The general linear equation on open connected sets

TL;DR

This paper characterizes functions on open connected sets in topological vector spaces that satisfy a general linear equation, showing they are affine functions with a unique linear part over a specific field.

Contribution

It extends classical results by providing a characterization of solutions to the general linear equation on open connected sets in topological vector spaces, including existence and uniqueness of affine solutions.

Findings

Solutions are affine functions with a unique linear part over a specified field.

Existence and uniqueness of solutions are established using a generalized extension theorem.

The results apply to functions on open convex cones and similar sets.

Abstract

Fix non-zero reals $\alpha_1,\ldots,\alpha_n$ with $n\ge 2$ and let $K$ be a non-empty open connected set in a topological vector space such that $\sum_{i\le n}\alpha_iK\subseteq K$ (which holds, in particular, if $K$ is an open convex cone and $\alpha_1,\ldots,\alpha_n>0$). Let also $Y$ be a vector space over $\mathbb{F}:=\mathbb{Q}(\alpha_1,\ldots,\alpha_n)$. We show, among others, that a function $f: K\to Y$ satisfies the general linear equation $$ \textstyle \forall x_1,\ldots,x_n \in K,\,\,\,\,\, f\left(\sum_{i\le n}\alpha_i x_i\right)=\sum_{i\le n}\alpha_i f(x_i) $$ if and only if there exist a unique $\mathbb{F}$-linear $A:X\to Y$ and unique $b\in Y$ such that $f(x)=A(x)+b$ for all $x \in K$, with $b=0$ if $\sum_{i\le n}\alpha_i\neq 1$. The main tool of the proof is a general version of a result Rad\'{o} and Baker on the existence and uniqueness of extension of the solution on the classical Pexider equation.

Equations

$$\forall x_1,\dots ,x_n\in K,f\left({\sum }_{i\le n}{\alpha }_i{x}_i\right)={\sum }_{i\le n}{\alpha }_{i}f(x_i)$$

$$\forall x,y\in {\mathbb{R}}^{+}\times {\mathbb{R}}^{+},\Phi \left(\alpha x+\beta y\right)=\alpha \Phi (x)+\beta \Phi (y),$$

$$\forall x_1,\dots ,x_n\in K,f\left({\sum }_{i\le n}{\alpha }_i{x}_i\right)={\sum }_{i\le n}{\beta }_{i}f(x_i).$$

$$\forall x\in X,\forall i=1,\dots ,n,A({\alpha }_{i}x)={\beta }_{i}A(x)$$

$$\forall x\in K,f(x)=A(x)+b,$$

$$\forall x_1,\dots ,x_n\in K,f\left({\sum }_{i\le n}{\alpha }_i{x}_i\right)={\sum }_{i\le n}{\alpha }_{i}f(x_i).$$

$$A_i:={\pi }_i(A)=\{b\in B:{\pi }_i(a)=b\text{ for some }a\in A\}$$

$$A_{+}:=\{a_1+\cdots +a_n:(a_1,\dots ,a_n)\in A\}.$$

$$\forall (x_1,\dots ,x_n)\in U,f\left({\sum }_{i\le n}{x}_i\right)={\sum }_{i\le n}{g}_i(x_i).$$

$$\forall x_1,\dots ,x_n\in X,F\left({\sum }_{i\le n}{x}_i\right)={\sum }_{i\le n}{G}_i(x_i).$$

$$\forall x\in X,\forall i=1,\dots ,n,F(x)=A(x)+u\text{ and }{G}_i(x)=A(x)+u_i.$$

$$\forall x_1,\dots ,x_n\in X,f\left({\sum }_{i\le n}{x}_i\right)={\sum }_{i\le n}{g}_i(x_i)$$

$$\forall x\in X,\forall i=1,\dots ,n,f(x)=A(x)+y\text{ and }{g}_i(x)=A(x)+y_i.$$

$$\tilde{f}\left({\sum }_{i\le n}{x}_i\right)={\sum }_{i\le n}{\tilde{g}}_i(x_i)$$

$$\forall x_1,\dots ,x_n\in X,f\left({\sum }_{i\le n}{\alpha }_i{x}_i\right)={\sum }_{i\le n}{g}_i(x_i)$$

$$3=f(2)-f(1)=(A(2)+y)-(A(1)+y)=A(1),$$

$$\forall z_1,\dots ,z_n\in V_x,f\left({\sum }_{i\le n}(x_i+z_i)\right)={\sum }_{i\le n}{g}_i(x_i+z_i).$$

$$\tilde{f}(z):=f\left(z+{\sum }_{i\le n}{x}_i\right)-f\left({\sum }_{i\le n}{x}_i\right)\text{ and }{\tilde{g}}_i(z):=g_i(x_i+z)-g_i(x_i)$$

$$\forall z_1,\dots ,z_n\in V_x,\tilde{f}\left({\sum }_{i\le n}{z}_i\right)={\sum }_{i\le n}{\tilde{g}}_i(z_i),$$

$$\forall z_1,z_2\in V_x,\tilde{f}(z_1+z_2)={\tilde{g}}_1(z_1)+{\tilde{g}}_2(z_2)$$

$$\forall z\in V_x,\forall i=1,\dots ,n,g_i(x_i+z)=A_x(z)+u_{x,i}$$

$$\forall z\in n\cdot V_x,f\left(z+{\sum }_{i\le n}{x}_i\right)=A_x(z)+u_x,$$

$$w+Z^n\subseteq (x+V_x^n)\cap (y+V_y^n).$$

$$\begin{array}{rl}{A}_w(z)& =g_i(w_i+z)-u_{w,i}=g_i(x_i+(w_i-x_i+z))-u_{w,i}\\ & =u_{x,i}+A_x(w_i-x_i+z)-u_{w,i}=(u_{x,i}-u_{w,i}+A_x(w_i-x_i))+A_x(z).\end{array}$$

$$u_{x,i}-u_{w,i}+A_x(w_i-x_i)=0,$$

$$\forall i=1,\dots ,n,\forall z\in (x_i+V_x)\cup (y_i+V_y),g_i(z)=A_x(z)+u_i$$

$$\forall z\in \left({\sum }_{i\le n}{x}_i+n\cdot V_x\right)\cup \left({\sum }_{i\le n}{y}_i+n\cdot V_y\right),f(z)=A_x(z)+u.$$

$$u_i:=u_{x,i}-A(x_i)=u_{w,i}-A(w_i)=u_{y,i}-A(y_i)$$

$$g_i(z)=g_i(x_i+z-x_i)=A(z-x_i)+u_{x,i}=A(z)+u_i,$$

$$\begin{array}{rl}f(z)& =f\left({\sum }_{i\le n}(x_i+v_i)\right)={\sum }_{i\le n}{g}_i(x_i+v_i)={\sum }_{i\le n}(A(v_i)+u_{x,i})\\ & ={\sum }_{i\le n}(A(v_i)+A(x_i)+u_i)=A\left({\sum }_{i\le n}(x_i+v_i)\right)+{\sum }_{i\le n}{u}_i=A(z)+u,\end{array}$$

Full text

[TABLE]

if and only if there exist a unique $\mathbb{F}$-linear $A:X\to Y$ and unique $b\in Y$ such that $f(x)=A(x)+b$ for all $x\in K$, with $b=0$ if $\sum_{i\leq n}\alpha_{i}\neq 1$. The main tool of the proof is a general version of a result Radó and Baker on the existence and uniqueness of extension of the solution on the classical Pexider equation.

P.L. was supported by the Austrian Science Fund (FWF), project F5512-N26.

1. Introduction

Motivated by the study of certainty equivalents in the theory of decision making under uncertainty, the authors in [6] solved a functional equation which, after some manipulations, led to the restricted general linear equation

[TABLE]

where $\alpha,\beta\in\mathbb{R}^{+}$ are given constants and $\Phi:\mathbb{R}^{+}\times\mathbb{R}^{+}\to\mathbb{R}^{+}$ is a continuous function such that $\lim_{x\to(0,0)}\Phi(x)=0$, with $\mathbb{R}^{+}:=(0,\infty)$, see [6, Equation (6)].

The aim of this article is to provide a characterization of the solutions of general linear equations as in (1), where the variables are restricted to an open connected set of a topological vector space. The main novelty of the proof is a general version of a result Radó and Baker [13] on the existence and uniqueness of extension of the solution of the classical Pexider equation (see Theorem 1.6 below). We refer to [9, Chapter 13.10] and [8, 11, 12] for the classical theory of general linear equations and references therein.

Our main result follows. The proof is given in Section 3.

Theorem 1.1**.**

Let $X$ be a topological vector space over $\mathbb{K}$, where $\mathbb{K}$ is the field of real or complex numbers, and let $Y$ be a vector space over a field $\mathbb{F}$ of characteristic zero. Fix also non-zero $\alpha_{1},\ldots,\alpha_{n}\in\mathbb{K}$ with $n\geq 2$, non-zero scalars $\beta_{1},\ldots,\beta_{n}\in\mathbb{F}$, and a non-empty open connected set $K\subseteq X$ such that $\sum_{i\leq n}\alpha_{i}K\subseteq K$. Finally, let $f:K\to Y$ be a function such that

[TABLE]

Then there exist a unique group homomorphism $A:X\to Y$ and a unique $b\in Y$ for which

[TABLE]

and

[TABLE]

where necessarily $b=0$ if $\sum_{i\leq n}\beta_{i}\neq 1$ (and $b\in Y$ arbitrary otherwise).

Conversely, if $A:X\to Y$ is a group homomorphism which satisfies (3), and $f:K\to Y$ is defined by (4) with $b=0$ if $\sum_{i\leq n}\beta_{i}\neq 1$, then $f$ satisfies (2).

Related results can found, e.g., in [8, Section 4]. We remark that, in the case $X=K=\mathbb{R}^{k}$, $Y=\mathbb{R}$, and $n=2$, it is possible to characterize all group homomorphisms $A:X\to Y$ satisfying (3), see e.g. [9, Theorem 13.10.5].

In the rather early paper [15] we can find an extension theorem for the classical Cauchy equation restricted to an arbitrary non-empty open, not necessarily connected, subset of $\mathbb{R}^{2}$. Other investigations by the same authors related to exponential polynomials and spectral analysis can be found in [16, 17, 18].

In some cases, the hypotheses of Theorem 1.1 are sufficiently easy to check. In this regard, given a real topological vector space $X$, a set $K\subseteq X$ is said to be a convex cone if $K+K\subseteq K$ and $\alpha K\subseteq K$ for all real $\alpha>0$. Note that a convex cone $K$ is connected and that $\alpha K=K+K=K$ for all $\alpha>0$. Therefore:

Corollary 1.2**.**

Theorem 1.1 holds if $K$ is a non-empty open convex cone and $\alpha_{1},\ldots,\alpha_{n}>0$, provided that $\mathbb{K}=\mathbb{R}$.

As another consequence, we have the following.

Corollary 1.3**.**

With the same hypothesis of Theorem 1.1, let us suppose that $\mathbb{K}=\mathbb{R}$, $\mathbb{F}:=\mathbb{Q}(\alpha_{1},\ldots,\alpha_{n})$, and $\alpha_{i}=\beta_{i}$ for all $i=1,\ldots,n$.

Then a function $f:K\to Y$ satisfies

[TABLE]

if and only if there exist a $\mathbb{F}$-linear $A:X\to Y$ and $b\in Y$ such that $f(x)=A(x)+b$ for all $x\in K$, with $b=0$ if $\sum_{i\leq n}\alpha_{i}\neq 1$.

Proof.

Thanks to Theorem 1.1, we just need to show that if $f$ satisfies (5) then $A$ is $\mathbb{F}$-linear. We obtained that $A$ is $\alpha_{i}$-homogeneous for all $i=1,\ldots,n$, i.e., $A(\alpha_{i}x)=\alpha_{i}A(x)$ for all $x\in X$. By a straightforward argument, $A$ is $p(\alpha_{1},\ldots,\alpha_{n})$-homogeneous, for each polynomial $p\in\mathbb{Q}[X_{1},\ldots,X_{n}]$, hence also to the corresponding field of rational functions. ∎

In particular, we obtain all the solutions of Equation (1) (we omit details).

Corollary 1.4**.**

Set $K:=\mathbb{R}^{+}\times\mathbb{R}^{+}$ and fix $\alpha,\beta\in\mathbb{R}^{+}$. Then a continuous function $\Phi:K\to\mathbb{R}^{+}$ such that $\lim_{x\to(0,0)}\Phi(x)=0$ satisfies Equation (1) if and only if there exists $y\in\mathbb{R}^{2}$ such that $\Phi(x)=\langle x,y\rangle$ for all $x\in K$.

Lastly, thanks to Theorem 1.1, we recover the characterization of the solutions of Jensen-like equations (again, we omit details). This may be used to shorten the proof of the main result in [10], cf. Equation (11) and (12) therein.

Corollary 1.5**.**

Fix positive reals $\alpha_{1},\ldots,\alpha_{n}$ with $n\geq 2$ and $\sum_{i\leq n}\alpha_{i}=1$. Then a continuous function $\Phi:[0,1]^{k}\to\mathbb{R}^{h}$ satisfies Equation (5) if and only if there exist a $h\times k$ real matrix $A$ and $b\in\mathbb{R}^{h}$ such that $\Phi(x)=Ax+b$ for all $x\in[0,1]^{k}$.

As we anticipated before, the main novelty in the proof of Theorem 1.1 is the use of (a natural extension of) a result of Radó and Baker [13] on the existence and uniqueness of extension of the solution of the classical Pedixer’s equation.

To this aim, we need to fix some notation. Given non-empty sets $A,B$ and $n\in\mathbb{N}^{+}$ with $A\subseteq B^{n}$, denote by $\pi_{i}$ the $i$-th projection, that is, $\pi_{i}(a):=a_{i}$ for all $a=(a_{1},\ldots,a_{n})\in A$, and define

[TABLE]

for each $i=1,\ldots,n$. Finally, set

[TABLE]

Theorem 1.6**.**

(Radó and Baker’s extension theorem.)* Let $X$ be a topological vector space over $\mathbb{K}$, where $\mathbb{K}$ is the field of real or complex numbers, and $Y$ be an abelian group. Moreover, given $n\geq 2$, let $U\subseteq X^{n}$ be a non-empty open connected set and fix functions $f:U_{+}\to Y$, $g_{1}:U_{1}\to Y,\ldots,g_{n}:U_{n}\to Y$ such that*

[TABLE]

Then there exists a unique extension $(F,G_{1},\ldots,G_{n})$ of $(f,g_{1},\ldots,g_{n})$ for which

[TABLE]

In fact, there exist a unique group homomorphism $A:X\to Y$ and unique $u,u_{1},\ldots,u_{n}\in Y$ such that $u=\sum_{i\leq n}u_{i}$ and

[TABLE]

The proof for the case $n=2$ can be found in [13, Theorem 1], cf. also [1, Theorem 4, p.80]. Moreover, see [5, Theorem 5] and [2, 3, 4] for related results. The proof of Theorem 1.6 follows in Section 2.

Lastly, if the Pexider equation (6) holds for all $x_{1},\ldots,x_{n}\in X$, then its analogue holds in a more general context; see [7, Proposition 1] for a related result.

Proposition 1.7**.**

Let $X,Y$ be abelian groups, written additively, and fix functions $f,g_{1},\ldots,g_{n}:X\to Y$, with $n\geq 2$. Then

[TABLE]

if and only if there exist a homomorphism $A:X\to Y$ and $y,y_{1},\ldots y_{n}\in Y$ with $y=\sum_{i\leq n}y_{i}$ such that

[TABLE]

Proof.

The if part is clear. Conversely, define $y_{i}:=g_{i}(0)$ for all $i=1,\ldots,n$, where [math] is the identity of $X$, and note that, setting $x_{1}=\cdots=x_{n}=0$ in (7), we obtain $f(0)=y:=\sum_{i\leq n}y_{i}.$ Hence, define the functions $\tilde{f},\tilde{g}_{1},\ldots,\tilde{g}_{n}:X\to Y$ by $\tilde{f}(x):=f(x)-y$ and $\tilde{g}_{i}(x)=g_{i}(x)-y_{i}$ for all $x\in X$ and $i=1,\ldots,n$. It follows by (7) that

[TABLE]

for all $x_{1},\ldots,x_{n}\in X$, and by construction $\tilde{f}(0)=\tilde{g}_{1}(0)=\cdots=\tilde{g}_{n}(0)=0$, where the latter [math] is the identity of $Y$. Given $x\in X$ and $i\in\{1,\ldots,n\}$, set $x_{i}=x$ and $x_{j}=0$ for all $j\in\{1,\ldots,n\}\setminus\{i\}$ in (8) so that $\tilde{f}(x)=\tilde{g}_{i}(x).$ By the arbitrariness of $x$, we conclude that $\tilde{f}=\tilde{g}_{i}$, hence $\tilde{f}\left(\sum_{i\leq n}x_{i}\right)=\sum_{i\leq n}\tilde{f}(x_{i})$ for all $x_{1},\ldots,x_{n}\in X$. Setting $x_{1}=\cdots=x_{n-2}=0$, we see that $\tilde{f}$ itself is a homomorphism. Therefore $A=\tilde{f}$, so that $f(x)=A(x)+y$ and $g_{i}(x)=A(x)+y_{i}$ for all $i=1,\ldots,n$ and $x\in X$. ∎

However, the analogous statement for the functional equation

[TABLE]

where $\alpha_{1},\ldots,\alpha_{n}$ are fixed non-zero integers does not hold. Indeed, consider the following example.

Example 1.8**.**

Set $n=2$, $X=Y=\mathbb{Z}_{4}$, $\alpha_{1}=\alpha_{2}=2$, and fix functions $f,g_{1},g_{2}:X\to Y$ such that $f(0)=f(2)=0$, $f(1)=1$, and $g_{1}(x)=g_{2}(x)=0$ for all $x\in X$. Then the functional equation (9) holds. However, if there exist a homomorphism $A:X\to Y$ and $y\in Y$ such that $f(x)=A(x)+y$ for all $x\in X$, then

[TABLE]

therefore $A(2)=2$, $y=f(1)-A(1)=2$, and $A(0)=f(0)-y=2$. This is impossible since we should have $A(0)=0$.

2. Proof of Theorem 1.6

Fix $x=(x_{1},\ldots,x_{n})\in U$. Since $U$ is open, there exists a neighborhood $G_{x}$ of $0\in X^{n}$ such that $x+G_{x}\subseteq U$. In particular, by the standing assumptions, we have $f\left(\sum_{i\leq n}(x_{i}+y_{i})\right)=\sum_{i\leq n}g_{i}(x_{i}+y_{i})$ for all $(y_{1},\ldots,y_{n})\in G_{x}$. Since each projection $\pi_{i}$ is an open map, it follows that $V_{x}:=\bigcap_{i\leq n}\pi_{i}(G_{x})$ is a non-empty open neighborhood of $0\in X$. In particular,

[TABLE]

At this point, define the functions $\tilde{f},\tilde{g}_{1},\ldots,\tilde{g}_{n}:V_{x}\to Y$ by

[TABLE]

for each $z\in V_{x}$ and $i=1,\ldots,n$. Since $f\left(\sum_{i\leq n}x_{i}\right)=\sum_{i\leq n}g_{i}(x_{i})$, it follows that

[TABLE]

and, in addition, $\tilde{f}(0)=\tilde{g}_{1}(0)=\cdots=\tilde{g}_{n}(0)=0$. As in the proof of Theorem 1.7, we obtain that

[TABLE]

and the restriction $\tilde{f}$ to $V_{x}$ coincides with each of $\tilde{g}_{1},\ldots,\tilde{g}_{n}$. Moreover, as in the proof of [13, Theorem 1], there exists a unique group homomorphism $A_{x}:X\to Y$ which extends $\tilde{f}$.

To sum up, this implies that, for each $x\in U$, there exist a neighborhood $V_{x}$ of $0\in X$, a unique group homomorphism $A_{x}:X\to Y$, and unique $u_{x},u_{x,1},\ldots,u_{x,n}\in Y$ with $u_{x}=\sum_{i\leq n}u_{x,i}$ such that

[TABLE]

and

[TABLE]

where $n\cdot V_{x}:=V_{x}+\cdots+V_{x}$, where $V_{x}$ is repeated $n$ times.

Claim** 1****.**

Fix $x,y\in U$ such that $(x+V_{x}^{n})\cap(y+V_{y}^{n})\neq\emptyset$. Then $A_{x}=A_{y}$.

Proof.

For each $i=1,\ldots,n$, we get by hypothesis that $(x_{i}+V_{x})\cap(y_{i}+V_{y})$ is a non-empty open set in $X$, hence there exists $w_{i}\in X$ and a neighborhood $W_{i}$ of $0\in X$ such that $w_{i}+W_{i}\subseteq(x_{i}+V_{x})\cap(y_{i}+V_{y})$. Set $w=(w_{1},\ldots,w_{n})$ and note that $\textstyle Z:=V_{w}\cap W_{1}\cap\cdots\cap W_{n}$ is a non-empty open neighborhood of $0\in X$. Hence

[TABLE]

For each $z\in Z$, we have $w_{i}+z\in x_{i}+V_{x}$ for each $i=1,\ldots,n$, so that

[TABLE]

Setting $z=0$ we obtain

[TABLE]

therefore $A_{w}(z)=A_{x}(z)$ for all $z\in Z$. Considering that $\bigcup_{n\geq 1}nZ=X$ by [14, Theorem 1.15.a] and both $A_{x}$ and $A_{w}$ are $\mathbb{Q}$-linear, we obtain that $A_{w}(z)=A_{x}(z)$ for all $z\in X$. Similarly $A_{w}(z)=A_{y}(z)$ for all $z\in X$, therefore $A_{x}=A_{y}$. ∎

Claim** 2****.**

With the same hypothesis of Claim 1, there exist unique $u,u_{1},\ldots,u_{n}\in Y$ with $u=\sum_{i\leq n}u_{i}$ such that

[TABLE]

and

[TABLE]

Proof.

Set $A:=A_{x}=A_{y}$. With the same notation of the proof of Claim 1, as it follows from Equation (10), we have

[TABLE]

for each $i=1,\ldots,n$. Hence, for each $z\in x_{i}+V_{x}$, it holds

[TABLE]

and similarly for $z\in y_{i}+V_{y}$.

To conclude, for each $z\in\sum_{i\leq n}x_{i}+n\cdot V_{x}$, there exist $v_{1},\ldots,v_{n}\in V_{x}$ such that $z=\sum_{i\leq n}(x_{i}+v_{i})$, hence

[TABLE]

and similarly for $z\in\sum_{i\leq n}y_{i}+n\cdot V_{y}$. ∎

Claim** 3****.**

Fix $x,y\in U$. Then there exists a finite sequence $t_{0},t_{1},\ldots,t_{k}\in U$ such that $t_{0}=x$, $t_{k}=y$, and $(t_{j}+V_{t_{j}}^{n})\cap(t_{j+1}+V_{t_{j+1}}^{n})\neq\emptyset$ for each $j=0,1,\ldots,k-1$.

Proof.

Note that $\{x+V_{x}^{n}:x\in U\}$ is a family of open sets in $U$ such that $\bigcup_{x\in U}(x+V_{x}^{n})=U$. Then, the claim follows by [4, Lemma 2.4]. ∎

Putting together Claims 1, 2, and 3, we conclude the proof.

3. Proof of Theorem 1.1

First, let us assume that $f:K\to Y$ is a function satisfying (2). For each $i=1,\ldots,n$, define the function $g_{i}:K\to Y$ by $g_{i}(x)=\beta_{i}f(x/\alpha_{i})$ for all $x\in K$. It follows that

[TABLE]

Setting $U:=\prod_{i\leq n}\alpha_{i}K$, it is readily seen that $U$ is a non-empty open connected subset of $X^{n}$ (with the usual product topology). Moreover, by construction $U_{i}=\alpha_{i}K$ for all $i=1,\ldots,n$ and $U_{+}=\sum_{i\leq n}\alpha_{i}K$. Thanks to (11) and the hypothesis $\sum_{i\leq n}\alpha_{i}K\subseteq K$, we obtain

[TABLE]

hence by Theorem 1.6 there exist a unique group homomorphism $A:X\to Y$ (hence, $A$ is $\mathbb{Q}$-linear) and unique $u,u_{1},\ldots,u_{n}\in Y$ such that

[TABLE]

with $u=\sum_{i\leq n}u_{i}$.

At this point, we claim that $A$ satisfies (3). To this aim, fix $i\in\{1,\ldots,n\}$. Taking into account (12) and the definition of $g_{i}$, we get

[TABLE]

However, since $A$ is $\mathbb{Q}$-linear, then also

[TABLE]

Calculating the differences of the above equations, we obtain that $A(\alpha_{i}x)=\beta_{i}A(x)$ for all $x\in K$. To conclude, fix $x\in X$, $y\in K$, and let $V$ be a neighborhood of [math] such that $y+V\subseteq K$ (which exists since $K$ is open). It follows by [14, Theorem 1.15.a] that there exists $n\in\mathbb{N}$ such that $x\in nV$. Therefore, considering that $A$ is $\mathbb{Q}$-linear, we obtain

[TABLE]

so that $A(\alpha_{i}x)=\beta_{i}A(x)$ for all $x\in X$.

Note that, thanks to (13), we have $u_{i}=\beta_{i}u$ for each $i=1,\ldots,n$. Summing these equations we obtain $u=\sum_{i\leq n}u_{i}=u\sum_{i\leq n}\beta_{i}.$ Considering that $\mathbb{F}$ has characteristic [math], it follows that $u=u_{1}=\cdots=u_{n}=0$ is the unique solution if $\sum_{i\leq n}\beta_{i}\neq 1$; finally, $u$ can be any value in $Y$ if $\sum_{i\leq n}\beta_{i}=1$. This concludes the proof of the first part.

Conversely, let us assume that $A:X\to Y$ is a group homomorphism which satisfies (3) and $f:K\to Y$ is a function defined by (4). If $\sum_{i\leq n}\beta_{i}\neq 1$ and $b=0$ then $f(x)=A(x)$ so that

[TABLE]

for all $x_{1},\ldots,x_{n}\in K$. On the other hand, if $\sum_{i\leq n}\beta_{i}=1$ and $b\in Y$ then

[TABLE]

Therefore, in both cases, $f$ satisfies (2).

4. Concluding Remark

In our Theorem 1.1 one could choose, e.g., $K=(-1,2)$ and fix non-zero reals $\alpha_{1},\ldots,\alpha_{n}$ with $\sum_{i\leq n}|\alpha_{i}|<1/2$ so that, in fact, $\sum_{i\leq n}\alpha_{i}K\subseteq(-1,1)\subseteq K$ (the same works, for instance, if $X=\mathbb{C}$, $K$ is the open circle with center $\nicefrac{{1}}{{2}}$ and radius $\nicefrac{{3}}{{2}}$, and $\alpha_{1},\ldots,\alpha_{n}$ are non-zero complex numbers such that $\sum_{i\leq n}|\alpha_{i}|<1/2$).

However, if $X$ is real topological vector space, one may ask whether it would be sufficient to require that $\alpha_{1},\ldots,\alpha_{n}$ are positive reals. More precisely:

Question 4.1**.**

Let $X$ be a real topological vector space, fix non-zero reals $\alpha_{1},\ldots,\alpha_{n}$, and let $K\subseteq X$ be a non-empty open connected set with $\sum_{i\leq n}\alpha_{i}K\subseteq K$. Does there exist a non-empty open connected set $K^{\prime}\subseteq K$ such that $\sum_{i\leq n}|\alpha_{i}|K^{\prime}\subseteq K^{\prime}$?

We can show that the answer is affirmative if $X=\mathbb{R}$ and $\sum_{i\leq n}|\alpha_{i}|\leq 1$. In such case, indeed, it would be sufficient to prove that $K^{\prime}:=K\cap(-K)$ is a non-empty neighborhood of [math] so that $\sum_{i\leq n}|\alpha_{i}|K^{\prime}$ is contained in both $\sum_{i\leq n}\alpha_{i}K\subseteq K$ and $-\sum_{i\leq n}\alpha_{i}K\subseteq-K$, hence also to the intersection $K^{\prime}$.

To this aim, let us assume that at least one $\alpha_{i}$ is negative, let us say $\alpha_{1},\ldots,\alpha_{k}<0$ and $\alpha_{k+1},\ldots,\alpha_{n}>0$, for some positive integer $k\leq n$. Define also $\alpha^{+}:=\sum_{i>k}\alpha_{i}$ and $\alpha^{-}:=\sum_{i\leq k}\alpha_{i}$ so that $\alpha^{-}\neq 0$, $\alpha^{+}\neq 1$, and $\alpha^{+}+\alpha^{-}\leq\alpha^{+}-\alpha^{-}\leq 1$. Since $K$ is a non-empty open connected set, there exist $a,b\in\mathbb{R}\cup\{\pm\infty\}$, with $a<b$, such that $K=(a,b)$. Note that $\alpha^{-}x+\alpha^{+}y\in K$ for all $x,y\in K$. Thus, if $a=-\infty$ then, given any $y_{0}\in K$ we have $\alpha^{-}(-n)+\alpha^{+}y_{0}\in K$ for all sufficiently large $n$, which implies $b=\infty$; thus $K=\mathbb{R}$. The case $b=\infty$ is similar.

Hence, let us assume hereafter that $a$ and $b$ are finite. Note that $\alpha K=(\alpha a,\alpha b)$ and $-\alpha K=(-\alpha b,-\alpha a)$ for all $\alpha>0$; in addition, $(x,y)+(x^{\prime},y^{\prime})=(x+x^{\prime},y+y^{\prime})$ for all non-empty intervals $(x,y)$ and $(x^{\prime},y^{\prime})$. Therefore

[TABLE]

Considering that $\sum_{i\leq n}\alpha_{i}K\subseteq K$, we obtain that $\alpha^{+}a+\alpha^{-}b\geq a$ and $\alpha^{+}b+\alpha^{-}a\leq b$, which can be rewritten as

[TABLE]

If $\alpha^{+}-\alpha^{-}=1$ then, by (14), $-\alpha^{-}(a+b)=0$, so that $a\leq 0$ and $b=-a\geq 0$ (and they cannot be equal since $(a,b)\neq\emptyset$). Otherwise $\alpha^{+}-\alpha^{-}<1$ and, in particular, $\alpha^{+}+\alpha^{-}<1$. Multiplying the second equation in (14) by $\frac{1-\alpha^{+}}{\alpha^{-}}<0$ and summing it to the first one, we obtain

[TABLE]

This implies that $b\geq 0$ and, similarly, $a\leq 0$.

To conclude, we claim that $0\in(a,b)$. Let us assume for the sake of contradiction that $a=0$. Then, choosing $x_{1}=\cdots=x_{k}=b-\varepsilon\in K$ and $x_{k+1}=\cdots=x_{n}=\varepsilon\in K$ with $\varepsilon>0$ sufficiently small, we obtain that $\sum_{i\leq n}\alpha x_{i}=\alpha^{-}(b-\varepsilon)+\alpha^{+}\varepsilon>0$, which is impossible. The case $b=0$ is similar.

4.1. Acknowledgments

The authors are grateful to an anonymous referee for suggestions that helped improving the overall presentation of the article.