Basis functions for a Schauder-Faber-Basis 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 $[0,1]$.
I found this proof here 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?
 A: Functions $f_n$ specified incorrectly. The correct ones formulas are
$$
f_n(x) = \begin{cases}
2^k ( x - (2^{-k}(2n - 2) - 1)) & \text{ if } \quad x \in I_n \\
1 - 2^k ( x - (2^{-k}(2n - 1) - 1)) & \text{ if } \quad x \in J_n \\
0 & \text{ otherwise }
\end{cases}.
$$
where 
$$
 I_n = [2^{-k}(2n-2)-1, 2^{-k}(2n-1)-1) \qquad
 J_n = [2^{-k}(2n-1)-1, 2^{-k}2n-1).
$$
In order to prove that $f_n=\Delta_{n-1}$ rewrite these formulas as
$$
f_n(x) = \begin{cases}
2^{j+1} ( x - (2^{-j-1}(2n - 2) - 1)) & \text{ if } \quad 2^{-j-1}(2n-2)-1\leq x<2^{-j-1}(2n-1)-1\\
1 - 2^{j+1} ( x - (2^{-j-1}(2n - 1) - 1)) & \text{ if } \quad 2^{-j-1}(2n-1)-1\leq x<2^{-j-1}2n-1 \\
0 & \text{ otherwise }
\end{cases}=
$$
$$
\begin{cases}
2^{j+1} x - n + 1 + 2^{j+1} & \text{ if } \quad (2n-2)\leq 2^{j+1}(x+1)<(2n-1)\\
2n-2^{j+1}x-2^{j+1} & \text{ if } \quad  2n-1\leq 2^{j+1}(x+1)<2n \\
0 & \text{ otherwise }
\end{cases}=
$$
$$
\begin{cases}
2(2^{j} x - (n - 1 - 2^{j})) & \text{ if } \quad 0\leq 2^{j} x - (n - 1 - 2^{j})<\frac{1}{2}\\
2-2(2^jx-(n-1-2^j)) & \text{ if } \quad  \frac{1}{2}\leq 2^{j} x - (n - 1 - 2^{j})<1 \\
0 & \text{ otherwise }
\end{cases}=
$$
$$
\Delta(2^jx-(n-1-2^j))=\Delta(2^jx-k)=\Delta_{n-1}(x)
$$
And now we are done.
