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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 $l^p$?

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  • These don't give the answer, but some information in that direction: https://ncatlab.org/nlab/show/type+(functional+analysis) ; https://ncatlab.org/nlab/show/isomorphism+classes+of+Banach+spaces – Chill2Macht Jun 04 '16 at 16:32
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    No. Look at the book on Banach spaces by Carothers to find examples. – Pedro Jun 04 '16 at 16:33
  • I think the counterexamples, by the way, are probably all non-separable; see for example: http://math.stackexchange.com/questions/112619/isometric-embedding-of-a-separable-banach-space-into-ell-infty; https://en.wikipedia.org/wiki/Banach%E2%80%93Mazur_theorem; Theorem 1.14 https://www0.maths.ox.ac.uk/system/files/coursematerial/2015/3119/26/Chapter_1.pdf; also http://mathoverflow.net/questions/77383/a-separable-banach-space-and-a-non-separable-banach-space-having-the-same-dual-s – Chill2Macht Jun 04 '16 at 16:35
  • I would bet that $\ell^p \times \ell^q$ with norm $|(u,v)| = |u|{\ell^p} + |v|{\ell^q}$ is not isometrically isomorph to some $\ell^r$ for $p \ne q$. And $\ell^p \times \ell^q$ is separable. – gerw Jun 04 '16 at 19:32

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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 ($1<p<\infty$).

Finite dimensions

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.