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In proving that if $f$ is a $C^3$ function from $\mathbb{R}$ to $\mathbb{R}$ and $f(p) = p$ and $f^{'}(p) = 1$ and $f^{''}(p) = 0$ and $f^{'''}(p) > 0$ then $p$ is a not Lyapunov stable fixed point of $f$, I have arrived at $(p,p+\delta) \subseteq W^u(p)$ for some $\delta > 0$. However, I can't prove the unstabillity. I know I should use the definition and have tackled with the problem but I can't solve it. Can anybody help? Thanks.

Emad
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1 Answers1

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So your map has a Taylor expansion of the form $$ f(x)=p+1(x-p)+0(x-p)^2+\frac16f'''(y)(x-p)^3=x+\frac16f'''(y)(x-p)^3 $$ for some $y$ close to $p$. Since $f$ is of class $C^3$ and $f''(p)>0$, also $f''(y)>0$ for any $y$ sufficiently close to $p$. This implies that for $x>p$ sufficiently close to $p$ we have $f(x)>x>p$. This gives you the instability.

You should avoid writing $W^u(p)$ since there is no hyperbolicity in this case.

John B
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  • I already know your final result $f(x)>x>p$. The question is how the unstability follows from that. And about $W^u(p)$, I mean the unstable set of $p$. This terminology is used in the book written by Devaney. – Emad Mar 13 '21 at 19:23
  • I explained quite clearly in the first paragraph how the instability follows. – John B Mar 13 '21 at 22:32
  • I understood it completely. The question is how we can deduce that: $$ \exists \varepsilon >0 ;; \forall \delta >0 ;; \exists x,n; ; ((|x-p|< \delta) \wedge (|f^n(x) - f^n(p)| >\varepsilon)) $$ which proves the unstabillity. – Emad Mar 14 '21 at 05:04
  • That's the trivial part, it follows immediately by continuity. – John B Mar 14 '21 at 10:46