Following [21, 22, 30], denote for with the admissible set

The main results of this paper are contained in the following three theorems. The arguments used in the proofs have common features with the ones developed in [22, 30, 34].

Theorem 3.1.

Suppose that for any given there exist subintervals and of such that

Let be an -tuple satisfying (2.2) in Lemma 2.1. If there exists a function , , such that

for , where

then (1.1) is oscillatory.

Proof.

To arrive at a contradiction, let us suppose that is a nonoscillatory solution of (1.1). First, we assume that is positive for all , for some .

Let , where is sufficiently large. Define

It follows that

and hence

By our assumptions (3.2) we have and for . Set

Then (3.7) becomes

In view of the arithmetic-geometric mean inequality, see [35],

and equality (3.9) we obtain

Multiplying both sides of inequality (3.11) by and then using the identity

result in

where

As demonstrated in [7, 12], we know that , and that if and only if

where stands for the inverse function. In our case, since , dynamic equation (3.15) has a unique solution satisfying . Clearly, the unique solution is . Therefore, on .

For the benefit of the reader we sketch a proof of the fact that . Note that if is a right-dense point, then we may write

Applying Young's inequality (Lemma 2.3) with

we easily see that holds. If is a right-scattered point, then can be written as a function of and as

Using differential calculus, see [7], the result follows.

Now integrating the inequality (3.13) from to and using on we obtain

which of course contradicts (3.3). This completes the proof when is eventually positive. The proof when is eventually negative is analogous by repeating the arguments on the interval instead of .

A close look at the proof of Theorem 3.1 reveals that one cannot take . The following theorem is a substitute in that case.

Theorem 3.2.

Suppose that for any given there exists a subinterval of such that

Let be an -tuple satisfying (2.5) in Lemma 2.2. If there exists a function such that

where

then (1.1) with is oscillatory.

Proof.

We proceed as in the proof of Theorem 3.1 to arrive at (3.7) with , that is,

Setting

and using again the arithmetic-geometric mean inequality

we have

The remainder of the proof is the same as that of Theorem 3.1.

As it is shown in [34] for the sublinear terms case, we can also remove the sign condition imposed on the coefficients of the sub-half-linear terms to obtain interval criterion which is applicable for the case when some or all of the functions , , are nonpositive. We should note that the sign condition on the coefficients of super-half-linear terms cannot be removed alternatively by the same approach. Furthermore, the function cannot take the value zero on intervals of interest in this case. We have the following theorem.

Theorem 3.3.

Suppose that for any given there exist subintervals and of such that

If there exist a function , , and positive numbers and with

such that

for , where

with

then (1.1) is oscillatory.

Proof.

Suppose that (1.1) has a nonoscillatory solution. We may assume that is eventually positive on when is sufficiently large. If is eventually negative, then one can repeat the proof on the interval . Rewrite (1.1) as follows:

with

Clearly,

where

Applying (2.8) and (2.9) to each summation on the right side with

we see that

where

From (3.32) and inequality (3.37) we obtain

where

Set

In view of inequality (3.39) it follows that

The remainder of the proof is the same as that of Theorem 3.1, hence it is omitted.