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The point is that for every fixed $k$, the Banach-Mazur distance between $E_k \simeq \ell_1^k$ and $E_{k,n}\simeq \ell_{p_n}^k$ goes to $1$ as $n\to \infty$. In particular, there is $n(k) >k$ such tha …

The answer is no: the projective tensor product is not left-exact on $\mathrm{Ban}_1$. There are several confusions in the question, that the following three points should hopefully clarify: For usua …

I cannot resist to provide an easily citable reference to Rudin's beautiful paper $L_p$-Isometries and equimeasurability, where Rudin proves the much stronger result that if $p$ is not an even integer …

This is not possible in general. The obstruction does not come from spaces that are not dual spaces, but from the spaces that appear in several different ways as dual spaces. Indeed, the restriction o …

It is not true that $\tilde A$ maps trace class operators to trace class operators in general. For a counterexample, consider the maps $A:X\mapsto \mathrm{Tr}(X) \vert 1\rangle \langle 1 \vert$. Then …

No, the completion of a strictly convex normed space can fail to be strictly convex. To put it differently, there are non strictly convex Banach spaces with a dense strictly convex subspace. Here is a …

I understand that you are happy with the case $y\geq 0$, so let's assume that for simplicity (I expect that small modifications should deal with arbitrary self-adjoint $y$). In that case, and if $1<p< …

The orbit structure is extremely simple. If $p=2$, there is one orbit (the isometry group of a Hilbert space acts transitively), whereas for $p\neq 2$ there are exactly $2$ orbits: the (classes of) fu …

[Copying here the content of the comments, for the question not to appear as unanswered] If by $L^1(0,T;X)$ you mean (as it is standard) the space of Bochner-measurable functions, then by definition a …

No, there are such continuous functions, which are continuous with values in $\mathcal{L}^p(H)$ for any $p$ but such that $\int_a^b f$ (which is well defined in the Banach space $\mathcal{L}^1(H)$) do …

The answer to (ii) is positive. Here is a construction, which relies on graphs with spectral gap. For simplicity I write things for trees, but one can probably do the same for more general graphs with …

There does not exist such a function $q$ if $\varepsilon<1/2$. Indeed, if $q$ is a positive measurable function on $[0,1]$ of integral $1$, pick $a \in [0,1]$ such that $\int_0^a q(x) dx = 1/2$. The …

I guess that the answer is no in general. More precisely what I consider as the discrete version of your question has a negative answer. I guess that one should be able to find a couterexample to your …

Here is a suggestion on how to obtain random unitaries from random gaussians. No idea on how to derandomize the construction. This is inspired by Pisier's remark 16.9 in his survey on Grothendieck's t …

The answer is no: there is no bounded projection from $B(H)$ onto $V$. For a proof, see for example Theorem 5.12 in Peller's book Hankel operators and their applications. If you replace $B(H)$ by the …