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As in, a function $f$ such that $f(u + v) = f(u) + f(v)$ for all $u,v \in \mathbb{R}^n$ but it doesn't hold that $f(tu) = tf(u)$ for all real $t$?

I know for rational $t$ it works, and if we had continuity of $f$ it would be done. I vaguely remember seeing a construction using the Choice Axiom of a weird function that involved linearity somehow, but I can't seem to find the link. (and I'm not even sure if it'll help with this problem, but maybe someone will know what I'm talking about)

violeta
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    Look up Cauchy's functional equation; given a Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$ (whose existence can be proven using choice), you can construct exotic, nonlinear, wildly discontinuous solutions to Cauchy's functional equation. – Ian Mar 07 '16 at 00:23
  • Here there are some links that considers the case $n=m=1$. (See the topic "There exists non-continuous solutions of (C)") – Pedro Mar 07 '16 at 00:31

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You make make such a thing for any $n$ and $m$ by considering both ${\mathbb R}^n$ and ${\mathbb R}^m$ as infinite dimensional vector spaces over ${\mathbb Q}$. My example will have $n=m=1$. Extend the linearly independent set $\{1,\pi\}$ to a basis of ${\mathbb R}$ as a ${\mathbb Q}$-vector space. Define $f$ on this basis by mapping $1$ to $1$ and every other basis element to $0$. Then $f$ is additive (since indeed it's ${\mathbb Q}$-linear), but $0=f(\pi 1) \neq \pi f(1) = \pi$.

P.S. It requires some form of the Axiom of Choice to prove that every vector space has a basis, and that every linearly independent set can be extended to a basis.

Jamie Radcliffe
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  • even with the axiom of choice, we will still have that the mapping used to compute the function at irrational values will be undefinable ? – reuns Mar 07 '16 at 00:41