Transformations preserving additivity

Question

I am interested in the space of real-valued additive function on $[0,1]^d$:
$\mathcal{F}= \{f:[0,1]^d \rightarrow R : f(x)= f_1(x_1)+...+f_d(x_d), f_i \in C^2, i=1,...,d \} $

Obviously there are mapping $H: [0,1]^d \rightarrow [0,1]^d$ such that $f \circ H \in \mathcal{F}$ for any $f\in \mathcal{F}$.

For example: $H(x)= (h_1(x_1),...,h_d(x_d))$ preserves additivity:
$(f\circ H) (x)= f_1(h_1(x_1))+...+f_d(h_d(x_d))$ is still additive for any $f$.

We could also consider permutations $\pi$ of $H$:
$(f\circ \tilde H) (x)= f_1(h_{\pi(1)}(x_{\pi(1)}))+...+f_d(h_{\pi(d)}(x_{\pi(d)}))$.

I would like to know if there are other maps that can preserve additivity?

Answer 1

I think I have a proof of the following result:

Proposition

The set of maps from $\mathbb{R}^n$ to $\mathbb{R}^n$ that preserve addidivity is the set $H$ of all $h: \mathbb{R}^n \to \mathbb{R}^n$ such that $\forall i\in \{1, ..., n\}$, $h_i$ (the $i$-th compenent of $h$) only depends on at most one $x_j$.

It is obvious that any $h \in H$ will preserve additivity, to prove the converse, I will use the following lemma:

Lemma

$f : \mathbb{R^n} \to \mathbb{R}$ is additive if and only if $(x, t) \mapsto f(x + t e_i) - f(x)$ does not depend on $x_j$ for $j \neq i$.

(One implication of this lemma is trivial the other a bit more tedious to show)

Proof of the proposition

Let's consider $h \notin H$, so there is $i_0 \in \{1, \dots, n\}$ such that $h_{i_0}$ depends on a least to components of $x$: $x_i$ and $x_j$ (with $i \neq j$). Without loss of generality we can assume that $i_0 = i = 1$. So $h_1$ depends on $x_1$ (meaning it is not constant with respect to $x_1$) and at least one other component of $x$.

Let's suppose that $h$ preserve additivity and get to a contradiction.

For $h$ to preserve additivity, we need that for all $f$ additive, $f ∘ h$ is additive, so using the lemma, for any $i \in \{1, ..., n\}$ and for any additive $f$ : \begin{equation}\forall (t,x) \in \mathbb{R}\times\mathbb{R}^n, f ∘ h(x + t.e_i) - f ∘ h(x) \text{ does not depend on } x_1, ..,x_{i-1}, x_{i + 1}, .., x_n \tag{$*$} \end{equation} If we write $(*)$ setting $f(x) = x_1$ (which is additive), we get that $h_1$ must be additive (using the lemma). So it means there exists functions $h_{11}, ..., h_{1n}$ such that $$h_1(x) = h_{11}(x_1) + h_{12}(x_2) + \dots +h_{1n}(x_n) = h_{11}(x_1) + h_{1+}(x_2, \dots, x_n)$$ with $h_{1+}(x_2, .., x_n) = h_{12}(x_2) + ..+h_{1n}(x_n)$. Neither $h_{11}$ nor $h_{1+}$ can be constant, otherwise it will violate the hypothesis that $h_1$ depends on $x_1$ and at least one other variable that $x_1$.

Now chosing in $(*)$ $f(x) = x_1 ^2$ (which is additive) and $e_i = e_1$ we get that

\begin{equation} f_{11}(x_1 + t) ^ 2 - f_{11}(x_1)^2 + 2f_{1+}(x_2, ..., x_n)(f_{11}(x_1 + t) - f_{11}(x_1)) \tag{$**$} \end{equation} does not depend on $x_2, ..., x_n$. But since $h_{11}$ is not constant with respect to $x_1$ (by hypthesis) we can fix one $t$ such that $h_{11}(x_1 + t) - h_{11}(x_1) \neq 0$, let's call that it $t(x_1)$ (I guess I'm using the choice axiom here). Setting $t$ to $t(x_1)$ we can write $(**)$ as $A(x_1) + B(x_1) h_{1+}(x_2, ..., x_n)$ with $B(x_1)\neq 0$ and $h_{1+}(x_2,...,x_n)$ not constant: this expression depends on at least one other variable than $x_1$. This is a contradiction since $(**)$ is supposed to depend only on $x_1$.

Hope it is clear enough (and correct), I tried to be as concise as I could...

So the set of additivity preserving maps is just a little bit bigger of the one you wrote, since for example $h(x) = (x_1, x_1, ..., x_1)$ is part of it but it is not a permutation of a function of the form $(h_1(x_1), ..., h_n(x_n))$.

Answer 2

You have described almost all of them, we show the following : Any funciton $H$ as above is of the form $H(x) = (h_1, \dots, h_d)$ where $h_i$ is a $\mathcal{C}^2$-function $[0,1]^d \rightarrow [0,1]$ that depends only on one variable which we will denote by $h_i(x) = h_i(x_j)$ and $x = (x_1, \dots, x_d)$.

Let $H$ be a function such that precomposition by $H$ preserve $\mathcal{F}$. Consider $f_i(x) = x_i$ then $f_i \in \mathcal{F}$ and $f_i \circ H = h_i$, then as $f_i \circ H$ is in $\mathcal{F}$ we can find $\mathcal{C}^2$ functions $h_i^j$ such that $h_i(x) = \sum_j h_i^j(x_j)$. We now want to see that there is only one $h_i^j$ that is non constant.

We fix $i =1$ (the argument is the same for all $i$), and suppose $h_1^1$ and $h_1^2$ are both non constant, fix $x_1$ and $x_2$ such that $(h_1^k)'(x_k) \neq 0$ for $k = 1,2$. Consider the function $g(x) = x_1^2$ then $g \in \mathcal{F}$ and thus $g \circ H \in \mathcal{F}$. We compute

$$g \circ H (x) = (h_1^1)^2(x_1) + 2h_1^1(x_1)h_1^2(x_2) + (h_1^2)^2(x_2) + \text{other terms}$$

and note that all terms are $\mathcal{C}^2$ and the "other terms" depend only on $x_1$ or $x_2$. Now as a function on $[0,1]^d$ it is $\mathcal{C}^2$, we are going to take partial derivative along $x_1$ and $x_2$, being $\mathcal{C}^2$ the order in which we do this doesn't matter. We get

$$\frac{\partial}{\partial x_1} \frac{\partial}{\partial x_2} g \circ H (x) = 2 (h_1^1)'(x_1)(h_1^2)'(x_2).$$

Now by hypothesis $g \circ H$ is in $\mathcal{F}$, it is easy to see that $\frac{\partial}{\partial x_1} \frac{\partial}{\partial x_2} t = 0$ for any $t \in \mathcal{F}$ so either $(h_1^1)'(x_1) = 0$ or $(h_1^2)'(x_2) = 0$ which contradicts our choice of $(x_1, x_2)$. Repeating the argument we get the claim and thus $h_i^j$ is non constant for only one $j$ which we denote by $j_i$, thus $h_i(x) = h_i^{j_i}(x_{j_i}) + C$ where $C$ is constant. Thus $H = (h_1, \dots, h_d)$ where each $h_i$ is a $\mathcal{C}^2$ function that depends only on one variable.

We now show that any such $H$ satisfies the condition that $f \circ H \in \mathcal{F}$ for any $f \in \mathcal{F}$. Let $f \in \mathcal{F}$ then $f(x) = \sum_i f_i(x_i)$ where $f_i \in \mathcal{C}^2$, then $f \circ H (x) = f(h_1(x_{j_1}), \dots, h_d(x_{j_d})) = \sum_i (f_i \circ h_i)(x_{j_i})$, by hypothesis $(f_i \circ h_i)$ is in $\mathcal{C}^2$ and $f \circ H$ is in $\mathcal{F}$.