# Topological vector space locally convex

Hello how to show the following:

Let $X$ be a vector space.

Then a norm on $X$ induces a topology under which $X$ is a locally convex topological
vector space.

Moreover, let $D$ be a nonempty subset of linear functionals $x’$. What is the neighborhood bases of the coarsest topology on $X$ making every $x’ \in D$ continous.

Thanks a lot!

#### Solutions Collecting From Web of "Topological vector space locally convex"

For the first:

Show that $+$ as a map from $X \times X$ to $X$ is continuous. As we have a vector space we only need to check this at $(0,0)$.

Show that that scalar multiplication $\cdot$ from $X \times K$ to $X$, where $K$ is the topological field we use (most likely $\mathbb{R}$ or $\mathbb{C}$) as scalars, is continuous.
Again we only need to do this at $(0,0)$.

show that the open balls around $0$, so sets like $\left\{x \in X: \|x\| < r \right\}$ where $r>0$ are convex. This will follow quite easily from the defining properties for a norm. This shows local convexity.

For the second: if every member of $D$ is to be continuous for a topology $\mathcal{T}$ on $X$, it had better be the case that for all open sets $O$ in $K$ (see above), and every member of $f$ of $D$, the set $f^{-1}[O]$ is in $\mathcal{T}$. So almost by definition, the topology that is generated by all sets of this form, call then $\mathcal{S}$, so $\mathcal{S} = \left\{ f^{-1}[O]: f \in D, O \subset K \mbox{ open} \right\}$, should be our candidate for the coarsest topology that makes all members of $D$ continuous. A base for this topology is the set of finite intersections of members of $S$, and a local base at $0$ is thus all those intersections that contain $0$, so $$\left\{ f_1^{-1}[O_1] \cap \ldots \cap f_n^{-1}[O_n]: n \in \mathbb{N}, \forall i \in \{1,\ldots,n\}: f_i \in D, O_i \subset K \mbox{ open }, 0 \in O_i \right\} .$$

Now you should try to verify that this is indeed a topology that makes $X$ also a topology vector space. For this you should have learnt some criteria, I suppose…