# Tag Info

## New answers tagged topological-vector-spaces

1

As mentioned in the answer above, it is quite elementary to prove that the existence a a metric inducing the topology of a topological vector space (t.v.s.) is equivalent to the existence of a translation invariant one. However, if you then want to speak about F-spaces, i.e. complete metrizable t.v.s., the equivalence of having a complete metric or a ...

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(I have withdrawn my earlier answer based on Tsang's justified criticism) Metrizable topological spaces always satisfy the first axiom of countability (take the open balls with radius $1/n$). In theorem 1.24 Rudin proves that if $X$ is a TVS with a countable local base then there is an invariant metric that is compatible with the topology. The proof ...

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As you found out, the two notions are not equivalent. It happens often that two different areas of mathematics use the same word to mean two different things. Think of all the meanings of "normal", "complete", or "regular"... The map $T:\ell_2\to\ell_2$ defined by $T(\{x_n\}) = \{x_n/n\}$ has a closed graph, but does not map closed sets to closed sets: ...

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Note that $S$ belongs to the one-dimensional subspace $\{f_c | c \in \mathbb{R}\}$. Then, $C[0,1]$ with the weak-$*$ topology is a locally convex, Hausdorff topological vector space. Hence, its one-dimensional subspaces are closed. Moreover, on finite-dimensional spaces, all Hausdorff vector topologies coincides. This should help to prove the closedness of ...

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Seperability and metriziability play a role. Note that every subset of a seperable metric space is second countable and hence LindelĂ¶f in the subspace topology. Hence, so is $X \setminus V$. Now, given $y \in X \setminus V$, sa $V$ is closed, there is an open convex neighbourhood $V_y$ of zero, such that $y + V_y \subseteq X \setminus V$. Now $\{y + V_y: y ... 0 I've added an answer to the original question on MSE, which can be seen here: http://math.stackexchange.com/questions/1539116/from-finite-to-sigma-finite-measure-space/1542206#1542206. I have left it un-copied since it's not research-level. 7 A general strategy in scenarios where you are trying to extend a proposition like this from a finite measure space to a$\sigma$-finite one is to break the$\sigma$-finite space in question into countably many finite-measure disjoint measurable pieces and apply the theorem to each chunk individually, and then try to stitch things back up again. To do this ... 2 The only point where the measures itself intervenes is in the definition of strong measurability, through simple convergence$\mu$-almost everywhere. A$\sigma$-finite measure is equivalent (in the sense has the same negligible sets) to a finite measure: partition$\Omega$into countably many sets of finite$\mu$-measures, and define$d\nu=\rho\ d\mu$where ... 1 Let's first assume convergence in measure. For$\epsilon > 0$, define the set$A_{n} := \{t \in [0,1] \; :\;|f_{n}(t) - f(t)| < \epsilon\}$. We obtain $$d(f_{n}, f) = \int_{A_{n}} \frac{|f_{n}(t) - f(t)|}{1 + |f_{n}(t) - f(t)|}\; dt + \int_{A_{n}^{c}} \frac{|f_{n}(t) - f(t)|}{1 + |f_{n}(t) - f(t)|}\; dt$$ Now the first integral is smaller than ... 1 It means that the range of the sequence ($Y_n$), i.e. {$Y_n$}, is a cofinal subset of$\Phi$. In other words, the set {$Y_n$} is a special cofinal subset of$\Phi$: It is indexed by the natural numbers. Moreover, the subset requirements$Y_1\subsetY_2\subset$... restrict how the indexes are assigned. 0 There is a theorem says that If$X$is a Hausdorff locally convex space, and$x,y$are two distinct points in$X$, there exists$\lambda\in X^*$such that$\varphi(x)\not=\varphi(y)$. The assumption "Hausdorff locally convex space" is used to make$(X^*,X)$as a dual pair. In general, one has the following statement: If$X$and$Y$are in ... 1 You need to prove that two maps $$m : X^{\ast} \times X^{\ast} \to X^{\ast} \text{ given by } (f,g) \mapsto f+g$$ and $$s : k\times X^{\ast} \to X^{\ast} \text{ given by } (\alpha, f) \mapsto \alpha f$$ are continuous. Let us prove it for$m$, and leave$s$for you to tackle. Suppose$W \subset X^{\ast}$is weak-$\ast$open, then we want to show that ... 2 Let's take a Cauchy-sequence$(a_{n})_{n}\subset V$. By the completeness of$V$with respect to the first norm$\Vert\cdot\Vert_{1}$, we have $$\forall\,\epsilon>0,\,\exists\,N\in\mathbb{N}:\forall\,n\geq N:\Vert a_{n}-a\Vert_{1}<\epsilon$$ for some$a\in V$. As$\Vert\cdot\Vert_{1}$and$\Vert\cdot\Vert_{2}$are equivalent, there exists$C>0$such ... 2 In the sequel,$E$is our Banach space and$E'$its dual. I consider$E$a real vector space. The "strong" topology is the one induced by some fixed norm$\Vert\cdot\Vert$. The "if" part: Let's denote$B_{w}:=B(a,u_{1},\dots,u_{n},R):=\left\{x\in E\,\vert\,|u_{i}(x-a)|<R,\,i=1,\dots,n\right\}$where$R>0$,$u_{i}\in E'$and$a\in E$(for some ... 2 The canonical projection$M\oplus N\to M$is continuous and as$0\in M$is closed ($M$is Hausdorff, too), its preimage$N$is closed. 1 Find$N\in\mathbb{N}$so that$x_n\neq 0$for$n\geq N$, and assume towards a contradiction that$x^{**}\in[x_n-x^{**}]_{n=N}^\infty$. Write $$x^{**}=\sum_{n=N}^\infty\lambda_n(x_n-x^{**}).$$ Set$a_k:=1+\sum_{n=N}^k\lambda_n$, and find a subsequence$(a_{k_j})$s.t. for all$j$, ... 2 Letting$f(z)= \langle z,x \rangle\$ like you said, we have that $$\|x_n-x\|^2 = \|x_n\|^2 -2 \langle x_n,x \rangle + \|x\|^2 = \|x_n\|^2 - 2f(x_n) +\|x\|^2$$$$\to \|x\|^2 - 2f(x) +\|x\|^2 =0$$

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