Simple functional equation: $f(x)=f(y) \implies f(ax)=f(ay)$

Question

Let $a$ be real number and variable $x,y\in\mathbb R^n$. Solve for all continuous function $f$ such that it is "proportional-invariance": $f(x)=f(y)\implies f(ax)=f(ay)$ for all $a$.

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Let's start at one dimension.

For $n=1$, it is straightforward strict monotonic function or constant are solutions. Edit: power-like functions are also solutions.

I have no idea how to deal with it when $n=2$, besides requiring that $f$ is also monotonic in both directions. Can anyone give me a hint for the simple case when $n=2$ and $a$ is positive?

Example: $f(x_1,x_2)=g(x_1+x_2)$ and $g$ is monotonic.

Edit: after reading the answers, my guessed solution is $f(x)=|x|g(\frac{x}{|x|})$

Answer 1

Define a cone of $\mathbb{R}^n$ as a set that can be written as the union the positive spans of vectors (we will call them positive rays) in $\mathbb{R}^n$. By positive span of a vector $\mathbf{x}$ I mean the set $\{a\mathbf{x} | a\geq 0\}$

Define a radial-function as one that can be written as a function of the length of a vector (i.e. $f$ is a radial function if there is some $g$ such that $f(\mathbb{x}) = g(||\mathbb{x}||)$ )

Then let $P$ be a partition of $\mathbb{R}^n$ into cones. It is easy to see that if the restriction of $f$ to every element of $P$ is a radial function and the one dimensional function that $f$ reduces to satisfies the one dimensional version of the problem, and moreover the images of the elements of $P$ do not intersect, then $f$ satisfies the condition.

The converse is maybe a bit more tricky to see, it entails constructing the partition from a given function that satisfies the condition.

Given a function $f$ that satisfies the condition in question, then define a relation $\sim$ on the set of all positive rays of $\mathbb{R}^n$ (call this set S), as follows: (we Identify every positive ray with the unit vector that positive spans that ray)

$ \mathbf{a} \sim \mathbf{b} \iff f(\mathbf{a}) = f(\mathbf{b}) $

Then $\sim$ is an equivalence relation, and the partition we are looking for is the partition induced by $\sim$. It is clear that the restriction of $f$ to some element of this partition is a radial function, and I leave it to you to check the intersections of the images.

Then finally you can add back the continuity condition.

This may not be as nice of a classification as you were looking for, but I don't think the solution set permits a much nicer classification.

Answer 2

Let us consider the following equivalence relation for $x,y \in \mathbb{R}^n$: $$ x\sim y \iff \frac{x}{\|x\|} = \frac{\pm y}{\|y\|} $$ where the $\pm$ should be read as one of the equations is true. Then the set of unsigned directions is given as the quotient set $$ D = \mathbb{R}^n/\sim $$ Now let $G_a = \{g_d: \mathbb{R}\to \mathbb{R} : d\in D, g_d(0)=a\}$ be a collection of functions originating in $a$. Then such a function set can be stiched together into a function on $\mathbb{R}^n$ by $$ f(x):= g_{[x]}(\|x\|) $$ where $[x]$ is the unsigned direction of $x$. Conversely you can split up a function $f$ into a set $G_{f(0)}$.

If $f$ was continuous, all the $g_d$ are continuous. And from $f(x)=f(y)\implies f(ax)=f(ay)$ we can deduce the same property for all $g_d$. Thus you can apply your reasoning in one dimension to deduce that the $g_d$ are either

• constant
• injective (and therefore strict monotonous due to continuity) or
• symmetric and injective on $\mathbb{R}_+$.

Let $C = \{d\in D : g_d \equiv c \in \mathbb{R}\}$ be the set of directions where the function is constant. Then $C \uplus C^\complement = D$. If both $C$ and its complement where not empty, we would have $d\in \partial C$. This implies that we can find arbitrarily close $d_c$, $d$ where $g_{d_c}$ is constant and $g_d$ is injective on $\mathbb{R}_+$ and therefore strictly monotonously increasing. But since $f$ is continuous this (probably) leads to a contradiction?

EDIT: No it doesn't. Example: $f(x,y) = x$ In the $y$ direction this is constant, in the $x$ direction it isn't.

So either $C=D$ or $C=\emptyset$. Now since all $g_d$ coincide in $0$, in the case they are constant $f$ is constant.

Now let us consider the $C=\emptyset$ case and let $g_d(0)=0$ w.l.o.g. If the intersection $g_d(\mathbb{R})\cap g_{d'}(\mathbb{R})$ of two $g_d$, $g_{d'}$ includes more than $0$ then you know that $g_d(x) = g_{d'}(y)$ for some $x,y\neq 0$. With $y= bx$ we therefore have $g_d(ax) = g_{d'}(abx)$ for all $a$ or in other words $g_d(t) = g_{d'}(bt)$ for all $t$. In other words then $g_d$ is just a rescaled version of $g_{d'}$.

You can probably rule out that the intersection only includes $0$ with continuity for $d$ close to each other. And similarly express $b$ as a continuous function of $d$. To ge the expression $g_d(x) = g(b(d) x)$.