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).)

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