As we mentioned above, our proof of Theorem 1.1 has a major part which is parallel to . Our matrix A is a little different from that in , we use a variable y instead of ε and δ. It simplifies the argument to an extent, though the improvement is not essential.
Throughout this paper, we assume that
. We will consider matrices
Then we have . The eigenvalues of A are , where .
Lemma 2.1 and .
If , then we would have or , which is contrary to the assumption. □
is unitary and
By the assumption and Theorem 1.3, A
satisfy the inequality (2). Then
hence we have
Lemma 2.2 Let p, q, r be positive real numbers. Then and .
, we have
hence we have . Thus .
It is obvious that and , and hence . □
Then it is easy to see that
is unitary and
The following lemma is one of the most important points in Tanahashi’s argument. Although the substance is presented in the whole proof of , Theorem], we should restate and prove it in our context for the readers’ convenience.
The formula (3) implies
Write the left-hand matrix as
Then, by the formula (5), we have
So, its determinant is also non-negative. We expand it to obtain
Hence, the formula (6) implies
Cancel the common positive factor
and substitute the definitions for
. Then a simple calculation shows that
By factorizing, we have
This completes the proof of Lemma 2.3. □
Now, we estimate each term of the inequality (4) with respect to . A key point in making use of the inequality (4) is that both estimations of the factor on the left-hand side and the factor on the right-hand side contain a common subfactor y. After the cancellation of this y, we will derive the desired functional inequality by letting , and applying l’Hopital’s rule. Terms in other factors can be roughly estimated.
In the following, o
, that is
and denotes a term such that ().
One can establish the following formulae:
Now, we have the estimation of the most delicate factor in the formula (4), whose constant term is canceled by subtraction.
Substitute these estimations for the inequality (4), cancel the positive factor y
, and let
, then we have
and applying l’Hopital’s rule, we have
This implies that, for arbitrary
For arbitrary , substitute for b in (7) and multiply by x, , , both sides. It is easy to see that x itself satisfies (7). This completes the proof of Theorem 1.1.