Give an example of a sequence $\{f_n\}$ of continous real valued functions on $[0, \infty)$ converging uniformly on $[0, \infty)$ to a function $f$ , such that for each $n\ge 1$, there exist $x_n \in [0, \infty)$ such that $f_n(x_n)=0$ , but satisfies $f(x) \neq 0, \forall x\in [0, \infty)$
My example(s) :-
I know for sure that $x_n$ has to be unbounded sequence otherwise there would exist $x_o \in [0,M]$ such that $f(x_o)=0$ where $M$ is an upper bound for $\{x_n\}$.
Let $f_n(x)=e^{-x} -\frac 1n, n\ge 1$ and $x\in [0,\infty)$
For $n=1$, clearly $f_1(0)=0$
Let $n\gt 1$ be fixed.
Then since $e^{-x}\to 0$ as $x\to \infty$ , there must exist $x_n \in (0, \infty)$ such that $e^{-x_n}=\frac 1n$ i.e $f_n(x_n)=0$.
Again for fixed $x\in [0, \infty)$ , we have
$f_n (x)\to e^{-x}=f(x) $ as $n\to \infty$
Now , For a given $\epsilon \gt 0$, we have
$|f_n(x)-f(x)|=\frac 1n \lt \epsilon $ if $n \gt \big[ \frac 1{\epsilon} \big] +1$ and for all $x\in [0, \infty)$ showing the uniform convergence.
Clearly $f(x) \gt 0, \forall x\in [0, \infty)$
I quickly realised, I could easily replaced $e^{-x}$ by any positive decreasing function converging to zero, $g(x)$ (say) and then take
$f_n(x)=g(x) -\frac 1n$
Is this okay? I still wonder whether other examples could be contructed.
Thanks for the time!