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This question already has an answer here:

Is there a function from $ \Bbb R^3 \to \Bbb R^3$ such that $$f(x + y) = f(x) + f(y)$$ but not $$f(cx) = cf(x)$$ for some scalar $c$?

Is there one such function even in one dimension? I so, what is it? If not, why?

I came across a function from $\Bbb R^3$ to $\Bbb R^3$ such that $$f(cx) = cf(x)$$ but not $$f(x + y) = f(x) + f(y)$$, and I was wondering whether there is one with converse.

Although there is another post titled Overview of the Basic Facts of Cauchy valued functions, I do not understand it. If someone can explain in simplest terms the function that satisfy my question and why, that would be great.

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marked as duplicate by levap, Jorge Fernández Hidalgo, Rohan, user91500, Brevan Ellefsen Jan 26 '17 at 7:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Hey! Future advice: backslashes () are used in Mathjax commands, not slashes (/). $\endgroup$ – AJY Jan 26 '17 at 3:49
  • $\begingroup$ There are functions like that. Search wiki page for cauchy functional equation, you will get loads of it. $\endgroup$ – Shobhit Jan 26 '17 at 3:53
  • $\begingroup$ I misread the problem and have deleted my answer. $\endgroup$ – Stella Biderman Jan 26 '17 at 4:02
  • $\begingroup$ I recommend this link for a comprehensive and entertaining exposition on the subject. $\endgroup$ – Fimpellizieri Jan 26 '17 at 6:25
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Take a $\mathbb Q$-linear function $f:\mathbb R\rightarrow \mathbb R$ that is not $\mathbb R$-linear and consider the function $g(x,y,z)=(f(x),f(y),f(z))$.

To see such a function $f$ exists notice that $\{1,\sqrt{2}\}$ is linearly independent over $\mathbb Q$, so there is a $\mathbb Q$-linear function $f$ that sends $1$ to $1$ and $\sqrt{2}$ to $1$. So clearly $f$ is not $\mathbb R$-linear. ( Zorn's lemma is used for this).

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    $\begingroup$ For those more interested in the details: the answer makes use of the fact that any vector space over any field (say, $\mathbb{R}$ over $\mathbb{Q}$) has a Hamel basis, which is equivalent to the axiom of choice. $\endgroup$ – lisyarus Jan 26 '17 at 4:10

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