Let $X$ be a compact Riemann surface and $\overline{L}$ be a Hermitian line bundle with curvature form $\Theta$. If $$ \int_X \log|f| \cdot \Theta = 0 $$ for every meromorphic function $f$ on $X$, I wonder if this implies the form $\Theta=0$, i.e. $\overline{L}$ is flat.
I know $\displaystyle \int_X g \cdot \Theta=0$ for every smooth function $g$ implies $\Theta=0$. But $\log |f|$ are only Green functions, i.e. $\Delta \log |f| = \delta_{\operatorname{div}(f)}$.
And we must note that $\big\{ \log|f|: \text{$f$ is meromorphic} \big\}$ does not exhaust all degree $0$ Green functions, since not every degree $0$ divisor is principal (This is true only when $g=0$. When $g>0$ there are so many degree $0$ divisors, so many in the sense that they form a dimension $g$ manifold).
I wonder if this small family of functions is already enough to test whether a form is zero. (But on the other hand, curvature form $\Theta$ is very special. Maybe $\log|f|$ is enough for curvature forms).