I am working through two proofs that there exists a Schauder Basis for $C([0,1])$.
One proof defines a basis $(f_n)_{n=0}^{\infty}$ with $ f_0(x) = 1 \qquad f_1(x) = x $ for $2^{k-1} < n \le 2^{k}$, where $k \ge 1$, we define $ f_n(x) = \left\{ \begin{array}{ll} 2^k ( x - (2^{-k}(2n - 2) - 1)) & \mathrm{if} ~ x \in I_n \\ 1 - 2^k ( x - (2^{-k}(2n - 1) - 1)) & \mathrm{if} ~ x \in J_n \\ 0 & \mathrm{otherwise} \end{array} \right. $ where $ I_n = [2^{-k}(2n-2), 2^{-k}(2n-1)) \qquad J_n = [2^{-k}(2n-1), 2^{-k}2n). $ The graphs of these functions form a sequence of "tents" of height one and width $2^{-k+1}$ that sweep across the interval $[0,1]$.
This proof is from this notes page 94.
Another proof i found on the internet goes like this, define the "triangle function" $ \Delta(x) = \begin{cases} 2x & \mathrm{if } \hspace{2mm} x \in \left[0, \frac{1}{2}\right] \\ \\ 2(1-x) & \mathrm{if } \hspace{2mm} x \in \left(\frac{1}{2},1\right] \\ \\ 0 & \mathrm{otherwise}. \end{cases}$ Then consider for $n > 0$ $ \Delta_n(x) = \Delta(2^j x - k) \quad \mathrm{for } \qquad n = 2^j + k, \quad j \ge 0,\quad 0 \le k < 2^j $ and $\Delta_{-1}(x) = 1, \Delta_0(x) = x$. Then the Sequence $\Delta_{-1}, \Delta_0, \Delta_1, \Delta_2, \ldots$ forms a Schauder basis for the Banach space of continuous functions on $[0,1]$.
This proof i found there and here. Now here's my question. I tried to prove that both, the $\Delta_n$ and the $f_n$'s define the same functions, but i am not able to convert the definitions to each other. Do you know how can i proof that these two functions are essentially the same "tent"-functions?