nLab Lebesgue space

Redirected from "L^p space".
Contents

Context

Functional analysis

Integration theory

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 pL^p space (or L pL_p space). Here we discuss the latter.

Lebesgue spaces L pL^p in this sense are normed vector spaces of functions on a measure space, equipped with the suitable version of the p-norm.

Beware that sometimes the notation ‘L pL_p’ is used as a synonym for L pL^p; sometimes it is used to mean L 1/pL^{1/p}.

Notation

For historical reasons (starting with the original paper by Riesz), the exponent pp is traditionally taken to be the reciprocal of the “correct” exponent.

If we take M p=L 1/pM^p=L^{1/p}, the spaces M pM^p form a C\mathbf{C}-graded algebra, where C\mathbf{C} denotes complex numbers.

This is a conceptual explanation for the appearance of formulas like 1/p+1/q=1/r1/p+1/q=1/r in Hölder's inequality.

In differential geometry, the notion of density does use the “correct” grading.

In the Tomita–Takesaki theory, the parameter tt for modular automorphism group is almost the “correct” grading, except that it is multiplied by the imaginary unit ii.

Definitions

If 1p<1 \leq p \lt \infty is a real number and (Ω,μ)(\Omega,\mu) is a measure space, one considers the L pL^p space L p(Ω)L_p(\Omega), which is the vector space of equivalence classes of those measurable (complex- or real-valued) functions f:Ω𝕂f\colon \Omega \to \mathbb{K} whose (absolute values of) ppth powers are integrable, in that the integral

f p( Ω|f| pdμ) 1/p< {\|f\|_p} \;\coloneqq\; \left( \textstyle{\int}_\Omega {|f|^p} \,d\mu \right)^{1/p} \;\lt\; \infty

exists. Two such are taken to be equivalent, fgf \sim g, if fg p=0{\|f-g\|_p} = 0. For p=2p = 2 this is the space L 2L^2 of square integrable functions.

On these spaces L p(X)L^p(X) of equivalence classes of pp-power integrable functions, the function f p{\|f\|_p} satisfies the triangle inequality (due to Minkowski's inequality, see below) and hence defines a norm, the p-norm, making them normed vector spaces.

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

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

For 0p<10 \leq p \lt 1, one can modify the definition to make L pL^p into an F-space (but not a Banach space). See the definitions at p-norm.

Minkowski’s inequality

We offer here a proof that f p{\|f\|_p} indeed defines a norm in the case 1<p<1 \lt p \lt \infty, in that it satisfies the triangle inequality. This is usually known as Minkowski's inequality.

(The cases p=1p = 1 and p=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 ff may be assumed to be real- or complex-valued.

Theorem

Suppose 1p1 \leq p \leq \infty, and suppose Ω\Omega is a measure space with measure μ\mu. Then the function |()| p:L p(Ω,μ){|(-)|_p} \,\colon\, L^p(\Omega, \mu) \to \mathbb{R} defined by

f p( Ω|f| pdμ) 1/p {\|f\|_p} \;\coloneqq\; \big( \textstyle{\int}_\Omega {|f|^p} \,d\mu \big)^{1/p}

defines a norm.

One must verify three things:

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

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

  3. Triangle inequality: f+g pf p+g p{\|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|x| px \mapsto {|x|^p} (as a function from real or complex numbers to nonnegative real numbers).

First, some generalities. Let VV be a (real or complex) vector space equipped with a function ():V[0,]{\|(-)\|}\colon V \to [0, \infty] that satisfies the scaling axiom: tv=|t|v{\|t v\|} = {|t|} \, {\|v\|} for all scalars tt, and the separation axiom: v=0{\|v\|} = 0 implies v=0v = 0. As usual, we define the unit ball in VV to be {vV|v1}.\{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{\|u\|} = {\|v\|} = 1, then tu+(1t)v1{\|t u + (1-t)v\|} \leq 1 for all t[0,1]t \in [0, 1].
Proof

Condition 1. implies condition 2. easily: if uu and vv are in the unit ball and 0t10 \leq t \leq 1, we have

tu+(1t)v tu+(1t)v = tu+(1t)v t+(1t)=1.\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(0,){\|v\|}, {\|v'\|} \in (0, \infty). Let u=vvu = \frac{v}{{\|v\|}} and u=vvu' = \frac{v'}{{\|v'\|}} be the associated unit vectors. Then

v+vv+v = (vv+v)vv+(vv+v)vv = tu+(1t)u\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=vv+vt = \frac{{\|v\|}}{{\|v\|} + {\|v'\|}}. If condition 3. holds, then

tu+(1t)u1{\|t u + (1-t)u'\|} \leq 1

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

v+vv+v1,\frac{{\|v + v'\|}}{{\|v\|} + {\|v'\|}} \leq 1,

which is the triangle inequality.

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

  • Condition 4: If |u| p p=1{|u|_{p}^{p}} = 1 and |v| p p=1{|v|_{p}^{p}} = 1, then |tu+(1t)v| p p1{|t u + (1-t)v|_{p}^{p}} \leq 1 whenever 0t10 \leq t \leq 1.

This allows us to remove the cumbersome exponent 1/p1/p in the definition of the pp-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 α,β\alpha, \beta be two complex numbers, and define

γ(t)=|α+βt| p\gamma(t) = {|\alpha + \beta t|^p}

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

Lemma

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

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

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

Proof of Minkowski’s inequality

Let uu and vv be unit vectors in L pL^p. By condition 4, it suffices to show that |tu+(1t)v| p1{|t u + (1-t)v|_p} \leq 1 for all t[0,1]t \in [0, 1]. But

Ω|tu+(1t)v| pdμ Ωt|u| p+(1t)|v| pdμ, \textstyle{\int}_\Omega {|t u + (1-t)v|^p} \,d\mu \;\leq\; \textstyle{\int}_\Omega t{|u|}^p + (1-t){|v|}^p \,d\mu \,,

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

References

Named after Henri Lebesgue.

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

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

  • Wikipedia (English): Lp space

category: analysis

Last revised on July 9, 2023 at 07:59:04. See the history of this page for a list of all contributions to it.