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We first start with following definitions.

Definition 1. A family $\mathcal{P}$ of seminorms on a real vector space $X$ is called filtering if for any $p_1,p_2\in \mathcal{P}$ there exsist $q\in \mathcal{P}$ and $r_1,r_2>0$ such that the two inequalities $r_1p_1\le q$ and $r_2p_2\le q$ hold on $X$.

Definition 2. A family $\mathcal{P}$ of seminorms on a real vector space $X$ is called separating if for every $x\in X$ with $x\neq 0$, there exists $p\in \mathcal{P}$ such that $p(x)>0$.

My question is this: Suppose that $(X,\tau)$ is a Hausdorff locally convex space. How can we construct a family of seminorms on $X$ which is both separating and filtering?

My idea. If $(X,\tau)$ is a Hausdorff locally convex space, then as pointed out in Rudin's Functional Analysis book, we can construct a separating family $\mathcal{P}$ of seminorms on $X$. At this point, I don't have any idea if $\mathcal{P}$ is filtering.

Tips are very much appreciated.

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  • $\begingroup$ The bottom line is that I want to verify if the topology of the Hausdorff locally space can be given by a filtering family of seminorms. I would be happy if this can be done.:) $\endgroup$
    – Juniven
    Feb 6 '13 at 4:43
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Since you only want a filtering family of seminorms you could simply take the family of all continuous seminorms $$ \mathcal{P} = \{p \colon X \to [0,\infty) \mid p \text{ is continuous}\} $$ this is clearly a filtering family and it is separating if and only if $X$ is Hausdorff.

More useful is the following construction: Let $\mathscr{U}$ a neighborhood basis of $0$ consisting of open, convex, balanced, circled and absorbing sets and for $U \in \mathscr{U}$ let $p_U(x) = \inf\{\lambda \geq 0 \mid x \in \lambda U\}$ be the Minkowski functional associated with $U$. Then $$ \mathcal{P} = \{p_U \mid U \in \mathscr{U}\} $$ is a family of continuous seminorms on $X$.

  • The family induces the given locally convex topology $\tau$ on $X$: Since all $p_U$ are continuous, the topology induced by $\mathcal{P}$ is weaker than $\tau$ and since $U = \{x \in X \mid p_U(x) \lt 1\}$, the topology induced by $\mathcal{P}$ is at least as strong as $\tau$.

  • If $X$ is Hausdorff, $\mathcal{P}$ is separating: if $x \neq 0$ there is $U \in \mathscr{U}$ such that $x \notin U$, so $p_U(x) \geq 1$.

  • The family $\mathcal{P}$ is filtering since for $U_1, U_2 \in \mathscr{U}$ there is $V \in \mathscr{U}$ such that $V \subseteq U_1 \cap U_2$ by the definition of a neighborhood basis, and therefore $p_V \geq p_{U_1}, p_{U_2}$.

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  • $\begingroup$ Are you still there? Thanks to your answer. Its nice. What guarantees that the family of all continuous seminorms is indeed filtering?:) $\endgroup$
    – Juniven
    Feb 6 '13 at 15:46
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    $\begingroup$ If $p_1$ and $p_2$ are continuous seminorms then so is $q = p_1 + p_2$ and $p_1,p_2 \leq q$ $\endgroup$
    – Martin
    Feb 6 '13 at 15:47
  • $\begingroup$ That's it. In the definition of filtering, do we really need the existence of $r_1$ and $r_2$?:) $\endgroup$
    – Juniven
    Feb 6 '13 at 15:52
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    $\begingroup$ It's slightly more general. For example, on $\mathbb R^2$ consider the semi-norms $|(x_1,x_2)|_i = |x_i|$ and $|(x_1,x_2)|_3 = \frac12 \max\{|x_1|, |x_2|\}$. This is filtering according to the definition with $r_i$, but not without the $r_i$. I can't think of a natural example off the top of my head. $\endgroup$
    – Martin
    Feb 6 '13 at 15:57
  • $\begingroup$ Now, I see.Thank you very much. Im now happy...:)God bless... $\endgroup$
    – Juniven
    Feb 6 '13 at 16:05

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