If $f: \mathbb R \to \mathbb R$ be a function such that for some $n_o \in \mathbb N$ , the $n_o$th iterate of $f$ has a unique fixed point $b$ , then how to prove that $f(b)=b$ ? I cant think of anything , please help . Thanks .


Hint: What is


  • $\begingroup$ Wow! so we don't actually need continuity ? $\endgroup$ – Souvik Dey Jul 2 '14 at 13:33
  • $\begingroup$ Not if the fixed point of $f^{n_0}$ is unique. $\endgroup$ – Daniel Fischer Jul 2 '14 at 13:34
  • $\begingroup$ If $b$ is a fixed point (not necessarily unique) and we assume that $f$ is continuous , then can we show $f(b)=b$ , or atleast that $f$ has a fixed point ? $\endgroup$ – Souvik Dey Jul 2 '14 at 13:40
  • $\begingroup$ If $b$ is a fixed point of $f^{n_0}$, then $f(b)$ is also one. To conclude that $f(b) = b$, some further conditions are required. One sufficient condition is that $f^{n_0}$ has only one fixed point. Other sufficient conditions can involve continuity, although I don't see any such that wouldn't impose a unique fixed point off-hand. However, if $b$ is a fixed point of $f^{n_0}$, and $f$ is continuous, we can deduce that $f$ must have a fixed point, since $b, f(b),\dotsc, f^{n_0-1}(b), f^{n_0}(b) = b$ form a cycle under $f$, and $f(x)-x$ must change sign between some two consecutive points of ... $\endgroup$ – Daniel Fischer Jul 2 '14 at 13:50
  • $\begingroup$ these (if $b$ itself is not yet a fixed point). $\endgroup$ – Daniel Fischer Jul 2 '14 at 13:51

Suppose that $f(b)=a$. Then $$f^{(n_0)}(a)=f^{(n_0)}(f(b))=f^{(n_0+1)}(b)=f(f^{(n_0)}(b))=f(b)=a\ .$$ Thus $a$ is a fixed point of $f^{(n_0)}$, and by assumption $a=b$.


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