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!

  • $\begingroup$ 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... $\endgroup$ – David C. Ullrich Jun 2 '16 at 14:05
  • $\begingroup$ @David C. Ulrich. I can't understand your last statement ! $\endgroup$ – Albert Jun 2 '16 at 14:20
  • $\begingroup$ en.wikipedia.org/wiki/Locally_convex_topological_vector_space Locally convex topological vectors spaces are determined by families of semi-norms. $\endgroup$ – user42761 Jun 2 '16 at 14:22
  • $\begingroup$ @DavidC.Ullrich do you know any proof for my question? $\endgroup$ – Albert Jun 2 '16 at 14:22
  • $\begingroup$ Can you tell me the definition of the weak topology on $X$? $\endgroup$ – David C. Ullrich Jun 2 '16 at 14:23

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.


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