Home > Algebra, Analysis, Geometry > Miklos Schweitzer 2013 Problem 7

## Miklos Schweitzer 2013 Problem 7

Problem 7. Suppose that ${f: \Bbb{R} \rightarrow \Bbb{R}}$ is an additive function (that is ${f(x+y) = f(x)+f(y)}$ for all ${x, y \in \Bbb{R}}$) for which ${x \mapsto f(x)f(\sqrt{1-x^2})}$ is bounded of some nonempty subinterval of ${(0,1)}$. Prove that ${f}$ is continuous.

Miklos Schweitzer 2013

Solution: (Lukas Geyer, Math-Stackexchange)

The idea is that we can go into two dimensions by defining ${F(x,y)=f(x)f(y)}$. 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 ${f}$ is continuous it is enough to prove that it is bounded on an interval. Since ${f}$ is an arbitrary solution of the Cauchy functional equation, we can use only the fact that ${f}$ is linear over ${\Bbb{Q}}$, so ${f(px+qy)=pf(x)+qf(y)}$ for all ${x,y \in \Bbb{R}}$ and all ${p,q \in \Bbb{Q}}$. On the other hand, we would like to work mostly on the arc of circle on which we know that ${f(x)f(y)}$ is bounded. To achieve that we work on a smaller subinterval ${J}$ which we can rotate with the help of a rotation matrix with rational coefficients a small enough angle so that we don’t leave ${J}$. This allows us to prove the boundedness of ${f}$ on ${J}$, and therefore that ${f}$ is continuous. Here are the details:

First observe that for every ${\epsilon > 0}$ there exist rational numbers ${p,q > 0}$ with ${p<\epsilon}$ and ${p^2 + q^2 = 1}$. This is just a consequence of well-known formulas for construction of Pythagorean triples. We can choose

$\displaystyle p = \frac{2n+1}{n^2 + (n+1)^2} \quad \text{ and } q = \frac{2n(n+1)}{n^2 + (n+1)^2}$

for ${n \in \mathbb{N}}$ sufficiently large. For such pairs we define an associated rotation matrix

$\displaystyle A_p = \begin{bmatrix} q & p \\ -p & q \end{bmatrix}.$

The assumption in the problem means that the function

$\displaystyle F(x,y) = f(x) f(y)$

is bounded on some arc ${I}$ of the unit circle ${x^2 + y^2 = 1}$.

Now if ${(x,y)}$ is arbitrary and ${A_p}$ is a rational rotation matrix as above, then (identifying the matrix and the induced linear map)

$\displaystyle F(A_p(x,y)) = F(qx + py, -px + qy) = f(qx+py) f(-px+qy)$

$\displaystyle = (qf(x)+ pf(y)) (-pf(x) + qf(y))$

$\displaystyle = qp(f(y)^2 - f(x)^2) + (q^2 - p^2) f(x)f(y),$

where we used the fact that ${f}$ is linear over the rational numbers.

Now we fix a small subarc ${J}$ of ${I}$, and a rational rotation matrix ${A_p}$ with ${p>0}$ such that ${A_p(J) \subset I}$. (This is possible because by the argument in the first paragraph we can find ${A_p}$ arbitrarily close to the identity.) For ${(x,y) \in J}$ the above calculation then implies that

$\displaystyle f(y)^2 - f(x)^2 = \frac{F(A_p(x,y))}{qp} - \frac{(q^2 - p^2) F(x,y)}{qp}$

so that ${f(y)^2 - f(x)^2}$ is bounded on ${J}$. Since ${f(x)f(y)}$ is also bounded on ${J}$, it follows that ${(f(y)+if(x))^2 = f(y)^2 - f(x)^2 + 2i f(x)f(y)}$ is bounded on ${J}$, so that ${f}$ is bounded on the projections of ${J}$ to both coordinate axes. This implies that ${f}$ 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 ${g: I \rightarrow \Bbb{R}}$ if ${x\mapsto f(x)f(g(x))}$ is bounded we also have that ${f}$ is bounded? This, of course is very broad, but note that in the solution we haven’t used at all bases of ${\Bbb{R}}$ over ${\Bbb{Q}}$ or any fact about how unbounded may ${f(x)f(g(x))}$ be if ${f}$ is not continuous.
• Maybe it is possible to go to more dimensions, by changing the hypothesis to

$\displaystyle (x_1,x_2,..,x_n) \mapsto f(x_1)f(x_2)...f(\sqrt{1-x_1^2-...-x_n^2}), \text{ is bounded}$

for ${x_i \in I \subset [0,1]}$ such that ${\sum_{i=1}^n x_i^2\leq 1}$.