If $f_n\to f$ pointwise on $[0,1]$, then $\sup_{[0,1]}f_n\to\sup_{[0,1]}f$ pointwise?

Question

Let $\{f_n(x)\}_{n=1}^\infty$ be a sequence of bounded functions that converge pointwise on $[0,1]$ to some bounded function $f(x)$.

I am trying to determine whether it is true that $\sup_{x\in[0,1]}f_n(x)\to\sup_{x\in[0,1]}f(x)$ pointwise on $[0,1]$. I think it is, and here is my proof:

Let $x\in[0,1]$. Fix $\epsilon>0$. Obtain some $N$ so that for any $n>N$, $|f_n(x)-f(x)|<\epsilon$. Then, $f_n(x)-f(x)<\epsilon$ and $f(x)-f_n(x)<\epsilon$. Then, we have

$$f_n(x)-\sup_{x\in[0,1]}f(x)\le f_n(x)-f(x)<\epsilon\implies \sup_{x\in[0,1]}f_n(x)-\sup_{x\in[0,1]}f(x)\le f_n(x)-f(x)<\epsilon$$ $$f(x)-\sup_{x\in[0,1]}f_n(x)\le f(x)-f_n(x)<\epsilon\implies \sup_{x\in[0,1]}f(x)-\sup_{x\in[0,1]}f_n(x)\le f(x)-f_n(x)<\epsilon$$

It proves $|\sup_{x\in[0,1]}f_n(x)-\sup_{x\in[0,1]}f(x)|<\epsilon$.

Did I make any mistakes?

Answer 1

Presumably the intended question was

• If $f_n\to f$ pointwise on $[0,1]$, must $\sup_{[0,1]}f_n\to\sup_{[0,1]}f$?

If so, the answer is "no".

For a counterexample, if we let $$ f_n(x)= \begin{cases} 1&\text{if}\;x={\large{\frac{1}{n}}}\\[4pt] 0&\text{otherwise}\\ \end{cases} $$ then $f_n\to 0$ pointwise, but $\sup_{[0,1]} f_n=1$ for all $n$.