example of construction of a Schauder basis

Consider an uniformly continuous function $f:[0,1]\\rightarrow\\mathbb{R}$ . A Schauder basis $\\{f_{n}(x)\\}_{0}^{\\infty}\\in C[0,1]$ is constructed. For this purpose we set $f_{0}(x)=1$ , $f_{1}(x)=x$ . Let us consider the sequence of semi-open intervals in $[0,1]$

I_{n}=[2^{-k}(2n-2),2^{-k}(2n-1)),\\qquad J_{n}=[2^{-k}(2n-1),2^{-k}2n),

where $2^{k-1}<n\\leq 2^{k}$ , $k\\geq 1$ . Define now

\\displaystyle f_{n}(x)

\\displaystyle=

\\displaystyle\\left\\{\\begin{array}[]{ll}2^{k}[x-(2^{-k}(2n-2)-1)]&\\text{if}\\,\\,%x\\in I_{n},\\\\1-2^{k}[x-(2^{-k}(2n-1)-1)]&\\text{if}\\,\\,x\\in J_{n},\\\\0&\\text{otherwise.}\\end{array}\\right.

Geometrically these functions form a sequence of triangular functions of height one and width $2^{-(k-1)}$ , sweeping $[0,1]$ . So that if $f\\in C([0,1])$ , it is expressible in Fourier series $f(x)\\sim\\sum_{n=0}^{\\infty}c_{n}f_{n}(x)$ and computing the coefficients $c_{n}$ by equating the values of $f(x)$ and the series at the points $x=2^{-k}m$ , $m=0,1,\\ldots,2^{k}$ . The resulting series converges uniformly to $f(x)$ by the imposed premise.