Uniform convergence of sequence of functions with infinite roots to a limit with finite roots

Question

Consider a sequence of continuous functions $(f_n)$ defined over $[0,1]$ such that, for all $n$, the set:

$$A_n = \{x\in [0,1] : f_n(x) = 0\}$$

is infinite in cardinality. Can $(f_n)$ uniformly converge to some continuous limit $f$ which only has finitely many zeroes? What if $f$ only has finitely many crossings, and no other zeroes, and for each $f_n$ there are infinitely many crossings (and no other zeroes)?

Answer 1

For the first question. For $n>2$ let $$ f_n(x)=\begin{cases}\dfrac{1/2-1/n-x}{1/2-1/n} &0\le x\le1/2-1/n\\ 0 & 1/2-1/n<x<1/2+1/n\\ \dfrac{x-1/2-1/n}{1/2-1/n} & 1/2+1/n\le x\le1\end{cases} $$ Then $\{x\in[0,1]:f_n(x)=0\}=[1/2-1/n,1/2+1/n]$ and $f_n$ converges uniformly to $|2\,x-1|$.

For the second, you can use the same type of function changing its value on the interval $[1/2-1/n,1/2+1/n]$ to $$ \frac1n\,(1/2-1/n-x)(x-1/2-1/n)\sin\frac{1}{(1/2-1/n-x)(x-1/2-1/n)}. $$

Answer 2

For $n=3,4,\dots,$ let $g_n$ be the function whose graph consists of the three line segments connecting $(0,0),(1/n,0),(2/n,1),(1,1).$ Define $f_n(x) = xg_n(x).$ Then $f_n(x) \to x$ uniformly on $[0,1].$ Note each $f_n$ has uncountably many zeros, while $f(x)=x$ has only one zero.

I assume by "crossing" you mean a change of sign. Here you could let $f_n(x) = (1/n)x\sin(1/x).$ Each $f_n$ has infinitely many crossings, while the $f_n$'s converge uniformly to $0,$ which has no crossings.