Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
Assume the following:
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)?
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 $n$.
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.
