Discontinuous linear functionals and the axiom of choice

In lecture the other day, my functional analysis professor remarked that he could not think of any examples of a Banach space B and a linear functional f on B such that f was discontinuous which were not constructed without use of the axiom of choice, and said, “The logicians might have something to say about that.”

Indeed, Solovay’s model is a model of ZF (without the axiom of choice) in which every linear functional on every Banach space is continuous. Notably, this model thinks that every subset of every Polish space has the property of Baire (i.e. it differs from an open set by a meager set — this could never happen under ZFC, since a nonmeasurable set does not have the property of Baire). For the purposes of this problem, we might as well assume that B is Polish, since if not, and f is an unbounded functional, then there is a linearly independent sequence $x_n$ in B such that $|f(x_n)| \to \infty$ . The completion of the span of the sequence $x_n$ is a separable (hence Polish) Banach space on which f is discontinuous.

We now apply Pettis’s theorem:

Let $G$ be a Baire group and $H$ be a Polish group, and $\varphi: G \to H$ be a (possibly discontinuous) morphism of groups. If, for every open set $U \subseteq G$ , $\varphi^{-1}(U)$ has the property of Baire, then $\varphi$ is actually continuous.

Pettis’s theorem is, for example, Theorem 9.10 in Kechris’ book “Classical Descriptive Set Theory.” The proof is not especially difficult.

Since B is a Banach space, the Baire category theorem implies that B is an additive Baire group; while $\mathbb R$ is an additive Polish group and f is a morphism of groups. We can assume that Solovay’s model thinks that any $A \subseteq B$ has the property of Baire, so f is continuous by Pettis’s theorem.