4
$\begingroup$

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.

  • 0
    Hey! Future advice: backslashes (\) are used in Mathjax commands, not slashes (/). – 2017-01-26
  • 0
    There are functions like that. Search wiki page for cauchy functional equation, you will get loads of it. – 2017-01-26
  • 0
    I misread the problem and have deleted my answer. – 2017-01-26
  • 0
    I recommend [this link](http://www.cofault.com/2010/01/hunt-for-addictive-monster.html) for a comprehensive and entertaining exposition on the subject. – 2017-01-26

1 Answers 1

5

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
    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. – 2017-01-26