$f(x)=\sup\{|f_n(x)| \ :\ n=1,2,..\}<\infty$ find an example of the sequence $\{f_n\}$ such that, $\sup\{f(x)\ :\ 0<x<1\ \}=\infty$

Question

Let ${f_n}$ be a sequence of real- valued continuous functions on $[0,1]$ such that for every $x\in [0,1]$, we have

$f(x)=\sup\{|f_n(x)| \ :\ n=1,2,..\}<\infty$

find an example of the sequence $\{f_n\}$ such that,

$\sup\{f(x)\ :\ 0<x<1\ \}=\infty$.

My idea-

If $f(x)=\frac{1}{x}$ on $(0,1]$ and,

$f(x)=0$ when $x=0$

then $\sup(f(x))$ over $[0,1]$ is $\infty$.

But how do I create $f_n(x)$ on $[0,1]$ such that point wise supremum of $f_n(x)$ is $f(x)$.

Answer 1

For $n=1,2,\dots $ define $f_n(x)=n^2x , x \in [0,1/n],$ $f_n(x) = 1/x, x\in [1/n,1].$

Answer 2

I guess $\displaystyle f_n(x)=\frac{1}{\frac 1 n +x}$ should work.