Let $f: [0,\infty)\to [0,\infty)$ be continuous function and $\underline{f}(m):=\min\{f(y): {\frac{1}{4} m\le y\le m}\}$.
I'd like to prove that $(i)$ if $f_0=\infty,$ then $(\underline{f})_0=\infty$ and $(ii)$ if $f_\infty=\infty,$ then $(\underline{f})_\infty=\infty.$ Here, $\displaystyle f_0:=\lim_{s\to 0^+}\frac{f(s)}{s}$ and $\displaystyle f_\infty:=\lim_{s\to \infty}\frac{f(s)}{s}$.
For example, Let $f(s):=\left\{ \begin{array}{ll} s^{\frac{1}{2}}, & \hbox{for}~s\in [0,1) \\ s^{2}, & \hbox{for}~s\in [1,\infty), \end{array} \right.$ Then it can be easily proved that the above assertions are true, since $f$ is increasing in $[0,\infty)$ and $\underline{f}(m)=f(\frac{m}{4}).$
I was wondering if the above assertions are true for any continuous function $f$ or for $f$ with suitable assumptions. Please let me know if you have any idea or comment for my question.
Thanks in advance.
Let $f:[0,\infty)\to [0,\infty)$ be continuous. For each $m\in \mathbb{R}_+,$ by extreme value theorem, there exists $m_* \in [\frac{m}{4},m]$ satisfying $\underline f(m)=f(m_*),$ and $\frac{\underline f(m)}{m^{p-1}}=\frac{f(m_*)}{m_*^{p-1}}(\frac{m_*}{m})^{p-1}\ge \frac{1}{4^{p-1}}\frac{f(m_*)}{m_*^{p-1}}.$ As $m \to c\in\{0,\infty\},$ $m_* \to c$, and thus $(\underline f)_c=\infty$, provided $f_c=\infty$.