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  1. Examples of linear maps from $\phi : \mathbb R^2 \to \mathbb R$ that has homogeneity degree $1$ but is not linear.

  2. Example of a function $\phi : \mathbb C \to \mathbb C$ that is additive but is not linear.

All the examples I have found for these have gone over my head, can someone help me find a simple example for these?

Progress

I figured out a solution for the second part, can anyone help some more with the first part?

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  • What degree of homogeneity? (I imagine you want degree one, since otherwise the first question has obvious solutions.) – Semiclassical Jul 18 '14 at 02:32
  • For the first, try some cubic polynomial divided by $x^2+y^2$. For the second: something with $\bar z $ in it. (I assume $C$ means complex numbers). –  Jul 18 '14 at 05:04
  • So I figured out a solution for the second part, can anyone help some more with the first part? – Soaps Jul 23 '14 at 03:12
  • Another duplicate: http://math.stackexchange.com/q/540876/ – Jonas Meyer Aug 15 '14 at 06:45

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(Expanding my comment) Any homogeneous cubic polynomial, such as $4x^3-7xy^2+y^3$ (taking a random example), when divided by $x^2+y^2$, will produce something with degree of homogeneity $1$. The reason to divide by $x^2+y^2$ is that this expression is not zero except at the origin (we obviously don't want to divide by zero). The function still needs to be defined at the origin; let it be zero there, in consistence with homogeneity.

If the fraction $p(x,y)/(x^2+y^2)$ agreed with some linear function $ax+by$, we would have $p(x,y) = (x^2+y^2)(ax+by)$. But I'm pretty sure this is not the case for my choice of $p$. An easy way to check this is to look at $p(1,i)$, where $i$ is the imaginary unit.

Yes, I know that $x,y$ are meant to be real; but if a polynomial identity holds over reals, it holds over complex numbers too. If you are suspicious about this approach, do long division instead.