# Proof that a TVS is seminormed iff it is locally convex

• A_B

#### 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

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.