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Suppose one has a schauder basis $\{f_n\}_{n\in\mathbb{N}}$ for $L^p([0,1])$ and we wish to expand a function $f \in L^p([0,1])$ in our basis to get the expression $$f(y)=\sum_{n=0}^{\infty} a_n f_n(y).$$ Let $\sigma: \mathbb{N}\rightarrow \mathbb{N}$ be a permutation of the natural numbers, then

1) Is $\{x_{\sigma(n)}\}_{n\in \mathbb{N}}$ still a basis for $L^p([0,1])$?

2) Is the function $$g(y)=\sum_{n=0}^{\infty} a_{\sigma(n)} f_{\sigma(n)}(y)$$ the same as $f$ except on a set of zero measure? Or is it in general different?If different can you provide an example?

3) If the answer to question 2 is no, then does there exist a basis (need not be schauder) which would retain the same function even after rearranging the sum is in question 2?

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In every Banach space that has a Schauder basis there is a Schauder basis that is not unconditional (meaning that there is a rearrangement of the basis that is not a basis). See

Pełczyński, A.; Singer, I. On non-equivalent bases and conditional bases in Banach spaces. Studia Math. 25 1964/1965 5–25.

I cannot make any sense out of question 3.

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  • $\begingroup$ I understand that the OP wants, for a given $f$, a new basis $(g_n)_n$ and a sequence of numbers $(b_n)_n$ such that 1) $f = \sum b_n g_n$ 2) $f = \sum b_{\sigma (n)} g_{\sigma(n)}$ for any permutation $\sigma$ of the natural numbers. In this process, both $(g_n)_n$ and $(b_n)_n$ depend on $f$. $\endgroup$
    – Hachino
    May 12, 2015 at 10:21
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    $\begingroup$ Maybe, but it is trivial that if $X$ has a basis, then for any $f$ in $X$ there is another basis s.t. $f$ is the first vector of the new basis. $\endgroup$
    – Bill Johnson
    May 12, 2015 at 11:21

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