I'm trying to understand the difference between two different definitions of the Nussbaum property for real functions.
Have a look at Definition 1 ( taken from https://doi.org/10.1016/0167-6911(83)90021-X):
An even an differentiable function $f: \mathrm{R} \to \mathrm{R}$ is said to have the Nussbaum property, if for every $y_0 \in \mathrm{R}$, $$ \sup_{y > y_0} \int_{y_0}^y h(s)\, ds = \infty $$ and $$ \inf_{y > y_0} \int_{y_0}^y h(s)\, ds = -\infty. $$
Now look at the more ad-hoc Definition 2 (taken from some lecture notes, p.22)
A function $f$ is said to have the Nussbaum property, if
$$ \sup_{y> 0} \frac 1 y \int_0^y f(s) s \,ds = \infty $$ and $$ \inf_{y>0} \frac 1 y \int_0^y f(s) s \, ds = - \infty. $$
Examples:
- $f(s) = s \cos(s)$ has the Nussbaum property only according to Definition 2 (plot for D2)(plot for D1), as $f$ is not even
- $f(s) = s \sin(s)$ has the Nussbaum property according to Definition 1 and Definition 2 (plot for D1)(plot for D2)
- $f(s) = s^2 \cos(s)$ has the Nussbaum property according to Definition 1 and Definition 2 (plot for D1)(plot for D2)
One more example for a non-differentiable function
- $f(s) = \begin{cases} 1, &\text{if } 0 \leq s < 1\\ -1, &\text{if } 1 \leq s < 3\\ 1, &\text{if } 3 \leq s < 6\\ -1, &\text{if } 6 \leq s < 10\\ 1, &\text{if } 10 \leq s < 15\\ \ldots \end{cases}$ is a Nussbaum function according to Definition 2, but not according to Definition 1, since it is clearly not differentiable.
Clearly, there exist functions which are Nussbaum according to Def.2, but not according to Def.1. Hypothesis: Are the Def-1-Nussbaum functions a subset to the Def-2-Nussbaum functions then?
On the other hand, requiring $y > 0$ in Def.1 is much more general than requiring $y > y_0$ for every $y_0 \in \mathrm{R}$ as in Def.2. However, since we're dealing with suprema and infima, we surely can neglect some compact intervals in most but the edgiest cases?
Can anyone give a counterexample to my Hypothesis or give some hints or ideas on how I should start the formal proof?
Lastly, I consider coming up with my own Definition 3 as a variant to Definition 2:
A function $f$ is said to have the Nussbaum property, if for every $y_0 \in \mathrm{R}$,
$$ \sup_{y> y_0} \frac 1 {y- y_0} \int_{y_0}^y f(s) s \,ds = \infty $$ and $$ \inf_{y> y_0} \frac 1 {y - y_0} \int_{y_0}^y f(s) s \, ds = - \infty. $$
If a function is a Def-3-Nussbaum function, it also is a Def-2-Nussbaum function per definition. Is the opposite also true, i.e. are Def-3 and Def-2 equivalent?