*2.3. Sufficient Conditions*

In the next result, we assume that the *k*th derivative of *U* exists and we assume that

$$h\_k(\mathbf{x}) = \mathbf{g}(\mathbf{x}) \frac{\mathbf{U}^{(k)}(\mathbf{x})}{\mathbf{U}^{(k-1)}(\mathbf{x})} \to 0,$$

where *<sup>U</sup>*(0)(*x*) = *<sup>U</sup>*(*x*).

$$\text{(a)}\quad \text{If } k=1 \text{, we have } \mathcal{U}'(\mathbf{x})/\mathcal{U}(\mathbf{x}) = \epsilon(\mathbf{x})/g(\mathbf{x}) \text{ with } \epsilon(\mathbf{x}) \to 0 \text{ and } \epsilon$$

$$\int\_{\infty}^{\infty + y\wp(x)} \frac{\mathcal{U}'(z)}{\mathcal{U}(z)} dz = \int\_{\varkappa}^{\varkappa + y\wp(x)} \frac{\epsilon(z)}{\mathcal{g}(z)} dz, \eta$$

so that

$$\log \frac{\mathcal{U}(\mathbf{x} + y\mathbf{g}(\mathbf{x}))}{\mathcal{U}(\mathbf{x})} = \int\_0^y \frac{\epsilon(\mathbf{x} + z\mathbf{g}(\mathbf{x})) \mathbf{g}(\mathbf{x})}{\mathbf{g}(\mathbf{x} + z\mathbf{g}(\mathbf{x}))} dz \to 0, 1$$

and hence

$$\frac{\mathcal{U}(x+y\mathbf{g}(x))}{\mathcal{U}(x)} \to 1.$$

(b) If *k* = 2, then we have

$$\frac{\mathcal{U}'''(x)}{\mathcal{U}'(x)} = \frac{\epsilon(x)}{\mathcal{g}(x)}$$

and

$$\int\_{x}^{x+y\csc(x)} \frac{\mathcal{U}^{\prime\prime}(z)}{\mathcal{U}^{\prime}(z)} dz = \int\_{x}^{x+y\csc(x)} \frac{\epsilon(z)}{\mathcal{g}(z)} dz,$$

so that

$$\begin{aligned} &\log \frac{\mathcal{U}'(\mathbf{x} + y\mathbf{g}(\mathbf{x}))}{\mathcal{U}'(\mathbf{x})} \\ &= \int\_0^y \frac{\epsilon(\mathbf{x} + z\mathbf{g}(\mathbf{x}))}{\mathbf{g}(\mathbf{x} + z\mathbf{g}(\mathbf{x}))} \mathbf{g}(\mathbf{x}) dz \to 0. \end{aligned}$$

We find that

$$\frac{\mathcal{U}'(x+y\mathcal{g}(x))}{\mathcal{U}'(x)} \to 1.$$

Now, consider

$$\begin{aligned} \mathcal{U}(\mathbf{x} + \mathcal{Y}\mathbf{g}(\mathbf{x})) - \mathcal{U}(\mathbf{x}) &= \int\_{\mathbf{x}}^{\mathbf{x} + \mathcal{Y}\mathbf{g}(\mathbf{x})} \mathcal{U}'(\mathbf{z}) d\mathbf{z} \\ &= \mathcal{g}(\mathbf{x}) \int\_{0}^{y} \mathcal{U}'(\mathbf{x} + z\mathbf{g}(\mathbf{x})) dz \\ &= \mathcal{g}(\mathbf{x}) \mathcal{U}'(\mathbf{x}) \int\_{0}^{y} \frac{\mathcal{U}'(\mathbf{x} + z\mathbf{g}(\mathbf{x}))}{\mathcal{U}'(\mathbf{x})} dz \\ &\sim \mathcal{g}(\mathbf{x}) \mathcal{U}'(\mathbf{x}) y. \end{aligned}$$

and then

$$\frac{\mathcal{U}(\mathfrak{x} + y\mathfrak{x}(\mathfrak{x}))}{\mathcal{U}(\mathfrak{x})} - 1 \sim h\_1(\mathfrak{x})y\_{\mathcal{H}}$$

and thus Equation (1) holds with *r*(*x*) = <sup>1</sup>4*<sup>h</sup>*1(*x*). (c) If *k* = 3, as before, we have

$$\frac{\mathcal{U}'''(x+y\mathfrak{g}(x))}{\mathcal{U}''(x)} \to 1$$

and

$$\begin{split} \mathcal{U}'(\mathbf{x} + \mathcal{Y}\mathbf{g}(\mathbf{x})) - \mathcal{U}'(\mathbf{x}) &= \ \mathbf{g}(\mathbf{x}) \mathcal{U}''(\mathbf{x}) \int\_0^y \frac{\mathcal{U}''(\mathbf{x} + z\mathbf{g}(\mathbf{x}))}{\mathcal{U}''(\mathbf{x})} dz \\ &\sim \ \mathbf{g}(\mathbf{x}) \mathcal{U}''(\mathbf{x}) y. \end{split}$$

Further, we have

$$\mathcal{U}(\mathbf{x} + y\mathbf{g}(\mathbf{x})) - \mathcal{U}(\mathbf{x}) = \mathbf{g}(\mathbf{x}) \int\_0^y \mathcal{U}'(\mathbf{x} + z\mathbf{g}(\mathbf{x})) dz$$

and

$$\begin{aligned} &\mathcal{U}(\mathbf{x}+\mathcal{Y}\mathbf{g}(\mathbf{x}))-\mathcal{U}(\mathbf{x})-\mathcal{g}(\mathbf{x})\mathcal{U}'(\mathbf{x})\mathbf{y} \\ &=\quad\mathcal{g}(\mathbf{x})\int\_{0}^{\mathcal{Y}}(\mathcal{U}'(\mathbf{x}+z\mathcal{g}(\mathbf{x}))-\mathcal{U}'(\mathbf{x}))dz \\ &\sim\quad\mathcal{g}^{2}(\mathbf{x})\mathcal{U}''(\mathbf{x})\frac{\mathbf{y}^{2}}{2}.\end{aligned}$$

We conclude that

$$\frac{\mathcal{U}(\mathbf{x} + y\mathbf{g}(\mathbf{x}))}{\mathcal{U}(\mathbf{x})} - 1 - h\_1(\mathbf{x})\mathbf{y} \sim h\_1(\mathbf{x})h\_2(\mathbf{x})\frac{\mathbf{y}^2}{2}.$$

(d) In general, we ge<sup>t</sup> a result of the type

$$\frac{\iota \mathcal{U}(\mathbf{x} + y\mathbf{g}(\mathbf{x}))}{\iota \mathcal{U}(\mathbf{x})} - 1 - \sum\_{i=1}^{k-1} \prod\_{j=1}^{i} h\_j(\mathbf{x}) \frac{y^i}{i!} \sim \prod\_{j=1}^{k} h\_j(\mathbf{x}) \frac{y^k}{k!}.$$

As a special case, we can take *g*(*x*) = 1: if *<sup>U</sup>*(*x*)4*<sup>U</sup>*(*x*) → 0, then

$$\frac{\mathcal{U}(x+y)}{\mathcal{U}(x)} - 1 - \frac{\mathcal{U}'(x)}{\mathcal{U}(x)} y \sim \frac{\mathcal{U}''(x)}{\mathcal{U}(x)} \frac{y^2}{2}.$$

## *2.4. More Results*

**Proposition 1.** *Suppose that <sup>F</sup>*(*x*) = *x*<sup>−</sup>*<sup>α</sup>L*(*x*) *where <sup>L</sup>*(·) *is a normalized slowly varying (SV) function (that is, xL*(*x*)4*L*(*x*) → 0*). Assume that g*(*x*) *and r*(*x*) *satisfy g*(*x*)4*x* → 0 *and r*(*x*)*g*(*x*)4*x* → *δ* > 0*. Then,*

$$r(\mathbf{x}) \left( \frac{\mathbb{F}(\mathbf{x} + y\mathbf{g}(\mathbf{x}))}{\mathbb{F}(\mathbf{x})} - 1 \right) \to -\alpha \delta y.$$

**Proof.** We have *<sup>F</sup>*(*x*) = *<sup>L</sup>*(*x*)*x*<sup>−</sup>*<sup>α</sup>* and then

$$\frac{\overline{F}(\mathbf{x} + y\mathbf{g}(\mathbf{x}))}{\overline{F}(\mathbf{x})} = \frac{L(\mathbf{x} + y\mathbf{g}(\mathbf{x}))}{L(\mathbf{x})} \times \left(1 + y\frac{\mathbf{g}(\mathbf{x})}{\mathbf{x}}\right)^{-\alpha}.$$

It follows that

$$\begin{aligned} \frac{\mathbb{F}(x+y\mathbf{g}(\mathbf{x}))}{\mathbb{F}(\mathbf{x})} - 1 &=& \frac{L(\mathbf{x}+y\mathbf{g}(\mathbf{x}))}{L(\mathbf{x})} \times \left( \left(1+y\frac{\mathbf{g}(\mathbf{x})}{\mathbf{x}}\right)^{-a} - 1 \right) + \frac{L(\mathbf{x}+y\mathbf{g}(\mathbf{x}))}{L(\mathbf{x})} - 1 \\ &=& I(a) + I(b). \end{aligned}$$

For *<sup>I</sup>*(*a*), we have

$$\frac{L(\mathbf{x} + y\mathbf{g}(\mathbf{x}))}{L(\mathbf{x})} = \frac{L\left(\mathbf{x}\left(1 + y\frac{\mathbf{g}(\mathbf{x})}{\mathbf{x}}\right)\right)}{L(\mathbf{x})} \to 1,$$

because *L* is SV and *g*(*x*)4*x* → 0. We also have

$$\left(1+y\frac{\mathbf{g}(\mathbf{x})}{\mathbf{x}}\right)^{-\alpha}-1 \sim -\alpha y \frac{\mathbf{g}(\mathbf{x})}{\mathbf{x}}.$$

so that

$$r(x) \left( \left( 1 + y \frac{\operatorname{g}(x)}{x} \right)^{-\alpha} - 1 \right) \sim -ay \frac{r(x)\operatorname{g}(x)}{x} \to -a\delta y.$$

For the second term, we have

$$\begin{split} r(\mathbf{x})(\frac{L(\mathbf{x}+y\mathbf{g}(\mathbf{x}))}{L(\mathbf{x})}-1) &=& \frac{r(\mathbf{x})}{L(\mathbf{x})} \int\_{\mathbf{x}}^{\mathbf{x}+y\mathbf{g}(\mathbf{x})} L'(\mathbf{t})dt \\ &=& \frac{r(\mathbf{x})}{L(\mathbf{x})} \int\_{\mathbf{x}}^{\mathbf{x}+y\mathbf{g}(\mathbf{x})} \frac{tL'(t)}{L(t)} \frac{L(t)}{t}dt \\ &=& o(1) \frac{r(\mathbf{x})g(\mathbf{x})}{L(\mathbf{x})} \int\_{0}^{y} \frac{L(\mathbf{x}+\theta g(\mathbf{x}))}{\mathbf{x}+\theta g(\mathbf{x})} d\theta \\ &=& o(1) \frac{r(\mathbf{x})g(\mathbf{x})}{\mathbf{x}}. \end{split}$$

We conclude that

$$r(\mathfrak{x}) \left( \frac{L(\mathfrak{x} + y\mathfrak{g}(\mathfrak{x}))}{L(\mathfrak{x})} - 1 \right) \to 0.$$

Combining these results, we obtain the desired result.

**Remark 2.** *The condition on <sup>L</sup>*(*x*) *in the previous theorem is equivalent to the requirement that*

$$\frac{\mathbf{x}f(\mathbf{x})}{\overline{F}(\mathbf{x})} \to a\_{\prime}$$

*where f*(*x*) = *<sup>F</sup>*(*x*) *is the density of F.*
