nLab
Lebesgue space

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 $1\le p<\mathrm{\infty }$ 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 $f:\Omega \to 𝕂$ whose (absolute values of) $p$th powers are integrable; i.e. whose norm

$$\parallel f{\parallel }_p={\left({\int }_{\Omega }\mid f{\mid }^{p}d\mu \right)}^{1/p}$${\|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, $l^p(S)$, for $S$ a set, is simply $L^p(S)$ if $S$ is equipped with counting measure?.)

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

$$\parallel f{\parallel }_p≔({\int }_{\Omega }\mid f{\mid }^{p}d\mu {)}^{1/p}$${\|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: ${\parallel tf\parallel }_p=\mid t\mid \parallel f{\parallel }_p$.

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 $x\mapsto \mid x{\mid }^p$ (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 $\parallel (-)\parallel :V\to [0,\mathrm{\infty }]$ that satisfies the scaling axiom: $\parallel tv\parallel =\mid t\mid \parallel v\parallel$ 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 $\{v\in V\mid \parallel v\parallel \le 1\}.$

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 $\parallel u\parallel =\parallel v\parallel =1$, then $\parallel tu+(1-t)v\parallel \le 1$ for all $t\in [0,1]$.

Proof

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

$$\begin{array}{ccc}\parallel tu+(1-t)v\parallel & \le & \parallel tu\parallel +\parallel (1-t)v\parallel \\ & =& t\parallel u\parallel +(1-t)\parallel v\parallel \\ & \le & t+(1-t)=1.\end{array}$$\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 $\parallel v\parallel ,\parallel v\prime \parallel \in (0,\mathrm{\infty })$. Let $u=\frac{v}{\parallel v\parallel }$ and $u\prime =\frac{v\prime }{\parallel v\prime \parallel }$ be the associated unit vectors. Then

$$\begin{array}{ccc}\frac{v+v\prime }{\parallel v\parallel +\parallel v\prime \parallel }& =& (\frac{\parallel v\parallel }{\parallel v\parallel +\parallel v\prime \parallel })\frac{v}{\parallel v\parallel }+(\frac{\parallel v\prime \parallel }{\parallel v\parallel +\parallel v\prime \parallel })\frac{v\prime }{\parallel v\prime \parallel }\\ & =& tu+(1-t)u\prime \end{array}$$\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

$$\parallel tu+(1-t)u\prime \parallel \le 1$${\|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 $\mid tu+(1-t)v{\mid }_p^p\le 1$ 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

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

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

Lemma

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

$$\mid tx+(1-t)y{\mid }^p\le t\mid x{\mid }^p+(1-t)\mid y{\mid }^p$${|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 $\mid tu+(1-t)v{\mid }_p\le 1$ for all $t\in [0,1]$. But

$${\int }_{\Omega }\mid tu+(1-t)v{\mid }^{p}d\mu \le {\int }_{\Omega }t{\mid u\mid }^p+(1-t){\mid v\mid }^{p}d\mu$$\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.

Related concepts

• canonical Hilbert space of half-densities

References

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

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

• Wikipedia (English): Lp space

category: analysis