$f:\mathbb{Q} \rightarrow \mathbb{R}$, with conditions on $f$ I would like to find all functions : $f:\mathbb{Q} \rightarrow \mathbb{R}$ such that :


*

*$f(x) \geq 0$, $\forall x \in \mathbb{Q}$ and the equality holds only if $x = 0$

*$f(x\cdot y) = f(x)\cdot f(y)$, $\forall x, y \in \mathbb{Q}$

*$f(x+y) \leq\max\{f(x), f(y)\}$


Any suggestions, ideas ? 
What I found : 


*

*$f$ is even

*$f(x) = 1/f(1/x)$

*$f(kx) \leq f(x)$ with $k \in \mathbb{N}$ and $f(kx) \leq 1$

 A: Not quite an answer, but some probably significant facts about $f(x)$:
From the statement that $f(x+y) \le \max \{f(x), f(y)\}$ we can derive the statement
$$f(2x) \le f(x)$$
and so
$$f(3x) \le \max \{f(2x), f(x)\}$$
$$f(3x) \le f(x)$$
and, without loss of generality,
$$f(kx) \le f(x)$$
for any positive integer $k$. Furthermore, since $f(x)$ is even,
$$f(-kx) \le f(x)$$
for any positive integer $k$. This means that, since $f(1)=f(-1)=1$, for all integer $x$,
$$f(x) \le 1$$
Can you extend this statement to all rational $x$ greater than one?
A: Some facts

  
*
  
*$f(n) \leq 1$ for all $n\in\mathbb{Z}$
  
*$f(x^{-1}) = f(x)^{-1}$
  
*$f(x)$ = $f(|x|)=|f(x)|$
  
*$f(1)= 1$ and $f(0)=0$
  
*If $x\neq 0$ and $f(x)\neq 1$ then there is a $p\in\mathbb{P}$ such that $f(p)<1$.
  

These facts aren't hard to varify.
functions which satisfy all conditions
Note that for every $x\in \mathbb{Q}$ and every $p\in \mathbb{P}$ we find $m,n,k \in \mathbb{Z}$such that
\begin{align}
x = p^k \frac{m}{n} \quad \text{and}\quad \gcd(m,n)=1
\end{align}
Now we define $f_{p,c}:\mathbb{Q} \to \mathbb{R},\; f(x) = f(p^k \frac{m}{n}) = c^k$ for $p\in\mathbb{P}$ and $c\in\mathbb{R}$. Note that if $x = \frac{m}{n}$ such that $p\nmid m$ and $p\nmid n$ then $f(x)=1$. 

The set of all functions which satisfy all your conditions is
  \begin{align} M := \big\{f_{p,c} : p\in\mathbb{P} \text{ and } c\in (0,1]\big\} \end{align}

It is easy to check that every function in $M$ satisfies all conditions. So I want to prove the more interesting fact that there aren't more functions:
There aren't more
If there is a $f:\mathbb{Q} \to \mathbb{R}$ which satisfies all conditions and $f\notin M$ then we can conclude from fact point 5. that there are at least $p_1, p_2\in\mathbb{P}$ such that $f(p_1)<1$, $f(p_2)<1$ and $p_1>p_2$.
Now we do some kind of recursion, take
\begin{align}k_{i+1} := \max\{n\in\mathbb{N} : p_i-np_{i+1} > 0\} \quad\text{and}\quad r_{i+2} := p_i - k_{i+1} p_{i+1}\end{align}
If $f(p_{i})<1$ and $f(p_{i+1})< 1$ then $f(r_{i+2}) \leq \max\{f(p_{i}),f(k_{i+1}p_{i+1})\}< 1$ so there is either


*

*a primfactor $q$ of $r_{i+2}$ such that $f(q)<1$ or

*$r_{i+2}=1$


but the second case leads to $f(1)< 1$ which contradicts fact point 4. So there will always happen case 1 and we can define
\begin{align}
p_{i+2} := q
\end{align}
Clearly we have $0<p_{i+2}\leq r_{i+2}<p_{i+1}<p_{i}$ and $p_i \in \mathbb{P}$. Now we can show by induction that there are infinitely many prime numbers smaller than $p_1$ which is false and gives us a contradiction to our assumption that $f \notin M$ and $f$ satisfies all conditions.
