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)?

  • $\begingroup$ 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


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 $n$.

  • $\begingroup$ 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

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.


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