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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)?

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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 –  Will Townes Nov 13 '12 at 19:59
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2 Answers

up vote 7 down vote accepted

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 $n$.

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Is this the same thing as Hurwitz's Theorem? –  Will Townes Nov 13 '12 at 22:29
    
That will do it too. –  Robert Israel Nov 13 '12 at 22:46
    
Lovely example Robert –  Adam Rubinson Nov 14 '12 at 13:59
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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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