I know that every Hilbert space is isometrically isomorphic to $l^2(\beta)$ where $\beta$ is a Hilbert basis for that space.
Do Banach spaces have the similar property? That is, is every Banach space isometrically isomorphic to some $l^p$?
I know that every Hilbert space is isometrically isomorphic to $l^2(\beta)$ where $\beta$ is a Hilbert basis for that space.
Do Banach spaces have the similar property? That is, is every Banach space isometrically isomorphic to some $l^p$?
The space $c_0$ (sequences converging to $0$, with supremum norm) is separable but is not even isomorphic to any $\ell_p$ (never mind isometrically isomorphic). Indeed, its dual is $\ell_1$ — a separable nonreflexive space. The dual of $\ell_p$ is either nonseparable ($p=1,\infty$) or reflexive ($1<p<\infty$).
Every $2$-dimensional $\ell_p$ space admits a $4$-fold symmetry, namely the map $T(x,y) = (-y,x)$ which is an isometry such that $T^4=\operatorname{id}$ but $T^k\ne \operatorname{id}$ for $k=1,2,3$. So, if the unit ball of a $2$-dimensional normed space does not admit such a symmetry, it is not isometric to any $\ell^p$ space. One example of such a unit ball is square with two half-disks attached to one pair of opposing sides.