I am trying to find solutions of functional equation $f(xy)=f(x)+f(y)$ for all $x,y\in\mathbb Q_p\setminus\{0\}=\mathbb Q_p^\star$. Where $f:\mathbb Q_p^\star\to\mathbb R$. I know some solutions:
1) $f(x)\equiv 0$.
2) $f(x)=\log |x|_p$
Here, $p$ is prime and $\mathbb Q_p$ is the field of all $p-$adic numbers.
My question is: Does there exist any other solution which of cause not similar to these solutions as above? (ie.. I mean $f(x)=\log |x|^n_p$ is "similar" to 2).)
You are asking for homomorphisms from the multiplicative group $\mathbb{Q}_p^*$ to the additive group of real numbers.
It is known that $\mathbb{Q}_p^*$ is a direct product of three subgroups. The infinite cyclic subgroup generated by $p$, the cyclic subgroup of order $p-1$ consisting of non-zero Teichmüller elements (generated by $\zeta$, a primitive root of unity of order $p-1$), the group $1+p\mathbb{Z}_p$ of $p$-adic integers congruent to $1$.
A homomorphism from a direct product of groups is determined, if we know its restriction to the constituent groups. The infinite cyclic group is easy to handle, as we can select the image of $p$ to by any real number $a$ we want, and define $f_a(p^n)=na$. The finite cyclic group is also easy to handle: the additive groups has darn few elements of a finite order, so such a homomorphism has to be trivial.
The last factor, $U:=1+p\mathbb{Z}_p$, is more interesting. It is actually a bit scary to think about classifying the homomorphisms from $U$ to $\mathbb{R}$. For one, I can't write down a single non-trivial one. Yet the following argument shows that they exist. Consider the subgroup $K=\langle 1+p\rangle \le U$. This is an infinite cyclic group, so we get a homomorphism $\phi_b$ from $K$ to $\mathbb{R}$ simply by sending $1+p$ to an arbitrary real number $b$, and extending that. Because the additive group of $\mathbb{R}$ is divisible, it is an injective object in the category of abelian groups (see e.g. Hilton-Stammbach). Therefore there is a homomorphism $g_b:U\to\mathbb{R}$ such that the restriction of $g_b$ to $K$ is equal to $\phi_b$. The proof of this involves a typical use of Zorn's lemma, so if you belong to the church that does not believe in the axiom of choice, then I cannot convince you that such a homomorphism exists.
Given the above there will other such functions. Namely those of the form $$ f(p^n\zeta^j u)= na + g_b(u). $$ If $g_b$ is the trivial homomorphism, then you get a scalar multiple of the logarithm of the $p$-adic value. Otherwise you get something else.
