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 $\Lambda \subseteq {\mathbb{R}}^{n}$ . And we have a function $p: X \times \Lambda \to \mathbb{R}$ which is positive and is integrable in $x$ for all $\lambda \in \Lambda$ . Also, $p(x,\lambda )\log p(x, \lambda')$ is integrable in $x$ for each pair $\lambda, \lambda' \in \Lambda$ .

Define $P: \Lambda \to \mathbb{R}$ by

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

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

by $$ Q(\lambda, \lambda' ) = \int p(x,\lambda ) \log p(x, \lambda') dm(x). $$

Lemma 1 (1) $P(\lambda ) \log \frac{P(\lambda')}{P(\lambda)} \ge Q(\lambda , \lambda' ) - Q(\lambda, \lambda) $
(2) if $Q(\lambda , \lambda') \ge Q(\lambda , \lambda) $ then $P(\lambda') \ge P(\lambda)$ . If equality holds then $p(x,\lambda) = p(x,\lambda')$ 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.

\begin{eqnarray*} Q(\lambda, \lambda')- Q(\lambda , \lambda ) &=& \int p(x,\lambda )[\log p(x, \lambda' ) - \log p(x, \lambda ) ] dm(x) \\ &=& \int p(x,\lambda ) \log \frac{p(x,\lambda')}{p(x,\lambda)} dm(x) \\ & =& P(\lambda) \int \log \frac{p(x,\lambda')}{p(x,\lambda)} d\nu(x) \\ & \le& P(\lambda) \log \int \frac{p(x,\lambda')}{p(x,\lambda)} d\nu(x) \\ & =& P(\lambda ) \log \int \frac{p(x, \lambda')}{P(\lambda)} dm(x) \\ & =& P(\lambda) \log \frac{P(\lambda')}{P(\lambda)}. \end{eqnarray*} The next lemma uses the notation of the parent entry.

Lemma 2 Suppose $r_i \ge 0$ for $i=1, \ldots, n$ and $\theta = (\theta_1, \ldots, \theta_n ) \in \sigma$ . 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 $\i=1, \ldots, n$ .

so that \begin{equation} \prod_i {x_i}^{\lambda_i} \le \sum_i \lambda_i x_i = \prod_i ( \sum_j \lambda_j x_j )^{\lambda_i} . \end{equation} 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:

\begin{equation} \prod_i {x_i}^{r_i} \le \prod_i (\sum_j \lambda_j x_j )^{r_i} \end{equation}so that \begin{equation} \prod_i (\frac{x_i}{\sum_j \lambda_j x_j })^{r_i} \le 1 \end{equation}Multiply (3) by $\prod (\frac{r_i}{\rho})^{r_i}$ to get:

\begin{equation} \prod_i ( \frac{r_i x_i}{\sum_j r_j x_j})^{r_i} \le \prod_i (r_i/\rho)^{r_i}. \end{equation} Claim: There exist $x_i>0 $ , $i=1,\ldots, n$ such that \begin{equation} \theta_i = \frac{r_i x_i}{\sum_j r_j x_j}. \end{equation}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 $\theta_i \sum_j r_j x_j = r_i x_i$ , $i=1, \ldots, n$ for $x_i$ . Rewriting this in matrix form, let $A=(a_{ij})$ , $R={diag}(r_1, \ldots, r_n)$ , and $x={diag}(x_1, \ldots, x_n)$ , where $a_{ii} = \theta_i-1$ and $a_{ij} = \theta_i$ if $i \not = j$ , $i,j=1,\ldots, n$ . 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 $Rx \not = 0$ . It follows that $r_i x_i \not = 0$ for some $i$ and therefore $\sum_j r_j x_j \not = 0$ . 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 differentiating under the integral sign. If $\lambda_i>0$ then $$ \frac{\partial P}{\partial \lambda_i} = r_i(\lambda)/\lambda_i . $$ Thus \begin{equation} \lambda_i \frac{\partial P}{\partial \lambda_i} = r_i(\lambda). \end{equation}If $\lambda_i =0$ then by writing $$ r_i(\lambda) = \int_E a_i(x) \ldots dm(x) + \int_{E^c} {\lambda_i}^{a_i(x)} \ldots dm(x) $$ 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 \begin{eqnarray*} Q(\lambda, \lambda') &=& \int \prod_j {\lambda_j}^{a_j(x)} \log \prod_i ( {\lambda_i}')^{a_i(x)} dm(x)\\ & =&\sum_i \log {\lambda_i}' \int a_i(x) \prod_j {\lambda_j}^{a_j(x)} dm(x) \\ &=& \sum_i r_i(\lambda) \log {\lambda_i}' \\ &=& \log \prod_i ({\lambda_i}')^{r_i(\lambda)} \\ & \le&\log \prod_i ( \frac{r_i(\lambda)}{\sum_j r_j(\lambda)})^{r_i(\lambda)} \\ & =& \log \prod_i ({\overline{\lambda_i}})^{ r_i(\lambda) }\\ & =& Q(\lambda , \overline{\lambda}). \end{eqnarray*} Now by Lemma 1, with $\overline{\lambda} = \lambda'$ we get $P(\overline{\lambda}) \ge P(\lambda)$ .