# Convergence of sequence of means in locally convex t.v.s.

In doing Ex 3.2 in Brezis's book of Functional Analysis, I found that it can be generalize to t.v.s.

Let $$E$$ be a locally convex t.v.s. and $$(x_n)$$ a sequence in $$E$$ such that $$x_n \to a\in E$$. Let $$y_{n}=\frac{1}{n}\left(x_{1}+x_{2}+\cdots+x_{n}\right).$$ Then $$y_n \to a$$.

Could you have a check on my attempt?

I posted my proof separately so that I can accept my own answer and thus remove my question from unanswered list. If other people post answers, I will happily accept theirs.

WLOG, we can assume $$a=0$$. Let $$U$$ be a neighborhood (nbh) of $$0$$. By continuity of $$E \times E \to E, (x,y) \mapsto x+y$$, there are nbhs $$V_1,V_2$$ of $$0$$ such that $$V_1+V_2 \subseteq U$$. Let $$V := V_1 \cap V_2$$. Then $$V$$ is also a nbh of $$0$$ such that $$V+V \subseteq U$$. Because $$E$$ is locally convex, we can assume $$V$$ is convex. It follows from $$x_n \to 0$$ that there is $$N$$ such that $$x_n \in V$$ for all $$n\ge N$$. We have \begin{aligned} y_{n} & = \frac{(x_1 +\cdots+x_N) + (x_{N+1} + \cdots +x_{n})}{n} \\ & = \frac{Ny_N}{n} + \frac{x_{N+1} + \cdots +x_{n}}{n}. \end{aligned}
By continuity of $$\mathbb R \times E \to E, (t,x) \mapsto tx$$, there is $$M$$ such that $$\frac{Ny_N}{n} \in V$$ for all $$n \ge M$$. Because $$V$$ is convex, $$0\in V$$, $$n \ge n-N$$, and $$x_n \in V$$ for all $$n\ge N$$, we get $$\frac{x_{N+1} + \cdots +x_{n}}{n} \in V$$. It follows that $$y_n \in V$$ for all $$n \ge \max\{M,N\}$$.