I am reading the fifth chapter ("Dual Spaces") from David Luenberger's Optimization by Vector Space Methods (1969). In Section 5.10, weak continuity for possibly nonlinear functionals on normed spaces is defined as follows.
(Note: $X^\ast$ is defined previously in the book as the normed dual of $X$, and $f$ is meant to be the possibly nonlinear functional being characterized.)
I am struggling to understand the motivation behind requiring the functional collection $\{x_1^\ast, x_2^\ast, \cdots, x_n^\ast\}$ to be finite. In particular, instead of requiring a finite collection, why does the definition not simply demand the existence of “at least one element”, say $x_1^\ast$ (from $X^\ast$), which abides by the $\varepsilon-\delta$ condition?
Would greatly appreciate help. Thanks.
This has to do with the way you define the weak topology on a normed vector space: Those finite intersections form a basis of neighborhoods of $x_0$, i.e. every weakly open set containing $x_0$ contains a set of that form for some finite collection. In this sense, those finite intersections are playing the role of $B_\delta(x_0)$ when you think of continuity on, say, $\mathbb{R}^n$.
The reason you can't just take one linear functional is that those sets are gigantic: Think of what happens when $X=\mathbb{R}^n$, then the set of $x$ for which $\langle x-x_0, x_1^*\rangle <\delta$ is a half-space! Requiring $|f(x)-f(x_0)|<\varepsilon$ for $x$ in half-space is way too much. On the other hand, if you allow for finite intersections, you can get polyhedra containing $x_0$ instead, and that is more reasonable to expect of a continuous function.