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

${\|f\|_p} = \left(\int_\Omega {|f|^p} \,d\mu\right)^{1/p}$

is finite.

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

For fixed $f$, the norm ${\|f\|_p}$ is continuous in $p$. Accordingly, for $p = \infty$, one may take the limit of ${\|f\|}_p$ as $p \to \infty$. However, this turns out to be the same as the essential supremum norm $\|f\|_\infty$. Therefore, $L^\infty(\Omega)$ 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 $0 \lt p \lt 1$, 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 ${\|f\|_p}$ defines a norm in the case $1 \lt p \lt \infty$; 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 $1 \leq p \leq \infty$, and suppose $\Omega$ is a measure space with measure $\mu$. Then the function ${|(-)|_p}\colon L^p(\Omega, \mu) \to \mathbb{R}$ defined by

${\|f\|_p} \coloneqq (\int_\Omega {|f|^p} \,d\mu)^{1/p}$

defines a norm.

One must verify three things:

1. Separation axiom: ${\|f\|_p} = 0$ implies $f = 0$.

2. Scaling axiom: ${\|t f\|}_p = {|t|} \, {\|f\|_p}$.

3. Triangle inequality: ${\|f + g\|_p} \leq {\|f\|_p} + {\|g\|_p}$.

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 $x \mapsto {|x|^p}$ (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 ${\|(-)\|}\colon V \to [0, \infty]$ that satisfies the scaling axiom: ${\|t v\|} = {|t|} \, {\|v\|}$ for all scalars $t$, and the separation axiom: ${\|v\|} = 0$ implies $v = 0$. As usual, we define the unit ball? in $V$ to be $\{v \in V \;|\; {\|v\|} \leq 1\}.$

###### 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 ${\|u\|} = {\|v\|} = 1$, then ${\|t u + (1-t)v\|} \leq 1$ for all $t \in [0, 1]$.

###### Proof

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

$\array{ {\|t u + (1-t)v\|} & \leq & {\|t u\|} + {\|(1-t)v\|} \\ & = & t {\|u\|} + (1-t) {\|v\|} \\ & \leq & t + (1-t) = 1.}$

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

$\array{ \frac{v + v'}{{\|v\|}+{\|v'\|}} & = & (\frac{{\|v\|}}{{\|v\|}+{\|v'\|}})\frac{v}{{\|v\|}} + (\frac{{\|v'\|}}{{\|v\|}+{\|v'\|}})\frac{v'}{{\|v'\|}} \\ & = & t u + (1-t)u'}$

where $t = \frac{{\|v\|}}{{\|v\|} + {\|v'\|}}$. If condition 3. holds, then

${\|t u + (1-t)u'\|} \leq 1$

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

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

which is the triangle inequality.

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

• Condition 4: If ${|u|_{p}^{p}} = 1$ and ${|v|_{p}^{p}} = 1$, then ${|t u + (1-t)v|_{p}^{p}} \leq 1$ whenever $0 \leq t \leq 1$.

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

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

for real $t$. Then $\gamma''(t)$ is nonnegative.

###### Lemma

Define $\phi\colon \mathbb{C} \to \mathbb{R}$ by $\phi(x) = |x|^p$. Then $\phi$ is convex, i.e., for all $x, y$,

${|t x + (1-t)y|^p} \leq t{|x|^p} + (1-t){|y|^p}$

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

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

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

$\int_\Omega {|t u + (1-t)v|^p} \,d\mu \leq \int_\Omega t{|u|}^p + (1-t){|v|}^p \,d\mu$

by Lemma 3. Using $\int {|u|^p} = 1 = \int {|v|^p}$, 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)