Miklos Schweitzer 2013 Problem 7
Problem 7. Suppose that is an additive function (that is for all ) for which is bounded of some nonempty subinterval of . Prove that is continuous.
Solution: (Lukas Geyer, MathStackexchange)
The idea is that we can go into two dimensions by defining . This new function is additive in each variable and the hypothesis implies that it is bounded on an arc of the unit circle. Let me state the general idea before going into the details. To prove that is continuous it is enough to prove that it is bounded on an interval. Since is an arbitrary solution of the Cauchy functional equation, we can use only the fact that is linear over , so for all and all . On the other hand, we would like to work mostly on the arc of circle on which we know that is bounded. To achieve that we work on a smaller subinterval which we can rotate with the help of a rotation matrix with rational coefficients a small enough angle so that we don’t leave . This allows us to prove the boundedness of on , and therefore that is continuous. Here are the details:
First observe that for every there exist rational numbers with and . This is just a consequence of wellknown formulas for construction of Pythagorean triples. We can choose
for sufficiently large. For such pairs we define an associated rotation matrix
The assumption in the problem means that the function
is bounded on some arc of the unit circle .
Now if is arbitrary and is a rational rotation matrix as above, then (identifying the matrix and the induced linear map)
where we used the fact that is linear over the rational numbers.
Now we fix a small subarc of , and a rational rotation matrix with such that . (This is possible because by the argument in the first paragraph we can find arbitrarily close to the identity.) For the above calculation then implies that
so that is bounded on . Since is also bounded on , it follows that is bounded on , so that is bounded on the projections of to both coordinate axes. This implies that is continuous.
This is a quite interesting solution: going from one dimension to two dimensions is not that intuitive. This leaves place for some questions though:
 Is this result valid for more general functions? For which functions if is bounded we also have that is bounded? This, of course is very broad, but note that in the solution we haven’t used at all bases of over or any fact about how unbounded may be if is not continuous.
 Maybe it is possible to go to more dimensions, by changing the hypothesis to
for such that .

November 22, 2013 at 8:22 pmMiklos Schweitzer 2013  Beni Bogoşel's blog