# Tag Info

## New answers tagged topological-vector-spaces

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"In Schwartz's Théorie des distributions, chapter III, Théorème VII : $\mathcal D$ is a Montel space, where bounded sets are relatively compact. Then the weak and strong topologies, restricted to bounded sets, coincide, and convergent sequences are the same in these two topologies (and also in weaker Hausdorff topologies). In more concrete terms: whenever a ...

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A topological vector space in which there is a bounded neighbourhood of $0$ is called locally bounded. Whether the TVS $X$ is $T_1$ or not doesn't play much of a role, $X$ is locally bounded if and only if its associated Hausdorff TVS $X/\overline{\{0\}}$ is locally bounded. If we're only considering vector spaces over $\mathbb{R}$ or $\mathbb{C}$, it ...

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$\mathscr D_K \cap W = \bigcap_{x_{j}\in K} \lbrace\varphi \in \mathscr D_K: |\varphi(x_j)| <c_j\rbrace$ for some finitely many $x_{m}$ in $K$ and corresponding constants $c_j>0$ (given explicitly in Rudin's proof). So all you need is the continuity of the evaluation $\mathscr D_K\to \mathbb C$, $\varphi\mapsto \varphi(x)$. By the way, I do not like ...

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Suppose that $(X,\mathcal{U})$ is sequentially compact. So there exists some $N \in \mathcal{U}$ (some entourage) such that $X$ cannot be covered by finitely many $B(x,N)$. So pick $x_0 \in X$, note that $B(x_0, N)$ does not cover $X$ so there is some $x_1 \notin B(x_0,N)$. By recursion, if we have constructed points $x_0,\ldots, x_n$ then we find some ...

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By definition, unitaries preserve the norm. So $$\|Cy-CSy\|=\|C(y-Sy)\|=\|y-Sy\|.$$ Thus, $CV=V$.

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Continuty is definitely needed： The quotient of X modulo kernel is a topological vector space only if the kernel (a singleton in quotient space) is closed. However, a linear functional is continuous if and only if its kernel is closed. Assume our linear functional is continuous. This follows from the uniqueness of vector topology: The image is either ...

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I believe you came across this in Rudin's Functional Analysis at the end of section 3.11. I think his intended argument is a bit unclear there. I was also confused, which is how I came across your post. He uses the fact that every neighborhood of zero in the weak topology contains an infinite dimensional subspace to prove that the weak topology is not ...

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Technically a metric space is not a topological space, and a topological space is not a metric space: a metric space is an ordered pair $\langle X,d\rangle$ such that $d$ is a metric on $X$, and a topological space is an ordered pair $\langle X,\tau\rangle$ such that $\tau$ is a topology on $X$. A metric on $X$ is a special kind of function from $X\times X$ ...

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Every metric space is a topological space. A subset $S$ of a metric space is open if for every $x\in S$ there exists $\varepsilon>0$ such that the open ball of radius $\varepsilon$ about $x$ is a subset of $S$. One can show that this class of set is closed under finite intersections and under all unions, and the empty set and the whole space are open. ...

2

The weak-* topology on $H^*$ is generated by the functionals $$\{h^* \mapsto h^*(x) \mid x \in H\}$$ If you pull-back this topology to $H$, you get the topology generated by $$\{y \mapsto \Phi(y)(x) \mid x \in H\} = \{y \mapsto (x,y) \mid x \in H\}$$ But this is the same as the weak topology which is generated by $$\{y \mapsto (y,x) \mid x \in H\}$$

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The map $\mathbf{R}\times X\to X$, $(r,x)\mapsto x$, is continuous by assumption, with $\mathbf{R}$ given the usual topology. As a consequence, the map $\mathbf{R}\to X$, $r\mapsto rx_0$, is continuous. Since the sequence $1/n$ converges to $0$, you have $$\lim_{n\to\infty}\frac{1}{n}x_0=0x_0=0$$ Well, if $X$ is not Hausdorff, $\lim$ is used in the sense ...

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Here is the idea. Suppose there are functionals $g_1,...,g_n$ and $g$ such that $\cap_k \ker g_k \subset g$ then we can write $g = \sum_k \eta_k g_k$ for some $\eta_k$. To see this, let $\phi:X \to \mathbb{C}^n$ be defined by $\phi(x) = (g_1(x),...,g_n(x))$ and note that $\phi(x) = \phi(y)$ means that $g(x) = g(y)$. In particular, if we define the ...

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Let $||\cdot||_0$ and $||\cdot||_1$ induce the same topology on a vector space $V$. Let $r>0$, since $B_r(0)_0$ is open in the topology induced by $||\cdot||_0$, it must be open in the one induced by $||\cdot||_1$. This means there exists a $k>0$ so that $B_k(0)_1 \subset B_r(0)_0$. Now any $x \in V, x \neq 0$ can be written as ...

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No. Take $X=C([0,1])$. Give $X$ the sup norm to make it a Banach space. But if we give $X$ the norm from $L^2([0,1])$ and $X$ is incomplete with this norm.

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