While studying functional analysis the following question came up.

Let $(E, P)$ be a locally convex space where $P$ is a family of seminorms. Also, let $F \subseteq E$ be a linear subspace endowed with the family of seminorms $P_F := \{p_{\vert F} \; : \; p \in P\}$. Show that a subset $V \subseteq F$ is open with regard to $P_F$ if and only if $V = U \cap F$ where $U \subseteq E$ and $U$ is open with regard to $P$.

What I tried so far is the following.

First, since $F$ is a linear subspace we know that $0 \in F$ and $E$ is locally convex, the intersection of convex sets is convex which makes $(F, P_F)$ a locally convex space as well, i.e. there is a balanced absorbing convex local base at zero. Also, $V$ being open with regard to the family of seminorms in the space $(F,P_F)$ is to say that

$$\forall x \in V, \ \exists \ \epsilon_x > 0, p_1^x, \ldots, p_{n_x}^x \in P_F \ : V(x, p_1^x, \ldots, p_{n_x}^x, \epsilon_x) = \bigcap_{i=1}^{n_x} \{y \in F : p_i^x(x - y) < \epsilon\} \subseteq V$$

For the "$\Rightarrow$" direction I now have to somehow show that, based on this notion of $V$ being open, we can infer that there has to be another set $U \subseteq E$ which is open to the entire seminorm family and for which it holds that $V = U \cap F$. But why is this the case?

The other "$\Leftarrow$" direction seems to be clearer. Since $U$ is open to the entire family of seminorms, we can restrict it to those sets being open to $P_F$ by intersecting it with $F$. But this seems heuristic at best, how would one use the notions of local convexity and openess regarding a seminorm system correctly to write it down?


Assume $V$ is open in $(F,P_F)$. So for all $x\in V$ there is $\epsilon_x > 0$ and $p_1^x,\ldots,p_{n_x}^x\in P_F$ such that $V(x,p_1^x,\ldots,p_{n_x}^x,\epsilon_x)\subseteq V$. For each $x\in V,$ choose $q_1^x,\ldots,q_{n_x}^x \in P$ such that $p^x_i={q_{i}^x}_{|F}$

Now define $$U=\bigcup_{x\in V}\bigcap_{i=1}^{n_x}\{y\in E \mid q_i^x(x-y) < \epsilon\}.$$ Note that for any $x\in V$ and $i=1,\ldots n_x,$ the set $U(x,\epsilon,q_i^x)=\{y\in E \mid q_i^x(x-y) < \epsilon\}$ is open in $(E,P)$. Because if $y\in U(x,\epsilon,q_i^x)$ then let $\delta=\epsilon - q_i^x(x-y) > 0$. By triangle inequality of $q_i^x$, $U(y,\delta,q_i^x)\subseteq U(x,\epsilon,q_i^x)$ and hence $U(x,\epsilon,q_i^x)$ is open in $(E,P)$. Since $U$ is arbitrary union of finite intersection of these set, $U$ is open in $(E,P)$. Now \begin{align} U\cap F & = \bigcup_{x\in V}\bigcap_{i=1}^{n_x}(U(x,\epsilon,q_i^x)\cap F)\\\\ & = \bigcup_{x\in V}\bigcap_{i=1}^{n_x}\{y\in F \mid p_i^x(x-y) < \epsilon\}\\\\ & = \bigcup_{x\in V}V(x,p_1^x,\ldots,p_{n_x}^x,\epsilon) = V \end{align}

$U$ need not be unique. Because for a particular semi-norm $p\in P_F$, there may exist several semi-norm in $P$ whose restriction in $F$ is $p$. So in the definition of $U$, you may replace $q_i^x$ with another semi-norm say $r_i^x \in P$ such that ${r_{i}^x}_{|F} = p_i^x$ but $r_i^x\neq q_i^x$ outside $F$.

For the other direction, your reasoning is correct. Try to formalize it mathematically.

  • $\begingroup$ Thank you. I got some clarificatory questions though: Why does it follow that your $U$ is open to $P$ and not just $P_F$? And how would I actually show that $V = U \cap F$ when I take your $U$? Also, is the $U$ you proposed the only $U$ possible or are there others possible? $\endgroup$ – Taufi Nov 13 '17 at 22:25
  • $\begingroup$ I think your definition of a set being open in $(F,P_F)$ is wrong. I have edited your question. Choice of $\epsilon$ and $p_i$ depends on each point, just like in metric space the radius of ball changes with each point in an open set. $\endgroup$ – S. Nandan Nov 14 '17 at 5:56
  • $\begingroup$ Thank you for the clarification! $\endgroup$ – Taufi Nov 14 '17 at 9:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.