Why weak and weak star topologies are locally convex? I searched for a basis that the open sets at the origin consisting of convex set but I did'nt reach any result!
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What basis at the origin do you have in mind? The one that follows immediately from the definition of the topology consists of convex sets... – David C. Ullrich Jun 02 '16 at 14:05
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@David C. Ulrich. I can't understand your last statement ! – Albert Jun 02 '16 at 14:20
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https://en.wikipedia.org/wiki/Locally_convex_topological_vector_space Locally convex topological vectors spaces are determined by families of semi-norms. – Jun 02 '16 at 14:22
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@DavidC.Ullrich do you know any proof for my question? – Albert Jun 02 '16 at 14:22
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Can you tell me the definition of the weak topology on $X$? – David C. Ullrich Jun 02 '16 at 14:23
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@DavidC.Ullrich topology that consist of all finite intersection of sets like this : – Albert Jun 02 '16 at 14:25
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@DavidC.Ullrich. {x;a<l(x)<b} – Albert Jun 02 '16 at 14:25
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Right, for real spaces anyway. Where $l$ is linear. That shows that ${x:a<l(x)<b}$ is convex! And the intersection of convex sets is convex. – David C. Ullrich Jun 02 '16 at 14:29
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Hint: A basis of neighbourhoods of $x_0\in X$ for the weak topology is obtained by varying $\epsilon$, $k$, and the $f_i$'s in $E^\ast$ in the following expression:
$$V(f_1,\cdots,f_k,\epsilon)=\{x \in E:\forall i=1,\cdots,k:|f_i(x-x_0)|<\epsilon\}$$ (see Proposition 3.4 of Brezis' "Functional Analysis, Sobolev Spaces and Partial Differential Equations")
Proof: Take $x,y\in V(f_1,\cdots,f_k,\epsilon)\equiv V$ for some $\epsilon$, $k$, and some $f_i$'s. To show that $z=\alpha x+(1-\alpha)y\in V$, we have to show that for any $i\in\{1,\cdots,k\}$, $|f_i(z-x_0)|<\epsilon$. Well, we have that $|f_i(x-x_0)|=|f_ix-f_ix_0|<\epsilon$ and $|f_iy-f_ix_0|<\epsilon$. Now use linearity of $f_i$ and convexity of the set $\{w\in\mathbb{C}:|w|<\epsilon\}$.
The weak $\star$ topology is proven to be locally convex similarly.
Carucel
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