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I need to find a nonlinear function $f:\mathbb{R}^2\rightarrow\mathbb{R}^2$ such that $f(\alpha (a,b))=\alpha f(a,b)$ for all $(a,b)\in\mathbb{R}^2$ and $\alpha\in\mathbb{R}$.

I can't find anything.

Context

The requirement $f(\alpha (a,b))=\alpha f(a,b)$ says that $f$ respects the scalar multiplication, just as linear maps do. In particular, $f$ is homogeneous of degree $1$. To make it nonlinear, one has to somehow destroy the additive property $f(a+c,b+d)=f(a,b)+f(c,d)$.

Community
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yemino
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    A duplicate of http://math.stackexchange.com/q/870478/ but the answer there isn't upvoted, so I can't vote to close. (The example there works for negative scalar multiplication too, not just homogeneity). –  Aug 12 '14 at 17:34
  • Side note: over $\mathbb{R}$, every scalar multiple of $\vec{v}$ is a a limit of rational scalar multiples of $\vec{v}$. And if additivity holds for a transformation, then it's easy to show that preservation of scalar multiplication by a rational number holds. So if you had the opposite task (seeking a transformation with additivity but not scalar preservation) then it would be impossible over $\mathbb{R}$. – 2'5 9'2 Aug 12 '14 at 17:38
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    @alex.jordan See this. – David Mitra Aug 12 '14 at 17:40
  • @DavidMitra Oh yes, I should have added "continuous" to my note, which relies on reals being approached by a sequence of rationals. – 2'5 9'2 Aug 12 '14 at 19:55
  • Another duplicate: http://math.stackexchange.com/q/540876/ – Jonas Meyer Aug 15 '14 at 06:45

4 Answers4

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The hypothesis that $f(\alpha\vec x)=\alpha f(\vec x)$ is equivalent to requiring that $f$ is linear on each line through the origin. You can map these lines to other lines in nonlinear ways. E.g., you could rotate a vector with polar angle $\theta$ by an angle of $\sin^2\theta$ (or any other nonconstant $\pi$-periodic function of $\theta$):

$$(r\cos(\theta),r\sin(\theta))\mapsto (r\cos(\theta+\sin^2(\theta)),r\sin(\theta+\sin^2(\theta)).$$

Jonas Meyer
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  • Wow, I didn't notice all that happened in this thread in the last 20 minutes or so. There is some inadvertent similarity to JHance's answer, and I hadn't seen from 900sit-ups's comment that this is a duplicate. – Jonas Meyer Aug 12 '14 at 17:58
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Hint: find a function $g:\mathbb R^2 \to \mathbb R$ that is non linear and satisfies your condition. Then take $$f(a,b) = (g(a,b),0)$$

Try $$g(a,b) = \begin{cases}\frac{a^2}b & b \ne 0\\0&\text{otherwise}\end{cases}$$

Mathmo123
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You can define f as follows : $f(0,y)=(0,y)$ for any $y$, and $f(x,y)=(y,0)$ for any $y$,whenever $x\neq 0$.

Ewan Delanoy
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As observed by others, in order for this to happen the function must fail to be additive. Now, one way to think about such problems is to ask the following: given my hypotheses on the function, how much data determines the function?

For example, in a linear function it is a standard observation that specifying $f$'s values at a basis determines the function. Now, given this homogeneity constrain we observe that specifying $f(a,b)$ determines $f$ along the entire ray by $(a,b)$ but nowhere else. So let's choose a value of $f(a,b)$ for every $(a,b)$ on the unit circle and such that $f(-a,-b) = -f(a,b)$. Then by scaling we have a function on the whole plane satisfying your constraint.

Now, final problem: figure out which values on the unit circle give linear functions (or equivalently, what values do we get on the unit circle if we start with a linear function). Choose something that isn't that.

jxnh
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