4
votes
"Adding one dimension" to an infinite dimensional topological vector space
Sophie Morel on Zulip found this counterexample (by Gowers) to $V \cong V \oplus \mathbb{R}$ with $V$ a Banach space: A Solution to Banach's Hyperplane Problem, and there's a nice blog post about it; ...
- 668
2
votes
Accepted
Sobolev extension operator with dense range
No, such an extension operator does not exist.
To see this, let $v \in H^1(\mathbb R)$ such that $v \equiv 0$ in a neighbourhood of $[0,1]$ such that $v \neq 0$. Since the range of $E$ is dense by (3),...
- 6,989
1
vote
Accepted
Proof verification: $(H^m_0)' \cong \overline{(L^2, \|.\|_{-m})}$
Your proof is correct if you just want to show that $H'=(H^m_0(\Omega))'$ and $\overline{(L^2(\Omega),\lVert \cdot \rVert_{-m})}$ are isometrically isomorphic. For this, it suffices to observe that ...
- 6,989
1
vote
True or false? if $f:X\to \mathbb{R}$ is linear on normed space $X$ with closed graph, then $f$ is bounded.
Following the ideas in the link by @Ningxin we provide a proof that the answer is affirmative:
since $f$ is linear functional, it is enough to prove that $\mathrm{Ker}(f)$ is closed. Let $(x_n) \...
- 5,255
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