If $f_n$ is continuous for all $n\in \mathbb{N}$ is $f(x)=\sup(f_n(x))$ continuous?

Question

Suppose there is a set of continuous functions $\{ f_n \}$ for $n\in \mathbb{N}$ such that $0\leq f_n \leq 1$. Then is $f(x) = \sup(f_n(x))$ continuous for all $x\in [0,1]$?

Is there a counterexample to this? If not then how can I go about proving it?

So far I have been able to show that there cannot exist some removable discontinuity that is greater than the rest of $f$ on some interval around it as since the $f_n$ it belongs to is continuous this would imply that for some interval around the point $f_n>f$ leading to a contradiction.

Aside from this special case I am stuck as to how to formulate a rigorous proof for this.

Answer 1

Try $f_n(x) = 1-x^n$. There's a problem near $x = 1$.

Answer 2

It isn't necessarily continuous. Let

$$ f_n(x) = \frac{1}{1 + e^{-nx}} \,. $$

Then $\sup(f_n(0)) = 1/2$ but $\sup(f_n(x)) = 1$ for $x>0$.

Answer 3

No, if there is no additional hypothesis then $f$ is not necessarily continuous.

Consider $f_n$ defined by: $$f_n(x)=\left\lbrace \begin{array}{l}0 \text{ if } x \in [0,1/2)\\n \left(x-1/2 \right) \text{ if } x \in [1/2,1/2+1/n)\\1 \text{ if } x \in [1/2+1/n,1] \end{array} \right.$$ then $\forall n$, $f_n$ is continuous but with $f=\sup_n f_n$ you have: $$f(x)=\left\lbrace \begin{array}{l}0 \text{ if } x \in [0,1/2]\\1 \text{ if } x \in (1/2,1] \end{array} \right.$$ which is not continuous.

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Edit: with pictures