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

## Sierpinski’s Theorem for Additive Functions

We say that ${f: \Bbb{R} \rightarrow \Bbb{R}}$ is an additive function if

$\displaystyle f(x+y)=f(x)+f(y),\ \forall x,y \in \Bbb{R}.$

1. Prove that there exist additive functions which are discontinuous with or without the Darboux Property.

2. Prove that for every additive function ${f}$ there exist two functions ${f_1,f_2:\Bbb{R} \rightarrow \Bbb{R}}$ which are additive, have the Darboux Property, and ${f=f_1+f_2}$.

The second part is similar to Sierpinski’s Theorem which states that every real function can be written as the sum of two real functions with Darboux property.

(A function ${g:I \rightarrow \Bbb{R}}$ has the Darboux property if for every ${[a,b]\subset I}$, ${g([a,b])}$ is an interval.)

Proof: We know that every additive function ${f: \Bbb{R} \rightarrow \Bbb{R}}$ is uniquely determined by its values over a basis of ${\Bbb{R}}$ regarded as a vector space over ${\Bbb{Q}}$. Consider ${\mathcal{B}}$ such a basis. In the following, the vector spaces are always considered over ${\Bbb{Q}}$.

1. (i) Examples of additive function without Darboux property Pick a basis ${\mathcal{B}}$ and ${b \in \mathcal{B}}$. Every real number ${x}$ can be expressed uniquely by

$\displaystyle x=\lambda b+\sum \lambda_ib_i,$

and we can define ${f}$ on the basis ${\mathcal{B}}$ by ${f(b)=2b,\ f(b')=b'}$ for ${b'\neq b \in \mathcal{B}}$. This generates an additive function ${f}$ which is injective. Indeed if ${t=\lambda b+\sum\lambda_i b_i}$, then ${f(t)=0}$ implies

$\displaystyle 2\lambda b+\sum \lambda_ib_i=0$

and since ${\{b,b_1,..,b_n\}}$ is linearly independent, all coefficients are zero, and finaly ${t=0}$.

Suppose ${f}$ has Darboux property. Then injectivity implies that ${f}$ is monotone which finally gets us to ${f(x)=cx}$ for some constant ${c}$. But this contradicts ${f(b)=2b}$ and ${f(b')=b'}$ for ${b'\neq b}$. Therefore ${f}$ does not have the Darboux property.

It is possible to construct another example by simply choosing all ${f(b) \in \Bbb{Q}}$ for ${b \in \mathcal{B}}$. This implies that ${f}$ takes only rational values, and therefore it does not have the Darboux property.

(ii) Example of additive function with Darboux property

We know that ${\mathcal{B}}$ is uncountable and ${\text{card}\ \mathcal{B}=\text{card}\ \Bbb{R}}$. We need a way to produce additive functions which vanish on a dense subset of ${\Bbb{R}}$ without being continuous. This is because any non-continuous function is not locally upper/lower blunded, which means that the image of any interval is dense in $\Bbb{R}$. To build an additive Darboux function we need it to take every real value in every interval, and in particular it takes the value zero in every interval. Furthermore, if $f$ is additive and $\ker f$ is dense in $\Bbb{R}$ then the function has the Darboux property. Indeed, for  an arbitrary interval $I$ and a real number $y$ such that $y=f(x)$ we know that $x+\ker f$ is dense in $\Bbb{R}$ so there exists $x_0\in I$ such that $x_0=x+z$ where $f(z)=0$. Therefore $f(x_0)=f(x)+f(z)=y$. This proves that $f(I)=\Bbb{R}$ and since the interval was arbitrary, $f$ has the Darboux property.

Choose ${b \in \mathcal{B}}$ and look at ${\Bbb{R}/b\Bbb{Q}}$ which has

$\displaystyle \mathcal{B}^*=\{b'+b\Bbb{Q} : b' \in \mathcal{B} \setminus \{b\}\}$

as a basis. Since ${\mathcal{B}}$ is infinite, it has the same cardinal as ${\mathcal{B}^*}$ (“removing” one element does not modify its cardinality) and therefore we can find an isomorphism ${\varphi : \Bbb{R}/b\Bbb{Q} \rightarrow \Bbb{R}}$.

Define now the function ${f:\Bbb{R} \rightarrow \Bbb{R}}$,

$\displaystyle f(t)= \varphi(t+b\Bbb{Q})$

which is additive and ${f(t)=0}$ for every ${t \in b\Bbb{Q}}$. Moreover, it is surjective (since ${\varphi}$ is surjective) and for every interval ${I}$ we have ${f(I)=\Bbb{R}}$ (if ${y \in \Bbb{R}}$ there exists ${t_y}$ such that ${\varphi(t_y+b\Bbb{Q})=y}$ so ${f(t_y+bq)=y}$ for every ${q \in \Bbb{Q}}$). Therefore ${f}$ has the Darboux property.

Another construction is possible starting from a basis ${\mathcal{B}}$ which contains ${1}$ and defining ${f(1)=0}$. We choose ${f}$ on the rest of the basis ${\mathcal{B}}$ such that ${f:\mathcal{B}\setminus \{1\} \rightarrow \Bbb{R}}$ is bijective. Furthermore, consider the equivalence relation ${x\sim y \Leftrightarrow x-y \in \Bbb{Q}}$. We find that ${f}$ is constant on every equivalence class of ${\sim}$ and furthermore, every interval contains one representative of every class (since all classes are dense in ${\Bbb{R}}$). Therefore ${f(I)=\Bbb{R}}$ and ${f}$ has the Darboux property.

Note that if an additive function is locally bounded (above/below) then it is continuous. Therefore, a discontinuous additive function with Darboux property always satisfies ${f([a,b])=\Bbb{R}}$ for every interval ${[a,b]\subset \Bbb{R}}$.

2. Pick ${b_1\neq b_2 \in \mathcal{B}}$. As above, we have two bases of ${\Bbb{R}/b_1\Bbb{Q}}$ and ${\Bbb{R}/b_2\Bbb{Q}}$ given by

$\displaystyle \mathcal{B}_i^*=\{ b+b_i \Bbb{Q} : b \in \mathcal{B}\setminus\{b_i\}\}.$

We can find a bijection ${g: \Bbb{R} \rightarrow \mathcal{B}\setminus \{b_1,b_2\}}$ and define ${P=g((-\infty,0)),Q=g([0,\infty))}$ (this is just a way to split ${\mathcal{B}\setminus \{b_1,b_2\}}$ into two parts of the same cardinality). Furthermore, we define the bijections ${g_1: P \rightarrow \Bbb{R},\ g_2: Q \rightarrow \Bbb{R}}$ and define ${\psi_i}$ on ${\mathcal{B}_i^*}$ by

$\displaystyle \psi_1 :\mathcal{B}_1^* \rightarrow \Bbb{R},\ \psi_1(b+b_1\Bbb{Q})=\begin{cases} g_1(b) & b \in P \\ f(b)-g_2(b) & b \in Q \\ f(b_2) & b=b_2\end{cases}$

$\displaystyle \psi_2 :\mathcal{B}_2^* \rightarrow \Bbb{R},\ \psi_2(b+b_2\Bbb{Q})=\begin{cases} f(b)-g_1(b) & b \in P \\ g_2(b) & b \in Q \\ f(b_1) & b=b_1\end{cases}$

Because of the definitions of ${g_1,g_2}$ we find that ${\psi_i(\mathcal{B}_i^*)=\Bbb{R},\ i=1,2}$.

We can define morphisms ${\varphi_i :\Bbb{R}/b_i\Bbb{Q} \rightarrow \Bbb{R}}$ such that ${{\varphi_i}_{\mid \mathcal{B}_i^*}=\psi_i}$ (just extend them from the definition on the basis) and these morphisms are surjective.

Define now ${f_i:\Bbb{R} \rightarrow \Bbb{R}}$ by ${f_i(t)=\varphi_i(t+b_i \Bbb{Q}),\ i=1,2}$. ${f_i,\ i=1,2}$ is of course additive and surjective. Moreover since every ${t+b_i\Bbb{Q}}$ is dense in ${\Bbb{R}}$, ${f_i(t)}$ takes every real value in every interval ${I}$. To prove this, choose ${r \in \Bbb{R}}$, fix ${i\in \{1,2\}}$ and by surjectivity of ${\varphi_i}$ there exists ${t_r}$ such that ${\varphi_i(t_r+b_i \Bbb{Q})=r}$. But ${t_r +b_i\Bbb{Q}}$ is dense in ${\Bbb{R}}$ so there exists ${y \in I}$ such that ${f_i(t)=\varphi_i(y+b_i\Bbb{Q})=\varphi_i(t_r+b_i\Bbb{Q})=r}$. Therefore ${f_i}$ has the Darboux property (since ${f(I)=\Bbb{R}}$ for every interval ${I}$) and is additive.

Now let’s look at ${f_1+f_2}$:

• ${b \in P}$: ${f_1(b)+f_2(b)=\psi_1(b+b_1\Bbb{Q})+\psi_2(b+b_2\Bbb{Q})=g_1(b)+f(b)-g_1(b)=f(b)}$.
• ${b \in Q}$: ${f_1(b)+f_2(b)=\psi_1(b+b_1\Bbb{Q})+\psi_2(b+b_2\Bbb{Q})=f(b)-g_2(b)+g_2(b)=f(b)}$.
• ${b=b_1}$: ${f_1(b_1)+f_2(b_1)=\psi_1(b_1+b_1\Bbb{Q})+\psi_2(b_1+b_2\Bbb{Q})=0+f(b_1)=f(b_1)}$.
• ${b=b_2}$: ${f_1(b_2)+f_2(b_2)=\psi_1(b_2+b_1\Bbb{Q})+\psi_2(b_2+b_2\Bbb{Q})=f(b_2)+0=f(b_2)}$.

Therefore the additive functions ${f_1+f_2}$ and ${f}$ coincide on ${\mathcal{B}}$ which proves that ${f=f_1+f_2}$ on ${\Bbb{R}}$.

(The last part was inspired from M. Megan, A.L. Sasu, B. Sasu Differential Calculus through Exercices and Problems (in Romanian).)