
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
for all measurable real- or complex-valued functions f and g on S.
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
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
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 ∈(1,∞).
Dividing f and g by ‖f‖p and ‖g‖q, respectively, we can assume that
for all nonnegative a and b, where equality is achieved if and only if ap = bq. Hence
Integrating both sides gives
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
such that
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:
Proof
The function on (0,∞) is convex because , so by Jensen's inequality,
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 , i.e.
Hence we have, using , hence , and letting ,
Finally, we get
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 are neither null nor infinity, and that : 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
where and is -measurable (real or complex) function on . To prove this, we must bound by . There is no constant that will make for all . Hence, we seek an inequality of the form
for suitable choices of and .
We wish to obtain on the right-hand side after integrating this inequality. By trial and error, we see that the inequality we wish should have the form
where are non-negative and . Indeed, the integral of the right-hand side is precisely . So, it remains to prove that such an inequality does hold with the right choice of
The inequality we seek would follow from:
which, in turn, is equivalent to
It turns out there is one and only one choice of , subject to , that makes this true: and, necessarily, . (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 and multiplying both sides by .
Assume that 1 ≤ p < ∞ and let q denote the Hölder conjugate. Then for every f ∈ Lp(μ),
where max indicates that there actually is a g maximizing the right-hand side. When p = ∞ and if each set A in the σ-fieldΣ with μ(A) = ∞ contains a subset B ∈ Σ with 0 < μ(B) < ∞ (which is true in particular when μ is σ-finite), then
Proof of the extremal equality:
Proof
By Hölder's inequality, the integrals are well defined and, for 1 ≤ 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
By checking the cases p = 1 and 1 < p < ∞ separately, we see that ‖g‖q = 1 and
It remains to consider the case p = ∞. For ε ∈(0, 1) define
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-fieldΣ if necessary, there exists a subset B ∈ Σ of A with 0 < μ(B) < ∞. Define g on S by
Then g is well-defined, measurable and |g(x)| ≤ 1/μ(B) for x ∈ B, hence ‖g‖1 ≤ 1. Furthermore,
The equality for fails whenever there exists a set of infinite measure in the -field with that has no subset that satisfies: (the simplest example is the -field containing just the empty set and and the measure with ) Then the indicator function satisfies but every has to be -almost everywhere constant on because it is -measurable, and this constant has to be zero, because is -integrable. Therefore, the above supremum for the indicator function is zero and the extremal equality fails.
For the supremum is in general not attained. As an example, let and the counting measure. Define:
Then For with let denote the smallest natural number with Then
The extremal equality is one of the ways for proving the triangle inequality ‖f1 + f2‖p ≤ ‖f1‖p + ‖f2‖p for all f1 and f2 in Lp(μ), see Minkowski inequality.
Hölder's inequality implies that every f ∈ Lp(μ) defines a bounded (or continuous) linear functional κf on Lq(μ) by the formula
The extremal equality (when true) shows that the norm of this functional κf as element of the continuous dual spaceLq(μ)* coincides with the norm of f in Lp(μ) (see also the Lp-space article).
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,
If
then the reverse Hölder inequality is an equality if and only if
Note: The expressions:
and
are not norms, they are just compact notations for
Proof of the reverse Hölder inequality (hidden, click show to reveal.)
Note that p and
are Hölder conjugates. Application of Hölder's inequality gives
Raising to the power p gives us:
Therefore:
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.
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 be vectors with positive entries and such that for all . If are nonzero real numbers such that , then:
if all but one of are positive;
if all but one of 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 be vectors in with positive entries and such that for all . If are nonzero real numbers such that , then:
if all but one of the numbers are positive;
if all but one of the numbers are negative.
As in the standard Hölder inequalities, there are corresponding statements for infinite sums and integrals.
Let (Ω, F, ) 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 on Ω,
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
and note that they are measurable with respect to the sub-σ-algebra. Since
it follows that |X| = 0 a.s. on the set {U = 0}. Similarly, |Y| = 0 a.s. on the set {V = 0}, hence
and the conditional Hölder inequality holds on this set. On the set
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
This is done by verifying that the inequality holds after integration over an arbitrary
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
Let S be a set and let be the space of all complex-valued functions on S. Let N be an increasing seminorm on meaning that, for all real-valued functions we have the following implication (the seminorm is also allowed to attain the value ∞):
Remark: If (S, Σ, μ) is a measure space and is the upper Lebesgue integral of then the restriction of N to all Σ-measurable functions gives the usual version of Hölder's 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.
^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, MR1492911
^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, MR3069109, S2CID253600514, under the additional assumption that exists, this inequality was already obtained by Hölder in 1889
^Beckenbach, E. F. (1980). General inequalities 2. International Series of Numerical Mathematics / Internationale Schriftenreihe zur Numerischen Mathematik / Série Internationale d'Analyse Numérique. Vol. 47. Birkhäuser Basel. pp. 145–150. doi:10.1007/978-3-0348-6324-7. ISBN978-3-7643-1056-1.
^For a proof see (Trèves 1967, Lemma 20.1, pp. 205–206).
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, vol. 25, New York, London: Academic Press, MR0225131, Zbl0171.10402.