seph
Turian
Mentor: Arteom ZvavitchDepartment of Theoretical Mathematics, |
Supervisor: Gideon SchechtmanWeizmann Institute of Science |
constant in the Khintchine Inequality for p > 3 using only elementary
calculus.
where 
and, 
, for all i such
that 
and 
are independent random variables that each assume the values 
with probabilities 
.
The actual inequality (in which 
and 
are not necessarily
the best) was proved by Khintchine. But we are looking for the best constants 
and 
that satisfy the inequality
for all natural numbers n. For all p, a solution and proof
was given in [1]. Here we present another proof, for p > 3, which
uses only elementary calculus.
.
Therefore, we can write the inequality in the following way:
Using the definition of mathematical expectation, we arrive to the following formula:
And finally:
As changing the signs of the 
does not change the sum, we may assume that every 
is non-negative. For n=1, we obtain 
.
If 
and 
for some k, without loss of generality let k=n. Then note
that:
and that the inequality is therefore
Thus, the equality for n reduces to the inequality for n-1.
We would like to show that to find the best value of 
,
it suffices to find the upper bound of
for all valid ns.
We now define the following function f:
with 
.
It follows from Theorem 1
that f is only maximal when the non-zero 
s
are equal.
Proof of Theorem 1:
We prove Theorem 1
by contradiction. Let us suppose that f is maximal for some 
,
such that, without loss of generality, 
.
We define following function:
with 
as a constant
dependent upon the 
sequence, 
:
and with
and 
. Note that 
, 
,
and 
, for all 
.
Because 
, 
and 
are not zero-functions.
If we can show that 
increases at 
, then we have
a contradiction because f increases with 
and is therefore not maximal for 
as assumed.
Proof of Lemma 1:
and therefore 
and 
.
Note that 
for any 
.
Define the following constant: 
.
Thus, c > 0.
Note that
and
Also note that
From this point forward, k (and 
)
shall be understood as 
(and 
) and l (and 
)
as 
(and 
).
Replacing 
and 
with explicit values, we obtain the following equation:
Note that 
, there is
an 
such that 
.
The inverse is also true. Therefore, the equation can be condensed:
Now we compare every non-negative 
with its corresponding 
.
Note that when the summation constraint is 
,
the sum is greater than or equal to the sum when the summation constraint
is 
, that is
Let us prove that each summand is greater than 0, which will prove that 
.
For each summand, define the constants: 
and 
. Note that 
.
Then:
Therefore, it is enough to prove that for all x > y > 0
Note that 
and 
.
Thus, if we can prove that 
is less than 0, we will prove that 
is greater than 0 for valid values of x and y.
Case 1: y-1=0 The question that arises is if 
is differentiable, or, more specifically, if 
is differentiable at y = 1.
for y-1 > 0
for y-1 < 0
By the definition of the derivative,
(if the limit exists).
To prove that the limit exists, we take the limit from both sides.
So thus 
and 
.
Otherwise, when 
,
Case 2: y-1>0
Case 3: y-1<0
Therefore, 
is greater
than 0 for x > y > 0. Note that this inequality does not
hold for 
Thus, 
and Lemma
1
has been proven.
Thus, Theorem 1 has been proven.
Q.E.D.
constant for the Khintchine Inequality has been derived for p > 3.
implies 
, then we could
also get the contradiction with which we proved Theorem 1.
Comments to: Joseph Turian / < jude@ai.mit.edu
>
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