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If you only require isomorphism in the sense of an invertible, continuous linear bijection, then the answer is yes. If you require isometric linear isomorphism, the answer is no (because the unique is …

Choose finite sets $A_n\subset{\bf N}$ with $\max(A_n)<\min(A_{n+1})$, and then define $T:c_{00}\to c_0$ by $T(e_n)=\chi_{A_n}$. It is easy to check that $T$ extends to a norm 1 linear map $\ell^2\to …

I'm not a Banach-space specialist, and don't have a copy of Lindenstrauss-Tzafriri to hand (though maybe the Albiac-Kalton book would be a friendlier read, if your library has a copy?) but it seems to …

Contrary to my original muddled guess, the answer is no: the problem is that your `extension operators' don't give enough control over what happens in the gap between $U_\infty$ and $U_r$. For a conc …

Does this paper of Kalton do the trick? (disclaimer: I haven't read through the details)

(Posting this comment as an answer, just so the question doesn't still show up as "unanswered".) Your first claim is not justified (as you yourself suspected). If $T:X\to Y$ has separable range, th …

No. Take $X=\ell_\infty({\bf N})$ and take $E_n = \operatorname{span}(e_1,\dots, e_n)$. Then $Y=c_0({\bf N})$ which is well-known – by a non-trivial argument – to be uncomplemented in $X$ (in the sens …

The answer to Q2 is negative. One way to see this (which might not be the first, and might not be the easiest) is to combine the following results. A separable dual (Banach) space has the Radon-Niko …

Examples were constructed (about two years ago?) by Argyros and Haydon. See this blog post for some non-technical discussion. It seems worth noting, as one is almost obliged to, that the space origina …

The first counterexample that comes to my mind, which is probably overkill (memory is easier than thought!) is J. Bourgain, A counterexample to a complementation problem. Compositio Mathematica, Volu …

My internet access at the moment is limited & sluggish, so I haven't been able to look up all the details; but I think your reasoning is correct. Certainly my impression is that duals in the sense bel …

This is a special case of a much more general phenomenon, so I'm writing an answer which deliberately takes a slightly high-level functional-analytic POV; I think (personally) that this makes it easie …

Edit: it seems that I probably misunderstood the question, see Bill Johnson's comments below. No. The identity map factors through $\ell^2$, so it is weakly compact (no doubt one can also see that …

In a different direction from Yves Cornulier's suggestion: how about something based on Cauchy kernels as elements of Hardy space? I'm writing this off the top of my head so I might not be taking the …

First of all, I am not sure what you mean by $L^\infty(X)$ for a general complete metric space $X$. Don't you want $C_b(X)$? Secondly: when $X$ is compact and $0<\alpha<1$, the result you want is The …