Let $\left\{r_{i}: i \geq 1\right\}$ be an enumeration of dyadic rationals in [0,1]: $0,1, \frac{1}{2}, \frac{1}{4}, \frac{3}{4}, \frac{1}{8}, \frac{3}{8}, \frac{5}{8}, \frac{7}{8}, \frac{1}{16}, \frac{3}{16}, \cdots$ Let $f_{1}(t) \equiv 1, f_{2}(t)=t ;$ and for $n>2$ define $f_{n}$ as follows. Let $f_{n}\left(r_{j}\right)=0$ if $j<n, f_{n}\left(r_{n}\right)=1,$ and let $f_{n}$ be linear between any two neighbours among the first $n$ dyadic rationals. Draw the graphs of $f_{3}, f_{4}$ and $f_{5}$ Show that every element $g$ of $C[0,1]$ has a unique representation $g=\sum a_{i} f_{i}$.
I can draw the graphs of $f_{3}, f_{4}$ ,$f_{5}$ and $f_{6}$ ,... .how we can show that every element $g$ of $C[0,1]$ has a unique representation $g=\sum a_{i} f_{i}$ ? i find :
(i)$a_{1}$ must be $g(0)$
(ii)$a_{2}$ must be $g(1)-a_{1}$
(iii)$a_{n}=g\left(r_{n}\right)-\sum_{i=1}^{n-1} a_{i} f_{i}\left(r_{n}\right)$
Note that for j>n, $f_j(r_n) = 0$, so by your choices of the $a_n$, you actually have
$$g(r_n) = \sum_{i=1}^\infty a_if_i(r_n)$$ for every $n$. I.e., the infinite sum and $g$ agree on every diadic rational.
So now you have to show it is also true for all the other numbers in $[0,1]$. For that, you need to know something about the behavior of $g$. That bit of information is given in the problem. Can you spot it? You will also need to prove the same thing is true for the infinite sum.
Lastly, there is a certain fact about the diadic rationals you are expected to know.
Obviously, I could tell you what these things are, but I think it would be better for you to find them yourself. Once you know them, how to proceed should be clear. If you are having trouble, we can discuss it in the comments.