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

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.

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, $$U := V \cap M$$.