# Epsilon tensor product of locally convex spaces

I want to understand the definition of the $$\varepsilon$$-tensor product of two locally convex vector spaces. (Mainly as a hobby.)

Let $$X,Y$$ be locally convex vector spaces and let $$B(X^{\ast},Y^{\ast})$$ denote the space of separately continuous bilinear forms on $$X^{\ast} \times Y^{\ast}$$, where $$X^{\ast}$$, and $$Y^{\ast}$$ carry the weak$$\ast$$ topology. Let $$A \subset X^{\ast}$$ and $$B \subset Y^{\ast}$$ be two equicontinuos sets of functionals and let $$\beta: X^{\ast} \times Y^{\ast} \rightarrow \mathbb{C}$$ be an element of $$B(X^{\ast}, Y^{\ast})$$. Why is

$$\sup_{(a,b) \in A \times B}{|\beta(a,b)|} < \infty\, ?$$

Here are my thoughts on this: I understand that, by using the Alaoglu-Bourbaki Theorem on polars and the assumption of equi-continuity, we can replace $$A$$ and $$B$$ by polars of neighbourhoods of zero in $$X$$ and $$Y$$ respectevely to obtain weak* compact, convex subsets of $$X^{\ast}$$ and $$Y^{\ast}$$ respectively. We may write the compact Hausdroff space $$A$$ as the union $$A = \bigcup_{n \geq 1}{A_n}, \qquad A_n :=\{a \in A \,:\, \sup_{b \in B}{|\beta(a,b)|} \leq n\}.$$

Using that (locally) comapct Hausdorff spaces are Baire spaces, we can fix $$n \geq 1$$ and $$x_0 \in A_n$$ and an open zero-neighborhood $$U \subset X^{\ast}$$ such that $$x_0 + U \subset A_n$$. Now choose a finite subcover $$A \subset \cup_{i=1}^n{(a_i + U)}$$ of $$\cup_{a \in A}{(a + U)}$$ and define $$C:= \max_{1 \leq i \leq n}{\sup_{b \in B}{|\beta(a_i -x_0, b)|}}.$$ Then for arbitrary $$b \in B$$, $$a \in A$$, we write $$a = a_i + x$$ where $$x \in U$$, $$1 \leq i \leq n$$ and bound $$|\beta(a,b)| =|\beta(a_i+x, b)| \leq |\beta(a_i-x_0, b)|+|\beta(x_0 + x, b)| \leq C+n.$$

Is this a valid argument?

The notes I am reading mention that one can use a "Banach-Steinhaus Theorem for compact convex sets", but my argument above doesn't use the convexity of $$A$$ or $$B$$. Which Theorem could the author of these notes (J.L. Taylor) be referring to?