**1. Introduction**

The notion of ultimately monotony introduced by Zygmund says that a function *U* ≥ 0 is slowly varying if for each > 0 the function *xU*(*x*) is ultimately increasing and *x*<sup>−</sup>*U*(*x*) is ultimately decreasing ([1], p. 186). A different kind of slowly varying functions was defined by Karamata [2] known as simply the class of slowly varying functions (KSV). It is known that any ZSV function is a KSV function (see [1], p. 186 and, e.g., [3], p. 49).

Recently, Seneta [4] found that the slowly varying functions *L* in the sense of Zygmund are characterized by the relation:

$$\lim\_{x \to \infty} x \left( \frac{L(x+y)}{L(x)} - 1 \right) = 0, \forall y.$$

More recently, Omey and Cadena's [5] functions extended the results of Seneta, and they considered functions for which the following relation holds:

$$\lim\_{\alpha \to \infty} r(x) \left( \frac{L(x + y\mathcal{g}(x))}{L(x)} - 1 \right) = 0, \quad \forall y.$$

Here, the function *g*(*x*) is self-neglecting (notation: *g* ∈ *SN*) and *r* is in the class <sup>Γ</sup>0(*g*) with *r*(*x*) → ∞. The class <sup>Γ</sup>0(*g*) is deeply studied in [6]. Recall that *g* ∈ *SN* if it satisfies

$$\lim\_{x \to \infty} \frac{g(x + yg(x))}{g(x)} = 1,$$

locally uniformly in *y*. In addition, recall that, for *g* ∈ *SN*, we have *f* ∈ <sup>Γ</sup>*α*(*g*) if *f* satisfies

$$\lim\_{\mathfrak{x}\to\infty} \frac{f(\mathfrak{x} + y\mathfrak{g}(\mathfrak{x}))}{f(\mathfrak{x})} = \mathfrak{e}^{\alpha y}, \forall y.$$

Now, we study more general relations of the form

$$\lim\_{x \to \infty} r(x) \left( \frac{\mathcal{U}(x + y\mathcal{g}(x))}{\mathcal{U}(x)} - \varepsilon^{ay} \right) = \theta(y), \quad \forall y.$$

where we assume that the convergence is l.u. in *y*. As before, we assume that *r* ∈ <sup>Γ</sup>0(*g*), *r*(*x*) → ∞ and that *g* ∈ *SN*.

Throughout this paper, we use the notation *f*(*x*) ∼ *g*(*x*) for representing *f*(*x*)4*g*(*x*) → 1 as *x* → ∞.

We study in detail the two cases: *α* = 0 and *α* -= 0. The case *α* = 0 can be considered as the class *SN* with a rate of convergence in the definition. This case is presented in the following section. The case where *α* -= 0 can be considered as the class <sup>Γ</sup>*α*(*g*) with a rate of convergence in the definition. This case is presented in Section 3. For each case, characterizations of the involved functions are provided. Concluding remarks are presented in the last section.

#### **2. The Case** *α* **= 0**

#### *2.1. The Limit Function*

Suppose that *U*, *g*,*<sup>r</sup>* > 0 are measurable functions and suppose that the following relation holds:

$$\lim\_{x \to \infty} r(x) \left( \frac{\mathcal{U}(x + yg(x))}{\mathcal{U}(x)} - 1 \right) = \theta(y), \tag{1}$$

and we assume that Equation (1) holds locally uniformly in *y*. As before, we assume that *r*(*x*) → <sup>∞</sup>, *r* ∈ <sup>Γ</sup>0(*g*) and that *g* ∈ *SN*.

Clearly, Equation (1) holds if and only if

$$\lim\_{\mathbf{x}\to\mathbf{x}} r(\mathbf{x}) \left( \mathcal{W}(\mathbf{x} + y\mathbf{g}(\mathbf{x})) - \mathcal{W}(\mathbf{x}) \right) = \theta(\mathbf{y}),\tag{2}$$

where *<sup>W</sup>*(*x*) = log *<sup>U</sup>*(*x*).

> Now, we replace *x* by *x* = *t* + *zg*(*t*). Note that *g*(*t*)4*<sup>t</sup>* → 0 so that *x*4*<sup>t</sup>* → 1 l.u. in *z*. We find

$$\lim\_{t \to \infty} r(t + z\emptyset(t)) \left( \mathcal{W}(t + z\emptyset(t) + y\emptyset(t + z\emptyset(t))) - \mathcal{W}(t + z\emptyset(t)) \right) = \theta(y).$$

 Using *r* ∈ <sup>Γ</sup>0(*g*), we have

$$\lim\_{t \to \infty} r(t) \left( \mathcal{W}(t + z\mathfrak{g}(t) + y\mathfrak{g}(t + z\mathfrak{g}(t))) - \mathcal{W}(t + z\mathfrak{g}(t)) \right) = \theta(y),$$

and then it follows that

$$\lim\_{t \to \infty} r(t) \left( \mathcal{W}(t + z\mathfrak{g}(t) + y\mathfrak{g}(t + z\mathfrak{g}(t))) - \mathcal{W}(t) \right) = \theta(y) + \theta(z).$$

Now, we have

$$\mathcal{W}(t+z\mathbf{g}(t)+y\mathbf{g}(t+z\mathbf{g}(t)))-\mathcal{W}(t) = \mathcal{W}\left(t+\left(z+y\frac{\mathbf{g}(t+z\mathbf{g}(t))}{\mathbf{g}(t)}\right)\mathbf{g}(t)\right)-\mathcal{W}(t).$$

Using l.u. convergence, we obtain that

$$\lim\_{t \to \infty} r(t)(\mathcal{W}(t + z\emptyset(t) + y\emptyset(t + z\emptyset(t))) - \mathcal{W}(t)) = \theta(y + z).$$

We conclude that

$$
\theta(z+y) = \theta(z) + \theta(y),
$$

and (since *θ* is measurable) hence also that *<sup>θ</sup>*(*y*) = *θy* for some constant *θ*.

Conversely, we have the following (cf. [6]): if

$$\lim\_{\chi \to \infty} r(\mathbf{x}) \left( \frac{\mathcal{U}(\mathbf{x} + y\mathbf{g}(\mathbf{x}))}{\mathcal{U}(\mathbf{x})} - 1 \right) = \theta y\_{\mathbf{x}}$$

then this relation holds l.u. in *y*.

> To conclude, we have the following theorem.

**Theorem 1.** *Assume that g* ∈ *SN and that r* ∈ <sup>Γ</sup>0(*g*) *with r*(*x*) → ∞*.*


Three different ways to represent the functions satisfying Equation (1) follow.

## 2.2.1. First Form

For further use, let *<sup>A</sup>*(*x*) = *xa* 1/*g*(*t*)*dt*. Clearly, we have

$$A(\mathbf{x} + y\mathbf{g}(\mathbf{x})) - A(\mathbf{x}) = \int\_0^y \frac{\mathbf{g}(\mathbf{x})}{\mathbf{g}(\mathbf{x} + z\mathbf{g}(\mathbf{x}))} dz \to y$$

l.u. in *y*. Note that *<sup>A</sup>*(*x*) is an increasing function so that *fx*(*y*) = *<sup>A</sup>*(*x* + *yg*(*x*)) − *<sup>A</sup>*(*x*) is an increasing function of *y* for which *fx*(*y*) → *y* as *x* → ∞. As a consequence, the inverse function also satisfies *f* −1 *x* (*y*) → *y*. To calculate the inverse, we set

$$f\_x(y) = A(x + y\mathbb{g}(x)) - A(x) = t$$

so that *x* + *yg*(*x*) = *<sup>A</sup>*−<sup>1</sup>(*<sup>t</sup>* + *<sup>A</sup>*(*x*)) and

$$y = f\_x^{-1}(t) = \frac{A^{-1}(t + A(x)) - A^{-1}(A(x))}{g(x)}.$$

We conclude that

$$\frac{A^{-1}(t+A(x)) - A^{-1}(A(x))}{g(x)} \to t\_r$$

so that (replacing *<sup>A</sup>*(*x*) by *x* and *t* by *y*)

$$\frac{A^{-1}(x+y) - A^{-1}(x)}{g(A^{-1}(x))} \to y,$$

l.u. in *y*.

> Now, let *<sup>K</sup>*(*x*) := *<sup>W</sup>*(*<sup>A</sup>*−<sup>1</sup>(*x*)). We have (using l.u. convergence in the last step):

$$\begin{aligned} r(\mathbf{r})(\mathbf{K}(A(\mathbf{x}) + \mathbf{y}) - \mathbf{K}(A(\mathbf{x}))) &= \quad r(\mathbf{x})(\mathcal{W}(A^{-1}(A(\mathbf{x}) + \mathbf{y})) - \mathcal{W}(\mathbf{x})) \\ &= \quad r(\mathbf{x}) \left( \mathcal{W} \left( \mathbf{x} + \mathbf{g}(\mathbf{x}) \frac{A^{-1}(A(\mathbf{x}) + \mathbf{y}) - \mathbf{x}}{\mathbf{g}(\mathbf{x})} \right) - \mathcal{W}(\mathbf{x}) \right) \\ &\to \quad \theta \mathbf{y}. \end{aligned}$$

It follows that

$$r(A^{-1}(\mathbf{x}))(\mathcal{K}(\mathbf{x}+\mathbf{y}) - \mathcal{K}(\mathbf{x})) \to \theta \mathbf{y} \tag{3}$$

l.u. in *y*. Taking the integral 1*<sup>y</sup>*=<sup>0</sup>(.)*dy* in Equation (3) we have

$$\int\_{y=0}^{1} r(A^{-1}(x))(K(x+y) - K(x))dy \to \int\_{0}^{1} \theta y dy$$

or

$$r(A^{-1}(\mathbf{x})) \int\_{\mathbf{x}}^{\mathbf{x}+1} K(z)dz - r(A^{-1}(\mathbf{x}))K(\mathbf{x}) \to \frac{\theta}{2}.$$

We see that *<sup>K</sup>*(*x*) is of the form

$$\begin{aligned} K(\mathbf{x}) &= \, ^C \mathbf{C} + \int\_{\mathbf{x}}^{\mathbf{x}+1} K(\mathbf{z}) d\mathbf{z} + \frac{\mathbf{C}(\mathbf{x})}{r(A^{-1}(\mathbf{x}))} \\ &= \, ^L L(\mathbf{x}) + \frac{\mathbf{C}(\mathbf{x})}{r(A^{-1}(\mathbf{x}))} \end{aligned}$$

where *<sup>C</sup>*(*x*) → *C*(= *θ*/2) and *<sup>L</sup>*(*x*) = *x*+1 *x <sup>K</sup>*(*z*)*dz*. Note that

$$r(A^{-1}(\mathbf{x})L'(\mathbf{x}) = r(A^{-1}(\mathbf{x})(K(\mathbf{x}+1) - K(\mathbf{x})) \to \theta.)$$

Using *<sup>W</sup>*(*x*) = *<sup>K</sup>*(*A*(*x*)), we find that

$$\mathcal{W}(\mathbf{x}) = T(\mathbf{x}) + \frac{\mathbf{C}^{\diamond}(\mathbf{x})}{r(\mathbf{x})},$$

where *<sup>C</sup>*◦(*x*) = *<sup>C</sup>*(*A*(*x*)) → *C* and *<sup>T</sup>*(*x*) = *<sup>L</sup>*(*A*(*x*)). Note that

$$r(\mathbf{x})\,\mathbf{g}(\mathbf{x})T'(\mathbf{x}) = r(\mathbf{x})L'(A(\mathbf{x}))\mathbf{g}(\mathbf{x})A'(\mathbf{x}) = r(\mathbf{x})L'(A(\mathbf{x})) \to \theta.$$

We prove the following result:

**Theorem 2.** *Assume that g* ∈ *SN and that r* ∈ <sup>Γ</sup>0(*g*)*, r*(*x*) → ∞*.*

*(a) If Equation (1) holds with <sup>θ</sup>*(*x*) = *θx, then <sup>W</sup>*(*x*) = log *<sup>U</sup>*(*x*) *is of the form*

$$W(\mathbf{x}) = T(\mathbf{x}) + \frac{C(\mathbf{x})}{r(\mathbf{x})},$$

*where <sup>C</sup>*(*x*) → *C and r*(*x*)*g*(*x*)*<sup>T</sup>*(*x*) → *θ.*

*(b) If <sup>W</sup>*(*x*) = *<sup>T</sup>*(*x*) + *<sup>C</sup>*(*x*)4*r*(*x*)*, where <sup>C</sup>*(*x*) → 0 *and r*(*x*)*g*(*x*)*<sup>T</sup>*(*x*) → *θ, then Equation (1) holds with <sup>θ</sup>*(*y*) = *θy.*

**Proof.** The proof of (a) is given above. To prove (b), we have

$$\begin{aligned} &W(\mathbf{x} + y\mathbf{g}(\mathbf{x})) - W(\mathbf{x}) \\ &= \quad T(\mathbf{x} + y\mathbf{g}(\mathbf{x})) - T(\mathbf{x}) + \frac{\mathbf{C}(\mathbf{x} + y\mathbf{g}(\mathbf{x}))}{r(\mathbf{x} + y\mathbf{g}(\mathbf{x}))} - \frac{\mathbf{C}(\mathbf{x})}{r(\mathbf{x})}. \end{aligned}$$

Clearly, we have

$$r(\mathbf{x})(T(\mathbf{x} + y\mathbf{g}(\mathbf{x})) - T(\mathbf{x})) = yr(\mathbf{x})\mathbf{g}(\mathbf{x})T'(\mathbf{x} + \beta\mathbf{g}(\mathbf{x})) $$

for some *β* ∈ (0, *y*). It follows that

$$\begin{split} r(\mathbf{x})(T(\mathbf{x} + y\mathbf{g}(\mathbf{x})) - T(\mathbf{x})) &= \quad \mathcal{Y}(\boldsymbol{\theta} + o(1)) \frac{r(\mathbf{x})\mathbf{g}(\mathbf{x})}{r(\mathbf{x} + \beta \mathbf{g}(\mathbf{x}))r(\mathbf{x} + \beta \mathbf{g}(\mathbf{x}))} \\ &\to \quad \mathcal{Y}\boldsymbol{\theta}. \end{split}$$

For the second term, we have

$$r(\mathbf{x}) \left( \frac{\mathbb{C}(\mathbf{x} + y\mathbf{g}(\mathbf{x}))}{r(\mathbf{x} + y\mathbf{g}(\mathbf{x}))} - \frac{\mathbb{C}(\mathbf{x})}{r(\mathbf{x})} \right) = \frac{r(\mathbf{x})}{r(\mathbf{x} + y\mathbf{g}(\mathbf{x}))} \mathbb{C}(\mathbf{x} + y\mathbf{g}(\mathbf{x})) - \mathbb{C}(\mathbf{x}) \to 0.$$

The result follows.
