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.

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.