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bio website maths.lancs.ac.uk/~kania
location Warsaw / Lancaster
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visits member for 2 years, 10 months
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2h
comment If $\varphi f\in L^1(\mu)$ for every $f\in L^1(\mu),$ then $\varphi \in L^\infty$
It is enough to assume that $\mu$ is localisable. (It covers the case of the counting measure on an uncountable set which is a quite important example.)
16h
comment Banach Spaces: Totally Bounded Subsets
The unit ball is bounded by definition. Maybe you mean totally bounded in the non-Banach-space context?
16h
comment Continuous functional such that $f(x_0)\ne 0$
The map $x\mapsto x+x_0$ is a homeomorphism. Just take a suitable neighbourhood of the origin and translate it by $x_0$.
16h
comment Continuous projections in $\ell_1$ with norm $>1$
This is a result of Helemskii arxiv.org/pdf/1112.5750v1.pdf
17h
comment A group with five elements is Abelian
My feeling is that this is the best solution.
17h
comment Continuous projections in $\ell_1$ with norm $>1$
Sorry, what is $\ell_1^0(M)$?
23h
comment Continuous projections in $\ell_1$ with norm $>1$
Every Banach space which is not isometric to a Hilbert space contains a two-dimensional subspace which is not 1-complemented. On the other hand, every 2-dimensional Banach space embeds isometrically into $\ell_1$ (Herz–Lindenstrauss), so I think that even at the level of two-dimensional subspaces the situation can be quite difficult. Maybe you should somehow narrow yourself and ask a more specific question?
1d
revised $L^{\infty} (X, \mu)$ is not separable?
retract
1d
revised $L^{\infty} (X, \mu)$ is not separable?
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1d
revised $L^{\infty} (X, \mu)$ is not separable?
more
1d
answered $L^{\infty} (X, \mu)$ is not separable?
1d
comment Continuous functional such that $f(x_0)\ne 0$
Use the Minkowski functional of some convex, balanced, open and proper neighbourhood of $x_0$.
1d
revised Continuous functional such that $f(x_0)\ne 0$
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1d
answered Continuous functional such that $f(x_0)\ne 0$
1d
comment Example of locally metrizable Lindelöf Hausdorff space that is not metrizable
A good question. I think it'd get more attention at Mathoverflow, though.
2d
comment Proof that the set of all functions from $\mathbb N$ to $\mathbb N$ is not enumerable
Can you elaborate on "a set of functions such that $g_n\colon \mathbb{N}\to \{n\}$..."? I don't understand your notation.
Aug
24
revised In a Hilbert space $H$, if the closed unit ball is compact, then how can it be proved that $H$ is finite-dimensional?
improvements
Aug
24
comment Topological spaces vs. metric spaces
Zariski topologies usually are not even Hausdorff. I am also not sure how natural the Moore spaces are.
Aug
24
suggested suggested edit on In a Hilbert space $H$, if the closed unit ball is compact, then how can it be proved that $H$ is finite-dimensional?
Aug
24
answered In a Hilbert space $H$, if the closed unit ball is compact, then how can it be proved that $H$ is finite-dimensional?