# Solution of a pair of equations converging to another solution.

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.

• Are all the variables one-dimensional, and all the functions from $\mathbb{R}$ to $\mathbb{R}$? Nov 12, 2019 at 0:44
• I think you're also going to have to assume that the curves $\{(h_1(t),k_1(t)):t \in \mathbb{R}\}$ and $\{(h_2(t),k_2(t)):t \in \mathbb{R}\}$ cross past each other at their intersection point, otherwise I'm pretty sure the statement you want to prove is wrong. Nov 12, 2019 at 1:16
• More precisely, I mean by this that the function $s \mapsto k_2(h_2^{-1}(s)) - k_1(h_1^{-1}(s))$ has neither a local minimum nor a local maximum at $h_1(a)$. Nov 12, 2019 at 1:32
• @JulianNewman Yes, all are one-dimensional and real valued. I’m not entirely sure what you mean by your last two comments? How would the proof look like with that assumption and why it us required?
– NPHA
Nov 12, 2019 at 7:26

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 $$L(b) and $$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, $$l_n(b) and $$l_n(c)>c.$$ Therefore $$x_n\in [b,c]$$.

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

• I think it's also possible to replace the condition "$L(z)-z$ takes both positive and negative values" with the condition "the sequence $|x_n|$ doesn't tend to $\infty$ as $n \to \infty$", provided the convergence of $l_n$ to $L$ is something a little bit stronger than pointwise convergence (e.g. uniform convergence on compact sets). This enables one to include some cases where $L(z)-z$ doesn't take both positive and negative values. Moreover, I think this condition will be not only sufficient but also necessary (again assuming the same stronger form of convergence of $l_n$ to $L$). Nov 13, 2019 at 18:15

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 $$a=0$$.

• That is something I also tried but couldn’t find an counterexample. However, you are not using the limiting conditions that I assume to hold for the functions $f$ and $g$?
– NPHA
Nov 12, 2019 at 11:56
• Thus I’m not sure if such an $l_n$ and $L$ could be constructed that those limits hold. Also, do you know what we should assume more from $f$, $g$, $h$, $k$ so that the result holds (or even from $l_n$ or $L$)?
– NPHA
Nov 12, 2019 at 11:59
• I am using the limiting conditions. Just set all functions to the identity function except $g_{2,n}=l_n, k_2=L$.
– user502266
Nov 12, 2019 at 12:06
• Your second query is relevant to why I set the proof out as above (rather than just stating the simplest counterexample). To make the proof work requires (and only requires) preventing this type of counterexample.
– user502266
Nov 12, 2019 at 12:17
• The 'simplest' condition I can find to ensure the truth of your conjecture is that $L(z)-z$ must take both positive and negative values.
– user502266
Nov 12, 2019 at 13:13