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 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,

‖fg‖1≤‖f‖p‖g‖q.{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 iff |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. 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 L¹(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 Rogers (1888), and discovered independently by Hölder (1889).

Contents

• 
  1 Remarks
  
  
  
  • 
    1.1 Conventions
    
  
  
  • 
    1.2 Estimates for integrable products
    
  
  
  • 
    1.3 Generalization for probability measures
    
  
  
  
  


• 
  2 Notable special cases
  
  
  
  • 
    2.1 Counting measure
    
  
  
  • 
    2.2 Lebesgue measure
    
  
  
  • 
    2.3 Probability measure
    
  
  
  • 
    2.4 Product measure
    
  
  
  • 
    2.5 Vector-valued functions
    
  
  
  
  


• 
  3 Proof of Hölder's inequality
  


• 
  4 Extremal equality
  
  
  
  • 
    4.1 Statement
    
  
  
  • 
    4.2 Remarks and examples
    
  
  
  • 
    4.3 Applications
    
  
  
  
  


• 
  5 Generalization of Hölder's inequality
  
  
  
  • 
    5.1 Interpolation
    
  
  
  
  


• 
  6 Reverse Hölder inequality
  


• 
  7 Conditional Hölder inequality
  


• 
  8 Hölder's inequality for increasing seminorms
  


• 
  9 See also
  


• 
  10 Citations
  


• 
  11 References
  


• 
  12 External links
  

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 ||f ||_p and ||g||_q stand for the (possibly infinite) expressions

(∫S|f|pdμ)1p(∫S|g|qdμ)1q{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

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

|∫Sfg¯dμ|≤∫S|fg|dμ=‖fg‖1{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

|⟨f,g⟩|≤‖f‖2‖g‖2,{displaystyle |langle f,grangle |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 ||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

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

‖fg‖1≤‖f‖p‖g‖q{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

∑k=1n|xkyk|≤(∑k=1n|xk|p)1p(∑k=1n|yk|q)1q for all (x1,…,xn),(y1,…,yn)∈Rn or Cn.{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:

∑k=1∞|xkyk|≤(∑k=1∞|xk|p)1p(∑k=1∞|yk|q)1q for all (xk)k∈N,(yk)k∈N∈RN or CN.{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})_{kin mathbb {N} },(y_{k})_{kin 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

∫S|f(x)g(x)|dx≤(∫S|f(x)|pdx)1p(∫S|g(x)|qdx)1q.{displaystyle int _{S}{bigl |}f(x)g(x){bigr |},mathrm {d} xleq {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 (Ω,F,P),{displaystyle (Omega ,{mathcal {F}},mathbb {P} ),} let E{displaystyle mathbb {E} } denote the expectation operator. For real- or complex-valued random variables X{displaystyle X} and Y{displaystyle Y} on Ω,{displaystyle Omega ,} Hölder's inequality reads

E[|XY|]⩽(E[|X|p])1p(E[|Y|q])1q.{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 0<r<s{displaystyle 0<r<s} and define p=sr.{displaystyle p={tfrac {s}{r}}.} Then q=pp−1{displaystyle q={tfrac {p}{p-1}}} is the Hölder conjugate of p.{displaystyle p.} Applying Hölder's inequality to the random variables |X|r{displaystyle |X|^{r}} and 1Ω{displaystyle 1_{Omega }} we obtain

E[|X|r]⩽(E[|X|s])rs.{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

For two σ-finite measure spaces (S₁, Σ₁, μ₁) and (S₂, Σ₂, μ₂) define the product measure space by

S=S1×S2,Σ=Σ1⊗Σ2,μ=μ1⊗μ2,{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 σ-algebra Σ 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

∫S1∫S2|f(x,y)g(x,y)|μ2(dy)μ1(dx)≤(∫S1∫S2|f(x,y)|pμ2(dy)μ1(dx))1p(∫S1∫S2|g(x,y)|qμ2(dy)μ1(dx))1q.{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, ..., f_n) and g = (g₁, ..., 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

∫S∑k=1n|fk(x)gk(x)|μ(dx)≤(∫S∑k=1n|fk(x)|pμ(dx))1p(∫S∑k=1n|gk(x)|qμ(dx))1q.{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

α(|f1(x)|p,…,|fn(x)|p)=β(|g1(x)|q,…,|gn(x)|q),{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.

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 also assume p, q ∈ (1, ∞).

Dividing f  and g by ||f ||_p and ||g||_q, respectively, we can assume that

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

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

ab≤app+bqq{displaystyle ableq {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

|f(s)g(s)|≤|f(s)|pp+|g(s)|qq,s∈S.{displaystyle |f(s)g(s)|leq {frac {|f(s)|^{p}}{p}}+{frac {|g(s)|^{q}}{q}},qquad sin S.}

Integrating both sides gives

‖fg‖1≤‖f‖ppp+‖g‖qqq=1p+1q=1,{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

α=‖g‖qq,β=‖f‖pp,{displaystyle alpha =|g|_{q}^{q},qquad beta =|f|_{p}^{p},}

such that

α|f|p=β|g|q{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 (*).

Alternate proof using Jensen's inequality

Recall the Jensen's inequality for the convex function xp{displaystyle x^{p}} (it is convex because obviously p≥1{displaystyle pgeq 1}):

∫hdν≤(∫hpdν)1p{displaystyle int hdnu leq left(int h^{p}dnu 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 gq{displaystyle g^{q}}, i.e.

dν=gq∫gqdμdμ{displaystyle dnu ={frac {g^{q}}{int g^{q},dmu }}dmu }

Hence we have, using 1p+1q=1{displaystyle {frac {1}{p}}+{frac {1}{q}}=1}, hence p(1−q)+q=0{displaystyle p(1-q)+q=0}, and letting h=fg1−q{displaystyle h=fg^{1-q}},

∫fgdμ=(∫gqdμ)∫fg1−q⏟hgq∫gqdμdμ⏟dν≤(∫gqdμ)(∫fpgp(1−q)⏟hpgq∫gqdμdμ⏟dν)1p=(∫gqdμ)(∫fp∫gqdμdμ)1p{displaystyle int fg,dmu =left(int g^{q},dmu right)int underbrace {fg^{1-q}} _{h}underbrace {{frac {g^{q}}{int g^{q},dmu }}dmu } _{dnu }leq left(int g^{q}dmu right)left(int underbrace {f^{p}g^{p(1-q)}} _{h^{p}}underbrace {{frac {g^{q}}{int g^{q},dmu }},dmu } _{dnu }right)^{frac {1}{p}}=left(int g^{q},dmu right)left(int {frac {f^{p}}{int g^{q},dmu }},dmu right)^{frac {1}{p}}}

Finally, we get

∫fgdμ≤(∫fpdμ)1p(∫gqdμ)1q{displaystyle int fg,dmu leq left(int f^{p},dmu right)^{frac {1}{p}}left(int g^{q},dmu right)^{frac {1}{q}}}

This assumes f,g{displaystyle f,g} real and non negative, but the extension to complex functions is straightforward (use the modulus of f,g{displaystyle f,g}). It also assumes that ‖f‖p,‖g‖q{displaystyle |f|_{p},|g|_{q}} are neither null nor infinity, and that p,q>1{displaystyle 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(μ),

‖f‖p=max{|∫Sfgdμ|:g∈Lq(μ),‖g‖q≤1},{displaystyle |f|_{p}=max left{left|int _{S}fg,mathrm {d} mu right|:gin L^{q}(mu ),|g|_{q}leq 1right},}

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

‖f‖∞=sup{|∫Sfgdμ|:g∈L1(μ),‖g‖1≤1}.{displaystyle |f|_{infty }=sup left{left|int _{S}fg,mathrm {d} mu right|:gin L^{1}(mu ),|g|_{1}leq 1right}.}

Proof of the extremal equality

By Hölder's inequality, the integrals are well defined and, for 1 ≤ p ≤ ∞,

|∫Sfgdμ|≤∫S|fg|dμ≤‖f‖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

g(x)={‖f‖p1−p|f(x)|p/f(x)if f(x)≠0,0otherwise.{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

∫Sfgdμ=‖f‖p.{displaystyle int _{S}fg,mathrm {d} mu =|f|_{p}.}

It remains to consider the case p = ∞. For ε ∈ (0, 1) define

A={x∈S:|f(x)|>(1−ε)‖f‖∞}.{displaystyle A=left{xin 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

g(x)={1−εμ(B)‖f‖∞f(x)if x∈B,0otherwise.{displaystyle g(x)={begin{cases}{frac {1-varepsilon }{mu (B)}}{frac {|f|_{infty }}{f(x)}}&{text{if }}xin 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,

|∫Sfgdμ|=∫B1−εμ(B)‖f‖∞dμ=(1−ε)‖f‖∞.{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 p=∞{displaystyle p=infty } fails whenever there exists a set A{displaystyle A} of infinite measure in the σ{displaystyle sigma }-field Σ{displaystyle Sigma } with that has no subset B∈Σ{displaystyle Bin Sigma } that satisfies: 0<μ(B)<∞.{displaystyle 0<mu (B)<infty .} (the simplest example is the σ{displaystyle sigma }-field Σ{displaystyle Sigma } containing just the empty set and S,{displaystyle S,} and the measure μ{displaystyle mu } with μ(S)=∞.{displaystyle mu (S)=infty .}) Then the indicator function 1A{displaystyle 1_{A}} satisfies ‖1A‖∞=1,{displaystyle |1_{A}|_{infty }=1,} but every g∈L1(μ){displaystyle gin L^{1}(mu )} has to be μ{displaystyle mu }-almost everywhere constant on A,{displaystyle A,} because it is Σ{displaystyle Sigma }-measurable, and this constant has to be zero, because g{displaystyle g} is μ{displaystyle mu }-integrable. Therefore, the above supremum for the indicator function 1A{displaystyle 1_{A}} is zero and the extremal equality fails.


• For p=∞,{displaystyle p=infty ,} the supremum is in general not attained. As an example, let S=N,Σ=P(N){displaystyle S=mathbb {N} ,Sigma ={mathcal {P}}(mathbb {N} )} and μ{displaystyle mu } the counting measure. Define:

{f:N→Rf(n)=n−1n{displaystyle {begin{cases}f:mathbb {N} to mathbb {R} \f(n)={frac {n-1}{n}}end{cases}}}

Then ‖f‖∞=1.{displaystyle |f|_{infty }=1.} For g∈L1(μ,N){displaystyle gin L^{1}(mu ,mathbb {N} )} with 0<‖g‖1⩽1,{displaystyle 0<|g|_{1}leqslant 1,} let m{displaystyle m} denote the smallest natural number with g(m)≠0.{displaystyle g(m)neq 0.} Then

|∫Sfgdμ|⩽m−1m|g(m)|+∑n=m+1∞|g(n)|=‖g‖1−|g(m)|m<1.{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 + f₂||_p ≤ ||f₁||_p + ||f₂||_p for all f₁ and f₂ 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

κf(g)=∫Sfgdμ,g∈Lq(μ).{displaystyle kappa _{f}(g)=int _{S}fg,mathrm {d} mu ,qquad gin 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

Assume that r ∈ (0, ∞] and p₁, …, p_n ∈ (0, ∞] such that

∑k=1n1pk=1r{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, …, f_n defined on S,

‖∏k=1nfk‖r≤∏k=1n‖fk‖pk{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,

fk∈Lpk(μ)∀k∈{1,…,n}⟹∏k=1nfk∈Lr(μ).{displaystyle f_{k}in L^{p_{k}}(mu );;forall kin {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.

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 p₁ ≤ … ≤ p_n.

Case 1: If p_n = ∞, then

∑k=1n−11pk=1r.{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

‖f1⋯fn‖r≤‖f1⋯fn−1‖r‖fn‖∞≤‖f1‖p1⋯‖fn−1‖pn−1‖fn‖∞.{displaystyle {begin{aligned}left|f_{1}cdots f_{n}right|_{r}&leq |f_{1}cdots f_{n-1}|_{r}|f_{n}|_{infty }\&leq |f_{1}|_{p_{1}}cdots |f_{n-1}|_{p_{n-1}}|f_{n}|_{infty }.end{aligned}}}

Case 2: If p_n < ∞, then necessarily r < ∞ as well, and then

p:=pnpn−r,q:=pnr{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

‖|f1⋯fn−1|r|fn|r‖1≤‖|f1⋯fn−1|r‖p‖|fn|r‖q.{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 1/r and rewriting,

‖f1⋯fn‖r≤‖f1⋯fn−1‖pr‖fn‖qr.{displaystyle |f_{1}cdots f_{n}|_{r}leq |f_{1}cdots f_{n-1}|_{pr}|f_{n}|_{qr}.}

Since qr = p_n and

∑k=1n−11pk=1r−1pn=pn−rrpn=1pr,{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, ..., p_n ∈ (0, ∞] and let θ₁, ..., θ_n ∈ (0, 1) denote weights with θ₁ + ... + θ_n = 1. Define p as the weighted harmonic mean, i.e.,

1p=∑k=1nθkpk.{displaystyle {frac {1}{p}}=sum _{k=1}^{n}{frac {theta _{k}}{p_{k}}}.}

Given measurable real- or complex-valued functions fk{displaystyle f_{k}} on S, then the above generalization of Hölder's inequality gives

‖|f1|θ1⋯|fn|θn‖p≤‖|f1|θ1‖p1θ1⋯‖|fn|θn‖pnθn=‖f1‖p1θ1⋯‖fn‖pnθn.{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 f1=⋯=fn=:f{displaystyle f_{1}=cdots =f_{n}=:f} gives

‖f‖p⩽∏k=1n‖f‖pkθk.{displaystyle |f|_{p}leqslant prod _{k=1}^{n}|f|_{p_{k}}^{theta _{k}}.}

Specifying further θ₁ = θ and θ₂ = 1-θ, in the case n = 2, we obtain the interpolation result (Littlewood's inequality)

‖f‖pθ⩽‖f‖p1θ⋅‖f‖p01−θ,{displaystyle |f|_{p_{theta }}leqslant |f|_{p_{1}}^{theta }cdot |f|_{p_{0}}^{1-theta },}

for θ∈(0,1){displaystyle theta in (0,1)} and

1pθ=θp1+1−θp0.{displaystyle {frac {1}{p_{theta }}}={frac {theta }{p_{1}}}+{frac {1-theta }{p_{0}}}.}

An application of Hölder gives Lyapunov's inequality: If

p=(1−θ)p0+θp1,θ∈(0,1),{displaystyle p=(1-theta )p_{0}+theta p_{1},qquad theta in (0,1),}

then

‖|f0|p0(1−θ)p⋅|f1|p1θp‖pp≤‖f0‖p0p0(1−θ)‖f1‖p1p1θ{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

‖f‖pp⩽‖f‖p0p0(1−θ)⋅‖f‖p1p1θ.{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 f∈Lp0∩Lp1,{displaystyle fin L^{p_{0}}cap L^{p_{1}},} then f∈Lp{displaystyle fin L^{p}} for all p0<p<p1.{displaystyle p_{0}<p<p_{1}.}

Reverse Hölder inequality

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,

‖fg‖1⩾‖f‖1p‖g‖−1p−1.{displaystyle |fg|_{1}geqslant |f|_{frac {1}{p}},|g|_{frac {-1}{p-1}}.}

If

‖fg‖1<∞and‖g‖−1p−1>0,{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

∃α⩾0|f|=α|g|−pp−1μ-almost everywhere.{displaystyle exists alpha geqslant 0quad |f|=alpha |g|^{frac {-p}{p-1}}qquad mu {text{-almost everywhere}}.}

Note: The expressions:

‖f‖1pand‖g‖−1p−1,{displaystyle |f|_{frac {1}{p}}quad {text{and}}quad |g|_{frac {-1}{p-1}},}

are not norms, they are just compact notations for

(∫S|f|1pdμ)pand(∫S|g|−1p−1dμ)−(p−1).{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

q:=pp−1∈(1,∞){displaystyle q:={frac {p}{p-1}}in (1,infty )}

are Hölder conjugates. Application of Hölder's inequality gives

‖|f|1p‖1=‖|fg|1p|g|−1p‖1⩽‖|fg|1p‖p‖|g|−1p‖q=‖fg‖11p‖|g|−1p−1‖1p−1p{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:

‖|f|1p‖1p⩽‖fg‖1‖|g|−1p−1‖1p−1.{displaystyle left||f|^{frac {1}{p}}right|_{1}^{p}leqslant |fg|_{1}left||g|^{frac {-1}{p-1}}right|_{1}^{p-1}.}

Therefore:

‖|f|1p‖1p‖|g|−1p−1‖1−(p−1)⩽‖fg‖1.{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

Let (Ω, F, 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 on Ω,

E[|XY||G]≤(E[|X|p|G])1p(E[|Y|q|G])1qP-almost surely.{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

E[Z|G]=supn∈NE[min{Z,n}|G]a.s.{displaystyle mathbb {E} [Z|{mathcal {G}}]=sup _{nin 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

U=(E[|X|p|G])1p,V=(E[|Y|q|G])1q{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

E[|X|p1{U=0}]=E[1{U=0}E[|X|p|G]⏟=Up]=0,{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

E[|XY||G]=0a.s. on {U=0}∪{V=0}{displaystyle mathbb {E} {bigl [}|XY|{big |},{mathcal {G}}{bigr ]}=0qquad {text{a.s. on }}{U=0}cup {V=0}}

and the conditional Hölder inequality holds on this set. On the set

{U=∞,V>0}∪{U>0,V=∞}{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

E[|XY||G]UV≤1a.s. on the set H:={0<U<∞,0<V<∞}.{displaystyle {frac {mathbb {E} {bigl [}|XY|{big |},{mathcal {G}}{bigr ]}}{UV}}leq 1qquad {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∈G,G⊂H.{displaystyle Gin {mathcal {G}},quad Gsubset 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

E[E[|XY||G]UV1G]=E[E[|XY|UV1G|G]]=E[|X|U1G⋅|Y|V1G]≤(E[|X|pUp1G])1p(E[|Y|qVq1G])1q=(E[E[|X|p|G]Up⏟=1 a.s. on G1G])1p(E[E[|Y|q|G]Vp⏟=1 a.s. on G1G])1q=E[1G].{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}}}

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Hölder's inequality for increasing seminorms

Let S be a set and let F(S,C){displaystyle F(S,mathbb {C} )} be the space of all complex-valued functions on S. Let N be an increasing seminorm on F(S,C),{displaystyle F(S,mathbb {C} ),} meaning that, for all real-valued functions f,g∈F(S,C){displaystyle f,gin F(S,mathbb {C} )} we have the following implication (the seminorm is also allowed to attain the value ∞):

∀s∈Sf(s)⩾g(s)⩾0⇒N(f)⩾N(g).{displaystyle forall sin Squad f(s)geqslant g(s)geqslant 0qquad Rightarrow qquad N(f)geqslant N(g).}

Then:

∀f,g∈F(S,C)N(|fg|)⩽(N(|f|p))1p(N(|g|q))1q,{displaystyle forall f,gin 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 p{displaystyle p} and q{displaystyle q} are Hölder conjugates.^[1]

Remark: If (S, Σ, μ) is a measure space and N(f){displaystyle N(f)} is the upper Lebesgue integral of |f|{displaystyle |f|} then the restriction of N to all Σ-measurable functions gives the usual version of Hölder's inequality.

See also

• Cauchy–Schwarz inequality


• Minkowski inequality


• Jensen's inequality


• Young's inequality for products


• Clarkson's inequalities

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.mw-parser-output cite.citation{font-style:inherit}.mw-parser-output q{quotes:"""""""'""'"}.mw-parser-output code.cs1-code{color:inherit;background:inherit;border:inherit;padding:inherit}.mw-parser-output .cs1-lock-free a{background:url("//upload.wikimedia.org/wikipedia/commons/thumb/6/65/Lock-green.svg/9px-Lock-green.svg.png")no-repeat;background-position:right .1em center}.mw-parser-output .cs1-lock-limited a,.mw-parser-output .cs1-lock-registration a{background:url("//upload.wikimedia.org/wikipedia/commons/thumb/d/d6/Lock-gray-alt-2.svg/9px-Lock-gray-alt-2.svg.png")no-repeat;background-position:right .1em center}.mw-parser-output .cs1-lock-subscription a{background:url("//upload.wikimedia.org/wikipedia/commons/thumb/a/aa/Lock-red-alt-2.svg/9px-Lock-red-alt-2.svg.png")no-repeat;background-position:right .1em center}.mw-parser-output .cs1-subscription,.mw-parser-output .cs1-registration{color:#555}.mw-parser-output .cs1-subscription span,.mw-parser-output .cs1-registration span{border-bottom:1px dotted;cursor:help}.mw-parser-output .cs1-hidden-error{display:none;font-size:100%}.mw-parser-output .cs1-visible-error{font-size:100%}.mw-parser-output .cs1-subscription,.mw-parser-output .cs1-registration,.mw-parser-output .cs1-format{font-size:95%}.mw-parser-output .cs1-kern-left,.mw-parser-output .cs1-kern-wl-left{padding-left:0.2em}.mw-parser-output .cs1-kern-right,.mw-parser-output .cs1-kern-wl-right{padding-right:0.2em}
  


• 
  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, 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.

External links

• 
  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).


• 
  Tisdell, Chris (2012), Holder's Inequality, Online video on Dr Chris Tisdell's YouTube channel.

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