schauder basis vs isomorphisms My doubt is about a space with schauder basis. If a space has schauder basis so can i say that it has an isometric isomorphism with some space? For an exemple: $c$ has a schauder basis, can i considere another set a dual space who it is isomorphic?
I am thinking this because i know that $(c_0)'=l_1$ and can i see something like this to $c_0$?
ps: I know that every separate space has an isomorphism with some subspace of $l_{\infty}$. But in this case i think that i can considerate is just trivial isomorphism.
plus: I know that $l_1'=l_{\infty}$ and $l_p=l_q'$, with $\frac{1}{p}+\frac{1}{q}=1$. Is there another set that can i make this easily?
 A: In general the Schauder basis won't give you a way to make an isometric isomorphism between different Banach spaces. For instance $\ell^2$ is not isometrically isomorphic to any $\ell^p$ if $p \neq 2$. One way to see this is that $\ell^2$ is a Hilbert space but $\ell^p$ for $p \neq 2$ isn't.
You may be interested in the theorem which states that any two infinite-dimensional separable Hilbert spaces are isometrically isomorphic (in particular isometrically isomorphic to $\ell^2$).
A: The existence of a Schauder basis does not in general guarantee that one Banach space is isometrically isomorphic to another Banach space, as was noted in the previous answer. However, it is perhaps worth noting that there do exist some Banach spaces beyond those mentioned so far which have Schauder bases and are isometrically isomorphic to some other Banach spaces.
For example, the James space (see here), $J$, is a rather curious Banach space in the sense that the canonical embedding $\iota:J\to J^{**}$ is not an isomorphism (so $J$ is not reflexive), yet $J$ is still isometrically isomorphic to $J^{**}$ by means of a non-canonical linear function.
As to your question, $J$ is well-known to have a conditional Schauder basis, whereas the bases for some previous Banach spaces mentioned in this thread such as the $\ell_{p}$ spaces are unconditional.
