integration

# Contents

## Idea

The term ‘Lebesgue space’ can stand for two distinct notions: one is the general notion of measure space (compare the Springer Encyclopaedia of Mathematics) and another is the notion of ${L}^{p}$ space (or ${L}_{p}$ space). Here we discuss the latter.

Sometimes ‘${L}_{p}$’ is used as a synonym for ${L}^{p}$; sometimes it is used to mean ${L}^{1/p}$.

## Definitions

If $1\le p<\infty$ is a real number and $\left(\Omega ,\mu \right)$ is a measure space, one considers the ${L}^{p}$ space ${L}_{p}\left(\Omega \right)$ of (equivalence classes of) measurable (complex- or real-valued) functions $f:\Omega \to 𝕂$ whose (absolute values of) $p$th powers are integrable; i.e. whose norm

$\parallel f{\parallel }_{p}={\left({\int }_{\Omega }\mid f{\mid }^{p}\phantom{\rule{thinmathspace}{0ex}}d\mu \right)}^{1/p}${\|f\|_p} = \left(\int_\Omega {|f|^p} \,d\mu\right)^{1/p}

is finite.

The ${L}^{p}$ spaces are examples of Banach spaces; they are continuous analogues of ${l}^{p}$ spaces of $p$-summable series. (Indeed, ${l}^{p}\left(S\right)$, for $S$ a set, is simply ${L}^{p}\left(S\right)$ if $S$ is equipped with counting measure?.)

For fixed $f$, the norm $\parallel f{\parallel }_{p}$ is continuous in $p$. Accordingly, for $p=\infty$, one may take the limit of ${\parallel f\parallel }_{p}$ as $p\to \infty$. However, this turns out to be the same as the essential supremum norm $\parallel f{\parallel }_{\infty }$. Therefore, ${L}^{\infty }\left(\Omega \right)$ makes sense as long as $\Omega$ is a measurable space equipped with a family of null sets (or full sets); the measure $\mu$ is otherwise irrelevant.

For $0, we can define the ${L}^{p}$ space using the same formula; however, it is no longer a Banach space (but still an F-space?). For $p=0$, we can again take the limit, or equivalently use the essential infimum; every measurable function on $\Omega$ belongs to ${L}^{0}$ (which is no longer even an $F$-space).

## Minkowski’s inequality

We offer here a proof that $\parallel f{\parallel }_{p}$ defines a norm in the case $1; the cases $p=1$ and $p=\infty$ follow by continuity and are easy to check from first principles. The most usual textbook proofs involve a clever application of Hölder's inequality?; the following proof is more straightforwardly geometric. All functions $f$ may be assumed to be real- or complex-valued.

###### Theorem

Suppose $1\le p\le \infty$, and suppose $\Omega$ is a measure space with measure $\mu$. Then the function $\mid \left(-\right){\mid }_{p}:{L}^{p}\left(\Omega ,\mu \right)\to ℝ$ defined by

$\parallel f{\parallel }_{p}≔\left({\int }_{\Omega }\mid f{\mid }^{p}\phantom{\rule{thinmathspace}{0ex}}d\mu {\right)}^{1/p}${\|f\|_p} \coloneqq (\int_\Omega {|f|^p} \,d\mu)^{1/p}

defines a norm.

One must verify three things:

1. Separation axiom: $\parallel f{\parallel }_{p}=0$ implies $f=0$.

2. Scaling axiom: ${\parallel tf\parallel }_{p}=\mid t\mid \phantom{\rule{thinmathspace}{0ex}}\parallel f{\parallel }_{p}$.

3. Triangle inequality: $\parallel f+g{\parallel }_{p}\le \parallel f{\parallel }_{p}+\parallel g{\parallel }_{p}$.

The first two properties are obvious, so it remains to prove the last, which is also called Minkowski’s inequality.

Our proof of Minkowski’s inequality is broken down into a series of simple lemmas. The plan is to boil it down to two things: the scaling axiom, and convexity of the function $x↦\mid x{\mid }^{p}$ (as a function from real or complex numbers to nonnegative real numbers).

First, some generalities. Let $V$ be a (real or complex) vector space equipped with a function $\parallel \left(-\right)\parallel :V\to \left[0,\infty \right]$ that satisfies the scaling axiom: $\parallel tv\parallel =\mid t\mid \phantom{\rule{thinmathspace}{0ex}}\parallel v\parallel$ for all scalars $t$, and the separation axiom: $\parallel v\parallel =0$ implies $v=0$. As usual, we define the unit ball? in $V$ to be $\left\{v\in V\phantom{\rule{thickmathspace}{0ex}}\mid \phantom{\rule{thickmathspace}{0ex}}\parallel v\parallel \le 1\right\}.$

###### Lemma

Given that the scaling and separation axioms hold, the following conditions are equivalent:

1. The triangle inequality is satisfied.
2. The unit ball is convex.
3. If $\parallel u\parallel =\parallel v\parallel =1$, then $\parallel tu+\left(1-t\right)v\parallel \le 1$ for all $t\in \left[0,1\right]$.
###### Proof

Condition 1. implies condition 2. easily: if $u$ and $v$ are in the unit ball and $0\le t\le 1$, we have

$\begin{array}{ccc}\parallel tu+\left(1-t\right)v\parallel & \le & \parallel tu\parallel +\parallel \left(1-t\right)v\parallel \\ & =& t\parallel u\parallel +\left(1-t\right)\parallel v\parallel \\ & \le & t+\left(1-t\right)=1.\end{array}$\array{ {\|t u + (1-t)v\|} & \leq & {\|t u\|} + {\|(1-t)v\|} \\ & = & t {\|u\|} + (1-t) {\|v\|} \\ & \leq & t + (1-t) = 1.}

Now 2. implies 3. trivially, so it remains to prove that 3. implies 1. Suppose $\parallel v\parallel ,\parallel v\prime \parallel \in \left(0,\infty \right)$. Let $u=\frac{v}{\parallel v\parallel }$ and $u\prime =\frac{v\prime }{\parallel v\prime \parallel }$ be the associated unit vectors. Then

$\begin{array}{ccc}\frac{v+v\prime }{\parallel v\parallel +\parallel v\prime \parallel }& =& \left(\frac{\parallel v\parallel }{\parallel v\parallel +\parallel v\prime \parallel }\right)\frac{v}{\parallel v\parallel }+\left(\frac{\parallel v\prime \parallel }{\parallel v\parallel +\parallel v\prime \parallel }\right)\frac{v\prime }{\parallel v\prime \parallel }\\ & =& tu+\left(1-t\right)u\prime \end{array}$\array{ \frac{v + v'}{{\|v\|}+{\|v'\|}} & = & (\frac{{\|v\|}}{{\|v\|}+{\|v'\|}})\frac{v}{{\|v\|}} + (\frac{{\|v'\|}}{{\|v\|}+{\|v'\|}})\frac{v'}{{\|v'\|}} \\ & = & t u + (1-t)u'}

where $t=\frac{\parallel v\parallel }{\parallel v\parallel +\parallel v\prime \parallel }$. If condition 3. holds, then

$\parallel tu+\left(1-t\right)u\prime \parallel \le 1${\|t u + (1-t)u'\|} \leq 1

but by the scaling axiom, this is the same as saying

$\frac{\parallel v+v\prime \parallel }{\parallel v\parallel +\parallel v\prime \parallel }\le 1,$\frac{{\|v + v'\|}}{{\|v\|} + {\|v'\|}} \leq 1,

which is the triangle inequality.

Consider now ${L}^{p}$ with its $p$-norm $\parallel f\parallel =\mid f{\mid }_{p}$. By Lemma 1, this inequality is equivalent to

• Condition 4: If $\mid u{\mid }_{p}^{p}=1$ and $\mid v{\mid }_{p}^{p}=1$, then $\mid tu+\left(1-t\right)v{\mid }_{p}^{p}\le 1$ whenever $0\le t\le 1$.

This allows us to remove the cumbersome exponent $1/p$ in the definition of the $p$-norm.

The next two lemmas may be proven by elementary calculus; we omit the proofs. (But you can also see the full details.)

###### Lemma

Let $\alpha ,\beta$ be two complex numbers, and define

$\gamma \left(t\right)=\mid \alpha +\beta t{\mid }^{p}$\gamma(t) = {|\alpha + \beta t|^p}

for real $t$. Then $\gamma ″\left(t\right)$ is nonnegative.

###### Lemma

Define $\varphi :ℂ\to ℝ$ by $\varphi \left(x\right)=\mid x{\mid }^{p}$. Then $\varphi$ is convex, i.e., for all $x,y$,

$\mid tx+\left(1-t\right)y{\mid }^{p}\le t\mid x{\mid }^{p}+\left(1-t\right)\mid y{\mid }^{p}${|t x + (1-t)y|^p} \leq t{|x|^p} + (1-t){|y|^p}

for all $t\in \left[0,1\right]$.

###### Proof of Minkowski’s inequality

Let $u$ and $v$ be unit vectors in ${L}^{p}$. By condition 4, it suffices to show that $\mid tu+\left(1-t\right)v{\mid }_{p}\le 1$ for all $t\in \left[0,1\right]$. But

${\int }_{\Omega }\mid tu+\left(1-t\right)v{\mid }^{p}\phantom{\rule{thinmathspace}{0ex}}d\mu \le {\int }_{\Omega }t{\mid u\mid }^{p}+\left(1-t\right){\mid v\mid }^{p}\phantom{\rule{thinmathspace}{0ex}}d\mu$\int_\Omega {|t u + (1-t)v|^p} \,d\mu \leq \int_\Omega t{|u|}^p + (1-t){|v|}^p \,d\mu

by Lemma 3. Using $\int \mid u{\mid }^{p}=1=\int \mid v{\mid }^{p}$, we are done.

## References

• W. Rudin, Functional analysis, McGraw Hill 1991.

• L. C. Evans, Partial differential equations, Amer. Math. Soc. 1998.

• Wikipedia (English): Lp space

category: analysis

Revised on July 14, 2013 02:09:42 by Urs Schreiber (89.204.139.156)