basic question on normed algebra

Question

Prove the multiplication map $m:A\times A\rightarrow A$, sending $(x,y)\rightarrow x*y$ is jointly continuous in a normed algebra. i can't understand what's jointly cont.? this is a problem from evan's monumental book on quantum symmetries and operator algebra

Answer 1

There have been very nice answers using product topology. But since you are talking about normed algebras, I guess there is an easier answer.

The product $\cdot:A\times A\to A$ is jointly continuous if there is $C<+\infty$ such that \begin{equation} \|a\cdot b\|\le C\|a\|\|b\| \end{equation} for all $a,b\in A$.

This is to be distinguished from separately continuous, which says for each $a\in A$, there is $C_a<+\infty$ such that \begin{equation} \|a\cdot b\|\le C_a\|b\|, \end{equation} and \begin{equation} \|b\cdot a\|\le C_a\|b\| \end{equation} for all $b\in A$.

Note that the difference is that this constant $C_a$ depends on $a$. But if your are talking about Banach algebra then these two are the same as a consequence of Banach-Stenhauss.

Also as far as I know this is a standard terminology.

Answer 2

Jointly continuous appears to mean continuous on the product topology. For example, the first (non-PDF) page that came up when I Googled the term was http://topospaces.subwiki.org/wiki/Jointly_continuous_map

The fact that the first Google result was a PDF suggests that it is not a very standard term...