I'm researching topological vector spaces for my dissertation, and have been reading Functional Analysis by Walter Rudin. In the definition of a TVS, he (and others) state the following.

"Multiplication is continuous means that the mapping $(\alpha, x) \to \alpha x$ is continuous; if $x \in X$, $\alpha$ is a scalar, and $V$ is a neighbourhood of $\alpha x$, then for some $r > 0$ and some neighbourhood $W$ of $x$, it follows that $\beta W \subset V$ whenever $|\beta - \alpha| < r$."

My question is: why doesn't the definition just state "...then for some neighbourhood $W$ of $x$, $\alpha W \subset V$" --- why are the variables $r$ and $\beta$ necessary?

  • $\begingroup$ Multiplication is continuous as function $K\times X\to X$ where $K$ denotes the scalar field. $\endgroup$ – user251257 Apr 1 '17 at 17:39
  • $\begingroup$ @user251257 I know, I'm asking why this is the definition of multiplicative continuity though. $\endgroup$ – Bill Wallis Apr 1 '17 at 17:43
  • $\begingroup$ He does not only say that the map $x \mapsto \alpha x $ is continuous for a fixed $\alpha$. He states that multiplication is jointly continuous. $\endgroup$ – Olivier Apr 1 '17 at 17:47
  • $\begingroup$ @Olivier I have not come across the term jointly continuous before. I gave it a quick search and I'm not sure what relevance that has here --- is there a chance you may be able to elaborate on that for me please? $\endgroup$ – Bill Wallis Apr 1 '17 at 17:54
  • 3
    $\begingroup$ @BillWallis You certainly have come across the concept before. A function $f: \mathbb{R}^2 \to \mathbb{R}$ being continuous is not equivalent to $f_x: \mathbb{R} \to \mathbb{R}$ given by $f_x(y)=f(x,y)$ being continuous for every $x \in \mathbb{R}$. This is essentially the issue here: the former is much stronger, and its parallel on this question is needed in a lot of applications. $\endgroup$ – Aloizio Macedo Apr 1 '17 at 18:34

What Rudin is saying is that the function $f: K \times X \rightarrow X: (\alpha, x) \rightarrow \alpha x $, where $K$ is the scalar field is continuous when $K \times X$ has the product topology (generated by all sets of the form $U \times V$, $U$ open in $K$, $V$ open in $X$). He doesn't seem to want to assume that the reader knows about topological products.

He then gives the local continuity for this condition: if $W$ is a neighbourhood of $f(\alpha,x)$ we need a basic neighbourhood $U \times V$ of $(\alpha,x)$ in the product such that $f[U \times V] \subseteq W$, and we can assume that $U$ (as open balls form a base in any metric space) is of the form $B_r(\alpha)$, the $r$-ball around $\alpha$, which are all $\beta \in K$ such that $|\beta - \alpha| < r$.

So the existence of $U \times V$ is just the existence of $r>0$ and $V$ a neighbourhood of $x$ such that for all $(\beta,v) \in B_r(\alpha) \times V$, so for all $\beta \in K$ with $|\beta - \alpha | < r$ and all $v \in V$, we have that $f(\beta, v) = \beta v \in W$.

So Rudin just reformulates continuity in the product topology on $K \times V$ without explicitly mentioning products.

  • $\begingroup$ Perfect answer --- thank you! $\endgroup$ – Bill Wallis Apr 2 '17 at 10:48
  • $\begingroup$ @BillWallis nothing's perfect. But thx. Glad I could help $\endgroup$ – Henno Brandsma Apr 2 '17 at 11:34

In any useful application of TVS I believe you need the property that if $x_n\rightarrow x\in X$ and $\alpha_n\rightarrow \alpha\in K$ then $\alpha_n x_n \rightarrow \alpha x$, whence this joint continuity property (which is btw very strong). You could create a topology which does not verify joint continuity but I can't think of any use in applications.


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