Let $(X,\mathcal{O}_X)$ be an analytic space, and let $L$ be a line bundle on $X$. Intuitively, a metric $||\cdot||$ is a continuous choice of a metric for each fiber of the line bundle, which is a complex vector space of dimension $1$. The formal definition is as follows:
Given a line bundle $L$ over $X$, we define a (continuous) metric $||\cdot||$ on $L$ as being the datum, for any open set $U \subseteq X$ and any section $s \in \Gamma(U,L)$, of a continuous function $||s||_{U}:U\to \mathbb{R}^{\ge 0}$, satisfying the following properties:
(1) For any open set $V \subseteq U$, $||s||_V$ is the restriction to $V$ of the function $||s||_U$;
(2) For any function $f \in \mathcal{O}_X(U)$, $||fs||=|f|\cdot||s||$;
(3) If $s$ is a local frame on $U$, then $||s||$ doesn’t vanish at any point of $U$.
We would expect, by our intuition, that $||\cdot||$ induces a metric on each fiber $L\mid_x=L_x\otimes_{\mathcal{O}_{X,x}} \mathfrak{m}_x$. Using (1), it is easy to see that $||\cdot||$ induces a metric on the stalk $L_x$. However, it is not clear to me if this metric is also well-defined on the fibers. For example, let $[(s,U)]_x\in L_x$ be such that it is zero in $L\mid_x$. I cannot see why we would have $||s||_U(x)=0$.
Is it possible to conclude that only with the metric properties (1),(2),(3) given above? Or would we have to add another property that gives conditions for the vanishing of the $||s||$, in order to obtain a definition of metric coherent with our intuition?