## Cauchy’s functional equation II

In this post I will finish what I began in my previous post. I have collected the two posts in a pdf-file, which might be a bit easier to read. Again you are welcome to comment on both math and language.

##### 4: More wild facts about non-linear solutions

A function could have a graph that is dense in the plane, but still be ‘nice’ on a large part of the domain. E.g. there exist functions that are $0$ on all irrational numbers, but still have a graph that is dense in the plane. Inspired by this, here is a list of things that would make a function wild even in a measure theory sense. Again, the list is ordered, and again all these statements are true for any non-linear solution to Cauchy’s functional equation.

• $f$ is not measurable. (When I write measurable function and measurable set, I always mean Borel-measurable.)
• $|f|$ is not dominated by any measurable function.
• If $A$ is measurable set such that $f$ is bounded above on $A$, then $A$ has measure $0$.
• If $A$ is measurable set such that $f$ is bounded above on $A$, then $A$ has measure $0$.
• If $B\subset \mathbb{R}$ is open and non-empty and $A\subset \mathbb{R}$ is measurable and $f(A)\cap B= \emptyset$, then $A$ have measure zero.

I will prove that last one.

Proof: Assume for contradiction that there is a non-linear function $f$ satisfying Cauchy’s functional equation and a non-empty open set $B\subset \mathbb{R}$ and a measurable set $A\subset \mathbb{R}$ with positive measure, and $f(A)\cap B=\emptyset$. Any open set contains an open interval, so without loss of generality, we can assume that $B$ is an open interval. We assumed that $m(A)>0$, where $m(X)$ denotes the measure of a set $X\subset \mathbb{R}$, and since measures are countably additive, there must be an interval of length $1$ with $m(A\cap I)>0$, so without loss of generality, we assume that $A$ is contained in an interval of length $1$. To reach a contradiction, I will show that under these assumptions, there exist two sequences of sets, $A_n$ and $B_n$ of subsets of $\mathbb{R}$ satisfying:

• Each $A_n$ is a subset of a interval of length $1$.
• $m(A_n)\geq m(A)>0$ for each $n$.
• Each $B_n$ is a open interval.
• The length of $B_n$ tends to infinity as $n\to\infty$.
• $f(A_n)\cap B_n=\emptyset$.

Since $f(A)\cap B = \emptyset$ and $f(nx)=nf(x)$ for $n\in\mathbb{N}$ we have $f(nA)\cap nB = \emptyset$ for $n\in\mathbb{N}$. We assumed that $A$ is contained in an interval of length $1$, so $nA$ is contained in an interval of length $n$. Furthermore, $m(nA)=n\cdot m(A)$ so using the pigeonhole principle we can find an interval $I_n$ of length $1$ such that $m(nA\cap I_n)\geq m(A)$. We see that $f(nA\cap I_n)\cap nB =\emptyset$ so I choose $A_n=nA\cap I_n$ and $B_n=nB$.

I will now use the existence of $A_n$ and $B_n$ to reach a contradiction. For each $n$ let $a_n$ be a number such that $A_n\subset [a_n,a_n+1]$ and let $b_n$ and $c_n$ be numbers such that $B_n=(b_n,c_n)$. We know that the graph of $f$ is dense in the plane, so we can find some $x_n\in (a_n-1,a_n)$ with $\frac{3b_n+c_n}{4}<\frac{b_n+3c_n}{4}$. Now the sequences $C_n=A_n-x_n$ and $D_n=B_n-f_n$ satisfy the above five requirements for $A_n$ and $B_n$, and furthermore $C_n\subset [0,2]$ and the lower and upper bound of $D_n$ will tend to minus infinity resp. plus infinity. Now for all $x\in \mathbb{R}$ there is some $N\in \mathbb{N}$ such that $x\in D_n$ for all $n\geq N$. But $f(C_n)\cap D_n=\emptyset$, so the sequence of indicator functions $1_{C_n}$ converge pointwise to the $0$-function. It is dominated by $1_{[0,2]}$ which have integral $2$, so the dominated convergence theorem tells us that $\int 1_{C_n}=m(C_m)$ tends to $0$ as $n\to\infty$. But this contradicts $m(C_n)\geq m(A)>0$. QED.

Assuming the axiom of choice we can find a discontinuous solution to Cauchy’s functional equation, and if we let $B$ in the above be the set of positive real numbers, we see that any measurable set $A$ with

$A\subset \{x|f(x)\leq 0\}$

must have measure $0$. But similarly, any measurable set $A$ with

$A\subset \mathbb{R}\setminus \{x|f(x)\leq 0\}=\{x|f(x)> 0\}\subset \{x|f(x)\geq 0\}$

must also have measure $0$. Thus we have found a set, such that neither the set nor its complement contains a set with positive measure.

##### 5: Extra wild functions

We have seen that the graph of a non-linear solution is in many ways ‘spread out all over the plane’. But there are some ways to interpret ‘spread out all over the plane’ for which this is not true for all solutions. E.g. let $(x_i)_{i\in I}, x_i\in \mathbb{R}$ be a Hamel basis. Now the function

$f\left(\sum_{i\in I}q_ix_i\right)=\sum_{i\in I}q_i, \quad q_i\in \mathbb{Q}, |\{i\in I|q_i\neq 0\}|<\infty$

is a solution to Cauchy’s functional equation, but $f(x)$ is always rational. However, there exist surjective solutions, so in some ways, some solution functions are `wilder’ than others. As usual, I will give a list of wild properties a function can have, and as usual the list is ordered such that any of the properties imply the one above. Unlike for the other lists, there are some solution functions that do not satisfy any of the properties and some that satisfy all of them. Moreover, for any two properties on the list, there exist solutions, that satisfy the upper of the two, but not the lower one.

• $f$ is surjective.
• The graph of $f$ intersects any line in the plane.
• For any continuous function $c:\mathbb{R}\to \mathbb{R}$ there is a $x\in \mathbb{R}$ with $f(x)=c(x)$.
• For any continuous function $c:\mathbb{R}\to \mathbb{R}$ the set $\{x\in\mathbb{R}|f(x)=c(x)\}$ is dense in $\mathbb{R}$.
• any continuous function $c:\mathbb{R}\to \mathbb{R}$, any measurable set $A\subset \mathbb{R}$ with $\{x\in A|f(x)= c(x)\}=\emptyset$ have measure $0$.

First I will prove that there is a solution $f$ satisfying the third property. The proof of the existence of solutions satisfying the two last properties are similar, and I will sketch those proofs afterwards.

Proof: We begin by choosing a Hamel basis $(x_i)_{i\in I}$ and well-order $I$. That is, we find an ordering on $I$ such that any subset of $I$ has a least element. The existence of such an ordering (on any set) is equivalent to the axiom of choice. The set $\{x_i|i\in I\}$ is a subset of $\mathbb{R}$, so the cardinality of this set is not greater than the cardinality of $\mathbb{R}$. Since $x_i\neq x_j$ for $i\neq j$ we get that $|I|\leq |\mathbb{R}|$. Assume for contradiction that $|I|<|\mathbb{R}|$. Using the rules for calculations with cardinality we know that $|I\times \mathbb{Q}|=\max(|I|,|Q|)$ and more generally $|(I\times \mathbb{Q})^n|=\max(|I|,|\mathbb{Q}|,|I|,\dots,|\mathbb{Q}|)=\max(|I|,|\mathbb{Q}|)$. Since any element in $\mathbb{R}$ is a linear combination of the $x_i$‘s over $\mathbb{Q}$ so for any real number $y$ here is a $n\in \mathbb{N}$ and $((i_1),(q_1),\dots, (i_n,q_n)), i_k\in I,q_k\in\mathbb{Q}$ such that

$y=\sum_{k=1}^nq_kx_{i_k}$

Hence

$|\mathbb{R}|\leq \left|\bigcup_{n=1}^{\infty}(I\times \mathbb{Q})^n\right|=|\mathbb{N}|\cdot |\max(|I|,|\mathbb{Q}|)|=\max(|\mathbb{N}|,|I|,|\mathbb{Q}|)<|\mathbb{R}|$

To reach this contradiction, we assumed that $|I|<|\mathbb{R}|$, so $|I|=|\mathbb{R}|$.

Now, let’s see how many continuous functions $c:\mathbb{R}\to\mathbb{R}$ there is. A continuous function is uniquely determined by its value on the rational numbers, so $|C|\leq |\mathbb{R}^{\mathbb{Q}}|=|\mathbb{R}|$, where $|C|$ is the set of continuous functions. On the other hand, the constant functions are continuous, and there are $\mathbb{R}$ of them, so $|C|\geq |\mathbb{R}|$. Hence $|C|=|\mathbb{R}|=|I|$, so we can index the set of continuous functions with the set $I$, so that $C=\{c_i|i\in I\}$. We can now define $f(x_i)=\lambda_i=c_i(x_i)$ to make sure that the equation $f(x)=c_i(x)$ have a solution. Now $f$ is given by

$f\left(\sum_{i\in I}q_ix_i\right)=\sum_{i\in I}q_i\lambda_i \quad q_i\in \mathbb{Q}, |\{i\in I|q_i\neq 0\}|<\infty.$

QED.

If we want the set of solutions to $f(x)=c(x)$ to be dense in $\mathbb{R}$, it is a bit more complicated. The idea is, that instead using the set $I$ to index the set of continuous functions, we use it to index the set of continuous functions times the set of open intervals. Unfortunately we cannot be sure that $x_i$ is in the open interval corresponding to $i$. Instead we start by defining $f(1)=0$. Now we know that for each $i$ there is a $q_i\in\mathbb{Q}$ such that $q+x_i$ is in the open interval corresponding to $i$, and we define $f(x_i)=c_i(q_i+x_i)$.

If we want to show that there exist solution functions with the last property, it is much more complicated: Here we need transfinite induction, because we need to choose the elements of the Hamel basis one at a time. We know that the set of (Borel-)measurable set has the same cardinality as $\mathbb{R}$, thus the set of functions $c:A\to \mathbb{R}$, that can be defined as a continuous function $\mathbb{R}\to\mathbb{R}$ restricted to a measurable set $A$ with positive measure, has the cardinality $|\mathbb{R}\times \mathbb{R}|=|\mathbb{R}|$. Now we index this set by a set $I$. Using axiom of choice, we can well-order this set, and we can even choose the ordering such that $|\{j\in I|j<|I|=|\mathbb{R}|$ for all $i\in I$. Now for each $i\in I$ we choose $x_i$ and $\lambda_i$ such that $x_i$ is in the domain of $c_i$ and $f(x_i)=c_i(x_i)$ and such that the $x_i$‘s to be linear independent over $\mathbb{Q}$.

To show that this is possible, I only need to show that when we have chosen $x_j$ for all $j < i$ we can choose $x_i$ such that $x_i$ is linearly independent of the $\{x_j|j over $\mathbb{Q}$ and in the domain of $c_i$. The rest follows by transfinite induction. We know that measurable sets with positive measure are uncountable, so if we assume the continuum hypothesis (a statement independent of ZFC: it states that a set cannot have a cardinality between $|\mathbb{N}|$ and $|\mathbb{R}|$), any measurable set have the same cardinality as $\mathbb{R}$. (It is still true without the continuum hypothesis, but it is more difficult to prove. See [BS].) We know that $|\{x_j|j<|\mathbb{R}|$ so the cardinality of the linear span over $\mathbb{Q}$ of this set is also less than $|\mathbb{R}|$, since you cannot reach $|\mathbb{R}|$ by taking countable union and finite products of sets of smaller cardinalities. (In general, still assuming axiom of choice, you cannot get a set with some infinite cardinality $\kappa$, by taking finite products of sets with smaller cardinality, or by taking union of $\mu<\kappa$ sets with smaller cardinality.) Since the domain of $c_i$ have the same cardinality as $\mathbb{R}$, we can choose an element $x_i$ in the domain of $c_i$ and not in the linear span of $\{x_j|j.

By transfinite induction, we have now chosen $x_i$‘s such that they are linearly independent over $\mathbb{Q}$. However, we cannot be sure that they span all of $\mathbb{R}$. So we end by using the axiom of choice once again to extend the set $\{x_i|i\in I\}$ to a Hamel basis, and we set the rest of the $\lambda$‘s to be zero. This gives us a solution to Cauchy’s functional equation, with a graph that intersects any continuous function on any measurable set with positive measure.

##### References

[BS]: James M. Briggs and Thomas Schaffter. Measure and Cardinality. The American Mathematical Monthly, Vol. 86, No. 10, pp. 852-855.

[EH]: Ernst Hansen. Measure Theory. Department of Mathematical Sciences University of Copenhagen. 2009.

[MO]: mathoverflow: Do sets with positive lebesgue measure have same cardinality as R?

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### 2 Comments on “Cauchy’s functional equation II”

1. […] I’ll post the first half of this, and I will post the rest in a few days (update: It’s here. Here you can also find the notes in pdf.). If you have anything to say about the mathematics in […]

2. Nice thanks for the pdf file.