continuity of the multiplication in a topological vector space This is a follow-up question related to this one.

I don't follow the part of (joint) continuity of multiplication. 

My question is: how is it done in the proof? 


Let $(\lambda_0,x_0)\in F\times X$ where $F$ is the underlying field $(\Bbb{R}$ or $\Bbb{C})$ and let $V\in\mathcal{N}$ such that $\lambda_0x_0\in \lambda_0x_0+V$. How should I find a neighborhood $W$ in $F\times X$ so that $\lambda x\in \lambda_0x_0+V$ for all $(\lambda,x)$?  [I was trying to follow the definition of continuity of a map between two topological spaces.]

The proof shows several things but I don't know how things are put together:


*

*For each $x\in X$, the map $\lambda\mapsto \lambda x$ is continuous at $\lambda=0$. [Fix $x\in X$. Let $V\in\mathcal{N}$ be a nbhd of $0$. Since $V$ is absorbing, there exists $\delta>0$ such that $(\lambda-0)x\in V$ for all $\lambda$ with $|\lambda-0|<\delta$. This gives the desired continuity.]

*For each $\lambda\in F$, the map $x\to\lambda x$ is continuous at $x=0$.[To get $\lambda U\subset V$ as in the proof, one should also use the fact that $U$ is balanced. For example, consider $\lambda=1.5$. Also, I think in the proof, one should consider $|\lambda|\leq 2^n$ instead of $\lambda\leq 2^n$.]

*(Joint) continuity at $(0,0)$[? I don't get this part.]

*How the information above give the continuity at any $(\lambda,x)\in F\times X$?

 A: Fix any $(x_0,\lambda_0)\in X\times\mathbb R$ and consider any $V\in\mathcal N$. By applying (2) twice, there is some $U\in\mathcal N$ such that $$U+U+U+U\subseteq V.\tag{i}$$ Next, take $n\in\mathbb N$ so large that $|\lambda_0|+1\leq 2^n$ (I added $+1$ to ensure positivity). Again, applying (2) $n$ times, this time on $U$, one can find some $N\in\mathcal N$ such that $$(|\lambda_0|+1)N\subseteq U.\tag{ii}$$ (As you have already observed, one does need to consider absolute values, indeed, and also exploit the fact that every neighborhood in $\mathcal N$ is balanced by assumption.) Now, since $U$ is absorbing, there exists some $\alpha_0>0$ such that $$0\leq|\alpha|<\alpha_0\Rightarrow \alpha x_0\in U.\tag{iii}$$ Let $$\varepsilon\equiv\min\{\alpha_0,|\lambda_0|+1\}>0$$ and let $$W\equiv N\times(-\varepsilon,\varepsilon),$$ which is an open neighborhood of $(0,0)$ in the product topology on $X\times\mathbb R$ (the corresponding argument is very similar if $\mathbb F=\mathbb C$). I claim that if $(x-x_0,\lambda-\lambda_0)\in W$, then $\lambda x\in\lambda_0x_0+V$, which yields that the multiplication operator is continuous at any point in $X\times\mathbb R$, indeed. I will prove this claim in four steps.
Step 1 Observe that if $(x-x_0,\lambda-\lambda_0)\in W$, then $x-x_0\in N$ and $|\lambda-\lambda_0|<\varepsilon$. Since $N$ is balanced, one has that $\beta(x-x_0)\in N$ whenever $|\beta|\leq 1$; so take $\beta\equiv\lambda_0/(|\lambda_0|+1)$. Then, (ii) implies that $$\lambda_0(x-x_0)\in(|\lambda_0|+1)N\subseteq U.$$
Step 2 By a similar argument, given that $$|\lambda-\lambda_0|<\varepsilon\leq|\lambda_0|+1,$$ one has that $$(\lambda-\lambda_0)(x-x_0)\in(|\lambda_0|+1)N\subseteq U.$$
Step 3 Since $$|\lambda-\lambda_0|<\varepsilon\leq\alpha_0,$$ (iii) implies that $$(\lambda-\lambda_0)x_0\in U.$$
Step 4 Obviously, $0\in U$.
These four steps and (i) together imply that $$\lambda x-\lambda_0 x_0=\lambda_0(x-x_0)+(\lambda-\lambda_0)(x-x_0)+(\lambda-\lambda_0)x_0+0\in U+U+U+U\subseteq V,$$ as claimed.
