Supremum (pairwise disjoint functions) Suppose $(f_n)$ is a sequence of pairwise disjoint continuous functions on $[0,1]$ so that 
$\underset{t\in[0,1]}{\sup}\left\vert f_n(t) \right\vert=1$ for all $n$, 
then $\underset{t\in[0,1]}{\sup}\left\vert \Sigma a_n f_n(t) \right\vert=\underset{n}{\sup}\left\vert a_n \right\vert$. 
I can see why this is true but if I am to write down some steps to explain this clearly, any suggestion?
 A: Since all the functions are continuous, the $\sup$ of each function is a $\max$ (that is, there is some argument in $[0,1]$ where it is reached by the function).
Let $S = \sum_n a_n f_n(t)$ and $x_n\in [0,1]$ so that $|f_n(x_n)| = \underset{t\in[0,1]}{\sup}\left\vert f_n(t) \right\vert = 1$
Note that $x_n$ might not be unique.
Let $x_S\in [0,1]$ so that $\underset{t\in[0,1]}{\sup}\left\vert \sum_n a_n f_n(t) \right\vert = \vert\sum_n a_n f_n(x_S)\vert \neq 0$ because $\underset{t\in[0,1]}{\sup}\left\vert f_n(t) \right\vert = 1$.
Then, $\exists ! k\in \mathbb{N}, f_k(x_S)\neq 0$ and $S(x_S)=a_kf_k(x_S)$.
Note that $|f_k(x_S)|=1$ otherwise $|S(x_k)| = |a_kf(x_k)|=|a_k| > |a_kf_k(x_S)| = \underset{t\in[0,1]}{\sup}\left\vert S(t) \right\vert$.
Then, $\underset{t\in[0,1]}{\sup}\left\vert S(t) \right\vert = |a_k\times 1|$
Now we have to prove that $|a_k| = \underset{n\in \mathbb{N}}{\sup}\left\vert a_n \right\vert$.
If $\exists l\in \mathbb{N}, |a_l|>|a_k|$, then $|S(x_l)|=|a_l|> \underset{t\in[0,1]}{\sup}\left\vert S(t) \right\vert$ using the same reasoning as before. Hence the result.
