This note is to concern a recent result by David G. Costa [here]. Here the statement
Theorem 1.1. For all and one has
where . In addition, if , then
where the constant is sharp.
Here’s the proof.
Proof. For all and one has, for all , the following
Expanding the above yields
If we denote the last integral by and write it as
an integration by parts gives
Keep in mind that
A second integration by parts on the first integral above yields
so that becomes
Therefore, we have
for any where is given above and
This is equivalent to , i.e.
This completes the first inequality. On the other hand, since
we know that
provided . This proves the second inequality.