7
$\begingroup$

Assume the following:

  • $f_n{(x)}$ is a sequence of continuous functions, each with a unique zero $x_n^*$
  • $f_n\to f$ uniformly
  • $f$ has a unique zero at $x$

Does it then follow that $x_n^*\to x$?

If this claim is false, what are the minimum additional assumptions needed in order to make it true (for example, do we need to assume that all of the $f_n$'s are analytic)?

$\endgroup$
2
  • $\begingroup$ I am posting this question on behalf of Mark Reid, with his consent. He already has an answer but I wanted to share the question more widely $\endgroup$ Nov 13, 2012 at 19:59
  • $\begingroup$ I've been reading about M-estimators and estimating equations in statistics. It seems like this theory actually comes up there too! Pretty cool to see the connection. $\endgroup$ Sep 23, 2015 at 5:08

2 Answers 2

8
$\begingroup$

Consider the functions $f_n(x) = \dfrac{(x^2 + 1/n)(x/n - 1)}{1 + x^4}$ on $\mathbb R$.

EDIT: These, and their limit $f(x) = \dfrac{-x^2}{1+x^4}$, are real-analytic.

On the other hand, if $f_n$ are analytic in a domain $D$ of the complex plane containing $x$ (the unique zero of $f$ in $D$) and converge uniformly to $f$ on compact subsets of $D$, then by the argument principle $f_n$ must have a zero in $D$ for sufficiently large $n$.

$\endgroup$
3
  • $\begingroup$ Is this the same thing as Hurwitz's Theorem? $\endgroup$ Nov 13, 2012 at 22:29
  • $\begingroup$ That will do it too. $\endgroup$ Nov 13, 2012 at 22:46
  • $\begingroup$ Lovely example Robert $\endgroup$ Nov 14, 2012 at 13:59
1
$\begingroup$

Robert Israel's answer is correct. I would like to add that it is easy to see that $$ \| f(x_n^*) \| = \| f(x_n^*) - f_n(x_n^*) \| \le \sup_y \| f(y) - f_n(y) \| < \epsilon $$ for large enough $n$. Hence every limit point of $x_n^*$ is a zero of $f$. For example, if we are on a compact domain, there exist convergent subsequences of $x_n^*$ with limit point, say $x^*$, and hence $f(x^*)=0$ for every such limit point. If $f$ has a unique zero on this compact domain, then $x^*=x$ and your proposition follows.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.