I'm currently reading Saber Elaydi's Introduction to Difference Equations. I am confused with the proof of theorem 1.15 which states the following:
Let $x^*$ be an equilibrium point of $x(n+1)=f(x(n))$ where $f$ is continuously differentiable at $x^*$. Then it follows that:
If $f''(x^*)\neq 0$, then $x^*$ is unstable.
If $f''(x^*)=0$ and $f'''(x^*)>0$, then $x^*$ is unstable.
If $f''(x^*)=0$ and $f'''(x^*)<0$, then $x^*$ is asymptotically stable.
The proof of the first item is given in the book (pp. 30), while the other two are not. Those are left as exercises.
The proof of the first item starts off by stating that $f$ must be either concave upward or downward depending on the sign of $f''(x^*)$. Then we assume $f''(x^*)> 0$, since this is the case, it follows that $f'$ is increasing on a neighborhood of $x^*$.
My first question comes from this part. The author states:
If $f''(x^*)>0$, then $f'(x)>1$ for all $x$ in a small interval $\rbrack x^*,x^*+\varepsilon\lbrack$.
Then it would follow by another stability criterion that $f$ is unstable at $x$.
Does that hold for the whole neighborhood of $x$? Including $x^*$, even if it is in the boundary of the neighborhood? I don't know how to justifiy this.
How does $f''(x^*)>0$, imply $f'(x)>1$? My first attempt to answer this question was to say: Since $f'$ grows near $x^*$ then in some neighborhood it must grow around 1.
I know it sounds wrong, but it didn't when I thought of it the first time. The problem with my idea is that $f'$ can be bounded and I don't know if $\varepsilon $ can be made as large as I want. This also threw me off when reading the next part.
(...) if $f''(x^*)<0$, then $f'(x)>1$ for $x\in\rbrack x^*-\varepsilon,x^*\lbrack$.
Once again by a previous criterion, $x^*$ is unstable.
How does $f'(x)>1$ follow from $f''(x^*)<0$? If the same conclusion followed from $f''(x^*)>0$, in which way is that hypothesis being used?
That $f''(x^*)<0$ implies that $f'$ is decreasing in a neighborhood of $x^*$. I know that from the left, the values of $f'$ must be bigger than $f'(x^*)$. But what if $f'$ is bounded, or always negative? I can't wrap my head around this.
These are all my questions for the first part of the proof. I tried gathering some ideas for the second statement. If $f'''(x^*)>0$ then by the same argument as before $f''(x)>1$ in a neighborhood of $x^*$, in particular $f''(x)>0$ Then I think that the argument should run as before.
For the third statement, I am trying to get to the condition $|f'(x^*)|<1$ to inmediately use the stability criterion and show that $x^*$ is asymptotically stable. However I can also try to get to the definition of being asymptotically stable, but the function with its third derivative at this point seems too convoluted to actuall try to use the definition.
Any lead for the second and third part is greatly appreciated.