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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 $a$.

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 $f(x)=|x|g(\frac{x}{|x|})$

High GPA
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    Any injective continuous function is a solution. – geetha290krm Jul 13 '23 at 07:34
  • @geetha290krm https://math.stackexchange.com/questions/170147/a-continuous-injective-function-f-mathbbr-to-mathbbr-is-either-strict – High GPA Jul 13 '23 at 07:37
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    you should probably consider polar coordinates for $\mathbb{R}^n$. Let $g: S^n \to \mathbb{R}$ be any function on a sphere. Then $f(x) := g\bigl(\frac{x}{|x|}\bigr)$ satisfies your requirement I think – Felix B. Jul 13 '23 at 07:40
  • @Gonçalo You are right I probably made a mistake; I'll rethink. $ax^b+c$ are also solutions for dimension-1 I think. – High GPA Jul 13 '23 at 07:41
  • @FelixB. I agree. Is it possible to prove that those functions are all solutions? – High GPA Jul 13 '23 at 07:42
  • sorry my ignorance, but how can a function be linear and translational invariant? If it's linear, then $f(ax+by)=af(x)+bf(y)$, but if it's translational invariant, then $f(x+y)=f(x)$ – Daniel Muñoz Jul 13 '23 at 07:44
  • actually - I just noticed that the opposite side is an issue with $a=-1$. I mean in $n=1$ you still need both points at $+1$ and $-1$ to coincide for the function to be constant. Unfortunately the half-sphere has a border and the annoying thing is that you also have opposite points on the border. The border is an $S^{n-1}$ sphere, so I guess you have to take half of the border recursively to define you function $g$ on a half sphere – Felix B. Jul 13 '23 at 07:44
  • @DanielMuñoz Sorry, I agree with that my nomenclature is most-likely be sloppy. Could you please suggest an alternative naming? – High GPA Jul 13 '23 at 07:46
  • or you simply require $g$ to satisfy $g(x) = g(-x)$ – Felix B. Jul 13 '23 at 07:47
  • @FelixB. You are right. Let's first think about positive $a$ to get a structure for $f$. The issue with negative $a$ can easily be fixed later, I guess. – High GPA Jul 13 '23 at 07:47
  • @HighGPA the thing is, if your eq is $f(ax)=f(ay)$, when $x\neq y$, then you have two cases. $f$ is periodic or is constant. I can't think of other cases. If it's constant, then it's not linear. If it's periodic, then it's not linear and translational invariant. – Daniel Muñoz Jul 13 '23 at 07:51
  • @HighGPA with you new edit, Are you reffering to "scale invariant"? – Daniel Muñoz Jul 13 '23 at 07:56
  • @DanielMuñoz We could check if $f=x^4+b$ meets the requirements – High GPA Jul 13 '23 at 07:56
  • @DanielMuñoz Yes something like "scale invariant". Though from my personal perspectives scale invariant means $f(ax+b)=f(ay+b)$ – High GPA Jul 13 '23 at 07:57
  • @HighGPA scale invariance means $f(ax)=f(x)$, which is the property you are defining for. – Daniel Muñoz Jul 13 '23 at 07:59
  • @HighGPA your eq implies for all $x,y\in\Bbb R^n$ or some? – Daniel Muñoz Jul 13 '23 at 08:01
  • @DanielMuñoz For all $x,y$. – High GPA Jul 13 '23 at 08:03
  • @HighGPA then for $\Bbb R$, your solution is the constant function, because is the only function which is scale invariant. Note that $f(x)=f(y)$, $y$ can be wrote as $kx$ where $k\in \Bbb R$. If you choose a polynomial as solution, then don't satisfize the first condition $f(x)=f(y)$ – Daniel Muñoz Jul 13 '23 at 08:10
  • Any radially symmetric continuous function works in any dimension – Carlyle Jul 13 '23 at 08:11
  • @Carlyle I agree. I also want to find all solutions. – High GPA Jul 13 '23 at 08:26
  • @FelixB. I think your expression works but it excludes many other simple solutions like $f(x,y)=x+y$. – High GPA Jul 13 '23 at 08:28
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    @HighGPA I believe my answer captures all solutions, but could be mistaken, I also don't know if there is a simpler classification – Carlyle Jul 13 '23 at 08:42

2 Answers2

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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.

Carlyle
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  • If you have a more general solution without continuity, then it is even more exciting for me! – High GPA Jul 13 '23 at 09:24
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    Let $f(x,y)=x+2y$. How can you rewrite it into the radial function $g(|x|)$? – High GPA Jul 13 '23 at 15:11
  • @HighGPA You can't, but it isn't the entire function that must be rewritten in terms of a radial function, it is its restriction to the cones, so when we define $\sim$, we relate those unit vectors that get sent to the same thing under $f$ but no two unit vectors get mapped to the same number under this $f$ so the relation just relates every unit vector only to itself, the set of cones is then precisely the set of positive rays, and $f$ restricted to any such ray is radial. – Carlyle Jul 13 '23 at 17:00
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    But your function does in fact cause problems, because the third condition, that the intersections of the images of the cones must be empty, is not satisfied, but the function satisfies the property, so this requirement is too strict, the images may intersect, but in that case, either they must intersect at $f(\mathbf{0})$, or they're associated radial functions must be scalar multiples of each other (I am guessing that this is the solution, but maybe look for another counter example) – Carlyle Jul 13 '23 at 17:04
  • I also think that each different radial function $g_1,g_2$ must be related by scale: $g_1=\lambda g_2$. If not, we can try to construct a counterexample! – High GPA Jul 13 '23 at 17:06
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    If the radial functions share a non trivial value then yes I believe you are right, but I think it is possible for two radial functions to be completely unrelated, provided their ranges don't overlap except at the origin, but this radial scaling requirement is an interesting restriction in its own right, and leads to "nicer looking" functions imo – Carlyle Jul 13 '23 at 17:17
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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 $g_d(x) = g(b(d) x)$.

Felix B.
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