Pointwise convergence and Continuity implies Convergence of Zeros Let $f_n, f$ be real-valued continous functions, such that $f_n(x) \to f(x)$ for all real $x$. Let $f(x_0)=0$ and $f_n$ have exactly one zero at $x_n$. Show that $x_n \to x_0$.
I assume you have to plug in various values of $x$, such as $x_0$, $x_n$, $x_0\pm\delta$, into the definitions of continuity and pointwise convergence. However, the main thing I cannot see is how to get from a statement involving convergence of functions to convergence of its argument. From what I see, the continuity definition only allows me to imply things from arguments to functions. (i.e. In the definition below, $\delta$ is a function of $\epsilon$.)
[ $f$ is continuous means for all $\epsilon>0$, there is a $\delta>0$ such that for all $|x-y|<\delta$, $|f_n(x)-f_n(y)|<\epsilon$ ]
 A: The claim is not true.
Let 


*

*$f_n(x)=x(x-1)^2+\frac xn$ for all $n$

*$f(x)=x(x-1)^2$

*$x_0=1$

*$x_n=0$ for all $n$


Then we immediatly verify all conditions:


*

*$f_n,f$ are real valued continuous functions

*$f_n\to f$ pointwise

*$f(x_0)=0$

*$f_n$ has exactly one zero at $x_n$


And yet $x_n\not\to x_0$.

Exercise: In the above example $f$ has a "spurious" root. Find a counterexample with domain $\Bbb R$ where $f$ also has exactly one root.

 One can achieve $x_n\to\infty$ while at the same time $x_0$ comes from a parabola "sitting down" as above, e.g., with $f_n(x)=\left(e^{-x}-\frac1n\right)\cdot\left(x^2+\frac1n\right)\to f(x)=e^{-x}x^2$.


On the other hand we have
Proposition. Let $D\subseteq \Bbb R$ be an interval and let  $f_n\colon D\to \Bbb R$ for $n\in \Bbb N$ be continuous functions converging pointwise to a function $f\colon D\to\Bbb R$. Let  $x_0\in \overline{f^ {-1}(\left]0,\infty\right[)}\cap \overline{f^{-1}(\left]{-\infty},0\right[)}$. Then there exists a sequence $(x_n)_n$ with $x_n\in D$ and $x_n\to x_0$ and $f_n(x_n)=0$ for almost all $n$.
Proof.
Given $\epsilon>0$ we find $\xi^+,\xi^-\in D$ such that $|\xi^\pm-x_0|<\epsilon$ and $f(\xi^\pm)\gtrless 0$. There exists $N_\epsilon$ such that $|f_n(\xi^\pm)-f(\xi^\pm)|<|f(\xi^\pm)|$ for all $n>N_\epsilon$. This implies $f_n(\xi^\pm)\gtrless0$ for these $n$. By the intermediate value theorem, for all $n>N_\epsilon$ there exists $x_{n,\epsilon}\in D$ with $f_n(x_{n,\epsilon})=0$ and $|x_{n,\epsilon}-x_0|<\epsilon$.
We may assume wlog. that the sequence $(N_{1/k})_k$ is strictly increasing.
Then for all $n>N_1$ there exists a maximal $k\in\Bbb N$ with $n>N_{1/k}$ and we let $x_n=x_{n,1/k}$.
By this choice, $f_n(x_n)=0$ and $x_n\to x_0$. $_\square$
A: Even with "continuous", the claim is false.  Let $f_n(x) = (x^2 + n^{-2})(x-1)$, $f(x) = x^2(x-1)$, and $x_0 = 0$.  The (real) zero of $f_n(x)$ is at $1$ for all $n$.
