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\noindent {\bf Question}

\noindent Suppose that $f$ is continuous and that the sequence
$c$, $f(c)$, $f(f(c))$, $f(f(f(c))),\ldots$ converges to $a$.
Prove that $f(a)=a$.


\medskip

\noindent {\bf Answer}

\noindent first, for the sake of notational clarity, define the
$n$-fold composition of $f$ with itself by $f^{\circ n}$, so that
$f^{\circ n} =f\circ f^{\circ (n-1)}$.  The hypothesis can then be
restated as saying that the sequence $\{ f^{\circ n}(c)\}$
converges to $a$.  Now, apply $f$ to both sides. Since $f$ is
continuous, the sequence $\{ f(f^{\circ n}(c))\}$ converges to
$f(a)$, by the note below.  However, since $f(f^{\circ n}(c))
=f\circ f^{\circ n}(c) = f^{\circ (n+1)}(c)$, the sequence $\{
f(f^{\circ n}(c))\}$ is the same as the sequence $\{ f^{\circ
n}(c)\}$ with the first term removed, and so $\{ f(f^{\circ
n}(c))\}$ converges to $a$ as well.  Hence, since $\{ f(f^{\circ
n}(c))\}$ converges to both $a$ and $f(a)$, we have that $a =
f(a)$.

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\noindent Note:

\noindent Suppose that $f$ is continuous and that the sequence $\{
a_n\}$ converges to $a$.  Then, the sequence $\{ f(a_n)\}$
converges to $f(a)$.
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