A problem on basic concepts of Real analysis Let $A$ be the set of all continuous functions from $\mathbb{R} $ to $\mathbb{R} $, satiesfying the following three conditions.
a)$f(x)=f(|x|)$
b)$f(x)=|f(x)|$
c)$f$ has no fixed point.
then,


*

*Identify the set. $\bigcap_{f\in A}\{c\in \mathbb{R} :f(0)\neq c\} $

*If $f\in A$ is bounded, then what is the value of $f(0)$?

*If $f\in A$ is differentiable, the what is the value of $f'(0)$?

*If $p\in A$ and $p$ is a polynomial then, what can you say about the degree of the polynomial?

*$p\in A $ and $p$ is a polynomial and $p(\alpha)=0,$then what can you say about $\alpha$.?


I approach this problem geometrically, from condition (a), the function is symmetric about y-axis. From condition (b) the graph of the function is on the upper half plane. From the condition (c), the graph of the function is above the region of the straight line $y=x$.
Using this we can conclude 


*

*$\{0\}$. I considering the function $|x|+c$.

*$f$ can not be bounded.

*$0$

*The graph can not touch $x$-axis and every odd degree polynomial passes through $x$-axis, hence it is even.

*$\alpha$ is a complex number considering the function $x^2$+1.


I want to verify whether the result is correct or not. I want theoretical proof for all these question, Can anyone help?
 A: *

*$S:=\cap_{f\in A}\{c\in {\mathbb R}:f(0)\neq c\}=(-\infty,0]$
Firstly $(-\infty,0]\subseteq S$: Note that $f(0)\neq 0$ by c), and $f(0)=|f(0)|>0$ by b). Since $f(0)>0$ for any $f\in A$, the inclusion is clear. It remains to show that for any $c>0$, $c\notin S$. This is clear since $g(x)=c+|x|$ lies in $A$ and $g(0)=c$, namely $c\notin S$.


*$f\in A\Rightarrow f$ is unbounded:


One already knows that $f(0)>0$. If $f$ is bounded, then there exist $M,N>0$ such that $f(x)<M,\forall x>N$. It follows that for $N_1:=\max(N+1,M),$ $$f(N_1)<M\leq N_1.$$ By the intermediate value theorem, $\exists 0<x_0<N_1$ such that $f(x_0)=x_0,$ which contradicts c).


*If $f(x)$ is even and differentiable, then $f'(x)$ is odd:


By a), $f$ is even, thus one has $$f(x)=f(-x)$$ $$\Rightarrow f'(x)=-f'(-x)$$ $$\Rightarrow f'(0)=0.$$


*$p=a_nx^n+\cdots+a_0$ must be of even degree:


If not, by 2) and b), the leading coefficient $a_n>0$, but then $\lim_{x\rightarrow -\infty}p(x)=-\infty,$ which gives a contradiction.


*$\alpha$ cannot be real:


As shown above, $p(0)>0$. If $\alpha\in{\mathbb R}$ and $p(\alpha)=0$, then by a) $p(|\alpha|)=0<|\alpha|,$ so by the intermediate value theorem, there exists $\beta\in (0,|\alpha|)$ such that $p(\beta)=\beta$, contradicting c).
