0

Let $X$ be a real Banach space and $f:X \rightarrow \mathbb{R}$ be a map such that $f(x+y)=f(x)+f(y)$ for all $x,y \in X$. If $f$ is continuous we know that $f(\lambda x)=\lambda f(x)$, for all $x \in \mathbb{R}$ and $\lambda \in \mathbb{R}$.

My question: I would like to know if someone could give me an example of a Banach space $X$ and a discontinuous map $f:X \rightarrow \mathbb{R}$ such that $f(x+y)=f(x)+f(y)$ for all $x, y \in X$ and $f(\lambda_0 x_0) \neq \lambda_0 f(x_0)$ for some $\lambda_0 \in \mathbb{R}$ and $x_0 \in X$. Would it be possible to obtain an map satisfying this property in any Banach space?

Math
  • 2,473
  • 1
    Just take a basis $B$ of $X$ as a $\mathbb{Q}$-vector space that extends the $\mathbb{Q}$-linearly independent set ${x\neq0, \sqrt{2}x}$. Define $f$ on $B$ by putting $f(x)=r\neq 0$ and $f(\sqrt{2}x)\neq \sqrt{2}r$ and on the rest of $B$ in any way you want. Extend the definition of $f$ to $X$ by $\mathbb{Q}$-linearity. – plop Sep 21 '22 at 14:29
  • 1
    No needs for $X$ to be complete. Take $f=g\circ h$ where $h:X\to\mathbb R$ is any nonzero linear functional and $g:\mathbb R\to\mathbb R$ is any discontinuous solution of Cauchy's functional equation. – Anne Bauval Sep 21 '22 at 14:30
  • 4
    It is possible to prove existence of such functions using the axiom of choice. But as I know, there is no way to find a concrete example even for the case $X=\mathbb{R}$. – Mark Sep 21 '22 at 14:32

1 Answers1

3

Actually it's in a certain sense impossible to explicitly give such an example. This is because it's consistent with ZF that every homomorphism between Polish groups is automatically continuous (see e.g. Theorem 2.2 here), and separable Banach spaces are Polish. So there's a model of ZF in which every group homomorphism between separable Banach spaces is automatically continuous!

Using the axiom of choice terrible constructions can be performed as described in the comments, by considering a $\mathbb{Q}$-vector space basis. This can be done for any $\mathbb{R}$-vector space. See Cauchy's functional equation for more details and background.

Qiaochu Yuan
  • 468,795