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a topological vector space that is not a metric space: take $V=C(\Bbb{R})$ where the topology is given by convergence on compact sets. A basis for this topology is given by sets of the form $$U_{K,f,\varepsilon} = \{ g : \sup_K |g-f| < \varepsilon \}$$ where $f \in V$ is continuous, $\varepsilon >0$ is a positive real number, $K \subset \Bbb{R}$ is a ...


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There are many equivalent ways to construct the topology on $C^\infty_c(\Bbb{R^d})$ that makes it an LF-space. I'm not familiar with Tao's method, but I've looked at a different construction that should be equivalent, and is in my opinion very intuitive. First, we set our goal to be to find a topology on $C^\infty_c(\Bbb{R^d})$ that makes it a locally ...


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In fact it turns out that the evaluation map is discontinuous for any locally convex topology on $E'$ (if $E$ is not normable). You can find a concise sketch of the argument on the top of p.2 of Kriegl and Michors book: The convenient setting of global analysis (available online 1): To prove your statement one argues by contradiction and assumes that $ev$ ...


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$\mathbb R^\mathbb N$, the space of real-valued sequences with the product topology. It has a tvs structure and is metrizable, but not normable. See Aliprantis–Border (2006, p. 207).


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Note that $h\colon \mathbb C\to\mathbb C$, $z\mapsto\begin{cases}0&\text{if }|z|\le \varepsilon\\\frac{|z|-\varepsilon}{|z|}\cdot z&\text{if }|z|\ge\varepsilon\end{cases} $ is continuous. Then $g=h\circ f$ seems to do what you want, i.e., the support of $g$ is the compact set $F$ and $|f-g|\le \varepsilon$. If desired, one can deform $h$ suitably so ...


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I will assume that $X$ is Hausdorff. We may assume that $F=\{x\in X:|g(x)|\ge\epsilon\}$, since the righthand side is closed. Let $K=\left\{x\in X:|g(x)|\ge\frac{\epsilon}2\right\}$; then $F\subseteq\operatorname{int}K$, and $K$ is a compact Hausdorff space, so $K$ is normal. Let $U$ be an open set in $X$ such that $F\subseteq ...


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We have $\alpha(t)=(\alpha_1(t),\alpha_2(t))$, where $\alpha_1(t)$ maps $[a,b]$ on the $x$-axis and $\alpha_2(t)$ maps $[a,b]$ on the $y$-axis. Write $$ \phi(t)=\ln\left(\sqrt{[\alpha_1(t)]^2+[\alpha_2(t)]^2}\right)+2\alpha_1(t)+3\alpha_2(t) $$ and use the chain rule to take $\frac{d}{dt}$ in the usual manner (since $\phi$ is not a vector valued ...


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There is a general fact about a TVS $X$ defined by a family of seminorms $\{\rho_{\alpha}\}_{\alpha \in J}$ (which is perhaps proved in the textbook before this): Theorem: If $\varphi : X\to \mathbb{C}$ is a continuous linear functional, then $\exists C>0$ and $\alpha_1, \alpha_2,\ldots, \alpha_n \in J$ such that $$ |\varphi(x)| \leq C \sum_{j=1}^n ...


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Let $A$ be an open subset of $X$. If $\,x\,$ is an element of the convex hull of $A$, then there exist $\,x_1,\dots,x_n$ in $A$ and $\lambda_k \ge 0$ with $\sum_{k=1}^n \lambda_k=1\,$ and $\,x=\sum_{k=1}^n \lambda_k x_k$. At least one $\lambda_k$ does not vanish: say $\lambda_1$. The function $f: X \rightarrow X$, defined by ...


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You want to apply the UBP to the functionals given by the point evaluations, say $Q_x (f)=f (x) $. So your $Y $ ( in the notation from the linked article in Wikipedia ) is $\mathbb C $. And the topology on $C $ plays no role in your argument: you are already assuming that, for each $f\in E $, $|Q_x (f)|\leq M_x$. So you apply the UBP to the functionals ...


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I guess I should explicit my comment here since it is a bit big. I insist that there are details missing in the question, as the current statement is false. (The question you referred to has an answer concerning normed vector spaces. We are talking about topological vector spaces here, this is different.) Think for instance of a field $K$ and the map $K_1 ...


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1)In the case we have a metric: Let $z \in [A]$ ($[A]$ denotes the convex hull of $A$). Therefore for some $n \in \mathbb{N}$ there are $x_1,x_2 \ldots, x_n \in A$ and $\lambda_1,\lambda_2 \ldots,\lambda_n \in (0,1)$ such that $\lambda_1 + \lambda_2 + \ldots + \lambda_n =1$ and $$z = \lambda_1 x_1 + \lambda_2 x_2 + \ldots +\lambda_n x_n$$ Now $\exists ...


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The topologies induced by the dual pairing $\sigma:L^1\times L^\infty \to \mathbb R$, defined by $$ \sigma(u,v) = \int_I uv\ dx $$ make them topological vector spaces with weak and weak-star topology: $L^1$ supplied with topology induced by seminorms $p_v(u):=|\sigma(u,v)|$ coincides with $L^1$ supplied with the weak topology. This follows from the fact ...


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Let $$\tau=\{U\subseteq\Bbb R:0\notin U\text{ or }\Bbb R\setminus U\text{ is finite}\}\;;$$ then $\langle\Bbb R,\tau\rangle$ is a compact Hausdorff space, so it’s regular. (In fact it’s the one-point compactification of a discrete space of cardinality $|\Bbb R|=\mathfrak{c}$.) It is not first countable, since the point $0$ has no countable base. To see ...


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Define a topology on $\mathbb N=\{1,2,3,\dots\}$ by calling a set $S\subseteq\mathbb N$ open if $$\text{either}\quad1\notin S\quad\text{or else}\quad\sum_{n\in\mathbb N\setminus S}\frac1n\lt\infty.$$ With this topology $\mathbb N$ is a regular Hausdorff space, also zero-dimensional and normal, but not first-countable and therefore not metrisable. Another ...


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I don't know what topological vector spaces are. But what I know is that any finite dimensional real vector space can be made into a topological space by choosing any norm on it which does not depend on the choice of the norm. Now if $V$ is any finite dimensional real vector space and $W$ is a linear subspace of $V$, the projection map $\pi:V\to V/W$ can be ...



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