Why is $L(x) = x^{o(1)}$ for any slowly varying function $L$?

Question

Question

I've heard that for any slowly varying function $L$, it's true that $L(x) = x^{o(1)}$. But why is this the case?

Note: Here I'm using the standard (Karamata) definition of slow variation, namely a measurable function $L : (0,\infty) \to (0,\infty)$ is called slowly varying if $$ \frac{L(\gamma x)}{L(x)} \underset{x \to \infty}{\longrightarrow} 1 \quad \text{for all $\gamma > 0$.} $$

Thoughts/Attempt

From the Karamata representation of $L$, we have $$ L(x) = \exp\left\{ \eta(x) + \int_B^x \frac{\varepsilon(t)}{t} dt \right\}, \quad \text{for all $x > B$,} $$ for some $B>0$ and some bounded measurable functions $\eta$ and $\varepsilon$ with $\lim_{x \to \infty} \varepsilon(x) = 0$ and $\lim_{x \to \infty} \eta(x) \in \mathbb R$.

So, $$ \begin{aligned} && L(x) &= x^{o(1)} \\ &\Leftrightarrow & \log L(x) &= \eta(x) + \int_B^x \frac{\varepsilon(t)}{t} dt = o(1) \cdot \log x \\ &\Leftrightarrow & \underbrace{\frac{\eta(x)}{\log x} }_{\to 0} + \frac{1}{\log x} \int_B^x \frac{\varepsilon(t)}{t} dt &= o(1) \\ &\Leftrightarrow & \frac{1}{\log x} \int_{\log B}^{\log x} \varepsilon\big(\text e^u \big) du &= o(1) \qquad \text{($u := \log t$)}. \end{aligned} $$ The term on the left-hand side of the last line is clearly $O(1)$. But why is it $o(1)$ -- i.e. why does it have to converge to $0$?

Answer 1

$\newcommand\ep\varepsilon\newcommand\de\delta$Take any real $\de>0$. Since $\ep(t)\to0$ as $t\to\infty$, there is some real $t_\de>B$ such that $|\ep(t)|\le\de$ for all $t\ge t_\de$. So, for $x>\max(1,t_\de)$, $$\Big|\int_{t_\de}^x\frac{\ep(t)}t\,dt\Big| \le\de\,\int_{t_\de}^x\frac1t\,dt=\de(\ln x-\ln t_\de)$$ and hence $$\limsup_{x\to\infty}\frac1{\ln x}\Big|\int_{t_\de}^x\frac{\ep(t)}t\,dt\Big| \le\de\limsup_{x\to\infty}\frac{\ln x-\ln t_\de}{\ln x}=\de.$$ So, $$\limsup_{x\to\infty}\frac1{\ln x}\Big|\int_B^x\frac{\ep(t)}t\,dt\Big| \\ \le\limsup_{x\to\infty}\frac1{\ln x}\Big|\int_B^{t_\de}\frac{\ep(t)}t\,dt\Big| +\limsup_{x\to\infty}\frac1{\ln x}\Big|\int_{t_\de}^x\frac{\ep(t)}t\,dt\Big| \\ \le0+\de=\de,$$ for any real $\de>0$. So, $$\limsup_{x\to\infty}\frac1{\ln x}\Big|\int_B^x\frac{\ep(t)}t\,dt\Big|=0,$$ that is,
$$\lim_{x\to\infty}\frac1{\ln x}\int_B^x\frac{\ep(t)}t\,dt=0,$$ as desired.