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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.


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

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  • $\begingroup$ 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

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).

  • 1
    $\begingroup$ 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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