Proof of Theorem 2.1.
-
(1)
We first prove that the point
where the maximum of
is attained lies on the boundary of the hypercube
and moreover, it is a vertex. This result is the subject of Lemma 3.1. We then find the configuration where the maximum is realized.
Lemma 3.1.
The function
attains its maximum at the point
if and only if
for all 
Proof of Lemma 3.1.
Since
is continuous on the compact interval
, there is a point
where
attains its maximum. If
is an interior point of
, then
for all
therefore
which implies
for all
However, if
, then
which clearly is not the maximum of
Consequently,
lies on the boundary of
. Due to symmetry and since
there exist
and
such that
If
then
For this case, consider the function
defined by
If the point
where the maximum of
is attained is interior to
, in virtue of Fermat's theorem, we deduce that
for all
This is equivalent to
hence
A simple computation shows that
and for this configuration we have
Let us define the function
as
and prove it is increasing. Indeed, one finds
where
, and
Since
it follows that
so
is increasing and the upper bound is
This finally proves that
of the numbers
are equal to
while the other
are equal to
as anticipated. This ends the proof of Lemma 3.1.
The only thing to be done is to find the value of
for which the expression
attains its maximum.
To do this, consider the function
defined by
and find the points where the maximum of
is attained in the interval
.
The critical points of
are found from the equation
so they satisfy
As seen in the definition of the Stolarsky mean for this case,
It is finally found that
has a single critical point
which (fortunately) is contained in the interior of 
Taking into account that the second derivative of
is
the extremal point
is a point of maximum for
, and also the function
is decreasing on the interval
. Because
, we obtain
for
, and
for
Finally, this means that
is increasing on
and decreasing on
.
We conclude that
The maximum of (3.13) is then attained when
takes one of the values
and
, where
The value of this
is to be called
from now on.
Remark 3.2.
Because in our case
the Stolarsky mean satisfies the strict inequality
, so 
-
(2)
Using the properties of the integer part
, we obtain
so
It is then enough to work out the limit
On the other hand we have
Due to symmetry the partial derivatives are equal, so the desired limit is
Taking the limit
in (3.23), we obtain that the limit of
as
is confined to the interval 
Proof of Theorem 2.3.
Considering
and
in Theorem 2.1, we obtain
Out of here, we can immediately obtain the best constant
for which
Following the steps mentioned before, the function gets the maximum only when
where
, or
.
This proves that the following inequality holds:
so the best constant
will be
Remark 3.3.
Although appealing, a result involving arbitrary powers
would depend on which the exact value of
is (out of the two possibilities). At the same time, the power
on the righthand-side can only be obtained for 
Proof of Theorem 2.4.
To ease the notations we write
and
The following relation holds:
Using the notation
the limit can be written as
Since the denominator converges to
it only remains to examine the limit
which can be written as
It can be proven that the two terms of (3.36) converge to finite limits, and analyze each. From the hypothesis
so the limit of the first term is
while second term can be written as
Since
the same argument as above can be used to obtain
where
In the end we obtain