# On every infinite-dimensional Banach space there exists a discontinuous linear functional.

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?

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.
An example for these models are Solovay's model, or models of $\mathsf{ZF+AD}$.
• model, $\mathsf{ZF + \lnot AC}$, Solovay's model, $\mathsf{ZF+AD}$. Commented Jan 27, 2013 at 14:30