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Find a function $f:V \rightarrow W$, where $V$ and $W$ are vector spaces (and V is defined on $\mathbb{K}$), such that $$f(x+y) = f(x) + f(y), \forall x,y \in V$$ but $$\exists a \in \mathbb{K}: f(ax)\neq af(x).$$ I have so long proved with some algebra that this is impossible for functions on $\mathbb{Q}$, which can be some nice exercise as well for who might want to do this exercise.

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  • $\begingroup$ But if $a=2$ and $x=y$, then according to first property $f(2x)=2f(x)$. Contradiction? $\endgroup$
    – Kaster
    Feb 26, 2013 at 20:44
  • $\begingroup$ @Kaster Presumably the OP means that there is SOME $a \in \mathbb{F}, x \in \mathbb{V}$ such that equality doesn't hold. $\endgroup$ Feb 26, 2013 at 20:47
  • $\begingroup$ As Kaster noted, allquantors for all variables are usually implied, i.e. $\forall x\in V\forall y\in V$ for the first condition. The second condition should shurely be the negation of $\forall x\in V\forall a\in K\colon f(ax)=af(x)$, which is $\exists x\in V\exists a\in K\colon f(ax)\ne af(x)$ and not (with implied allquantors) $\forall x\in V\forall a\in K\colon f(ax)\ne af(x)$. $\endgroup$ Feb 26, 2013 at 20:49
  • $\begingroup$ here we go, made it clearer, sorry but these are my first questions :) $\endgroup$
    – Kore-N
    Feb 26, 2013 at 21:00

3 Answers 3

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Hint: When working in $\mathbb{Z_p}, (x+y)^p = x^p + y^p$ (Where $p$ is a prime.)

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  • $\begingroup$ Maybe you think of a function better than mine, but did you consider that $a^{p}\equiv a$ in $\mathbb{Z}_{p}$ $\endgroup$
    – Kore-N
    Feb 26, 2013 at 21:21
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    $\begingroup$ @Cornelis, that's true, but what happens when you work in $V = \mathbb{Z_p}[X]$? $\endgroup$ Feb 26, 2013 at 21:42
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Let $K=\mathbb Q(\sqrt 2)$ be our ground field, $V=W=K$ onedimensional and define $f\colon V\to W$, $a+b\sqrt 2\mapsto a-b\sqrt 2$. Or, take $V=W=K=\mathbb C$ and $f(z)=\bar z$, which is even continuous.

If the ground field is $\mathbb Q$, you are right: Additivity implies $f(n\cdot x)=n\cdot f(x)$ and hence also $f(\frac xn)=\frac1nf(x)$ and ultimately for arbitrary fractions.

If the ground field is $\mathbb R$, it is a field extension of $K=\mathbb Q(\sqrt 2)$ above and the $f$ defined above can be extended (using the Axiom of Choice, admittedly) accordingly to give an example.

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  • $\begingroup$ Sorry for the real part, my error, it is impossible only if the function is must be continuous! $\endgroup$
    – Kore-N
    Feb 26, 2013 at 20:54
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For $f:\mathbb{R}\longrightarrow\mathbb{R}$ satisfying Cauchy's functional equation $f(x+y)=f(x)+f(y)$, we have $$ f(\alpha x)=\alpha f(x) \quad \forall \alpha,x\in \mathbb{R} $$ if and only $f$ is continuous.

There are lots of discontinuous solutions to this equation. They are known as Cauchy-Hamel functions. You can find here a way to construct them.

Note Hamel proved that their graph is dense in $\mathbb{R}^2$.

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  • $\begingroup$ you are right sorry, my error, was thinking of continuous functions! Thanks a lot, will cancel the real part right away! $\endgroup$
    – Kore-N
    Feb 26, 2013 at 20:51

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