A weighted infinite sum of functions attains its maximum?

Question

Let $f_n:[0,1]\to[0,1]$ be a sequence of continuous functions with $\sum_{n\ge 1}f_n(x)=1$ for all $x\in[0,1]$. Let $a_n$ be a sequence of numbers in $[0,1]$. Can we show that the function $$S:[0,1]\to [0,1],\quad x\mapsto\sum_{n\ge 1}a_nf_n(x)$$ attains its maximum? (Note that the series defining $S$ is pointwisely convergent.) This is used in a paper that I am reading, but I fail to prove it.

Answer 1

$S$ is the limit of an increasing sequence of continuous functions and therefore lower semi-continuous (see, for example, Properties of semi-continuous functions).

By the same argument, $$ 1 - S(x) = \sum_{n=1}^\infty (1-\alpha_n) f_n(x) $$ is also lower semi-continuous.

It follows that $S$ is continuous, and therefore attains its maximum on the compact set $[0, 1]$.

Remark: The compactness of the domain is not needed to show that $S$ is continuous, only to assert that $S$ has a maximum.

Answer 2

By the Dini theorem the series $\sum f_n$ is uniformly convergent. Therefore the series $\sum a_nf_n$ is uniformly convergent as well. Indeed $$0\le \sum_{k=n}^\infty a_kf_k\le \sum_{k=n}^\infty f_k$$Therefore the series $\sum a_nf_n$ represents a continuous function, hence it attains its maximum and minimum.