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The Lp distance for 0<p<1 is a complete translation-invariant metric

Statement

Let 0<p<1 and let Lp(μ) denote the set of almost-everywhere classes of functions in Lp(μ). Define

dp([f],[g]):=fgpdμ.

Then dp is a translation-invariant metric on Lp(μ), and (Lp(μ),dp) is complete.

Facts & Assumptions

Given: A measure space (X,A,μ) and an exponent 0<p<1.

[L1]

The class notation Lp(μ) means almost-everywhere equivalence classes of Lp(μ) representatives (The space Lp(μ) as the quotient by null functions).

[L3]
[L4]
[L5]

Sums, scalar multiples, absolute values, and pointwise limits of measurable functions are measurable (Closure properties of measurable functions used by the integral).

[L6]

Countable unions of measurable null sets are measurable and null (Finite and countable subadditivity of measures).

[L7]

For 0<p<1 and nonnegative reals a,b, (a+b)pap+bp. Indeed, if a+b=0 the claim is trivial. Otherwise set s:=a+b>0, u:=a/s, and v:=b/s. Then u,v[0,1] and u+v=1. If 0<u1, then logu0, so plogulogu because 0<p<1; strict increase of the exponential and the definition of real power therefore give upu. The same holds for v. Hence ap+bp=sp(up+vp)sp(u+v)=sp=(a+b)p. (Real powers for positive bases, with the zero-base positive-exponent convention, Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm, The exponential function is strictly increasing) [given]

Proof

Proof technique: Because (a+b)pap+bp for 0<p<1, dp([f],[g])=fgpdμ defines a metric on quotient classes and is translation invariant. Completeness follows by repeating the Riesz-Fischer telescoping argument without taking p-th roots.

1.1

If ff and gg, then ff and gg vanish almost everywhere. Outside the union of those two null sets, one has fg=fg, so fgp=fgp almost everywhere. Hence dp([f],[g]) is well defined. The same union-of-null-sets argument shows that addition and scalar multiplication descend to the quotient classes, and the inequality in [L7] shows that Lp(μ) is closed under those operations.

L1L5L6L7
1.2

Symmetry of dp is immediate. If dp([f],[g])=0, then fgpdμ=0, so f=g almost everywhere and hence [f]=[g]. For the triangle inequality, the pointwise inequality from [L7] gives [L2, L4, L7] fhp=(fg)+(gh)pfgp+ghp, and integrating yields dp([f],[h])dp([f],[g])+dp([g],[h]). Thus [L2] makes dp a metric.

1.3

Let (un) be Cauchy in dp. Choose by least indices a subsequence (unk) with [L3, L5, given, choose] dp(unk+1,unk)<2k. Choose representatives fk of unk and define hk:=fk+1fkp,gm:=j<mhj. Each hk is measurable and integrable, and [L3] gives a measurable pointwise limit g:=j=0hj with gdμ=limmgmdμj=02j<. Hence g< almost everywhere.

2.1

Translation invariance is pointwise: [step 1.1] dp([f]+[u],[g]+[u])=(f+u)(g+u)pdμ=dp([f],[g]).

2.2

Fix x outside the null set where g(x)=. Then jfj+1(x)fj(x)p<, so the terms tend to 0. Thus fj+1(x)fj(x)1 for all large j, and then [step 1.3, L3, L7] fj+1(x)fj(x)fj+1(x)fj(x)p=hj(x). So the real series jfj+1(x)fj(x) converges by comparison with jhj(x), which makes (fk(x)) converge to some real value f(x). By [L3], the resulting function f is measurable. Also f(x)fk(x)p(jkfj+1(x)fj(x))pjkhj(x). Integrating and using monotone convergence on the tails yields dp(unk,[f])jkdp(unj+1,unj)jk2j0.

3.1

Because (un) is Cauchy, given ε>0 choose K with dp(un,um)<ε/2 for m,nK, then choose k with nkK and dp(unk,[f])<ε/2 from step 2.2. The triangle inequality from step 1.2 gives dp(un,[f])<ε for all nK. Hence (Lp(μ),dp) is complete.

step 1.2step 2.2
4.1

Steps 1.2 and 2.1 prove that dp is a translation-invariant metric, and step 3.1 proves completeness.

step 1.2step 2.1step 3.1

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