# Injectivity of a certain operator

Consider a compact $K$ of the complex plane of interior void, $f$ a rational fraction leaving $K$ invariant (i.e. $f(K) \subset K$), and $\mu$ a borelian probability measure supported by $K$, and ergodic for $f$. To any function $g \in L^1(\mu,K)$ we associate the function $$h(z)=\int_K \frac{g(y) d\mu(y)}{z-f(y)}$$

$h$ is analytical outside $K$.

Question : if $h$=0, does it imply that $g=0$ $\mu$-a.e. ? This is the case for $f(y)=y$, as can be seen for example through Runge's theorem.

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We can simplify the setup by introducing a complex-valued measure $\nu$ such that $d\nu(y)=g(y)\,d\mu(y)$. Let $\tilde \nu=f_*\nu$, the pushforward of $\nu$ under $f$. Then $$h(z)=\int_K \frac{d\nu(y)}{z-f(y)} = \int_K \frac{d\tilde \nu(w)}{z-w} \tag1$$ By what you already proved, $\tilde \nu=0$. The converse is obviously true as well: if $\tilde \nu=0$ then $h\equiv 0$.
Is it possible to have $\tilde \nu=0$ when $\nu$ is not zero? Yes. For example, let $K=\{z:|z|=1\}$, $f(z)=z^2$, and define $d\nu (\zeta)=(\operatorname{sign}\operatorname{Im} \zeta) \,|d\zeta|$ on $K$.
But if $g$ (hence $\nu$) is nonnegative, then $\tilde \nu=0$ implies $\nu=0$.