Lebesgue space



Functional analysis

Integration theory



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}.


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(\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.


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 (\int_\Omega {|f|^p} \,d\mu)^{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\}.


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].

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.)


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.


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


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 May 4, 2020 at 09:28:58. See the history of this page for a list of all contributions to it.