A function $\ell$ is slowly varying (in Karamata's sense) if it is a positive measurable function, defined on some neighbourhood $[X,\infty)$ of infinity, and satisfying $$ \ell(\lambda x)/\ell(x)\to1\quad\text{as}\quad x\to\infty $$ for each $\lambda>0$.

Is the difference $|\ell(\lambda x)-\ell(x)|$ bounded for large values of $x$, i.e. is there an $x_0$ such that the inequality $$ |\ell(\lambda x)-\ell(x)|\le c(\lambda) $$ is true for $x>x_0$?

We have that $|\ell(\lambda x)-\ell(x)|=\ell(x)|\ell(\lambda x)/\ell(x)-1|$. If $\ell$ is bounded, we are done. Hence, we need to investigate the case when $\ell(x)\to\infty$ as $x\to\infty$. We have an indeterminate form $\infty\times0$ in this case and we need to do something else. I can show that the inequality is true for some particular slowly varying functions (for example, the logarithmic function), but I have no idea how to proceed in the general case.

Any help is much appreciated!


1 Answer 1


Let $\ell(x) = \log^2 x$. Then, for $\lambda > 0$, we have \begin{align*} \frac{\ell(\lambda x)}{\ell(x)} &= \frac{(\log \lambda + \log x)^2}{\log^2 x}\\ &= \frac{\log^2 \lambda + 2\log x \log \lambda + \log^2 x}{\log^2 x}\\ &\to 1. \end{align*} so $\ell$ is slowly varying. On the other hand \begin{align*} \ell(\lambda x) - \ell(x) &= 2 \log x \log \lambda + \log^2 \lambda \end{align*} is unbounded, so in general, your claim is not true.


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