Let $\Omega$ be a measure space. Then the set $L^2(\Omega)$ is a Hilbert space. Hence it has an orthonormal basis $B$, which yields an isometry of Banach spaces: \begin{align*} L^2(\Omega)\cong L^2(B)\end{align*} Now I was wondering: does there also exist an isomorphism $L^p(\Omega)\cong L^p(B)$, for $p\in [1,\infty]\setminus \{2\}$?
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The case $p=\infty$ (when such an isomorphism may exist) is discussed here. – May 28 '18 at 05:30
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If $H$ is separable then $L^{p}(B)$ is $l^{p}$. For $p=1$ this is a dual space $(c_0)^{*}=l^{1}$. $L^{1}[0,2\pi]$ is not a dual space so it is not isomorphic to $l^{1}$. So the assertion is false for $p=1$.

Kavi Rama Murthy
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