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 space $X$ with measure $m$.We have a subset $\mathrm{\Lambda }\subseteq {ℝ}^n$.And we have a function $p:X\times \mathrm{\Lambda }\to ℝ$ which ispositive and is integrable in $x$ for all $\lambda \in \mathrm{\Lambda }$.Also, $p(x,\lambda )\log p(x,{\lambda }^{\prime })$ is integrable in $x$ for eachpair $\lambda ,{\lambda }^{\prime }\in \mathrm{\Lambda }$.

Define $P:\mathrm{\Lambda }\to ℝ$ by

$$P(\lambda )=\int p(x,\lambda )𝑑m(x)$$

and $Q:\mathrm{\Lambda }\times \mathrm{\Lambda }\to ℝ$

by

$$Q(\lambda ,{\lambda }^{\prime })=\int p(x,\lambda )\log p(x,{\lambda }^{\prime })𝑑m(x).$$

Lemma 1 (1) $P(\lambda )\log \frac{P({\lambda }^{\prime })}{P(\lambda )}\ge Q(\lambda ,{\lambda }^{\prime })-Q(\lambda ,\lambda )$
(2) if $Q(\lambda ,{\lambda }^{\prime })\ge Q(\lambda ,\lambda )$ then$P({\lambda }^{\prime })\ge P(\lambda )$. If equality holds then $p(x,\lambda )=p(x,{\lambda }^{\prime })$ a.e [m].

Proof It is clear that (2) follows from (1), so we only need to prove (1). Define a measure $d\nu (x)=\frac{p(x,\lambda )dm(x)}{P(\lambda )}$.Then

$$\int 𝑑\nu (x)=1$$

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

$Q(\lambda ,{\lambda }^{\prime })-Q(\lambda ,\lambda )$$=$${\displaystyle \int p(x,\lambda )[\log p(x,{\lambda }^{\prime })-\log p(x,\lambda )]𝑑m(x)}$

$=$${\displaystyle \int p(x,\lambda )\log \frac{p(x,{\lambda }^{\prime })}{p(x,\lambda )}dm(x)}$

$=$$P(\lambda ){\displaystyle \int \log \frac{p(x,{\lambda }^{\prime })}{p(x,\lambda )}d\nu (x)}$

$\le$$P(\lambda )\log {\displaystyle \int \frac{p(x,{\lambda }^{\prime })}{p(x,\lambda )}𝑑\nu (x)}$

$=$$P(\lambda )\log {\displaystyle \int \frac{p(x,{\lambda }^{\prime })}{P(\lambda )}𝑑m(x)}$

$=$$P(\lambda )\log {\displaystyle \frac{P({\lambda }^{\prime })}{P(\lambda )}}.$

The next lemma uses the notation of the parent entry.

Lemma 2 Suppose $r_i\ge 0$ for $i=1,\mathrm{\dots },n$ and $\theta =({\theta }_1,\mathrm{\dots },{\theta }_n)\in \sigma$. If ${\sum }_j{r}_j>0$ then

$$\prod _{i=1}^n\theta _i{}^{r_i}\le \prod _i{(\frac{r_i}{{\sum }_j{r}_j})}^{r_i}.$$

Proof. Let $\lambda =({\lambda }_i)\in \sigma$. By the concavity of the$\log$ function we have

$$\sum _i{\lambda }_i\log x_i\le \log \sum _i{\lambda }_i{x}_i$$

where $x_i>0$ for $ı=1,\mathrm{\dots },n$.

so that

$$\prod _{i}x_i{}^{{\lambda }_i}\le \sum _i{\lambda }_i{x}_i=\prod _i{(\sum _j{\lambda }_j{x}_j)}^{{\lambda }_i}.$$(1)

It is enough to prove the lemma for the case where $r_i>0$ for all $i$.We can also assume ${\theta }_i>0$ for all $i$, otherwise the result is trivial.

Let $\rho ={\sum }_j{r}_j>0$ and ${\lambda }_i=\frac{r_i}{\rho }$ so that$\rho {\lambda }_i=r_i$.

Raise each side of (1) to the $\rho$ power:

$$\prod _{i}x_i{}^{r_i}\le \prod _i{(\sum _j{\lambda }_j{x}_j)}^{r_i}$$(2)

so that

$$\prod _i{(\frac{x_i}{{\sum }_j{\lambda }_j{x}_j})}^{r_i}\le 1$$(3)

Multiply (3) by $\prod {(\frac{r_i}{\rho })}^{r_i}$ to get:

$$\prod _i{(\frac{r_i{x}_i}{{\sum }_j{r}_j{x}_j})}^{r_i}\le \prod _i{(r_i/\rho )}^{r_i}.$$(4)

Claim: There exist $x_i>0$, $i=1,\mathrm{\dots },n$ such that

$${\theta }_i=\frac{r_i{x}_i}{{\sum }_j{r}_j{x}_j}.$$(5)

If so, then substituting into (4)

$$\prod _i\theta _i{}^{r_i}\le \prod _i{(\frac{r_i}{\rho })}^{r_i}=\prod _i{(\frac{r_i}{{\sum }_j{r}_j})}^{r_i}$$

So it remains to prove the claim. We have to solve the system of equations${\theta }_i{\sum }_j{r}_j{x}_j=r_i{x}_i$, $i=1,\mathrm{\dots },n$ for $x_i$.Rewriting this in matrix form, let $A=(a_{ij})$, $R=\text{diag}(r_1,\mathrm{\dots },r_n)$,and $x=\text{diag}(x_1,\mathrm{\dots },x_n)$,where $a_{ii}={\theta }_i-1$ and $a_{ij}={\theta }_i$ if $i\ne j$, $i,j=1,\mathrm{\dots },n$.The columns sums of $A$ are $0$, since$\theta \in \sigma$. Hence $A$ is singular and the homogenous system $ARx=0$has a nonzero solution, say $x$. Since $R$ is nonsingular, it follows that$Rx\ne 0$. It follows that $r_i{x}_i\ne 0$ for some $i$ and therefore${\sum }_j{r}_j{x}_j\ne 0$. If necessary, we can replace $x$ by $-x$ so that${\sum }_j{r}_j{x}_j>0$. From (5) it follows that $x_j>0$ for all $j$.

Now we can prove the PTAH inequality.Let $r_i(\lambda )=\int a_i(x){\prod }_j\lambda _j{}^{a_j(x)}dm(x)$.

We calculate $\frac{\partial P}{\partial {\lambda }_i}$ by differentiatingunder the integral sign. If ${\lambda }_i>0$ then

$$\frac{\partial P}{\partial {\lambda }_i}=r_i(\lambda )/{\lambda }_i.$$

Thus

$${\lambda }_i\frac{\partial P}{\partial {\lambda }_i}=r_i(\lambda ).$$(6)

If ${\lambda }_i=0$ then by writing

$$r_i(\lambda )={\int }_E{a}_i(x)\mathrm{\dots }𝑑m(x)+{\int }_{E^c}\lambda _i{}^{a_i(x)}\mathrm{\dots }𝑑m(x)$$

where $E=\{x\in X|a_i(x)=0\}$it is clear that each integral is 0, so that $r_i(\lambda )=0$.So again, (6) holds. Therefore,

$$\frac{r_i(\lambda )}{{\sum }_j{r}_j(\lambda )}=\frac{{\lambda }_i\partial P/\partial {\lambda }_i}{{\sum }_j{\lambda }_j\partial P/{\lambda }_j}=\overline{{\lambda }_i}.$$

Then

$Q(\lambda ,{\lambda }^{\prime })$$=$${\displaystyle \int \prod _j\lambda _j{}^{a_j(x)}\log \prod _i{(\lambda _i{}^{\prime })}^{a_i(x)}dm(x)}$

$=$${\displaystyle \sum _i}\log \lambda _i{}^{\prime }{\displaystyle \int a_i(x)\prod _j\lambda _j{}^{a_j(x)}dm(x)}$

$=$${\displaystyle \sum _i}{r}_i(\lambda )\log \lambda _i{}^{\prime }$

$=$$\log {\displaystyle \prod _i}{(\lambda _i{}^{\prime })}^{r_i(\lambda )}$

$\le$$\log {\displaystyle \prod _i}{({\displaystyle \frac{r_i(\lambda )}{{\sum }_j{r}_j(\lambda )}})}^{r_i(\lambda )}$

$=$$\log {\displaystyle \prod _i}{(\overline{{\lambda }_i})}^{r_i(\lambda )}$

$=$$Q(\lambda ,\overline{\lambda }).$

Now by Lemma 1, with $\overline{\lambda }={\lambda }^{\prime }$ we get$P(\overline{\lambda })\ge P(\lambda )$.