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 $n \ge \max\{M,N\}$.
