There exists an function $f$ with this properties?

Question

Let $N >2$ be an fixed natural number. There exist an function $f$ such that?

$(1)$ $f \in C^{1}(\mathbb{R};\mathbb{R}),$

$(2)$ $\lim_{s \to 0} \frac{f(s)}{s}=0$

$(3)$ There exists $C>0$ and $p \in \left(2,\frac{2N}{N-2}\right)$ such that

$$|f(s)| \leq C(|s|+|s|^{p-1})\,\,\ \forall s \in \mathbb{R}, $$

$(4)$ There exists $\mu>2$ such that $$\mu F(t) \leq t f(t)\,\,\, \forall t \in \mathbb{R}, $$ where $F(t)=\int_{0}^{t} f(s)ds, $

$(5)$ There exists $t_0>0$ such that $F(t_0)>0.$

I was studying a elliptic partial differential equation where the non linearity satisfies all these properties, but the text does not show an example of an function satisfying all of this, anyone knows any reference or how to construct functions satisfying properties $(1)$ to $(5)$ ? Any help will be very useful, thank you.

Answer 1

For any $p \in \left(2,\frac{2N}{N-2}\right)$ you can take $$f(s) = \begin{cases} s^{p-1} & \text{if $s \geq 0,$} \\ 0 & \text{if $s < 0.$} \\ \end{cases}$$

As $p > 2$ this is continuously differentiable at $0$, so $(1)$ is satisfied. $(2)$ is just the fact that this function has derivative $0$ at $0$. $(3)$ is satisfied with $ C = 1 $ and with $\mu = p$ both sides of $(4)$ are either $0$ for negative $t$ or $t^{p}$ for nonnegative $t$. $(5)$ is obvious with any positive $t_0$.

Answer 2

Edit: of course, with the new condition (2), the reasoning below completely fails and there are simple solutions, as another answer shows.

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Old answer: Such a function cannot exist. In fact, conditions (2) and (4) (assuming that $f$ is continuous) already imply that $F(s) \leq 0$ for all $s>0$, which contradicts (5).

By (4), the derivative of $g: s \in (0,\infty) \longmapsto s^{-\mu}F(s)$ is $-\mu s^{-mu-1}F(s)+s^{-\mu}f(s)=s^{-\mu-1}(sf(s)-\mu F(s)) \geq 0$, so this function is nondecreasing.

Because $\frac{f(s)}{s} \rightarrow 0$ when $s\rightarrow \infty$ (condition (2)), it follows that $F(s)=o(s^2)$ (as $s \rightarrow \infty$). But since $\mu >2$, $g(s) \rightarrow 0$ as $s \rightarrow \infty$.

Since $g$ is nondecreasing, $g(s) \leq 0$ for all $s>0$, so $F(s) \leq 0$ for all $s>0$.