Search for two Real Valued functions.

Question

Can we have two real valued functions $f_1$ and $f_2$ defined on $[a,b]$ such that $f_1(x)=f_2(x)$ for infinitely many points and $f_1(x)\neq f_2(x)$ for infinitely many points. ?

Answer 1

$$f_1(x)=|x|$$ $$f_2(x)=x$$ where $x\in (-1,1)$.

$\forall x\in [0,1) \ f_1(x)=f_2(x)$ but $\forall x \in (-1,0) \ f_1(x)\neq f_2(x)$. Both intervals are dense, therefore have infinitely many points.

Answer 2

Consider $$f_1:[a,b]\to \mathbb{R}:x \mapsto \left\{\begin{array}{l l} 1 & \text{if } a \leq x \leq \frac{a+b}{2} \\ 0 & \text{else}\end{array}\right.$$ and $$f_2:[a,b]\to \mathbb{R}:x \mapsto \left\{\begin{array}{l l} 1 & \text{if } a \leq x \leq \frac{a+b}{2} \\ 2 & \text{else}\end{array}\right.$$ then $f_1 ([a,\frac{a+b}{2}])= f_2 ([a,\frac{a+b}{2}])$ and $f_1((\frac{a+b}{2},b]) \neq f_2((\frac{a+b}{2},b])$.

Answer 3

How about $f_1(x)=0$ and $f_2(x)=\begin{cases}x\sin\frac1x &\mbox{if } x\neq 0 \\ 0 &\mbox{if } x=0 \end{cases}$, on the interval $[-1,1]$?

In this case, the functions are both continuous, which is kind of cool.