Let $f$ and $g$ be real-valued continuous functions on $\Bbb R^2$ that satisfy the following condition:
$$
x<y \implies x< f(x,y) < g(x,y) <y
$$
Assume that there are two sequences $\{a_n\}$ and $\{b_n\}$ that satisfy the following:
$$ a_1 < b_1, a_1,b_1 \in \Bbb R,\quad a_{n+1}=f(a_n, b_n),\quad b_{n+1}=g(a_n,b_n).
$$
It's easily seen that
$$a_1 < a_2 < …< a_n < …< b_n < b_{n-1} < … < b_1.$$
Hence $\{a_n\}$ and $\{b_n\}$ are monotonic and bounded thus convergent.
Now I want to prove that $\lim_{n \to \infty} a_n= \lim_{n \to \infty} b_n$.
My attempt: Assume that $\lim_{n \to \infty} a_n=a, \lim_{n \to \infty} b_n=b$.
Since $a_n < b_n$, $a \le b$. Since $a_{n+1}=f(a_n, b_n)$, we have $a=f(a,b)$.
If $a< b$ then $a < f(a,b)$, a contradiction.
It follows that $a=b$.
Does the proof above look fine?
Best Answer
It looks fine to me.
$\lim f(a_n, b_n) = f(a,b)$ follows from the continuity of $f(x,y)$, for if $f(x,y)$ is continuous at $(a,b)$ then for any sequence converging to $(a,b)$ the sequence of the images of the sequence converges to $f(a,b)$. This is known as the sequential criterion of continuity.