Neutral Fixed Point Theorem

Question

I was looking at Coming Up With A Neutral Fixed Points Theorem

when I came across

What I don't understand is how using change of coordinates we can assume $ F(x)=x+\frac{kx^2}{2} $

How can this be done for any function?

Answer 1

First, the statement is missing an assumption on $F$ which I'm guessing is implicit in the context of the chapter - namely, that $F$ is a quadratic polynomial. Lacking that assumption, the functions $$F_1(x) = k + x - k\cos(x)$$ and $$F_2(x) = x + k x^2/2 - k x^4/24,$$ for example, also satisfy the stated conditions for the fixed point $x_0=0$ but are not conjugate to $F(x)=x+kx^2/2$.

So, let's assume that $f(x)=ax^2+bx+c$ satisfies $f(x_0)=x_0$, $f'(x_0)=1$, and $f''(x_0)=k>0$ or, written out long hand: \begin{align} ax_0^2 + bx_0 + c &= x_0 \\ 2ax_0+b &= 1 \\ 2a &= k \end{align}

Solving these for $a$, $b$, and $c$, we find that \begin{align} a&=k/2\\ b&=1-kx_0\\ c&=kx_0^2/2 \end{align} so that $f(x)=kx^2/2+(1-kx_0)x+kx_0^2/2$.

Finally, we can conjugate with $\varphi(x)=x+x_0$ to effectively shift the fixed point to lie at zero. The easy way to do this is to simply set $x_0=0$ in the above expression for $f$ to get $f(x)=kx^2/2+x$.