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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?

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@Martin Argerami, may be you meant Hamel basis? –  Norbert Jun 18 '12 at 6:36
    
Of course I did. Thanks! –  Martin Argerami Jun 18 '12 at 14:29
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1 Answer 1

up vote 2 down vote accepted

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.

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There's something wrong in your code for $f_n$, as the functions given in the OP definitely are functions on $[0,1]$ while there's something wiggling over $[1,2]$ in your plots. –  t.b. Jun 21 '12 at 22:05
    
Of course because those functions was incorrect. Now I've written correct formulas. –  Norbert Jun 21 '12 at 22:09
    
I see. Indexing problems. +1 for wading through that horrible mess :) –  t.b. Jun 21 '12 at 22:15
    
@t.b. Thanks! ${}{}{}$ –  Norbert Jun 21 '12 at 22:19
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