Continuity of finite number of functions

Question

Assume that $f_n(x):[0,1]\rightarrow [0,1], n \in \mathbb{N}$ are continuous. Define $h_n(x) = \max(f_1(x), \cdots f_n(x))$. I want to show that $h_n(x)$ is continuous on $[0,1]$.

I have tried using induction and showing that the only points of possible discontinuity of $h_{k+1}$ are when $h_k(x) = f_{k+1}(x)$ (by using the intermediate value theorem).

I am looking for how to proceed from here (and a nicer, more elegant solution if possible)

Answer 1

Hint:

$$\text {max} \{f,g\}=\frac{f+g}{2}+\frac{|f-g|}{2}$$

hence we only need to proof that $|x|$ is continuous.

Answer 2

Let $\phi(x) = \max(x_1,...,x_n)$, then $\phi$ is Lipschitz with rank $1$.

$x_k = y_k + (x_k-y_k) \le \phi(y) + \|x-y\|$, and so $\phi(x) \le \phi(y) + \|x-y\|$. Swapping the roles of $x,y$ gives the desired result, $|\phi(x)-\phi(y)| \le \|x-y\|$. In particular, $\phi$ is continuous.

Note that $F(x) = (f_1(x),...,f_n(x))$ is continuous and $h_n = \phi \circ F$.