Find all real functions $f(x)$ that satisfy the following conditions:

a) there exists a finite interval in which $f(x)$ is bounded,

b) for all $x_{1}, x_{2}$,

$$
f\left(x_{1}+x_{2}\right)=f\left(x_{1}\right)+f\left(x_{2}\right)
$$

c) $f(1)=1$.
Suppose the function $f(x)$ satisfies the conditions of the problem. Then for any integer $k$,

$$
f(k \cdot x)=k \cdot f(x)
$$

since the condition b) written for the pairs $x, n \cdot x$ (where $n$ is a natural number) yields $f((n+1) \cdot x)=f(x)+f(n \cdot x)$, from which it follows by complete induction that $f(n \cdot x)=n \cdot f(x)$.

It is easy to show that $f(0)=0$. Applying b) to the numbers $x_{1}$ and $x_{2}=0$: $f\left(x_{1}+0\right)=f\left(x_{1}\right)+f(0)$, hence $f(0)=0$ follows. Knowing this, applying the b) condition to the numbers $n \cdot x$ and $-n \cdot x$ we get that $f(-n \cdot x)=-n \cdot f(x)$.

If $r$ is a rational number: $r=\frac{k}{m}$, where $k, m$ are integers and $m \neq 0$, then applying (1) twice

$$
k \cdot f(x)=f(k \cdot x)=f\left(m \cdot\left(\frac{k}{m} x\right)\right)=m \cdot f\left(\frac{k}{m} x\right)
$$

from which dividing by $m$

$$
f(r \cdot x)=r \cdot f(x)
$$

From this, by condition c), for every rational number

$$
f(r)=r
$$

Using condition a), we show that (3) also holds for irrational numbers. Suppose $f(x)$ is bounded in the interval $(a, b)$ and let $x_{0}$ be an irrational number. Then for any rational number $r$, there exists a real number $t$ such that $tr$ falls into the interval $(a, b)$, and $rx_{0}+tr$ is rational. Thus, by (3), b), and (2)

$$
r x_{0}+t r=f\left(r x_{0}+t r\right)=f\left(r x_{0}\right)+f(t r)=r f\left(x_{0}\right)+f(t r)
$$

from which

$$
tr-f(t r)=r\left(f\left(x_{0}\right)-x_{0}\right)
$$

Since here $r$ can be any arbitrarily large rational number, and the quantity on the left side is bounded, this can only be true if $f\left(x_{0}\right)=x_{0}$, so the function $f(x)=x$ is the only one that satisfies the problem's conditions.

Remarks. 1. From our solution, it can also be seen that the boundedness can be replaced by boundedness from above or below.

2. It can be shown that if condition a) is omitted, there are solutions other than the function $f(x)=x$ 1[^0]

[^0]:    ${ }^{1}$ Mathematical Journals, problem 166. Solution in the 20th volume (1969) on page 415.