**2. Preliminary Lemmas**

**Notation 1.** *The class of all eventually positive solutions of DDE (4) is denoted by the symbol* P*s.*

**Notation 2.** *To facilitate the presentation of the results, we define the function* Q *as*

$$\mathcal{Q} := q \cdot \left[ F \circ \left( 1 - [p \circ \mathcal{g}] \cdot \frac{[\mathcal{E}\_{n-2} \circ h \circ \mathcal{g}]}{[\mathcal{E}\_{n-2} \circ \mathcal{g}]} \right) \right].$$

**Lemma 1** (Lemma 3, [32])**.** *Suppose that x* ∈ P*s. Then,*

$$\frac{\mathbf{d}}{\mathbf{d}u} \left( a \cdot \frac{\mathbf{d}^{n-1}}{\mathbf{d}u^{n-1}} \omega \right) \le 0,$$

*eventually. Moreover, one of the following conditions is satisfied eventually:*


(D3) (−1) *<sup>m</sup>ω*(*m*) *are positive, for m* <sup>=</sup> 0, 1, . . . , *<sup>n</sup>* <sup>−</sup> 1.

**Lemma 2.** *Suppose that x* ∈ P*<sup>s</sup> and ω satisfies case* (D3) *in Lemma 1. Then, eventually,*

$$(-1)^{s+1} \frac{\mathbf{d}^s}{\mathbf{d}u^s} \omega \le \left[ a \cdot \frac{\mathbf{d}^{n-1}}{\mathbf{d}u^{n-1}} \omega \right] \cdot \mathcal{E}\_{n-s-2s} \tag{5}$$

*for s* = 0, 1, . . . , *n* − 2*.*

**Proof.** Since *<sup>a</sup>* · *<sup>ω</sup>*(*n*−1) <sup>≤</sup> 0, *<sup>ω</sup>*(*n*−1) <sup>≤</sup> 0 and *<sup>ω</sup>*(*n*−2) <sup>&</sup>gt; 0 for all *<sup>u</sup>* <sup>≥</sup> *<sup>u</sup>*1, where *<sup>u</sup>*<sup>1</sup> <sup>≥</sup> *<sup>u</sup>*0, we conclude that

$$
\omega^{(n-2)}(\boldsymbol{u}) \ge -\int\_{\boldsymbol{u}}^{\infty} \left[ a(\boldsymbol{\upsilon}) \omega^{(n-1)}(\boldsymbol{\upsilon}) \right] \boldsymbol{a}^{-1}(\boldsymbol{\upsilon}) \mathrm{d}\boldsymbol{\upsilon} \ge -a(\boldsymbol{u}) \omega^{(n-1)}(\boldsymbol{u}) \mathcal{E}\_0(\boldsymbol{u}).
$$

Integrating this inequality *n* − 2 times from *u* to ∞, we obtain

$$\begin{aligned} \left(\omega^{(n-3)}(\mathfrak{u})\right) &\leq \int\_{\mathfrak{u}}^{\infty} \left[a(\upsilon)\omega^{(n-1)}(\upsilon)\right] \mathcal{E}\_{0}(\upsilon) \mathsf{d}\upsilon \\ &\leq \left. a(\mathfrak{u})\omega^{(n-1)}(\mathfrak{u}) \int\_{\mathfrak{u}}^{\infty} \mathcal{E}\_{0}(\upsilon) \mathsf{d}\upsilon \right. \\ &= \left. a(\mathfrak{u})\omega^{(n-1)}(\mathfrak{u}) \mathcal{E}\_{1}(\mathfrak{u}) \right. \end{aligned}$$

and so on until we get

$$(-1)^{s+1} \omega^{(s)} \le \left[ a \cdot \omega^{(n-1)} \right] \cdot \mathcal{E}\_{n-s-2s}$$

for *s* = 0, 1, . . . , *n* − 2. The proof is complete.

**Lemma 3.** *Suppose that x* ∈ P*<sup>s</sup> and ω satisfies case* (D3) *in Lemma 1. Then,*

$$(-1)^{s} \frac{\mathrm{d}}{\mathrm{d}\mu} \left( \frac{1}{\mathcal{E}\_{\mathrm{n}-s-2}} \cdot \frac{\mathrm{d}^{s}}{\mathrm{d}\mu^{s}} \omega \right) \geq 0,\tag{6}$$

*for s* = 0, 1, . . . , *n* − 2*.*

**Proof.** From Lemma 2, we have that (5) holds. Using (5) with *<sup>s</sup>* <sup>=</sup> *<sup>n</sup>* <sup>−</sup> 2, we find *<sup>ω</sup>*(*n*−2) <sup>≥</sup> − *<sup>a</sup>* · *<sup>ω</sup>*(*n*−1) · E0. Thus,

$$\frac{\mathbf{d}}{\mathbf{d}\mu} \left( \frac{1}{\mathcal{E}\_0} \cdot \boldsymbol{\omega}^{(n-2)} \right) = \frac{1}{\mathcal{E}\_0^2} \left[ \mathcal{E}\_0 \cdot \boldsymbol{\omega}^{(n-1)} + \boldsymbol{a}^{-1} \cdot \boldsymbol{\omega}^{(n-2)} \right] \ge 0.$$

Furthermore, we find

$$-\omega^{(n-3)}(u) \ge \int\_u^\infty \left[\frac{1}{\mathcal{E}\_0(\upsilon)} \omega^{(n-2)}(\upsilon)\right] \mathcal{E}\_0(\upsilon) d\upsilon \ge \frac{1}{\mathcal{E}\_0(u)} \omega^{(n-2)}(u) \mathcal{E}\_1(u),$$

and so, −E<sup>0</sup> · *<sup>ω</sup>*(*n*−3) ≥ E<sup>1</sup> · *<sup>ω</sup>*(*n*−2). This implies

$$\frac{\mathbf{d}}{\mathbf{d}\mu} \left( \frac{1}{\mathcal{E}\_1} \cdot \boldsymbol{\omega}^{(n-3)} \right) = \frac{1}{\mathcal{E}\_1^2} \left[ \mathcal{E}\_1 \cdot \boldsymbol{\omega}^{(n-2)} + \mathcal{E}\_0 \cdot \boldsymbol{\omega}^{(n-3)} \right] \le 0.1$$

Proceeding in this manner, we arrive at

$$\frac{\mathbf{d}}{\mathbf{d}u} \left( \frac{1}{\mathcal{E}\_{n-2}} \cdot \boldsymbol{\omega} \right) \ge 0.$$

The proof is complete.

**Lemma 4.** *Suppose that x* ∈ P*<sup>s</sup> and ω satisfies case* (D3) *in Lemma 1. Then,*

$$\alpha \ge \left(1 - p \cdot \frac{[\mathcal{E}\_{n-2} \circ h]}{\mathcal{E}\_{n-2}}\right) \cdot \omega\_n$$

*and*

$$\frac{\mathbf{d}}{\mathbf{d}u} \left( a \cdot \frac{\mathbf{d}^{n-1}}{\mathbf{d}u^{n-1}} \omega \right) + \mathcal{Q} \cdot [F \circ \omega \circ \mathcal{g}] \le 0. \tag{7}$$

**Proof.** From Lemma 3, we have that (6) holds. Using (6) with *s* = 0, we obtain

$$\left[\left[\mathfrak{x}\circ h\right]\leq\left[\omega\circ h\right]\leq\frac{\left[\mathcal{E}\_{n-2}\circ h\right]}{\mathcal{E}\_{n-2}}\cdot\omega.\right]$$

This implies

$$\begin{aligned} \|\mathbf{x}\| &\geq \quad \left(1 - p \cdot \frac{[\boldsymbol{\omega}\circ h]}{\omega}\right) \cdot \omega \\ &\geq \quad \left(1 - p \cdot \frac{[\boldsymbol{\mathcal{E}}\_{n-2}\circ h]}{\boldsymbol{\mathcal{E}}\_{n-2}}\right) \cdot \omega .\end{aligned}$$

Thus, from (4) and (H3), we get

$$\frac{\mathbf{d}}{\mathbf{d}u} \left( a \cdot \frac{\mathbf{d}^{n-1}}{\mathbf{d}u^{n-1}} \omega \right) + q \cdot \left[ F \diamond \left( 1 - \left[ p \diamond g \right] \cdot \frac{\left[ \mathcal{E}\_{n-2} \diamond h \circ g \right]}{\left[ \mathcal{E}\_{n-2} \diamond g \right]} \right) \right] \cdot \left[ F \diamond \omega \circ g \right] \le 0, \quad \mathbf{d}u = \mathbf{0}, \quad \mathbf{d}u = \mathbf{0}$$

or

$$\frac{\mathbf{d}}{\mathbf{d}u} \left( a \cdot \frac{\mathbf{d}^{n-1}}{\mathbf{d}u^{n-1}} \omega \right) + \mathcal{Q} \cdot [F \diamond \omega \circ \mathbf{g}] \le 0.$$

The proof is complete.

**Lemma 5.** *Suppose that x* ∈ P*<sup>s</sup> and ω satisfies case* (D3) *in Lemma 1. If there is a constant α* > 0 *such that, eventually,*

$$\mathbb{Q} \cdot \mathcal{E}\_{n-2}^2 \ge \kappa \mathcal{E}\_{n-3\prime} \tag{8}$$

*then ω converges to zero as u* → ∞*.*

**Proof.** Since *ω* is positive and decreasing (from case (D3)), we have that *ω*(*u*) → *L* as *u* → ∞, where *L* ≥ 0.

Suppose that *L* > 0. Then, there is *u*<sup>1</sup> ≥ *u*<sup>0</sup> such that [*ω* ◦ *g*] ≥ *L* for *u* ≥ *u*1. Hence, from Lemma 4, we get

$$\frac{\mathbf{d}}{\mathbf{d}\boldsymbol{\mu}}\Big(a(\boldsymbol{\mu})\frac{\mathbf{d}^{n-1}}{\mathbf{d}\boldsymbol{\mu}^{n-1}}\boldsymbol{\omega}(\boldsymbol{\mu})\Big) \leq -\mathcal{Q}(\boldsymbol{\mu})\boldsymbol{F}(\boldsymbol{L}).$$

Integrating this inequality from *u*<sup>1</sup> to *u* and using (8), we find

$$\begin{split} a(u)\omega^{(n-1)}(u) &\leq & -a(u\_1)\omega^{(n-1)}(u\_1) - F(L) \int\_{u\_1}^{u} \mathcal{Q}(\upsilon) \mathrm{d}\upsilon \\ &\leq & -aF(L) \int\_{u\_1}^{u} \frac{\mathcal{E}\_{n-3}(\upsilon)}{\mathcal{E}\_{n-2}^2(\upsilon)} \mathrm{d}\upsilon \\ &=& -aF(L) \left(\frac{1}{\mathcal{E}\_{n-2}(u)} - \frac{1}{\mathcal{E}\_{n-2}(u\_1)}\right). \end{split} \tag{9}$$

We note that <sup>E</sup>*n*−2(*u*) <sup>→</sup> 0 as *<sup>u</sup>* <sup>→</sup> <sup>∞</sup>. Then, for all <sup>∈</sup> (0, 1), we have that <sup>E</sup>−<sup>1</sup> *<sup>n</sup>*−2(*u*) <sup>−</sup> <sup>E</sup>−<sup>1</sup> *<sup>n</sup>*−2(*u*1) <sup>≥</sup> E−<sup>1</sup> *<sup>n</sup>*−2(*u*), eventually. Therefore, (9) becomes

$$a \cdot \omega^{(n-1)} \le -\epsilon \mathfrak{a} F(L) \frac{1}{\mathcal{E}\_{n-2}}.\tag{10}$$

From Lemma 2, we have that (5) holds. Combining (5) at *s* = 1 and (10), we conclude that

$$\frac{\omega'}{\mathcal{E}\_{n-3}} \le a \cdot \omega^{(n-1)} \le -\epsilon \alpha F(L) \frac{1}{\mathcal{E}\_{n-2}}.$$

*ω* ≤ −*αF*(*L*) E*n*−3 E*n*−2 .

Integrating this inequality from *u*<sup>1</sup> to *u*, we find

$$
\omega(\mathfrak{u}) \le \omega(\mathfrak{u}\_1) - \epsilon \alpha F(L) \ln \frac{\mathcal{E}\_{\mathfrak{u}-2}(\mathfrak{u}\_1)}{\mathcal{E}\_{\mathfrak{u}-2}(\mathfrak{u})}.
$$

Thus, *ω*(*u*) → −∞ as *u* → ∞, a contradiction. Then, *ω*(*u*) → 0 as *u* → ∞. The proof is complete.
