$\sigma(M, M')$ coincides with the subspace topology that $\sigma(E, E')$ induces on $M$

Question

I'm trying to prove below result stated without proof in my textbook. Could you have a check on my attempt?

Let $E$ be a locally convex t.v.s. and $M$ its linear subspace. Let $\sigma(E, E')$ and $\sigma(M, M')$ be the weak topologies on $E$ and $M$ respectively. Then $\sigma(M, M')$ coincides with the subspace topology that $\sigma(E, E')$ induces on $M$.

PS: I posted my proof separately so that I can accept my own answer to remove my question from unanswered list. Surely, if other people post answers, then I will happily accept theirs.

Answer 1

Lemma: Let $g:E \to F$ be a continuous function between topological spaces. We endow $X \subseteq E$ the subspace topology. Then the map $g \restriction X: X \to F$ is also continuous.

Fix $x_0 \in M$.

• Let $U$ be a neighborhood (nbh) of $x$ in $\sigma(M, M')$. WLOG, we assume $U$ admits a form $$ U= \{x\in M \mid | \forall i = 1, \ldots,n : \langle f_i, x-x_0\rangle| < \varepsilon\}, $$ for some $f_1, \ldots, f_n \in M'$ and $\varepsilon>0$. We need to find a nbh $V$ of $x$ in $\sigma(E,E')$ such that $V \cap M \subseteq U$. WLOG, we assume $V$ admits a form $$ V = \{x\in E \mid | \forall i = 1, \ldots,m : \langle f'_i, x-x_0\rangle| < \varepsilon'\}, $$ for some $f'_1, \ldots, f'_n \in E'$ and $\varepsilon'>0$. In fact, such $V$ can be obtained easily be picking a linear continuous extension of $f_i$ as $f_i'$ and $\varepsilon' := \varepsilon$. Such extension is guaranteed by Hahn-Banach theorem. In this way, $V \cap M = U$.

• Let $V$ be a nbh of $x$ in $\sigma(E, E')$. WLOG, we assume $V$ admits a form $$ U= \{x\in E \mid | \forall i = 1, \ldots,n : \langle f'_i, x-x_0\rangle| < \varepsilon'\}, $$ for some $f'_1, \ldots, f'_n \in E'$ and $\varepsilon'>0$. We need to find a nbh $U$ of $x$ in $\sigma(M, M')$ such that $U \subseteq V \cap M$. WLOG, we assume $U$ admits a form $$ U= \{x\in M \mid | \forall i = 1, \ldots,n : \langle f_i, x-x_0\rangle| < \varepsilon\}, $$ for some $f_1, \ldots, f_n \in M'$ and $\varepsilon>0$. In fact, such $U$ can be obtained easily be picking $f_i := f_i' \restriction M$ and $\varepsilon' := \varepsilon$. By our Lemma, $f_i$ indeed belongs to $M'$. In this way, $U := V \cap M$.