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The weak topology has the property that the dual is the same as the dual with respect to the original topology, if $(E,\tau)$ is a topological vector space, and $E^\ast$ its (topological) dual, then $$(E,\sigma(E,E^\ast))^\ast = E^\ast.$$ For the topology of pointwise convergence on $C(S)$, the dual is easily described: if $\lambda \colon C(S) \to ...


2

The answer is no, regardless of the field. For example, the set of pairs $(x,y)$ such that $x+y \neq 0$ is not open as it should be if $+$ were a continuous operation. To see this, just note that any cocountable $A \subseteq \mathbb{R}$ contains a pair $\{x,-x\}\subseteq A$.


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There is a positive result of Rolewicz for locally bounded F-spaces: S. Rolewicz, A generalization of the Mazur–Ulam theorem, Studia Math., 31 (1968), 501–505. There is an extension of this result due to W. Jian: W. Jian, On the Generalizations of the Mazur–Ulam Isometric Theorem, Journal of Mathematical Analysis and Applications, 263 (2001), ...


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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's an answer here. I'm going to reproduce it here to fill in a couple of details. Let $$T : l^2 \rightarrow l^2 : (x_n) \mapsto \left(\frac{x_n}{n}\right).$$ Then $T$ is linear, and $$\|T(x_n)\|^2 = \left\|\left(\frac{x_n}{n}\right)\right\|^2 = \sum_{n=1}^\infty \frac{|x_n|^2}{n^2} \le \sum_{n=1}^\infty |x_n|^2 = \|(x_n)\|^2,$$ hence $T$ is bounded. ...


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You can do it without any calculations: Let $T: X\to Y$ be a continuous linear map between two Banach spaces with dense range (e.g., $\ell^2\to\ell^2$, $(x_n)_n\mapsto (x_n/n)_n$ as in Theo's answer). Then take $Z=X\times Y$, $V_1=\lbrace (x,T(x)): x\in X\rbrace$ the graph of $T$ and $V_2= X\times \lbrace 0 \rbrace$. $V_1$ and $V_2$ are closed but their sum ...


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Your proof is fine. In addition, looking at the proof of Th. 3.20 (c) of Papa Rudin's Functional Analysis, he just mentioned for metric spaces, closure of totally bounded sets are compact. Applying this to a compact set in a Frechet set, you conclude that the closed convex hull of a compact set is again compact. For the conclusion he uses the property (b) ...


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I think your intuition is right: you already have the structure in place to talk about fuzzy tangent bundles (Definition 18). So you can talk about (fuzzy) sections of bundles, hence fuzzy vector fields, fuzzy forms and all that. So, if your question is "does the notion of fuzzy $C^1$ manifold you've defined lead to a natural notion of fuzzy differential ...


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We have$$||f||_1^n =\int_K |f^{(n)} (u)| du =\int_K 1\cdot |f^{(n)} (u)| du \leq \left(\int_K 1^2 du \right)^{\frac{1}{2}} \left(\int_K |f^{(n)} (u)|^2 du \right)^{\frac{1}{2}} =||f||^n_2$$ and $$f^{(n)} (t) =\int_0^t f^{(n+1)}(u) du \leqslant \int_K |f^{(n+1)}(u)| du =||f||^{n+1}_1$$ hence $$||f||^n_{\infty} \leq ||f||^{n+1}_1$$ and $$||f||^n_2 ...


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First, even though you don't say which book it is, I bet the book doesn't just say the $K_n$ are nested, I bet it assumes the stronger condition $$K_{n}\subset K_{n+1}^o,$$where $A^o$ is the interior of $A$. People sometimes says that compact sets satisfying this condition form an "exhaustion" of $\Omega$. If you assume that stronger condition then ...



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