Let $K$ be a number field, $V=V(K)$ be the set of valuations of $K$, $V_0$ be the set of non-archimedean valuations of $K$, $V_1$ be the set of archimedean valuations of $K$. For any $x\in K^\times$, we have the product formula $$\prod_{v\in V}|x|_v = 1 .$$
Define the height funcetionfunctions $h_0(x)=\sum_{v\in V_0} \log(\max{|x|_v,|x^{-1}|_v)$, $h_1(x)=\sum_{v\in V_1} \log(\max{|x|_v,|x^{-1}|_v)$$h_0(x)=\sum_{v\in V_0} \log(\max(|x|_v,|x^{-1}|_v))$ and $h_1(x)=\sum_{v\in V_1} \log(\max(|x|_v,|x^{-1}|_v))$.
Then for $K=\mathbb{Q}$ and rational number $x=\pm\frac{a}{b}\neq 0$ (where $a,b$ are coprime positive integers), we have $$0 \le h_1(x) = |\log(a)-\log(b)|\le h_0(x)=\log(a)+\log(b).$$ That is to say, the archimedean height funcetion $h_1(x)$ is "controlled" by the non-archimedean height funcetion $h_0(x)$.
I‘m wondering for general number fields $K$, does there exists real number $C=C(K)>0$, such that for any $x\in K^\times$, we have $0\le h_1(x) \le C\cdot h_0(x)$? If exists, how to find this $C=C(K)$ from $K$? Thank you for reading this post.