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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.

NPHA
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  • Are all the variables one-dimensional, and all the functions from $\mathbb{R}$ to $\mathbb{R}$? – Julian Newman Nov 12 '19 at 00:44
  • 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. – Julian Newman Nov 12 '19 at 01:16
  • 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 $h_1(a)$. – Julian Newman Nov 12 '19 at 01:32
  • @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 '19 at 07:26

2 Answers2

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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 $a=0$.

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  • 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 '19 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 '19 at 11:59
  • I am using the limiting conditions. Just set all functions to the identity function except $g_{2,n}=l_n, k_2=L$. –  Nov 12 '19 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. –  Nov 12 '19 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. –  Nov 12 '19 at 13:13
  • I still fail to see intuitevely why that condition helps even though I understand the idea of the proof. I’ll go trough this carefully tomorrow once more. Somehow in my mind that condition doesn’t add too much as $L$ already has a fixed point and as it is monotonic it ’almost’ satisfies the additional condition already. – NPHA Nov 12 '19 at 16:16
  • Yes, it's a good idea to think it through carefully and see how the proof in the second answer fails in the original counterexample. –  Nov 12 '19 at 16:20
  • I used Mathematica to plot the piecewise functions and I see the problem now. However, I have one last question: Why is it the case that necessarily $\lim_{n \to \infty} l_n(z) = L(z)$? Because we didn't directly assume anything about the limits of the inverse of the functions $f_{1,n}$ and $g_{2,n}$ which are used in the construction of $l_n(z)$. – NPHA Nov 13 '19 at 10:04
  • I'm travelling at the moment and cannot give you a reference but if all the functions are continuous then pointwise convergence of the functions implies that of the inverses. –  Nov 13 '19 at 13:43
  • Yup. Found one here https://math.stackexchange.com/questions/1106324/convergence-of-a-sequence-of-functions-and-their-inverses – NPHA Nov 13 '19 at 16:27
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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, $x_n\to a$.

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  • 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 $L$). – Julian Newman Nov 13 '19 at 18:15