On every infinite-dimensional Banach space there exists a discontinuous linear functional.
Assuming the axiom of choice, every vector space has a basis. With an infinite basis, I can define on a countable subset $\{e_n:n\in\mathbb{N}\}$ a function $f(e_n)=n\|e_n\|$ and let $f(x)=1$ for all other basis vectors.
Then this determines an unbounded linear functional, which is therefore discontinuous.
But this argument, a, applies to any infinite-dimensional normed spaces, b, relies on the assumption of the axiom of choice.
Is there a smart answer that does make use of the condition that the space in question is a Banach space, and even better, avoids the use of axiom of choice?
Best Answer
No. There are models of $\mathsf{ZF+\lnot AC}$ in which every linear transformation from a Banach space to a normed space is automatically continuous. In particular this is true for linear functionals.
In such models, it follows, every linear functional has to be continuous.
An example for these models are Solovay's model, or models of $\mathsf{ZF+AD}$.