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 and that are defined for the PTAH inequality.
The setup for the first lemma is as follows:
Lemma 1 (1)
(2) if then . If equality holds then a.e [m].
Proof It is clear that (2) follows from (1), so we only need to prove (1). Define a measure . Then
The next lemma uses the notation of the parent entry.
Lemma 2 Suppose for and . If then
Proof. Let . By the concavity of the function we have
where for .
It is enough to prove the lemma for the case where for all . We can also assume for all , otherwise the result is trivial.
Let and so that .
Raise each side of (1) to the power:
Multiply (3) by to get:
Claim: There exist , such that
If so, then substituting into (4)
So it remains to prove the claim. We have to solve the system of equations , for . Rewriting this in matrix form, let , , and , where and if , . The columns sums of are , since . Hence is singular and the homogenous system has a nonzero solution, say . Since is nonsingular, it follows that . It follows that for some and therefore . If necessary, we can replace by so that . From (5) it follows that for all .
Now we can prove the PTAH inequality. Let .
We calculate by differentiating under the integral sign. If then
If then by writing
where it is clear that each integral is 0, so that . So again, (6) holds. Therefore,
Now by Lemma 1, with we get .