Solution of a pair of equations converging to another solution.

Question

Assume that a unique solution $(x,y)$ to a pair of equations \begin{align} f_{1,n}(x)=f_{2,n}(y) \\ g_{1,n}(x)=g_{2,n}(y) \end{align} exists. And also that similarly a unique solution $(a,b)$ to the pair \begin{align} h_1(a)=h_2(b)\\ k_1(a)=k_2(b) \end{align} exists. Further, $\lim_{n\to \infty}f_{1,n}(z)=h_1(z)$, $\lim_{n\to \infty}f_{2,n}(z)=h_2(z)$, $\lim_{n\to \infty}g_{1,n}(z)=k_1(z)$ and $\lim_{n\to \infty}g_{2,n}(z)=k_2(z)$. One can also assume that all the functions are continuous and have a well-defined inverse.

How I can argue based on this that the solution of the first pair converges to the solution of the second, i.e. $x \to a$ and $y \to b$ as $n$ goes to infinity?

Maybe it is straightforward but I can’t put it on paper even though it seems obvious.

Answer 1

Proof of the modified conjecture.

Suppose that $l_n$ and $L$ are continuous and invertible;

$\lim_{n\to \infty}l_n(z)=L(z)$;

$l_n(z)=z$ has unique solution $z=x_n$ and $L(z)=z$ has unique solution $z=a$;

$L(z)-z$ takes both positive and negative values.

Then $x_n\to a$.

Consider any interval $[b,c]$, strictly containing $a$. Without loss of generality we can suppose $L(b)<b$ and $L(c)>c.$

Now $\lim_{n\to \infty}l_n(b)=L(b)$ and $\lim_{n\to \infty}l_n(c)=L(c)$. So, for $n$ sufficiently large, $l_n(b)<b$ and $l_n(c)>c.$ Therefore $x_n\in [b,c]$.

Let $b\to a$ and $c\to a$. Then, as required, $x_n\to a$.

Answer 2

First we can reduce the problem to a simpler but equivalent configuration.

Let $l_n=f_{1,n}^{-1}f_{2,n}g_{2,n}^{-1}g_{1,n}$ and $L=h_1^{-1}h_2k_2^{-1}k_1$.

$l_n$ and $L$ are continuous and invertible and are therefore strictly monotonic;

$\lim_{n\to \infty}l_n(z)=L(z)$;

$l_n(z)=z$ has unique solution $z=x_n$ and $L(z)=z$ has unique solution $z=a$.

As a counterexample to the conjecture that $x_n\to a$, define $l_n$ and $L$ as follows:

$l_n(z)=2z-\frac{1}{n}$ for $z\le 0$,

$l_n(z)=\frac {z}{2}-\frac{1}{n}$ for $0\le z\le n$,

$l_n(z)=2z-\frac {3n}{2}-\frac{1}{n}$ for $z\ge n$.

$L(z)=2z$ for $z\le 0$, $L(z)=\frac {z}{2}$ for $z\ge 0$.

Then $x_n>n$ but $a=0$.