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Banach spaces, function spaces, real functions, integral transforms, theory of distributions, measure theory.
5
votes
Accepted
Orthonormal basis for non-separable inner-product space
This is Problem 54 in Halmos' "A Hilbert Space Problem Book".
However, I think this is a concrete counterexample. [Please let me know if not viewable.]
3
votes
Accepted
separability of a certain space of continuous functions
The answer is negative. For, pick some non-zero $e$ in $E$, and
choose a surjection $\rho\in C\left(O,\mathbb{R}\right)$ (there exists
!).
Next, consider the (uncountable, uniformly discrete) family …
11
votes
2
answers
860
views
Monotone Lipschitz embedding ?
In 1974, Aharoni proved that every separable metric space (X, d) is Lipschitz isomorphic to a subset of the Banach space c_0.
Thus, for some constant L, there is a map K: X --> c_0 that satisfies the …
6
votes
1
answer
723
views
The "ultimate" indefinite inner product space
This can be considered as a relative of Splitting a space into positive and negative parts.
Is there a real (non-trivial) vector space $V$, endowed with a nondegenerate symmetric bilinear pairing $\l …
6
votes
1
answer
413
views
Subspaces of $L^{2}$
[In what follows $0^{0}$= 1 by convention.]
Is there some closed infinite dimensional linear subspace $F$ of $L^{2}(0,1)$
such that $\left\lvert f\right\rvert^{\left\lvert f\right\rvert}$ belongs to $ …
5
votes
1
answer
494
views
On the failure of the infinite dimensional Brouwer Theorem
Let $K$ be the closed unit ball of some infinite dimensional Banach
space, and let $H$ be an autohomeomorphism of $K$, having fixed
points. Can $H/2$ be fixed point free ?
Also, let ${\mathcal{F}}$ : …
1
vote
Compact Convex sets and Extreme Points
[Just a historical remark.] AFAIK, the fact that the set of all extreme points of a compact convex
subset of $\mathbb{R}^{2}$ must be closed is due to the legendary American
mathematician G. Baley Pri …
5
votes
1
answer
401
views
Nonlinear Nuclear Operators ?
Is there a "right" definition of the nuclear
operator in the nonlinear framework ? Of course, such an operator
must be compact, while a linear operator should be "nonlinearly"
nuclear iff it is nuclea …
3
votes
Show a linear operator is not compact
Let { $ L_{n} $ } be the sequence of Laguerre polynomials,
and let us define
$e_{n}(t)=\dfrac{L_{n}(\ln t)}{t}$ $\cdot\chi_{\left(1,\infty\right)}\left(t\right)$
$(n\in\mathbb{N\textrm{, t > 0}})$ …
2
votes
0
answers
197
views
Generating cones having no surjections [in operator spaces]
Is this little toy known ?
Let $E$ be some Banach space, and let $K$ be the closed unit ball
of its dual, endowed with the weak-star topology. Also, let $j:E$ $\rightarrow$ $C(K)$
be the natural embe …
11
votes
Accepted
Is a subspace with a certain property dense in the dual of a vector space?
The answer is negative. Since the linear span of the Dirac masses is not a dense subspace of the dual of $C[0,1]$.
2
votes
Borsuk pairs of Banach spaces
This is a negative result, heavily relying on P. Dodos'
answer to Boundedness of nonlinear continuous functionals.
Let $\kappa$ be a measurable cardinal, let $K$ be the closed unit
ball of $\ell_{\i …
4
votes
Radii and centers in Banach spaces
I think that http://www.ams.org/journals/tran/1982-271-02/S0002-9947-1982-0654848-2/S0002-9947-1982-0654848-2.pdf [together with its references] provides us with several counterexamples [as well as wi …
11
votes
When is a Banach space a Hilbert space?
Just two isometric/isomorphic characterizations:
A Banach space $X$ is [isometric to] a Hilbert
space if and only if there exists a Banach space $Y$ and a symmetric
bilinear mapping $f:X\times X\righ …
9
votes
2
answers
1k
views
Borsuk pairs of Banach spaces
Given $X$, $Y$ two real Banach spaces, let's say that $(X,\ Y)$
is a Borsuk pair if for any continuous mapping $T$ : {$x$ $\in$
$X$ ; $||x||\leq1$} $\rightarrow$ $Y$ s.t. $T$ is odd on {$x$
$\in$ $X$ …