# Tag Info

0

It seems the following. Put $\mathcal L={\oplus}_{p=1}^\infty \ell_p$ and $\widehat{\mathcal L}=\Pi_{p=1}^\infty \ell_p$. There can be many norms on $\mathcal L$, but at least I expect that the restriction $\|\cdot\||_{\ell_p}$ of the norm $\|\cdot\|$ which you consider on $\mathcal L$, coincides with the standard norm $\|\cdot\|_p$ of $\ell_p$, which is ...

0

It seems the following. The pseudo-interior $I$ is a (dense) $G_\delta$ subset of the Hilbert cube $Q$, so it is a (completely) metrizable space, and hence it is Baire. The pseudo-interior $Q\setminus I$ is meager in itself. So $I$ and $Q\setminus I$ are not homeomorphic.

0

Suppose $u_1(x),\dots , u_p(x) : O_1 \to O_2$ are continuous and satisfy $|u_j(x)| = 1, x \in O_1, j = 1,\dots , p.$ If $u_1(x_0),\dots , u_p(x_0)$ are linearly independent (LI) for some $x_0\in O_1,$ then $u_1(x),\dots , u_p(x)$ are LI for $x$ in a neighborhood of $x_0.$ Proof: Consider the $p$ functions u_j(x) - \sum_{k\ne j}\langle u_j(x), u_k(x) ...

1

first reduce to the case where $m=p.$ To do this rotate so that the image of the differential map is the first $p$ coordinates and then project onto the first $p$coordinates. If this map has rank $p$ nearby that is enough. If $m =p$ then it follows from the inverse function theorem. If $m>p,$ then rotate the domain to make the differential applied to ...

2

Lemma 1 If $X$ is first countable and $x_{n}\rightarrow 0$, then there exist positive scalars $\lambda_{n}\rightarrow\infty$ such that $\lambda_{n}x_{n}\rightarrow 0$. Proof: Let $\left\{U_{k}\right\}$ be a balanced local base for $0\in X$ such that $U_{k+1}\subset U_{k}$. Since $x_{n}\rightarrow 0$, there exists an increasing sequence $(N_{k})$ of ...

2

This isn't a full answer, just a simple observation that spectral theory gives a partial answer of sorts (as you may already know). See here or any introduction to Banach algebra/C*-algebra theory for supporting details. A character of a unital commutative Banach algebra $A$ is a unital algebra homomorphism $\phi: A \to \mathbb{C}$. The set $\Phi_A$ of all ...

2

Yes, any locally convex space has a neighborhood basis around zero consisting of balanced convex sets (Wikipedia seems to take this as the definition but as far as I know including balanced" in the definition is non-standard, and in any case can be derived from the more general definition requiring the existence of a neighborhood basis around zero ...

2

This is true. See corollary of theorem 1.14 in W. Rudin Functional analysis. This property doesn't imply LCS have a countable separating family of seminorms.

0

Not an answer, but too long for a comment: Matt Samuel's comment isn't quite true: here's an embedding (=map which is a homeomorphism onto its image) of $[0, 1]$ into a topological vector space over $\mathbb{R}$, whose image is linearly independent: The space is $V=\prod_\mathbb{R}\mathbb{R}$ - elements of $V$ are maps $\mathbb{R}\rightarrow\mathbb{R}$, ...

2

Any completely regular (Hausdorff) topological space can be imbedded in $\mathbb{R}^J$ for some index set $J$, which is a topological vector space. Indeed, this is true if and only if a topological space is completely regular. Thus any Hausdorff topological space that is not completely regular cannot be imbedded in a topological vector space over the reals.

Top 50 recent answers are included