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Sierpinski’s Theorem for Additive Functions – Simplification


1. If {f} is a solution of the Cauchy functional equation which is surjective, but not injective, then {f} has the Darboux property.

2. For every solution {f} of the Cauchy functional equation there exist two non-trivial solutions {f_1,f_2} of the same equation, such that {f_1} and {f_2} have the Darboux property and {f=f_1+f_2}.

These two results were proven in this post. The version presented here is a simplified one, identifying exactly what we need in order to obtain the desired results.

Proof: 1. First note that {f} must be non-trivial. If {f} would be trivial, then {f(x)=cx} for some {c \in \Bbb{R}}. The hypothesis {f} surjective proves that {c \neq 0}, but then {f} is injective. Contradiction.

Since {f} is not injective, there exists {b \neq 0} such that {f(b)=0}, and using the properties of {f} we have that {f(qb)=0} for every {q \in \Bbb{Q}}.

Pick an interval {I \subset \Bbb{R}} and a value {y_0 \in \Bbb{R}}. Since {f} is surjective, there exists {x_0 \in \Bbb{R}} such that {f(x_0)=y_0}. Since the set {\{x_0+bq : q \in \Bbb{Q}\}} is dense in {\Bbb{R}}, there exists {q_0} such that {x_0+bq_0 \in I}. Therefore {f(x_0+bq_0)=f(x_0)=y_0} and {y_0 \in f(I)}. This proves that {f(I)=\Bbb{R}}. Since this happens for every interval {I}, it follows that {f} has the Darboux property.

2. Consider a basis {\mathcal{B}} of {\Bbb{R}} considered as a vector space over {\Bbb{Q}} and {b_1,b_2 \in \mathcal{B}}. Since {card \ (\mathcal{B}\setminus \{b_1,b_2\})=card\ \Bbb{R}} there exists a bijection {g:\Bbb{R} \rightarrow B \setminus \{b_1,b_2\}}. Denote {P=g((-\infty,0))} and {Q=g([0,\infty))}. Therefore, we have just partitioned {B\setminus \{b_1,b_2\}} into two uncountable sets {P} and {Q}. This allows us to construct another two bijections {g_1 :P \rightarrow \Bbb{R}} and {g_2 :Q \rightarrow \Bbb{R}}.

Now we can define the two functions {f_1,f_2} with the needed properties. As we have seen in the introductory part of this section, a solution of the Cauchy functional equation is well and uniquely defined if we know the values of the solution on a basis of {\Bbb{R}} over {\Bbb{Q}}, which in our case is {\mathcal{B}}.

We define {f_1} and {f_2} by their values over {\mathcal{B}} as follows:

\displaystyle f_1(b)=\begin{cases} 0,& b=b_1 \\ f(b_2), & b=b_2 \\ g_1(b),& b \in P \\ f(b)-g_2(b),& b \in Q\end{cases}

and

\displaystyle f_2(b)=\begin{cases} f(b_1),& b=b_1 \\ 0, & b=b_2 \\ f(b)-g_1(b),& b \in P \\ g_2(b),& b \in Q\end{cases}.

The two functions {f_1,f_2} constructed above, are solutions to the Cauchy functional equation. Moreover, it is easy to see that {f_1(b)+f_2(b)=f(b)} for every {b \in \mathcal{B}}, which implies {f_1+f_2=f}.

Since {g_1} is a bijection from {P} to {\Bbb{R}} it follows that {f_1} is surjective, and because {f(b_1)=0,\ b_1 \neq 0} we see that {f_1} is not injective. Using 1. we conclude that {f_1} is non-trivial and has the Darboux property. By a similar argument {f_2} is non-trivial and has the Darboux property.  {\square}

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