I am trying to prove that the Faber-Schauder System is a basis for $C[0,1]$ but I am stuck at some point. I will define the relevant functions to my question and explain what I have done so far.
Let $s_{n,k}(x) = \max\{1-|2(2^nx-k)-1|, 0\}$. Let $g$ be a continous function defined on $[0,1]$ with $g(0) = g(1) = 0$. Let $g_{0} = g$ and $h_0 = g_0(\frac{1}{2}) s_{0,0}$. Then we inductively define for $n \ge 1$, $g_n = g_{n-1} - h_{n-1}$ and $h_n = \sum_{k = 0}^{2^n-1}g_n(\frac{2k+1}{2^{n+1}})s_{n,k}$. And finally, we define $t_n = \sum_{k=0}^{n} h_k$. I proved by a series of induction that $t_n(x) = g(x)$ for $x \in D_n := \{\frac{j}{2^{n+1}}: j \text{ integer}, 0 \le j \le 2^{n+1} \}$.
What I couldn't prove is that $t_n$ converges uniformly to $g$ on $[0,1]$. I believe it is sufficient to prove that $t_n$ converges uniformly to $g$ on $D = \bigcup_{n = 0}^{\infty} D_n$ since D is dense in $[0,1]$.
So: $t_n(x)$ is the piecewise-linear function that agrees with $g(x)$ on $D_n$. Given $\epsilon > 0$, use the fact $g$ is uniformly continuous to take $n$ sufficiently large that for $|x - y| \le 2^{n+1}$, $|g(x) - g(y)| < \epsilon$, and use that to show that $|g(x) - t_n(x)| < \epsilon$.