# Generalization of the fundamental theorem of duality

The "fundamental theorem of duality" states:

If $$X$$ is a real linear space and $$f, f_1,...,f_n$$ are linear functionals on $$X$$, then $$f$$ lies in the span of $$f_1,...,f_n$$ (i.e. $$f = \sum_{i=1}^n \lambda_i f_i$$, where $$\lambda_i \in \mathbb R$$) if and only if $$\bigcap_{i=1}^n ker \ f_i \subseteq ker \ f$$ (where $$ker \ g = \{x \in X: g(x)=0\}$$).

My question is:

Can someone please direct me to a reference that generalizes this result to arbitrary collections of linear functionals?

In other words, if $$I$$ is an arbitrary index set, I am wondering how to characterize $$\bigcap_{i \in I} ker \ f_i \subseteq ker \ f.$$ Intuitively, I suspect that the equivalent statement is something like: there exists a finitely additive signed measure $$\mu$$ on $$2^I$$ such that for all $$x \in X$$ $$f (x) = \int f_i (x) \ \mu(di).$$ But this will be tricky as it involves finitely additive integration of potentially unbounded functions. (Clearly, though, this condition is sufficient for the condition about kernels to hold. The difficulty is in showing that it is also necessary.)

Anyway, surely generalizations of the fundamental theorem of duality have been explored, and any references or pointers would be appreciated.

• Please do not delete questions after having received an answer. – quid Feb 15 at 13:54
• Won't the generalization be trivial, i.e. that $f$ is in the span of $\{ f_i : i \in I\}$ if and only if there is a finite $J \subseteq I$ with $f \in \text{span} \{ f_j : j \in J\}$ ? It should not be necessary to take infinitary intersections to talk about the span, just as it's not necessary to look at infinite series. If you are thinking of some other generalization, it would clarify the question if you stated the theorem that you are thinking about. – Carl Mummert Feb 15 at 14:36

This is not an answer, but it won't fit as a comment, so I am posting it here. I am hoping that those more expert than I (i.e. almost everyone) can weigh in on this, because I have had the same question since I first encountered this theorem, and have looked around for references, without finding any. An obvious idea is to mimic the usual proof in the finite case, i.e. try to map $$X$$ into a "nice" vector space, and use the induced map out of this space into $$\mathbb R$$ to write $$f$$ as some combination of the $$f_i.$$
So, take $$\pi: X\to \mathbb R^{I}:x\mapsto \prod_{i\in I}f_i(x).$$ Since $$\bigcap_{i \in I} \text {ker} \ f_i \subseteq \text{ker} \ f,$$ there is a properly defined linear map $$F$$ such that $$F\circ \pi=f.$$ But as far as I can tell, little else can be said, even if $$I$$ is countable. The problem is there is no way around the fact that each $$f_i\neq 0$$ has range all of $$\mathbb R$$ so $$\mathbb R^I$$ can not be reduced to a "nice" space, like $$\ell^p$$ for example.