proof of PTAH inequality

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

Λ

⊆

ℝ

n

.
And we have a

function

p

:

X

×

Λ

→

ℝ

which is

positive

and is

integrable

in

x

for all

λ

∈

Λ

.
Also,

p

⁢

(

x

,

λ

)

⁢

log

⁡

p

⁢

(

x

,

λ

′

)

is integrable in

x

for each
pair

λ

,

λ

′

∈

Λ

.

Define

P

:

Λ

→

ℝ

by

P

⁢

(

λ

)

=

∫

p

⁢

(

x

,

λ

)

⁢

𝑑

m

⁢

(

x

)

and

Q

:

Λ

×

Λ

→

ℝ

by

Q

⁢

(

λ

,

λ

′

)

=

∫

p

⁢

(

x

,

λ

)

⁢

log

⁡

p

⁢

(

x

,

λ

′

)

⁢

𝑑

m

⁢

(

x

)

.

Lemma 1

(1)

P

⁢

(

λ

)

⁢

log

⁡

P

⁢

(

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

d

⁢

ν

⁢

(

x

)

=

p

⁢

(

x

,

λ

)

⁢

d

⁢

m

⁢

(

x

)

P

⁢

(

λ

)

.
Then

∫

𝑑

ν

⁢

(

x

)

=

1

so we can use Jensen’s

inequality

for the

logarithm

.

Q

⁢

(

λ

,

λ

′

)

-

Q

⁢

(

λ

,

λ

)

=

∫

p

⁢

(

x

,

λ

)

⁢

[

log

⁡

p

⁢

(

x

,

λ

′

)

-

log

⁡

p

⁢

(

x

,

λ

)

]

⁢

𝑑

m

⁢

(

x

)

=

∫

p

⁢

(

x

,

λ

)

⁢

log

⁡

p

⁢

(

x

,

λ

′

)

p

⁢

(

x

,

λ

)

⁢

d

⁢

m

⁢

(

x

)

=

P

⁢

(

λ

)

⁢

∫

log

⁡

p

⁢

(

x

,

λ

′

)

p

⁢

(

x

,

λ

)

⁢

d

⁢

ν

⁢

(

x

)

≤

P

⁢

(

λ

)

⁢

log

⁢

∫

p

⁢

(

x

,

λ

′

)

p

⁢

(

x

,

λ

)

⁢

𝑑

ν

⁢

(

x

)

=

P

⁢

(

λ

)

⁢

log

⁢

∫

p

⁢

(

x

,

λ

′

)

P

⁢

(

λ

)

⁢

𝑑

m

⁢

(

x

)

=

P

⁢

(

λ

)

⁢

log

⁡

P

⁢

(

λ

′

)

P

⁢

(

λ

)

.

The next lemma uses the notation of the parent entry.

Lemma 2

Suppose

r

i

≥

0

for

i

=

1

,

…

,

n

and

θ

=

(

θ

1

,

…

,

θ

n

)

∈

σ

. If

∑

j

r

j

>

0

then

∏

i

=

1

n

θ

i

r

i

≤

∏

i

(

r

i

∑

j

r

j

)

r

i

.

Proof.

Let

λ

=

(

λ

i

)

∈

σ

. By the concavity of the

log

function we have

∑

i

λ

i

⁢

log

⁡

x

i

≤

log

⁢

∑

i

λ

i

⁢

x

i

where

x

i

>

0

for

ı

=

1

,

…

,

n

.

so that

∏

i

x

i

λ

i

≤

∑

i

λ

i

⁢

x

i

=

∏

i

(

∑

j

λ

j

⁢

x

j

)

λ

i

.

(1)

It is enough to prove the lemma for the case where

r

i

>

0

for all

i

.
We can also assume

θ

i

>

0

for all

i

, otherwise the result is

trivial

.

Let

ρ

=

∑

j

r

j

>

0

and

λ

i

=

r

i

ρ

so that

ρ

⁢

λ

i

=

r

i

.

Raise each side of (1) to the

ρ

power:

∏

i

x

i

r

i

≤

∏

i

(

∑

j

λ

j

⁢

x

j

)

r

i

(2)

so that

∏

i

(

x

i

∑

j

λ

j

⁢

x

j

)

r

i

≤

1

(3)

Multiply (3) by

∏

(

r

i

ρ

)

r

i

to get:

∏

i

(

r

i

⁢

x

i

∑

j

r

j

⁢

x

j

)

r

i

≤

∏

i

(

r

i

/

ρ

)

r

i

.

(4)

Claim: There exist

x

i

>

0

,

i

=

1

,

…

,

n

such that

θ

i

=

r

i

⁢

x

i

∑

j

r

j

⁢

x

j

.

(5)

If so, then substituting into (4)

∏

i

θ

i

r

i

≤

∏

i

(

r

i

ρ

)

r

i

=

∏

i

(

r

i

∑

j

r

j

)

r

i

So it remains to prove the claim. We have to solve the system of equations

θ

i

⁢

∑

j

r

j

⁢

x

j

=

r

i

⁢

x

i

,

i

=

1

,

…

,

n

for

x

i

.
Rewriting this in matrix form, let

A

=

(

a

i

⁢

j

)

,

R

=

diag

⁢

(

r

1

,

…

,

r

n

)

,
and

x

=

diag

⁢

(

x

1

,

…

,

x

n

)

,
where

a

i

⁢

i

=

θ

i

-

1

and

a

i

⁢

j

=

θ

i

if

i

≠

j

,

i

,

j

=

1

,

…

,

n

.
The columns sums of

A

are

0

, since

θ

∈

σ

. Hence

A

is

singular

and the homogenous system

A

⁢

R

⁢

x

=

0

has a nonzero solution, say

x

. Since

R

is

nonsingular

, it follows that

R

⁢

x

≠

0

. It follows that

r

i

⁢

x

i

≠

0

for some

i

and therefore

∑

j

r

j

⁢

x

j

≠

0

. If necessary, we can replace

x

by

-

x

so that

∑

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

⁢

(

λ

)

=

∫

a

i

⁢

(

x

)

⁢

∏

j

λ

j

a

j

⁢

(

x

)

⁢

d

⁢

m

⁢

(

x

)

.

We calculate

∂

⁡

P

∂

⁡

λ

i

by differentiating
under the

integral sign

. If

λ

i

>

0

then

∂

⁡

P

∂

⁡

λ

i

=

r

i

⁢

(

λ

)

/

λ

i

.

Thus

λ

i

⁢

∂

⁡

P

∂

⁡

λ

i

=

r

i

⁢

(

λ

)

.

(6)

If

λ

i

=

0

then by writing

r

i

⁢

(

λ

)

=

∫

E

a

i

⁢

(

x

)

⁢

…

⁢

𝑑

m

⁢

(

x

)

+

∫

E

c

λ

i

a

i

⁢

(

x

)

⁢

…

⁢

𝑑

m

⁢

(

x

)

where

E

=

{

x

∈

X

|

a

i

⁢

(

x

)

=

0

}

it is clear that each

integral

is 0, so that

r

i

⁢

(

λ

)

=

0

.
So again, (6) holds. Therefore,

r

i

⁢

(

λ

)

∑

j

r

j

⁢

(

λ

)

=

λ

i

⁢

∂

⁡

P

/

∂

⁡

λ

i

∑

j

λ

j

⁢

∂

⁡

P

/

λ

j

=

λ

i

¯

.

Then

Q

⁢

(

λ

,

λ

′

)

=

∫

∏

j

λ

j

a

j

⁢

(

x

)

⁢

log

⁢

∏

i

(

λ

i

′

)

a

i

⁢

(

x

)

⁢

d

⁢

m

⁢

(

x

)

=

∑

i

log

⁡

λ

i

′

⁢

∫

a

i

⁢

(

x

)

⁢

∏

j

λ

j

a

j

⁢

(

x

)

⁢

d

⁢

m

⁢

(

x

)

=

∑

i

r

i

⁢

(

λ

)

⁢

log

⁡

λ

i

′

=

log

⁢

∏

i

(

λ

i

′

)

r

i

⁢

(

λ

)

≤

log

⁢

∏

i

(

r

i

⁢

(

λ

)

∑

j

r

j

⁢

(

λ

)

)

r

i

⁢

(

λ

)

=

log

⁢

∏

i

(

λ

i

¯

)

r

i

⁢

(

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

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