So every $f_n : [a,b] \to \Bbb R$ is a sequence of continuous functions and $f_n \to f$ uniformly. If each $f_n$ has a zero, then we have to show that $f$ also has a zero.
To prove this first I list out the definitions I am gonna use:
- Uniform continuity of every $f_n$. (since $[a,b]$ is given compact.)
$\forall \epsilon \gt 0, \exists \delta_n \gt 0$ such that $\forall x,y \in [a,b]$ & $|x-y| \lt \delta_n$ we have $|f_n(x)-f_n(y)| \lt \epsilon$ $\forall n \in \Bbb N.$
- Uniform convergence of $f_n$.
$\forall \epsilon \gt 0, \exists N \in \Bbb N$ such that if $n \ge N$ then $|f_n(x)-f(x)| \lt \epsilon$ $\forall x \in [a,b]$.
Proof: Let $c_n$ be zero of $f_n$. Then $f_n(c_n)=0 \; \forall n\in \Bbb N.$
For a given $\epsilon \gt 0 \; \exists N \in \Bbb N$ such that if $n \ge N$ then $|f_n(c_n)-f(c_n)| \lt \epsilon \implies |0-f(c_n)| \lt \epsilon.$ (Using definition 2)
This means that $f(c_n)$ is sequence of real numbers converging to zero. Hence there exists $c \in [a,b]$ such that $f(c)=0.$
Now I am particularly suspicious at the last step! Somehow I am not content with it. Is it correct? Also I haven't used definition 1. Could it be key here in the last step?