TITLE: Show Weak Topology on a Banach Space is Locally Convex
QUESTION [1 upvotes]: I am trying to show that the weak topology on a Banach space $X$ is locally convex.
I have some doubts on how to show addition is (weakly) continuous, i.e. $f:X\times X\to X$, $f(x,y)=x+y$ is (weakly) continuous.
1) Is it sufficient to show $f$ is weakly sequentially continuous? I.e. if $(x_n,y_n)$ converges weakly to $(x,y)$, then $f(x_n,y_n)$ converges weakly to $f(x,y)$?
Or do we have to show:
2) Preimage of open set (or subbasis) in $X$ is open in $X\times X$ under weak topology.
It seems to me that (1) is easier to show, but I am not sure if it is sufficient.

REPLY [3 votes]: Let $x_0, y_0$ be two points of $X$ and $s_0 = x_0+y_0$. Let $W$ be a weak neighbourhood of $s_0$. It is enough to show that there are $U,V$ weak neighbourhoods of $x_0, y_0$ such that $U+V \subset W$. Its is even enough to show that for basis of neighbourhoods. If $W$ is of the form $$W =\{x \in X : |\langle f_i,x-s_0 \rangle| < \epsilon \quad \forall i \in I\},$$ where $I$ is finite, $f_i \in X'$ and $\epsilon >0$, then set $$U =\{x \in X : |\langle f_i,x-x_0 \rangle| < \epsilon /2 \quad \forall i \in I\}$$ and $$V =\{x \in X : |\langle f_i,x-y_0 \rangle| < \epsilon /2 \quad \forall i \in I\}.$$ Indeed, let $(x,y) \in U \times V$. For all $i \in I$ one has $$ |\langle f_i,(x+y)-s_0 \rangle| = |\langle f_i,(x+y)-(x_0+y_0) \rangle| \leq \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon,$$ which means that $(x,y) \in W$.
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