[proofplan] The recurrence $a_{n+1}=f(a_n)$ lets order preservation propagate the first comparison $a_1 \leq a_2$ or $a_1 \geq a_2$ through the whole sequence. Since every iterate remains in the interval $I=[\alpha,\beta]$, the resulting [monotone sequence](/page/Monotone%20Sequence) is bounded, so it converges by the monotone convergence criterion. The inequalities $\alpha \leq a_n \leq \beta$ pass to the limit, placing the limit in $I$. Finally, if $f$ is continuous, the recurrence may be passed to the limit to obtain $L=f(L)$. [/proofplan]

[step:Propagate the initial inequality through the iteration]Assume first that $a_1 \leq f(a_1)$. Since $a_2=f(a_1)$, this says $a_1 \leq a_2$. We prove by induction that \begin{align*} a_n \leq a_{n+1} \end{align*} for every $n \in \mathbb{N}$. The base case $n=1$ is the inequality $a_1 \leq a_2$. Suppose $a_n \leq a_{n+1}$ for some $n \in \mathbb{N}$. Since $a_n,a_{n+1} \in I$ and $f:I \to I$ is order-preserving, applying $f$ to both sides gives \begin{align*} f(a_n) \leq f(a_{n+1}). \end{align*} Using the recurrence $a_{n+1}=f(a_n)$ and $a_{n+2}=f(a_{n+1})$, we obtain \begin{align*} a_{n+1} \leq a_{n+2}. \end{align*} Thus $(a_n)_{n=1}^{\infty}$ is increasing in the non-strict sense.[/step]

[guided]We begin with the only comparison initially available: \begin{align*} a_1 \leq f(a_1). \end{align*} Because the sequence is generated by fixed point iteration, the recurrence gives $a_2=f(a_1)$, so the initial comparison is exactly \begin{align*} a_1 \leq a_2. \end{align*} The reason this single inequality controls the whole sequence is that $f$ preserves order. We prove the full monotonicity statement by induction. The assertion to prove is \begin{align*} a_n \leq a_{n+1} \end{align*} for every $n \in \mathbb{N}$. The base case is already established. For the induction step, assume that \begin{align*} a_n \leq a_{n+1} \end{align*} for some fixed $n \in \mathbb{N}$. Both points $a_n$ and $a_{n+1}$ lie in $I$, because $a_1 \in I$ and $f$ maps $I$ into itself at every iteration. Since $f$ is order-preserving on $I$, the inequality between the inputs gives the inequality between the outputs: \begin{align*} f(a_n) \leq f(a_{n+1}). \end{align*} Now we translate this back into the sequence using the recurrence. The left-hand side is $f(a_n)=a_{n+1}$, and the right-hand side is $f(a_{n+1})=a_{n+2}$. Hence \begin{align*} a_{n+1} \leq a_{n+2}. \end{align*} This completes the induction and proves that $(a_n)_{n=1}^{\infty}$ is increasing in the non-strict sense.[/guided]

[step:Use bounded monotone convergence to obtain a limit in the interval] For every $n \in \mathbb{N}$, the iterate $a_n$ lies in $I$ because $a_1 \in I$ and $f:I \to I$. Hence \begin{align*} \alpha \leq a_n \leq \beta \end{align*} for every $n \in \mathbb{N}$. The sequence is increasing and bounded above by $\beta$, so by the [[eventual monotone convergence criterion](/theorems/8268)][citetheorem:8268], applied with eventual monotonicity starting at $n=1$, there exists $L \in \mathbb{R}$ such that \begin{align*} \lim_{n \to \infty} a_n=L. \end{align*} It remains to check that $L \in I$. Since $\alpha \leq a_n$ for every $n$, taking limits gives $\alpha \leq L$. Since $a_n \leq \beta$ for every $n$, taking limits gives $L \leq \beta$. Therefore \begin{align*} L \in [\alpha,\beta]=I. \end{align*} [/step]

[step:Pass to the limit under continuity to identify the fixed point] Assume now that $f$ is continuous. Since $L \in I$ and $a_n \to L$, continuity of $f$ at $L$ gives \begin{align*} \lim_{n \to \infty} f(a_n)=f(L). \end{align*} Using $a_{n+1}=f(a_n)$ for every $n \in \mathbb{N}$, we also have \begin{align*} \lim_{n \to \infty} f(a_n)=\lim_{n \to \infty} a_{n+1}=L. \end{align*} The shifted sequence $(a_{n+1})_{n=1}^{\infty}$ has the same limit as $(a_n)_{n=1}^{\infty}$ because deleting the first term of a convergent sequence does not change its limit. Hence $f(L)=L$. [/step]

[step:Repeat the argument with the reversed initial inequality] Now assume instead that $a_1 \geq f(a_1)$. Since $a_2=f(a_1)$, this says $a_1 \geq a_2$. We prove by induction that \begin{align*} a_n \geq a_{n+1} \end{align*} for every $n \in \mathbb{N}$. The base case is $a_1 \geq a_2$. If $a_n \geq a_{n+1}$, then $a_{n+1} \leq a_n$. Since $f$ is order-preserving, \begin{align*} f(a_{n+1}) \leq f(a_n). \end{align*} Using the recurrence, this becomes \begin{align*} a_{n+2} \leq a_{n+1}, \end{align*} or equivalently $a_{n+1} \geq a_{n+2}$. Thus $(a_n)_{n=1}^{\infty}$ is decreasing in the non-strict sense. Again $a_n \in I$ for every $n \in \mathbb{N}$, so \begin{align*} \alpha \leq a_n \leq \beta \end{align*} for every $n \in \mathbb{N}$. The sequence is decreasing and bounded below by $\alpha$, so by the [eventual monotone convergence criterion][citetheorem:8268], applied with eventual monotonicity starting at $n=1$, there exists $L \in \mathbb{R}$ such that \begin{align*} \lim_{n \to \infty} a_n=L. \end{align*} Passing the inequalities $\alpha \leq a_n \leq \beta$ to the limit gives $\alpha \leq L \leq \beta$, hence $L \in I$. If $f$ is continuous, then continuity at $L$ gives \begin{align*} \lim_{n \to \infty} f(a_n)=f(L), \end{align*} while the recurrence gives \begin{align*} \lim_{n \to \infty} f(a_n)=\lim_{n \to \infty} a_{n+1}=L. \end{align*} Therefore $f(L)=L$. This proves the decreasing case and completes the proof. [/step]