Consider a vector space $V$ and a family $\mathcal{P}$ of seminorms on $V$. For $p\in \mathcal{P}, z \in V$, define

$$p_z: V \to [0,\infty[: v \mapsto p(v-z)$$

Consider the collection of maps

$$\{p_z:z \in V, p \in \mathcal{P}\}$$

and let $\mathcal{T}$ be the initial topology w.r.t. this collection. I managed to show that $(V, \mathcal{T})$ is a topological vector space. Now, I'm focused on showing the following:

$$\{\{v \in V: \max_{i=1}^n p_i(v) < \epsilon\}: p_i \in \mathcal{P}, \epsilon > 0, n \geq 1\}$$

is a local basis for this topological space (thus for every open containing $0$, I can find a set in the above collection contained in it).

What confuses me is that this does not depend on the translations of seminorms in $\mathcal{P}$, and I tried to apply reverse triangle inequality to get rid of these translation terms but I was unsuccesful.

Thanks in advance for any help.

  • 1
    $\begingroup$ The translations are not necessary for local bases at $0$. For other points yes. $\endgroup$ Feb 21 '20 at 22:05
  • $\begingroup$ @HennoBrandsma Would you be so kind to demonstrate why? $\endgroup$
    – user745578
    Feb 21 '20 at 22:11
  • $\begingroup$ It's rather obvious: all your basic open sets are neighbourhoods of $0$, as $p(0) =0$ for any seminorm. If you want a base at $v_0$ use finitely many $p_{v_0,i}$ instead; the translates of the $0$-neighbourhoods. $\mathcal{T}$ is also just the initial topology wrt $\mathcal{P}$. $\endgroup$ Feb 21 '20 at 22:14
  • $\begingroup$ I need to show that given an open set $O$ containing $0$, I can find seminorms $p_1, \dots, p_n \in \mathcal{P}$ and $\epsilon > 0$ such that $\{\max_i p_i < \epsilon\} \subseteq O$. This is not clear to me. Since finite intersections of the subbasis giving the initial topology can contain translates, this gives me difficulties... How to construct this $\epsilon$ and these seminorms. $\endgroup$
    – user745578
    Feb 21 '20 at 22:17

$\mathcal T$ is the coarsest topology making all functions $p_z$ continuous. A subbasis for $\mathcal T$ is given by the family of all $(p_z)^{-1}(U)$ with $p \in \mathcal P$, $z\in V$ and $U \subset [0,\infty)$ open. Obviously $p_0 = p$, thus certainly each set $$B(p_1,\ldots,p_n, \epsilon) = \{v \in V: \max_{i=1}^n p_i(v) < \epsilon\} = \bigcap_{i=1}^n p_i^{-1}([0,\epsilon)$$ is an open neighborhood of $0$.

Now let $O$ be any open neighborhood of $0$. Thus there exist $p_1,\ldots,p_n \in \mathcal P$, $z_1,\ldots,z_n \in V$ and open $U_1,\ldots,U_n \subset [0,\infty)$ such that $$0 \in \bigcap_{i=1}^n ((p_i)_{z_i})^{-1}(U_i) \subset O .$$ We have $(p_i)_{z_i}(0) = p_i(-z_i) \in U_i$. There exists $\epsilon > 0$ such that $(p_i(-z_i)+\epsilon,p_i(-z_i)+\epsilon) \cap [0,\infty) \subset U_i$ for $i =1,\ldots,n$. But then $(p_i)^{-1}([0,\epsilon) \subset ((p_i)_{z_i})^{-1}(U_i)$: In fact, $v \in (p_i)^{-1}([0,\epsilon)$ means $p_i(v) = p_i(-v) \in [0,\epsilon)$ and we conclude $$p_i(-z_i) - \epsilon < p_i(-z_i) - p_i(-v) = p_i(v-z_i - v) - p_i(-v) \le p_i(v-z_i) + p_i(-v) - p_i(-v)\\ = p_i(v-z_i) \le p_i(v) + p_i(-z_i) < p_i(-z_i) + \epsilon. $$ Thus $(p_i)_{z_i}(v) = p_i(v-z_i) \in (p_i(-z_i)+\epsilon,p_i(-z_i)+\epsilon) \cap [0,\infty) \subset U_i$, i.e. $v \in ((p_i)_{z_i})^{-1}(U_i)$.

  • $\begingroup$ Thanks! I will look into it this evening! $\endgroup$
    – user745578
    Feb 27 '20 at 6:50
  • $\begingroup$ Perfectly clear. Thank you! $\endgroup$
    – user745578
    Feb 27 '20 at 19:13

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