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A Banach space is a complete normed vector space: A vector space equipped with a norm such that every Cauchy sequence converges.
22
votes
Accepted
Is the space of Hankel operators complemented in B(H)?
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 …
8
votes
Accepted
An abstract characterisation of weak* topologies
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 …
7
votes
Existence of spectral gap
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 …
7
votes
Subspace of $L^2$ that lies in $L^\infty$
First note that by the closed graph theorem, there is a $C$ such that $\|f\|_\infty \leq C\|f\|_2$ for any $f$ iin $E$. (this can also be checked directly if you consider that the closed graph theorem …
6
votes
Accepted
Approximation of an integral over the unit ball of L_1
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 …
6
votes
Accepted
Is a completion of strictly convex normed space strictly convex?
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 …
5
votes
Accepted
Measurability of superposition operator with non-separable Banach space
[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 …
5
votes
Accepted
How complex is the orbit equivalence relation of $\mathrm{Iso}_0(X)\curvearrowright S_X$ for...
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 …
5
votes
Accepted
Are projective tensor products left-exact if one considers only maps of norm at most 1?
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 …
4
votes
Accepted
Integration on quasi-Banach spaces and Schatten ideals
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 …
4
votes
Accepted
Why do distributional isomorphisms preserve joint distribution?
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 …
4
votes
Accepted
Ribe's Theorem: finitely representability between two uniformly homeomorphic Banach spaces
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 …
3
votes
Accepted
Spectrum invariant under (generalised) transpose as operator on trace class operators
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 …
3
votes
Set of unitaries with "spread-like" properties
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 …
3
votes
Accepted
Projections in Banach spaces
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 …