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.

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