# Math Help - Finding a Non-linear Map That Satisfies af(x) = f(ax)

1. ## Finding a Non-linear Map That Satisfies af(x) = f(ax)

Linear maps satisfy both conditions L1 and L2 below. I have to find a map R^2 -> R that satisfies ONLY L2 and not L1. I have been thinking about this for quite a while, but I haven't made that much progress. Do I have to use absolute values in some way?

It would be nice to see the actual solution or at least get some clear hints.

x and y are vectors.

L1: f(x+y) = f(x) + f(y)
L2: af(x) = f (ax)

Linear maps satisfy both conditions L1 and L2 below. I have to find a map R^2 -> R that satisfies ONLY L2 and not L1. I have been thinking about this for quite a while, but I haven't made that much progress. Do I have to use absolute values in some way?

It would be nice to see the actual solution or at least get some clear hints.

x and y are vectors.

L1: f(x+y) = f(x) + f(y)
L2: af(x) = f (ax)
Think in terms of polar coordinates. What about the map $[r,\theta]\mapsto rf(\theta)$, for some suitable function $f(\theta)$ ?

3. Originally Posted by Opalg
Think in terms of polar coordinates. What about the map $[r,\theta]\mapsto rf(\theta)$, for some suitable function $f(\theta)$ ?
Your polar coordinate hint led me to try trigonometric functions, but I couldn't find a solution there. I have been trying pretty much every possible polar coordinate function I can think of, but the problem is that I just can't seem to find this "suitable" function. Whenever I find a map that satisfies L2, it always satisfies L1 too. More help would be apprectiated.

4. Have you tried the norm function?

Send $(x,y)\to |(x,y)|=\sqrt{x^2+y^2}$.

5. Originally Posted by roninpro
Have you tried the norm function?

Send $(x,y)\to |(x,y)|=\sqrt{x^2+y^2}$.
This doesn't work since $\|\alpha (x,y)\|=\sqrt{\alpha^2x^2+\alpha^2y^2}=|\alpha|\|( x,y)\|$.

6. It seems to be a good place to look though - perhaps it can be modified to include a sign.

How about $(x,y)\mapsto \sqrt[3]{x^3+y^3}$?

You need a function $f(\theta)$ that satisfies $f(\theta+\pi) = -f(\theta)$, so as to take care of multiplication by a negative scalar. But you can't use $f(\theta) = \cos\theta$ or $f(\theta) = \sin\theta$, because they lead to linear functions. The function I had in mind was $f(\theta) = \cos^3\theta$.