Holder's inequality

In mathematical analysis , Hölder's inequality, together with Minkowski's inequality and Jensen's inequality, are among the fundamental inequalities for L ^p spaces . It was first proven by Leonard James Rogers in 1888; it is named after Otto Hölder , who published it a year later.

statement

Hölder's inequality

Let a dimension space and measurable functions be given ${\displaystyle (X,{\mathcal {A}},\mu )}$

${\displaystyle f,g\colon X\to {\overline {\mathbb {R} }}}$

For and with the convention one defines ${\displaystyle p\in [1,\infty )}$ ${\displaystyle \infty ^{p}=\infty ^{\frac {1}{p}}=\infty }$

${\displaystyle H_{p}(f)=\left(\int _{X}|f|^{p}\mathrm {d} \mu \right)^{\tfrac {1}{p}}}$

and

${\displaystyle H_{\infty }(f)=\mathrm {ess} \sup _{x\in X}|f(x)|}$

the essential supremum . The Hölder inequality then reads: for with , where is agreed, applies ${\displaystyle 1\leq p,q\leq \infty }$ ${\displaystyle {\tfrac {1}{p}}+{\tfrac {1}{q}}=1}$ ${\displaystyle {\tfrac {1}{\infty }}=0}$

${\displaystyle H_{1}(fg)\leq H_{p}(f)\cdot H_{q}(g)}$

It is called the Hölder exponent to be conjugated . The inequality is formulated more specifically as follows: If the space of -fold Lebesgue integrable functions (see Lp space ) and is the Lp norm , then applies forever ${\displaystyle q}$ ${\displaystyle p}$ ${\displaystyle {\mathcal {L}}^{p}(X,{\mathcal {A}},\mu )}$ ${\displaystyle p}$ ${\displaystyle \|\cdot \|_{p}}$ ${\displaystyle f\in {\mathcal {L}}^{p}(X,{\mathcal {A}},\mu ),g\in {\mathcal {L}}^{q}(X,{\mathcal {A}},\mu )}$

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

Special cases

Schwarz's inequality

If you choose a measurement space , i.e. a real interval with the Lebesgue measure and two functions , then the Hölder inequality reads with ${\displaystyle ([a,b],{\mathcal {B}}([a,b]),\lambda )}$ ${\displaystyle f,g\in {\mathcal {L}}^{2}([a,b],{\mathcal {B}}([a,b]),\lambda )}$ ${\displaystyle p=q=2}$

${\displaystyle \int _{a}^{b}|fg|\mathrm {d} \lambda \leq \left(\int _{a}^{b}|f|^{2}\mathrm {d} \lambda \right)^{\tfrac {1}{2}}\cdot \left(\int _{a}^{b}|g|^{2}\mathrm {d} \lambda \right)^{\tfrac {1}{2}}}$

This is exactly the Schwarz inequality or the integral formulation of the Cauchy-Schwarz inequality.

Cauchy's inequality

If one chooses the finite set as the measure space , provided with the power set and equipped with the counting measure , the inequality is obtained as a special case ${\displaystyle \{1,\ldots ,n\}}$

${\displaystyle \sum _{k=1}^{n}|x_{k}y_{k}|\leq \left(\sum _{k=1}^{n}|x_{k}|^{p}\right)^{1/p}\left(\sum _{k=1}^{n}|y_{k}|^{q}\right)^{1/q},}$

valid for all real (or complex) numbers . For one obtains the Cauchy inequality (or the discrete formulation of the Cauchy-Schwarz inequality) ${\displaystyle x_{1},\ldots ,x_{n},y_{1},\ldots ,y_{n}}$ ${\displaystyle p=q=2}$

${\displaystyle |\langle x,y\rangle |\leq \|x\|_{2}\cdot \|y\|_{2}}$

Hölder's inequality for series

If one chooses the natural numbers as the basic set of the measure space , again provided with the power set and the counting measure, then one obtains Hölder's inequality for series ${\displaystyle \mathbb {N} }$

${\displaystyle \sum _{k=1}^{\infty }|a_{k}b_{k}|\leq \left(\sum _{k=1}^{\infty }|a_{k}|^{p}\right)^{1/p}\left(\sum _{k=1}^{\infty }|b_{k}|^{q}\right)^{1/q}}$.

for real or complex sequences . In the borderline case this corresponds ${\displaystyle (a_{k})_{k\in \mathbb {N} },(b_{k})_{k\in \mathbb {N} }}$ ${\displaystyle q=\infty }$

${\displaystyle \sum _{k=1}^{\infty }|a_{k}b_{k}|\leq \left(\sum _{k=1}^{\infty }|a_{k}|\right)\cdot \sup _{k\in \mathbb {N} }|b_{k}|}$.

generalization

There are as well and for all . ${\displaystyle p_{j}\in [1,\infty ],j=1,\ldots ,m}$ ${\displaystyle \textstyle {\frac {1}{r}}:=\sum _{j=1}^{m}{\frac {1}{p_{j}}}}$ ${\displaystyle f_{j}\in L^{p_{j}}(S)}$ ${\displaystyle j=1,\ldots ,m}$

Then follows

${\displaystyle \prod _{j=1}^{m}f_{j}\in L^{r}(S)}$

and the estimate applies

${\displaystyle \left\|\prod _{j=1}^{m}f_{j}\right\|_{r}\leq \prod _{j=1}^{m}\left\|f_{j}\right\|_{p_{j}}.}$

The corollary of this generalization is the following theorem.

If is a family of sequences of non-negative real numbers, and there are non-negative real numbers with , then ${\displaystyle (a_{i,j})_{i\in \{1,2,\dots ,n\},j\in \{1,2,\dots ,m\}}}$ ${\displaystyle m}$ ${\displaystyle (\lambda _{j})_{j\in \{1,\dots ,m\}}}$ ${\displaystyle \sum _{j=1}^{m}\lambda _{j}=1}$

${\displaystyle \sum _{i=1}^{n}\prod _{j=1}^{m}a_{i,j}^{\lambda _{j}}\leq \prod _{j=1}^{m}\left(\sum _{i=1}^{n}a_{i,j}\right)^{\lambda _{j}}.}$

Inverted Hölder's inequality

It is for almost everyone . ${\displaystyle g(x)\neq 0}$ ${\displaystyle x\in S}$

Then the reverse of Hölder's inequality applies to all of them ${\displaystyle r>1}$

${\displaystyle \int _{S}|f(x)g(x)|dx\geq \left(\int _{S}|f(x)|^{\frac {1}{r}}dx\right)^{r}\left(\int _{S}|g(x)|^{-{\frac {1}{r-1}}}dx\right)^{-(r-1)}.}$

proofs

Proof of Hölder's inequality

For (and vice versa) the statement of Hölder's inequality is trivial. We therefore assume that holds. Be and without restriction . According to Young's inequality : ${\displaystyle p=1,q=\infty }$ ${\displaystyle 1<p,q<\infty }$ ${\displaystyle \|f\|_{p}>0}$ ${\displaystyle \|g\|_{q}>0}$

${\displaystyle AB\leq {\frac {A^{p}}{p}}+{\frac {B^{q}}{q}}}$

for everyone . Use this specifically . Integration delivers ${\displaystyle A,B\geq 0}$ ${\displaystyle A:={\tfrac {|f(x)|}{\|f\|_{p}}},\,B:={\tfrac {|g(x)|}{\|g\|_{q}}}}$

${\displaystyle {\frac {1}{\|f\|_{p}\|g\|_{q}}}\int _{S}|fg|\mathrm {d} \mu \leq {\frac {1}{p}}+{\frac {1}{q}}=1,}$

what the Hölder's inequality implies.

Proof of generalization

The proof is by mathematical induction on out. The case is trivial. So be now and be without restriction . Then two cases have to be distinguished: ${\displaystyle m}$ ${\displaystyle m=1}$ ${\displaystyle m\geq 2}$ ${\displaystyle p_{1}\leq \cdots \leq p_{m}}$

Case 1: Then, by the induction hypothesis, then holds ${\displaystyle p_{m}=\infty .}$ ${\displaystyle \textstyle {\frac {1}{r}}=\sum _{j=1}^{m-1}{\frac {1}{p_{j}}}.}$

${\displaystyle \|f_{1}\cdots f_{m}\|_{r}\leq \|f_{m}\|_{\infty }\|f_{1}\cdots f_{m-1}\|_{r}\leq \|f_{m}\|_{\infty }\|f_{1}\|_{p_{1}}\cdots \|f_{m-1}\|_{p_{m-1}}.}$

Case 2: . According to the (usual) Hölder inequality for the exponents , ${\displaystyle p_{m}<\infty }$ ${\displaystyle {\tfrac {p_{m}}{p_{m}-r}},{\tfrac {p_{m}}{r}}}$

${\displaystyle \int _{S}|f_{1}\cdots f_{m-1}|^{r}|f_{m}|^{r}\mathrm {d} \mu \leq \left(\int _{S}|f_{1}\cdots f_{m-1}|^{\frac {rp_{m}}{p_{m}-r}}\mathrm {d} \mu \right)^{\frac {p_{m}-r}{p_{m}}}\left(\int _{S}|f_{m}|^{p_{m}}\mathrm {d} \mu \right)^{\frac {r}{p_{m}}},}$

so . Well is . The induction step thus results from the induction assumption. ${\displaystyle \textstyle \|f_{1}\cdots f_{m}\|_{r}\leq \|f_{1}\cdots f_{m-1}\|_{\tfrac {rp_{m}}{p_{m}-r}}\|f_{m}\|_{p_{m}}}$ ${\displaystyle \textstyle \sum _{j=1}^{m-1}{\frac {1}{p_{j}}}={\frac {1}{r}}-{\frac {1}{p_{m}}}={\frac {p_{m}-r}{rp_{m}}}}$

Proof of the reverse Hölder's inequality

The reverse Hölder inequality results from the (usual) Holder's inequality by as exponents and selected. This gives: ${\displaystyle p:=r}$ ${\displaystyle q:=r'={\tfrac {p}{p-1}}}$

${\displaystyle \int _{S}|f|^{\frac {1}{r}}\mathrm {d} \mu =\int _{S}\left(|fg|^{\frac {1}{r}}\cdot |g|^{-{\frac {1}{r}}}\mathrm {d} \mu \right)\leq \left(\int _{S}|fg|\mathrm {d} \mu \right)^{\frac {1}{r}}\left(\int _{S}|g|^{-{\frac {r'}{r}}}\mathrm {d} \mu \right)^{\frac {1}{r'}}.}$

Rearranging and exponentiating this inequality with yields the reversed Hölder inequality. ${\displaystyle r}$

Applications

Proof of the Minkowski inequality

With Hölder's inequality one can easily prove the Minkowski inequality (that is the triangle inequality im ). ${\displaystyle L^{p}}$

Interpolation inequality for Lebesgue functions

Let be and , then it follows and the interpolation inequality applies ${\displaystyle f\in L^{p}(S)\cap L^{q}(S)}$ ${\displaystyle 1\leq q\leq r\leq p}$ ${\displaystyle f\in L^{r}(S)}$

${\displaystyle \|f\|_{r}\leq \|f\|_{p}^{1-\theta }\|f\|_{q}^{\theta }}$

with or for . ${\displaystyle {\tfrac {1}{r}}=:{\tfrac {1-\theta }{p}}+{\tfrac {\theta }{q}}}$ ${\displaystyle \theta :={\tfrac {q}{r}}{\tfrac {p-r}{p-q}}}$ ${\displaystyle q\neq p}$

Proof: be without restriction . Fix with . Note that and are the conjugate Hölder exponents. It follows from Hölder's inequality ${\displaystyle q<r<p}$ ${\displaystyle t\in (0,1)}$ ${\displaystyle r=tp+(1-t)q}$ ${\displaystyle {\tfrac {1}{t}}}$ ${\displaystyle {\tfrac {1}{1-t}}}$

${\displaystyle \int _{S}|f|^{r}\mathrm {d} \mu =\int _{S}|f|^{tp}|f|^{(1-t)q}\mathrm {d} \mu \leq \left(\int _{S}|f|^{p}\mathrm {d} \mu \right)^{t}\left(\int _{S}|f|^{q}\mathrm {d} \mu \right)^{1-t}}$.

Raising the inequality by and calculating the exponents implies the interpolation inequality . ${\displaystyle {\tfrac {1}{r}}}$

Proof of Young's convolution inequality

Another typical application is the proof of the generalized Young's inequality (for convolution integrals )

${\displaystyle \|f\star g\|_{r}\leq \|f\|_{p}\|g\|_{q}}$

for and . ${\displaystyle {\tfrac {1}{p}}+{\tfrac {1}{q}}=1+{\tfrac {1}{r}}}$ ${\displaystyle p,q,r\geq 1}$

literature

• Herbert Amann, Joachim Escher : Analysis III . 1st edition. Birkhäuser-Verlag, Basel / Boston / Berlin 2001, ISBN 3-7643-6613-3 .
• Jürgen Elstrodt : Measure and integration theory . 6th, corrected edition. Springer-Verlag, Berlin Heidelberg 2009, ISBN 978-3-540-89727-9 , doi : 10.1007 / 978-3-540-89728-6 .
• Achim Klenke : Probability Theory . 3. Edition. Springer-Verlag, Berlin Heidelberg 2013, ISBN 978-3-642-36017-6 , doi : 10.1007 / 978-3-642-36018-3 .

Individual evidence

1. ↑ Elstrodt: Measure and Integration Theory. 2009, p. 277.