Let $\Omega \subset \mathbb{R}^N$ a bounded smooth domain and let $f:\Omega \times \mathbb{R} \to \mathbb{R}$ an Carathéodory function such that
$(f_1): |f(x,s)| \leq c|s|^{\sigma} + d$ for all $x \in \Omega$ and $s \in \mathbb{R}$, where $c,d>0$ and $\sigma>1.$
$(f_2): 0<\mu F(x,s) \leq sf(x,s)$ uniformly in $x$ for $|s| \geq r$, where $r>0$, $\mu>2$ and $F(x,s)=\int_{0}^{s} f(x,\tau)d\tau$.
I'm studying the book An invitation to variational methods in differential equations by David G. Costa and the author says that is easy to show from $(f_1)$ and $(f_2)$ that exists constants $k,l>0$ such that $$F(x,s) \geq k |s|^{\mu}-l\,\,\, \forall s \in \mathbb{R}, \forall x \in \Omega.$$
I'm trying to prove this for hours and I don't know how to see it. My attempt so far:
By the relation $(f_2)$ we have that
$$0<\frac{\mu}{s} \leq \frac{f(x,s)}{F(x,s)}\,\,\, \text{for}\,\,\,|s| \geq r.$$ Then $$\int_{r}^{s} \frac{\mu}{\tau} d \tau \leq \int_{r}^{s} \frac{f(x,\tau)}{F(x,\tau)} d \tau,$$ where the last integral we can calculate as follows: $$\int_{r}^{s} \frac{f(x,\tau)}{F(x,\tau)} d \tau = \int_{r}^{s} [\log F(x,\tau)]' d\tau = \log \frac{F(x,s)}{F(x,r)}. $$ Noticing that $\int_{r}^{s} \frac{\mu}{\tau} d\tau = \mu \log \frac{s}{r}$, we can elevate both sides by $e$ to conclude that $$\left(\frac{s}{r}\right)^{\mu} \leq \frac{F(x,s)}{F(x,r)}\,\,\ \text{for}\,\,\, x \in \Omega\,\,\, \text{and}\,\,\, |s|\geq r. $$
From now I don't know to proceed and use $(f_1)$ to conclude. Every help will be very much appreciated, thank you
I think it is important also to use the fact that in $(f_2)$ the bound $\mu F(x,s)>0$ is uniform in $x$, which means that $$F(x,r) > m > 0$$ for every $x \in \Omega$. From this we can conclude the proof for $s \geq r$: $$F(x,s) \geq F(x,r)\frac{|s|^\mu}{r^\mu} > \frac{m}{r^\mu} |s|^\mu$$ Similarly, if $s \leq -r$ we exploit the fact that $\forall x \in \Omega$ $$F(x,-r) > m > 0$$ Instead, we need to use $(f_1)$ for the case $|s|<r$. We get $$F(x,s)=\int_0^s f(x,\tau) d\tau \geq \int_0^s (-c|\tau|^\sigma-d) d\tau=-c\frac{|s|^{\sigma+1}}{\sigma+1}-sd=\frac{m}{r^\mu} |s|^\mu-\frac{m}{r^\mu} |s|^\mu-c\frac{|s|^{\sigma+1}}{\sigma+1}-sd \geq \frac{m}{r^\mu} |s|^\mu-m-c\frac{r^{\sigma+1}}{\sigma+1}-rd$$