Find a sequence of functionf $f_n$ on [0,1] that
$0\le f_n(x)\le1$ for all $n\in \Bbb N$ and for all $x \in [0,1]$
every function $f_n$ is continious
the function $x\rightarrow sup_{n\in\Bbb N}f_n(x)$ is not continous
I had something in mind: $$f_n(x)=\frac{n}{nx+1}$$ but this wouldn't fullfill the 3rd condition
Could use some hint
How about this?
$$f_n(x) := \sqrt[n]{x}$$
Then clearly $0 \leq f_n(x) \leq 1$ for all $n$ and all $x \in [0, 1]$.
And all of these functions are continuous on $[0, 1]$.
And notice $\limsup_{n\in\mathbb{N}} \big\{ \sqrt[n]{0} \big\}=0$, but for any $\varepsilon > 0$ we have $\limsup_{n\in\mathbb{N}} \big\{ \sqrt[n]{\varepsilon} \big\} = 1$.**
So for any $x\in (0,1]$, we have $\limsup_{n\in\mathbb{N}} \big\{ \sqrt[n]{x} \big\}=1$, but when $x=0$ then this function outputs $0$. Therefore, it is not continuous on $[0, 1]$.
**I have left the proof that this limit equals 1 as an exercise for you!