Let me define the N-functions first,
Definition 1: A mapping $\Phi:\mathbb{R^+}\to\mathbb{R^+}$ is said to be an N function if
(1) $\Phi$ is continuous on $\mathbb{R^+}$
(2) $\Phi$ is convex
(3) $\Phi(x)=0$ iff $x=0$
(4) $\lim_{x\to 0}\frac{\Phi(x)}{x}=0$
(5) $\lim_{x\to \infty}\frac{\Phi(x)}{x}=\infty$
$\textbf{1.}$ Let $\Phi:\mathbb{R^+}\to\mathbb{R^+}$ be an N-function, then it will have exactly one fixed point other than zero.
$\textbf{Proof:}$
Since $\Phi$ is an N-function then by definition, $0$ is a fixed point of $\Phi$.
To show that there exists exactly one $0\neq x\in \mathbb{R^+}$ such that $\Phi(x)=x$
$\textbf{Existence:}$ Since $\Phi$ is N-function so $lim_{x\to 0}\frac{\Phi(x)}{x}=0$ implies $\exists \hspace{0.2cm} x_1\in \mathbb{R^+}$ such that $\Phi(x)<x$ $\forall x\leq x_1$ similarly since $lim_{x\to\infty}\frac{\Phi(x)}{x}=\infty$ implies $\exists$ $\hspace{0.2cm}$ $x_2\in \mathbb{R^+}$ such that $\Phi(x)>x \forall x\geq x_2$
$$\mbox{ Let } g(x)=\Phi(x)-x \mbox{ where } x\in[x_1,x_2]$$
since $\Phi(x)$ and $x$ are continuous then $g(x)$ is also continuous.
$$g(x_1)=\Phi(x_1)-x_1<0 \mbox{ and } g(x_2)=\Phi(x_2)-x_2>0$$
hence by intermediate value theorem there exists $c\in[x_1,x_2]$ such that $g(c)=0$ implies $\Phi(c)=c$
$\textbf{Uniqueness:}$ If $\Phi(x_1)<x_1$ and $\Phi(x_2)=x_2$,$0<x_1<x_2$ then for $x>x_2$ convexity implies $\Phi(x)>x$ hence there is only one fixed point.
Here I have proved that the N functions has exactly two fixed points including $0$,I just wanted to know is my proof is correct,please if someone help if there is any mistake that will be great help.Thanks.