### Proof of Theorem 3.1

Since *q* satisfies the condition (1.1), we obtain that the equation (E) has a strongly increasing solution.

Let *x* be arbitrary strongly increasing solution of (E) defined on \([T,\infty ), T\geq a\). First, we show that there exist positive constants *m*, *M* such that

$$ m X_{2}(t)\leq x(t)\leq M X_{1}(t),\quad \text{for large } t, $$

(5.1)

where \(X_{1}\) and \(X_{2}\) are given by (4.1) and (4.2), respectively. Integrating \(x'\) on \([T,t]\), we get

$$ x(t)=x(T)+ \int _{T}^{t} x'(s)\,ds\leq x(T)+ x'(t) (t-T), \quad t\geq T,$$

because \(x'\) is increasing. Hence, we find \(K_{1} > 0\) such that

$$ x(t)\leq K_{1} t x'(t),\quad t\geq T. $$

(5.2)

Since *x* is increasing, integration of (E) from *T* to *t* gives

$$ x'(t)=x'(T)+ \int _{T}^{t} q(s) x(s)^{\gamma }\,ds \leq x'(T)+x(t)^{\gamma } \int _{T}^{t} q(s) \,ds, \quad t\geq T,$$

implying, due to the fact \(\int _{T}^{t} q(s)\,ds \rightarrow \infty \) as \(t\to \infty \), that we find \(K_{2}>0\) such that

$$ x'(t) \leq K_{2} x(t)^{\gamma } \int _{T}^{t} q(s)\,ds,\quad t\geq T. $$

(5.3)

By combining (5.2) and (5.3), we have

$$ x(t) \leq K_{1} K_{2} t x(t)^{\gamma } \int _{T}^{t} q(s)\,ds, \quad t \geq T. $$

Thus, there exists \(M>0\) such that

$$ x(t) \leq M \biggl(t \int _{T}^{t} q(s)\,ds \biggr)^{\frac{1}{1-\gamma }},\quad t\geq T. $$

The right-hand side of the inequality (5.1) is proved.

Next, we prove the left-hand side of the inequality (5.1). Set \(w(t)=x(t)x'(t)\) and

$$ \nu =\frac{\gamma +1}{\gamma +3},\qquad \mu =\frac{2}{\gamma +3},\qquad \kappa =\frac{1-\gamma }{\gamma +3}. $$

(5.4)

An application of Young’s inequality gives

$$\begin{aligned} w'(t) =&w(t) \biggl(\frac{q(t)x(t)^{\gamma }}{x'(t)}+\frac{x'(t)}{x(t)} \biggr)\geq \frac{ w(t)}{\mu ^{\mu }\nu ^{\nu }} \biggl( \frac{q(t)x(t)^{\gamma }}{x'(t)} \biggr)^{\mu } \biggl(\frac{x'(t)}{x(t)} \biggr)^{\nu } \\ =& \frac{w(t)}{\mu ^{\mu }\nu ^{\nu }} x(t)^{\gamma \mu -\nu } x'(t)^{ \nu -\mu }q(t)^{\mu }. \end{aligned}$$

Since, \(\gamma \mu -\nu =\nu -\mu =-\kappa \), we get

$$ w'(t)\geq \frac{1}{\mu ^{\mu }\nu ^{\nu }} w(t)^{1-\kappa }q(t)^{\mu }. $$

(5.5)

After dividing (5.5) by \(w(t)^{1-\kappa }\) and integrating the obtained inequality on \([T,t]\), we get that there is \(k_{1}>0\) such that

$$ w(t)^{\kappa }\geq k_{1} \int _{T}^{t} q(s)^{\mu }\,ds, \quad t\geq T $$

or

$$ x(t) x'(t)\geq k_{1}^{\frac{1}{\kappa }} \biggl( \int _{T}^{t} q(s)^{\mu }\,ds \biggr)^{\frac{1}{\kappa }},\quad t\geq T. $$

(5.6)

Integrating (5.6) from *T* to *t*, we find \(k_{2}>0\) and \(T^{*}\geq T\) sufficiently large such that

$$ \frac{x(t)^{2}}{2}\geq k_{2} \int _{a}^{t} \biggl( \int _{a}^{s} q(r)^{\frac{2}{\gamma +3}} \,dr \biggr)^{\frac{\gamma +3}{1-\gamma }} \,ds,\quad t \geq T^{*}. $$

(5.7)

From (5.7), we obtain that there exists \(m>0\) such that the left-hand side of the inequality (5.1) is satisfied.

Next, we prove that *x* is a rapidly varying function of index ∞. Fix arbitrary \(\lambda >1\). Indeed, from (5.1) for sufficiently large *t*, we have

$$ m X_{2}(\lambda t) \leq x(\lambda t)\leq M X_{1}(\lambda t), $$

(5.8)

and

$$ \frac{1}{M X_{1}(t)}\leq \frac{1}{ x( t)}\leq \frac{1}{m X_{2}( t)}. $$

(5.9)

From (5.8) and (5.9), we obtain

$$ \frac{m}{M} \frac{ X_{2}(\lambda t)}{ X_{1}(t)}\leq \frac{x(\lambda t)}{ x( t)}\leq \frac{M}{m} \frac{X_{1}(\lambda t)}{X_{2}( t)} $$

(5.10)

for sufficiently large *t*. By Lemma 4.1 and Lemma 4.2, we have \(X_{1}(t)\stackrel {r}{\sim }X_{2}(t)\), \(t\to \infty \), which means

$$ \lim_{t\to \infty } \frac{ X_{2}(\lambda t)}{ X_{1}(t)}=\lim _{t \to \infty } \frac{X_{1}(\lambda t)}{X_{2}( t)}=\infty . $$

(5.11)

Since *λ* was arbitrary, combining (5.10) and (5.11) gives us \(\lim_{t\to \infty }\frac{x(\lambda t)}{x(t)}=\infty \) for all \(\lambda >1 \), that is, \(x\in \operatorname{RPV}(\infty )\).

It remains to prove that the solution *x* satisfies the asymptotic relation (3.1). Fix arbitrary \(\lambda >1\). Let *m* and *M* be positive numbers, satisfying (5.1) for \(t\geq T_{1}\geq T\). By Lemma 4.1 and Lemma 4.2, we have (4.3) and (4.5), so there exists \(T_{2}=T_{2}(\lambda )\geq T_{1} \) such that

$$ M X_{1}(t)\leq X(\lambda t) \wedge X \biggl( \frac{t}{\lambda } \biggr)\leq m X_{2}(t), \quad t\geq T_{2}. $$

Therefore, from (5.1), we conclude that

$$ X \biggl( \frac{t}{\lambda } \biggr)\leq x(t)\leq X(\lambda t),\quad t\geq T_{2}, $$

(5.12)

implying \(x(t)\stackrel {\star }{\sim }X(t)\), \(t\to \infty \). This completes the proof of Theorem 3.1. □

### Proof of Theorem 3.2

Assumption (1.2) ensures the existence of strongly decreasing solution of (E). Assume that *x* is the arbitrary strongly decreasing solution of (E) defined on \([T,\infty ), T\geq a\). First, we show that there exist positive constants *m* and *M* such that

$$ m Y_{2}(t)\leq x(t)\leq M Y_{1}(t),\quad \text{for large } t, $$

(5.13)

where \(Y_{1}\) and \(Y_{2}\) are given by (4.8) and (4.9), respectively. Since \(x'(t)\to 0\), \(t\to \infty \), and *x* is decreasing, integrating (E) from *t* to ∞, we get

$$ -x'(t)= \int _{t}^{\infty }q(s) x(s)^{\gamma }\,ds\leq x(t)^{\gamma } \int _{t}^{\infty }q(s) \,ds,\quad t\geq T. $$

(5.14)

Dividing (5.14) by \(x(t)^{\gamma }\) and then integrating from *t* to ∞, since \(x(t)\to 0\), \(t\to \infty \), we have

$$ \frac{1}{1-\gamma } x(t)^{1-\gamma }\leq \int _{t}^{\infty } \int _{s}^{\infty }q(r) \,dr\,ds,\quad t\geq T $$

implying that there exists \(M>0\) such that the right-hand side of the inequality (5.13) is satisfied.

Next, we prove the left-hand side of the inequality (5.13). Setting \(w(t)=x(t)|x'(t)|\) and *ν*, *μ*, *κ* as in (5.4), application of Young’s inequality gives

$$\begin{aligned} -w'(t) =&w(t) \biggl(\frac{q(t)x(t)^{\gamma }}{ \vert x'(t) \vert } + \frac{ \vert x'(t) \vert }{x(t)} \biggr)\geq \frac{ w(t)}{\mu ^{\mu }\nu ^{\nu }} \biggl(\frac{q(t)x(t)^{\gamma }}{ \vert x'(t) \vert } \biggr)^{\mu } \biggl( \frac{ \vert x'(t) \vert }{x(t)} \biggr)^{\nu} \\ =& \frac{w(t)}{\mu ^{\mu }\nu ^{\nu }} x(t)^{\gamma \mu -\nu } \bigl\vert x'(t) \bigr\vert ^{ \nu -\mu }q(t)^{\mu }= \frac{1}{\mu ^{\mu }\nu ^{\nu }}w(t)^{1-\kappa }q(t)^{\mu }, \end{aligned}$$

afterwards multiplying by \(w(t)^{\kappa -1}\) and integrating from *t* to ∞, we find \(k_{1}>0\) such that

$$ w(t)^{\kappa }\geq k_{1} \int _{t}^{\infty }q(s)^{\mu }\,ds,\quad t\geq T, $$

or

$$ -x(t) x'(t)\geq k_{1}^{\frac{1}{\kappa }} \biggl( \int _{t}^{\infty }q(s)^{\mu }\,ds \biggr)^{\frac{1}{\kappa }},\quad t\geq T. $$

(5.15)

Since \(x(t)\to 0\), \(t\to \infty \), integrating (5.15) from *t* to ∞ yields that there is \(k_{2}>0\) such that

$$ \frac{x(t)^{2}}{2}\ge k_{2} \int _{t}^{\infty } \biggl( \int _{s}^{\infty }q(r)^{\frac{2}{\gamma +3}} \,dr \biggr)^{\frac{\gamma +3}{1-\gamma }} \,ds. $$

(5.16)

From (5.16), we obtain that there exists \(m>0\) such that the left-hand side of the inequality (5.13) is satisfied.

That \(x\in \operatorname{RPV}(-\infty )\) and satisfies the asymptotic relation (3.3) can be proved in the same way as in the proof of Theorem 3.1, using Lemma 4.3 and Lemma 4.4. □