Lemma 1 If , , (), () are decreasing continuous functions satisfying , , define a function as follows:
Then there exists
such that
where is a Bernoulli function of the first order. In particular, for , , we have and
(8)
for , , if , then it follows and
(9)
Proof Define a decreasing continuous function as
Then it follows
Since , is a non-constant decreasing continuous function with , by the improved Euler-Maclaurin summation formula (cf. [6], Theorem 2.2.2), it follows
and then in view of the above results and by simple calculation, we have (7). □
Lemma 2 If , , , and and are weight functions given by
then we have
(12)
Proof Substituting in (10), and by simple calculation, we have
For fixed , we find
By the Euler-Maclaurin summation formula (cf. [6]), it follows
-
(i)
For , we obtain , and
Setting , wherefrom , and
then by (7), we find
In view of (11) and the above results, since for , namely , it follows
-
(ii)
For , we obtain , and
Since for , , by the improved Euler-Maclaurin summation formula (cf. [6]), it follows
In view of (13) and the above results, for , we find
Hence, for , we have , and then (12) follows. □
Lemma 3 Let the assumptions of Lemma 2 be fulfilled and, additionally, let , , , , be a non-negative measurable function in . Then we have the following inequalities:
Proof Setting , by Hölder’s inequality (cf. [22]) and (12), it follows
Then by the Lebesgue term-by-term integration theorem (cf. [23]), we have
Hence, (14) follows. By Hölder’s inequality again, we have
By the Lebesgue term-by-term integration theorem, we have
and in view of (12), inequality (15) follows. □
Lemma 4 Let the assumptions of Lemma 2 be fulfilled and, additionally, let , , . Setting , ; , , and , , then we have
Proof We find
and then (16) is valid. We obtain
and so (17) is valid. □