In order to prove our Theorem 1.1, we need a lemma which we present in this section.
Lemma 2.1 Let α ∈ ℝ, β ∈ (0, 1) ∪ (1, ∞) and
Then the following statements are true:
-
(1)
If β ≤ α + 1 and β ≤ 2α + 1, then h(t) < 0 for t ∈ (0, ∞);
-
(2)
If α + 1 < β < 2α + 1, then there exists λ 1 ∈ (0, ∞) such that h(t) < 0 for t ∈ (0, λ 1) and h(t) > 0 for t ∈ (λ 1, ∞);
-
(3)
If β ≥ α + 1 and β ≥ 2α + 1, then h(t) > 0 for t ∈ (0, ∞);
-
(4)
If 2α + 1 < β < α + 1, then there exists λ 2 ∈ (0, ∞) such that h(t) > 0 for t ∈ (0, λ 2) and h(t) < 0 for t ∈ (λ 2, ∞).
Proof Let h1(t) = e(β + 1)th'(t) and . Then simple computations lead to
(2.2)
(2.3)
(2.4)
(2.5)
and
(2.6)
-
(1)
If β ≤ α + 1 and β ≤ 2α + 1, then we divide the proof into four cases.
Case 1 If 0 < β < 1 and α < β ≤ 2α + 1, then from (2.4) and (2.6) we clearly see that
Therefore, h(t) < 0 for t ∈ (0, ∞), which follows from (2.7) and (2.8) together with (2.1) and (2.2).
Case 2 If 0 < β < 1 and β ≤ α, then (2.5) and (2.6) lead to
(2.9)
Therefore, h(t) < 0 for t ∈ (0, ∞), which follows from (2.9) and (2.10) together with (2.1) and (2.2).
Case 3 If 1 < β ≤ α, then (2.4) and (2.6) lead to
From equations (2.1) and (2.2) together with inequalities (2.11) and (2.12), we clearly see that h(t) < 0 for t ∈ (0, ∞).
Case 4 If β > 1 and α < β ≤ α + 1, then we clearly see that
(2.13)
From (2.3)-(2.6), we know that
(2.14)
(2.16)
From (2.15)-(2.17), we clearly see that there exists t1> 0 such that h2(t) < 0 for t ∈ (0, t1) and h2(t) > 0 for t ∈ (t1, ∞). Hence, h1(t) is strictly decreasing in [0, t1] and strictly increasing in [t1, ∞).
From (2.2) and (2.14) together with the monotonicity of h1(t), we know that there exists t2> 0 such that h1(t) < 0 for t ∈ (0, t2) and h1(t) > 0 for t ∈ (t2, ∞). Hence, h(t) is strictly decreasing in [0, t2] and strictly increasing in [t2, ∞).
Therefore, h(t) < 0 for t ∈ (0, ∞) follows from (2.1) and (2.13) together with the monotonicity of h(t).
(2) If α + 1 < β < 2α + 1, then we clearly see that
(2.18)
and (2.14)-(2.17) hold again. From the proof of Case 4 in Lemma 2.1(1), we know that there exists λ > 0 such that h(t) is strictly decreasing in [0, λ] and strictly increasing in [λ, ∞).
Therefore, Lemma 2.1(2) follows from (2.1) and (2.18) together with the monotonicity of h(t).
(3) If β ≥ α + 1 and β ≥ 2α + 1, then we divide the proof into three cases.
Case I If β > 1 and β ≥ 2α + 1, then
and it follows from (2.4) that
Equation (2.6) and inequality (2.19) lead to
Therefore, h(t) > 0 for t ∈ (0, ∞) follows from (2.1), (2.2), (2.20), and (2.21).
Case II If 0 < β < 1 and α ≤ -1, then from (2.5) and (2.6) we clearly see that
(2.22)
Inequalities (2.22) and (2.23) imply that
for t ∈ (0, ∞).
Therefore, h(t) > 0 for t ∈ (0, ∞) follows from (2.1) and (2.2) together with (2.24).
Case III If 0 < α + 1 ≤ β < 1, then we clearly see that
(2.25)
It follows from (2.3)-(2.6) that
(2.26)
(2.27)
(2.28)
Inequalities (2.27)-(2.29) imply that there exists t3> 0 such that h2(t) > 0 for t ∈ (0, t3) and h2(t) < 0 for t ∈ (t3, ∞). Hence, h1(t) is strictly increasing in [0, t3] and strictly decreasing in [t3, ∞).
It follows from (2.2) and (2.26) together with the monotonicity of h1(t) that there exists t4> 0 such that h1(t) > 0 for t ∈ (0, t4) and h1(t) < 0 for t ∈ (t4, ∞). Hence, h(t) is strictly increasing in [0, t4] and strictly decreasing in [t4, ∞).
Therefore, h(t) > 0 for t ∈ (0, ∞) follows from (2.1) and (2.25) together with the monotonicity of h(t).
(4) If 2α + 1 < β < α + 1, then we clearly see that
(2.30)
and (2.26)-(2.29) hold again.
From the proof of Case III in Lemma 2.1(3) we know that there exists μ > 0 such that h(t) is strictly increasing in [0, μ] and strictly decreasing in [μ, ∞).
Therefore, Lemma 2.1(4) follows from (2.1) and (2.30) together with the monotonicity of h(t).