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On sort-of-linear functions

It is an interesting exercise to show that $\operatorname{Gal}(\mathbb R ~\colon \mathbb Q)$ is trivial. The only solution I know hinges on the fact that the automorphism is order-preserving, which in turn depends on the fact that $\theta(xy)=\theta(x)\theta(y)$ for $\theta \in \operatorname{Aut} \mathbb R$.

Now, a function $L:\mathbb R \to \mathbb R$ with just the property that $L(x+y)=L(x)+L(y)$ can be shown to preserve multiplication on the rationals. And, I have been unsuccessful in trying to extend this fact to the reals. This could be because such a function could be discontinuous, an example of which I also have failed to construct (it's a bad day).

My question:

Can you give me an example of a discontinuous function $L:\mathbb R \to \mathbb R$ with the property that $L(x+y)=L(x)+L(y)$ and $L|_{\mathbb Q}= \text{identity}$?

An idea: Perhaps we could consider a function that preserves order on rationals but reverses it on irrationals.


marked as duplicate by Qiaochu Yuan Feb 21 '11 at 15:22

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  • $\begingroup$ Why does $L$ with just that property preserve multiplication on the rationals? Can't you just take $L(x) = \lambda x$ to ensure $L(x+y)=\lambda(x+y)=\lambda x + \lambda y = L(x)+L(y)$, and then $L(x)\cdot L(y) = (\lambda x)\cdot(\lambda y) =\lambda^2 (x\cdot y) \neq L(x\cdot y)$ for $\lambda \notin \{0,1\}$? $\endgroup$ – joriki Feb 21 '11 at 14:14
  • $\begingroup$ @Dactyl: An additive function on the reals must be $\mathbb{Q}$-linear, but it need not "preserve multiplication on the rationals". That is, $L$ must satisfy $L(qx) = qL(x)$ for all $q\in\mathbb{Q}$, but it does not have to satisfy $L(pq) = L(p)L(q)$, as joriki points out. You may have misspoken there. $\endgroup$ – Arturo Magidin Feb 21 '11 at 14:26
  • $\begingroup$ Has this not come up before? I'm aghast. $\endgroup$ – Arturo Magidin Feb 21 '11 at 14:26
  • $\begingroup$ This has definitely come up before, although I'm bad at tracking down these kinds of questions. $\endgroup$ – Qiaochu Yuan Feb 21 '11 at 14:42
  • 1
    $\begingroup$ Oops, not only has it come up before, but I answered a variant of this question. This one was more specific to have it be the identity on the rationals. (And my previous answer gave a convoluted example because I was thinking of something else.) math.stackexchange.com/questions/16175/… $\endgroup$ – Jonas Meyer Feb 21 '11 at 15:09

This is impossible to do in ZF alone, but possible with a Hamel basis for $\mathbb{R}$ as a $\mathbb{Q}$ vector space. Hence no "explicit" example can be expected. Complete $\{1\}$ to a Hamel basis $B$, let $x\in B\setminus\{1\}$, and let $f:\mathbb{R}\to\mathbb{R}$ be the unique linear function such that $f(x)=0$ and $f(b)=b$ for all $b\in B\setminus\{x\}$; in particular, $f(1)=1$ implies that $f$ is the identity on the rationals. Because $x$ can be approximated by rationals (not to mention with rational multiples of other elements of $B$), $f$ is not continuous at $x$. In fact, $f$ is not continuous anywhere.


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