Does $L^p$ have a basis for which the Pythagorean identity with exponent $p$ holds? In the $\ell^p$ spaces with $1\leq p<\infty$, let $\{e_n\}$ be the standard basis.  If $x=\sum_{n=1}^\infty a_ne_n$ is in $\ell^p$, then for any $k$ we can write 
$$||x||^p=\sum_{n=1}^k |a_n|^p+\sum_{n=k+1}^\infty |a_n|^p=||h||^p+||g||^p$$ 
where $h\in$ span$\{e_1,\ldots,e_k\}$ and $g\in$ span$\{e_{k+1},e_{k+2},\ldots\}$. My question is whether a similar equality holds in the $L^p[0,1]$ spaces for $1\leq p<\infty$?
Some internet searching brought me to a sequence of functions $\{f_n\}$ called the Franklin System that turns out to be an orthonormal basis for $L^2[0,1]$. Also, the $\{f_n\}$ are a Schauder basis for $L^p[0,1]$. So can a similar computation be carried out for $f\in L^p[0,1]$? That is, for every $k$ do we have the equality
$||f||^p=||h||^p+||g||^p$, where $h\in$ span$\{f_1,\ldots,f_k\}$ and $g\in$ span$\{f_{k+1},f_{k+2},\ldots\}$?
Or is there another basis for $L^p[0,1]$ for which this equality holds? Thank you.
 A: Look at Lamperti's work on isometries of L^p (see book by Jamison and Fleming).  It is shown that when $p \neq 2$, parallelogram law holds in $L^p$ iff supports are disjoint.
A: Okay, the following is not a full answer, but a hunch that sadly is a bit to long for a comment:
I believe that in general this is false for $L^p$. The reason it works in $l^p$ is that your basis has in a sense "disjunct" support. Consider the toy example where you have the basis elements $f_1 = e_1 + e_2$ and $f_2 = e_1-e_2$. Then for $p\neq 2$ in general
$$\|a f_1 +bf_2\|_p^p = |a+b|^p+|a-b|^p \neq |a|^p + |b|^p$$
(apart from a few special cases like $a=0,b=0$). Basically the scaling of those $a+b$ and $a-b$ terms simply is not right. Your $l^p$ example works, since those basis elements never really add factors for the same $e_i$, so you do not get such terms.
For $L^p$ there is no basis disjunct support in this way. For any basis  element $f_i$, there will be (infinitely many) other basis elements $f_j$, such that $\{x\mid f_i(x) \neq 0\} \cap \{x \mid f_j(x) \neq 0\}$ is a set of positive measure, so you will run into the problem above.
There is probably an elegant way to turn this into a proof, but I am a bit too tired right now not to make any mistakes.
