Proof that a TVS is seminormed iff it is locally convex

[A_B]

Hi,

There is a guided exercise in the course of functional analysis that I am following where we have to prove that a topological vectors space is seminormed if and only if it is locally convex. There is one step in te proof that I can't figure out.

Let X be a topological vector space and \mathcal{U} a convex neighbourhood of 0.
Define
\mathcal{V} = \bigcap_{\lambda \in S_1} \lambda \mathcal{U}.
Where S_1 are the complex numbers of modulus one.

Use the compactness of S_1 to prove that \mathcal{V} is a balanced convex neighbourhood of zero.

Proving that \mathcal{V} is balanced an convex is straightforward and doesn't require the compactness of S_1. I haven't been able to prove that \mathcal{V} is still a neighbourhood of zero though.Thanks,

A_B

 

[Hawkeye18]

Try to use the continuity of multiplication by scalars.

 

[A_B]

OK, I think I've got it.

Denote by m : \mathbb{C} \times X \to X : (z, x) \to zx the scalar multiplication map. Since it is continuous and \mathcal{U} is a neighbourhoud of zero, the inverse m^{-1}(\mathcal{U}) is a neighbourhood of any pair (z, x) such that zx = 0. In particular, m^{-1}(\mathcal{U}) is a neighbourhood of (\lambda, 0) for any \lambda \in S^1. So we can find for every such \lambda an open neighbourhood \mathcal{O}_{\lambda} \subset \mathbb{C} of \lambda and an open neighbourhood \mathcal{W}_{\lambda} \subset X of zero such that m(\mathcal{O}_{\lambda} \times \mathcal{W}_{\lambda}) \subset \mathcal{U}. The \{ \mathcal{O}_{\lambda}\} are an open cover of the compact set S^1 so we can take a finite subcover \{\mathcal{O}_i \}_{i=1,\dots,k} where to each \mathcal{O}_i there still corresponds an open neighbourhood \mathcal{W}_i of zero such that m(\mathcal{O}_i \times \mathcal{W}_i) \subset \mathcal{U}. But \mathcal{W} = \bigcap_{i = 1, \dots, k} \mathcal{W}_i is also an open neighbourhood of zero, and for all i=1,\dots,k we have m(\mathcal{O}_i \times \mathcal{W}) \subset \mathcal{U}. Since \{\mathcal{O}_i \}_{i=1,\dots,k} covers S^1 we have that \lambda \mathcal{W} \subset \mathcal{U} for any \lambda \in S^1. Therefore, \mathcal{W} \subset \lambda \mathcal{U} for any \lambda \in S^1 whence \mathcal{W} \subset \mathcal{V}. Since \mathcal{W} is an open neighbourhood of zero, we conclude that \mathcal{V} is a neighbourhood of zero.Thanks!