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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?

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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.

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  • $\begingroup$ 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

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