proof of PTAH inequality

In order to prove the PTAH inequality two lemmas are needed. The first lemma is quite general and does not depend on the specific P and Q that are defined for the PTAH inequality.

The setup for the first lemma is as follows:

We still have a measure spaceMathworldPlanetmath X with measure m. We have a subset Λn. And we have a function p:X×Λ which is positive and is integrable in x for all λΛ. Also, p(x,λ)logp(x,λ) is integrable in x for each pair λ,λΛ.

Define P:Λ by


and Q:Λ×Λ



Lemma 1 (1) P(λ)logP(λ)P(λ)Q(λ,λ)-Q(λ,λ)
(2) if Q(λ,λ)Q(λ,λ) then P(λ)P(λ). If equality holds then p(x,λ)=p(x,λ) a.e [m].

Proof It is clear that (2) follows from (1), so we only need to prove (1). Define a measure dν(x)=p(x,λ)dm(x)P(λ). Then


so we can use Jensen’s inequalityMathworldPlanetmath for the logarithm.

Q(λ,λ)-Q(λ,λ) = p(x,λ)[logp(x,λ)-logp(x,λ)]𝑑m(x)
= p(x,λ)logp(x,λ)p(x,λ)dm(x)
= P(λ)logp(x,λ)p(x,λ)dν(x)
= P(λ)logp(x,λ)P(λ)𝑑m(x)
= P(λ)logP(λ)P(λ).

The next lemma uses the notation of the parent entry.

Lemma 2 Suppose ri0 for i=1,,n and θ=(θ1,,θn)σ. If jrj>0 then


Proof. Let λ=(λi)σ. By the concavity of the log function we have


where xi>0 for ı=1,,n.

so that

ixiλiiλixi=i(jλjxj)λi. (1)

It is enough to prove the lemma for the case where ri>0 for all i. We can also assume θi>0 for all i, otherwise the result is trivial.

Let ρ=jrj>0 and λi=riρ so that ρλi=ri.

Raise each side of (1) to the ρ power:

ixirii(jλjxj)ri (2)

so that

i(xijλjxj)ri1 (3)

Multiply (3) by (riρ)ri to get:

i(rixijrjxj)rii(ri/ρ)ri. (4)

Claim: There exist xi>0, i=1,,n such that

θi=rixijrjxj. (5)

If so, then substituting into (4)


So it remains to prove the claim. We have to solve the system of equations θijrjxj=rixi, i=1,,n for xi. Rewriting this in matrix form, let A=(aij), R=diag(r1,,rn), and x=diag(x1,,xn), where aii=θi-1 and aij=θi if ij, i,j=1,,n. The columns sums of A are 0, since θσ. Hence A is singularPlanetmathPlanetmath and the homogenous system ARx=0 has a nonzero solution, say x. Since R is nonsingular, it follows that Rx0. It follows that rixi0 for some i and therefore jrjxj0. If necessary, we can replace x by -x so that jrjxj>0. From (5) it follows that xj>0 for all j.

Now we can prove the PTAH inequality. Let ri(λ)=ai(x)jλjaj(x)dm(x).

We calculate Pλi by differentiating under the integral sign. If λi>0 then



λiPλi=ri(λ). (6)

If λi=0 then by writing


where E={xX|ai(x)=0} it is clear that each integral is 0, so that ri(λ)=0. So again, (6) holds. Therefore,



Q(λ,λ) = jλjaj(x)logi(λi)ai(x)dm(x)
= ilogλiai(x)jλjaj(x)dm(x)
= iri(λ)logλi
= logi(λi)ri(λ)
= logi(λi¯)ri(λ)
= Q(λ,λ¯).

Now by Lemma 1, with λ¯=λ we get P(λ¯)P(λ).

Title proof of PTAH inequality
Canonical name ProofOfPTAHInequality
Date of creation 2013-03-22 16:55:00
Last modified on 2013-03-22 16:55:00
Owner Mathprof (13753)
Last modified by Mathprof (13753)
Numerical id 29
Author Mathprof (13753)
Entry type Proof
Classification msc 26D15