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The proof Fabian alludes to in the book reference Mark gave is a modern one using the notions of cotype and type. One way to prove that a Banach space $X$ is not isomorphic to a Banach space $Y$ is t …

Ozawa, Schechtman, and I finally wrote up what we know on this question. The estimate is that for every $\epsilon > 0$ there is a constant $C_\epsilon$ so that for every $n$, $\lambda(n)\le C_\epsilon …

The duals of $C[0,1]$ and of $C[0,1]\oplus_\infty c_0(\Bbb{R})$ are isometrically isomorphic.

My question has a negative answer. Lemma. Suppose $X$ has the approximation property (AP), $Y$ is a subspace of $X$, and $X/Y$ fails the AP. Then there is a nuclear operator $T$ on $X$ s.t. $TX\subs …

The answer to the first question is "no". You can see this with specific examples, but here is a more conceptual approach: Take $Y$ an uncomplemented subspace of $X$ and in $Z:= X\oplus_1 X$ identify …

Write $L_\infty$ as the closure of a net (directed by inclusion) of finite dimensional $\ell_\infty$ spaces. Compose the operator into $L_\infty$ with norm one projections onto these subspaces and ex …

No. For complicated and important examples, consider any $\mathcal{L}_\infty$ space that does not contain a subspace isomorphic to $c_0$. The first such examples were constructed by Bourgain and Del …

While I do not have a complete answer to the OP’s question, I made enough observations that I think it is worthwhile to record them as an answer. (1) If $X$ and $Y$ are isomorphic (closed) subspaces …

For Lipschitz functions in finite dimensional spaces, Gateaux and Frechet differentiability are the same, but there are huge differences when the domain is infinite dimensional. Lipschitz functions a …

The answer is no. Let $X$ be a separable Pisier counterexample [P] to Grothendieck’s problem. That is, both $X$ and $X^*$ have cotype 2 and every operator from $X$ or $X^*$ into a Hilbert space i …

Yes. The reason is that the unit vector basis for $c_0$ is a shrinking basis, which means that the biorthogonal functionals to the basis are a Schauder basis for $c_0^* = \ell_1$. This implies that …

There is a beautiful result of Odell and Schlumprecht that gives an answer to this question for separable Banach spaces.Odell, E.(1-TX); Schlumprecht, Th.(1-TXAM) Asymptotic properties of Banach space …

No. The condition implies that the subspace is isomorphic to a Hilbert space. In fact, Kadec and Pelczynski proved that a subspace of $L_p$, $2<p<\infty$, is closed in $L_r$ for some $r<p$ if and …

Kevin, there are non reflexive spaces with non trivial type--even of type 2. James constructed the first one; his argument is very complicated. Later Pisier-Xu did it much more simply using interpol …

$K(H)$, the compact operators on $H$, is the only proper closed ideal in $B(H)$ when $H$ is a separable infinite dimensional Hilbert space, and $K(H)$ is not complemented in $B(H)$ (because if it were …