A characterization of the module function on a locally compact division ring References:
Weil's Basic Number Theory(written as BNT).
Bourbaki's Commutative Algebra(written as BCA).
Let $K$ be a topological ring with an identity.
Suppose every non-zero element of $K$ is invertible.
Let $K^* = K - \{0\}$ be the multipilcative group of $K$
If the map $x \rightarrow x^{-1}$ is continuous on $K^*$, we say $K$ is a topological division ring.
Suppose the topological space $K$ is non-discrete and locally compact(by this Hausdorff is implicitly assumed).
Then we say, by abuse of terminology,  $K$ is a locally compact division ring.
Let $K$ be a locally compact division ring.
Then the aditive group $K$ is a locally compact group.
Hence there exists a Haar measure $\mu$ on $K$.
Let $a$ be an element of $K^*$.
Then the map $x \rightarrow ax$ is an automorphism of the locally compact group $K$.
Hence the map $X \rightarrow \mu(aX)$ defines an invariant measure on $K$ where $X$ is any measurable subset of $K$.
Therefore there exists a constant $c \gt 0$ such that $\mu(aX) = c\mu(X)$ for every measurable subset $X$ such that $0 \lt \mu(X) \lt \infty$.
We denote $c$ by $mod(a)$.
We define $mod(0) = 0$.
$mod(a)$ can also be defined by the map $x \rightarrow xa$(see BNT or BCA).
Clearly $mod(ab) = mod(a)mod(b)$ for all $a, b \in K$.
The function $mod$ is continuous(see BNT or BCA).
The subset $\{x \in K|\ mod(x) \le d\}$ is compact for every real number $d \gt 0$(see BNT).
Locally compact division rings are classified as follows(see BNT or BCA).


*

*The field of real numbers $\mathbb{R}$.

*The field of complex numbers $\mathbb{C}$.

*The field of Hamilton's quaternions $\mathbb{H}$.

*Finite division algebras over the field of $p$-adic numbers.

*Finite division algebras over the field of formal Laurent series over a finite field.
Here is my question.
Is the following proposition true?
Proposition
Let $K$ be a locally compact division ring.
Let $\phi$ be a real valued function defined on $K$.
Suppose $\phi$ satisies the following conditons.


*

*$\phi$ is continuous.

*$\phi(x) \gt 0$ for all $x \neq 0$ and $\phi(0) = 0$.

*$\phi(xy) = \phi(x)\phi(y)$ for all $x, y$.
Then there exists a real number $c \gt 0$ such that $\phi(x) = mod(x)^c$ for all $x$.
Remark
Let $K$ be a locally compact division ring.
Let $x \in K^*$.


*

*If $K = \mathbb{R}$, then $mod(x) = |x|$.

*If $K = \mathbb{C}$, then $mod(x) = |x|^2$.

*If $K = \mathbb{H}$, then $mod(x) = |x|^4$.

*If $K = \mathbb{Q}_p$, then $mod(x) = |x|_p$ where $|x|_p$ is the canonical absolute value, i.e. $|p|_p = 1/p$.
 A: For the archimedean case, we can take advantage of the isomorphism 
$$f: K^\ast \cong M \times \mathbb{R}_{> 0}$$ 
$$x \mapsto (\frac{x}{|x|}, |x|)$$ 
where $M = \{x \in K^\ast: |x| = 1\}$ is a compact group. The image $\phi(M)$ of the summand $M$ is a compact subgroup of $\mathbb{R}_{>0}$, hence is trivial. The restriction of $\phi$ to the summand $\mathbb{R}_{>0}$ must be of the form $r \mapsto r^c$ for some real $c$ (being just a variation on the Cauchy functional equation, where the continuous solutions are well-known). Continuity at $0$ then forces $c > 0$. 
I believe the nonarchimedean case is similar. In characteristic zero, we have (see Matthew Emerton's answer here: reduced norm is proper) a topological group isomorphism 
$$K^\ast \cong \mathcal{O}_K^\times \times \pi^\mathbb{Z}$$ 
where $\mathcal{O}_K^\times$ is compact. Applying the same argument as before, this means that a continuous homomorphism $\phi: K^\times \to \mathbb{R}_{>0}$ is determined by its restriction to the summand $\pi^\mathbb{Z}$ (it is trivial on the other summand). A (continuous) homomorphism $\pi^{\mathbb{Z}} \to \mathbb{R}_{>0}$ is of the form $\pi^n \mapsto |\pi|^{nc}$ for some real $c$, and $c > 0$ is forced by the requirement of continuity at $0 \in K$. Summarizing: the modulus function on $K^\times$ (a division algebra of dimension $n^2$) is given by the composite 
$$K^\ast \stackrel{N_{K/k}^{1/n}}{\to} k^\ast \stackrel{|-|_\pi}{\to} |\pi|^\mathbb{Z}$$ 
(again referring to the thread in which Emerton answered), and so we have $\phi = \text{mod}^c$ for some $c > 0$. 
