Home > Analysis, Linear Algebra, Topology > Sierpinski’s Theorem for Additive Functions – Simplification

## 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.  ${\square}$