I am not sure if this question is too specific on notations (I think the question is intuitive, but basically the only reference I know with this kind of notations is Bingham, Goldie & Teugels book Regular Variation), so let me introduce some things.

Firstly, let $f$ be a measurable positive real function defined in $[a, +\infty)$ for some real $a \geqslant 0$.


Say that $f$ is almost increasing when there is $M\geqslant a$ and $m>0$ such that $$ f(y) \geqslant mf(x),\quad \forall y\geqslant x\geqslant M. $$

The definition of almost decreasing is similar. Now define the upper ($\alpha$) and lower ($\beta$) Matuszewska indices of $f$ as

$$ \alpha(f) := \inf\{\alpha \in \mathbb{R} : x^{-\alpha} f(x) \text{ is almost decreasing}\},$$ $$ \beta(f) := \sup\{\beta \in \mathbb{R} : x^{-\beta} f(x) \text{ is almost increasing}\}.$$

These indices basically bound $f$ between two growth orders "neglecting" some slowly varying functions, i.e. $$ \frac{x^{\beta(f)}}{L_1(x)} \ll f(x) \ll x^{\alpha(f)} L_2(x),$$ for some non-decreasing $L_1(x), L_2(x) = o(x^{\varepsilon})$ for all $\varepsilon > 0$.


$f$ is said to be $O$-regularly varying when $$ \forall \lambda > 0, f(\lambda x) = \Theta(f(x)),$$ that is, for all $\lambda >0$ there is $M=M(\lambda)>0$ such that: $$ \forall x > M, \exists c_1=c_1(\lambda),c_2 = c_2(\lambda)~: ~c_1 f(x) \leqslant f(\lambda x) \leqslant c_2f(x).$$

This is (un?)surprisingly equivalent to say that $f$ has both Matuszewska indices finite (in $(-\infty, +\infty)$).

My question is:

Q: Let $f$ be a measurable, positive, increasing real function in $[0,\infty)$. If $f$ is $O$-regularly varying and $f(x) \gg x^{1/n}$ for some integer $n \geqslant 1$, then the lower Matuszewska index $\beta(f)$ is $> 1/2n$?

To me it is a somewhat intuitive statement (at a first glance I would even say that $\beta(f) \geqslant 1/n$), but I could not prove it or find it explicitly stated on Bingham's et al. book and other references (Feller's Probability books and some papers). I was looking for some help to develop a better intuition on this statement (being it true or false!). Any tips and/or references would be welcome.


Disclaimer: What I tried: http://i.stack.imgur.com/qGooL.png (These calculations didn't help...)

(Here $A\equiv f$ and $h\equiv n$. Sorry, too lazy to fix!)


1 Answer 1


The answer is no. Take any real $a>0$. Let, e.g., $y_k:=e^{k^3}$ and $x_k:=y_k/e^k=e^{k^3-k}$, so that $0<x_1<y_1<x_2<y_2<\dots$. Here in this answer $k$ is any natural number, unless further specified. For $k\ge2$, let
\begin{equation} a_k:=a\,\frac{\ln(y_k/y_{k-1})}{\ln(x_k/y_{k-1})}=a\,\frac{k^3-(k-1)^3}{k^3-k-(k-1)^3}>a,\quad c_k:=y_{k-1}^a/y_{k-1}^{a_k}, \end{equation} so that \begin{equation} c_ky_{k-1}^{a_k}=y_{k-1}^a,\quad c_kx_k^{a_k}=y_k^a, \end{equation} $c_k>0$, and $a_k\to a$; any convergence here is as $k\to\infty$. For each $k\ge2$, let next \begin{equation} f(x):=c_kx^{a_k}\quad\text{if}\quad y_{k-1}\le x\le x_k,\quad f(x):=y_k^a\quad\text{if}\quad x_k\le x\le y_k. \end{equation}

This defines a positive nondecreasing function $f\colon[e,\infty)\to\mathbb R$ such that $f(x)\ge x^a$ for all $x\ge e$. Moreover, $f$ is $O$-regular, since it is a piecewise-power function with alternating exponents $a_k$ and $0$, where $a_k\to a$, so that the positive nondecreasing function $f$ grows no faster than the power function with exponent $2a$ (say). Here are the graphs $\{(\ln x,\ln f(x))\colon y_1\le x\le y_4\}$ and $\{(\ln x,\ln(x^a)\colon y_1\le x\le y_4\}$ for $a=1$:

enter image description here

Yet, the lower Matuszewska index of $f$ is $0$ (rather than $a$ or $a/2$), because for any real $b>0$ one has $y_k^{-b}f(y_k)=o(x_k^{-b}f(x_k))$.

  • $\begingroup$ Wow, this is a beautiful counterexample! It even rules out some additional hypotheses I've been thinking (like $x^{a-\varepsilon} \ll_{\varepsilon} f(x) \ll_{\varepsilon} x^{a+\varepsilon}$, i.e $\forall \varepsilon >0$). It also reminds me of an answer to a question I asked on M.SE last year, but the fact that $a_k\to a$ makes it even better to this particular case. I'll surely add this to my pocket counterexamples :) Thank you very much! $\endgroup$
    – Alufat
    Mar 23, 2016 at 2:13
  • 1
    $\begingroup$ I am glad this was useful. $\endgroup$ Mar 23, 2016 at 2:25

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