TITLE: Does $L^p(\Bbb R^n)$ have Schauder basis?
QUESTION [7 upvotes]: I can only find this result for compact subsets for some reason, but it should be true. Does $L^p(\Bbb R^n)$ have Schauder basis?

REPLY [4 votes]: The space $L_p(\mathbb{R}^n)$ (Lebesgue measure implied), $1\leq p<\infty$, is isometrically isomorphic to $L_p[0,1]$. (https://www.math.tamu.edu/~schlump/lecture24and25.pdf)

More generally, if $(X,\mathscr{F})$ countably generated, and $\mu$ is a $\sigma$-finite measure of $(X,\mathscr{F})$, $L_p(X,\mu)$, $1\leq p<\infty$,  is isometrically isomorphic to $L_p[0,1]\oplus\ell_p(I)$ for some countable set $I$.  If $\mu$ has no atoms then one can take $I=\emptyset$.
From this you may deduce that such $L_p(X)$ spaces admit Schauder bases.