Let $L$ be a hyperelliptic function field ($[L : K(x)]=2$), with $K$ an algebraically closed field. The norm is defined as $N(f)=f\bar{f}$, where $\bar{f}$ is the conjugate of $f$. If $y^2+h_1(x)=h(x)$ is the equation of the underlying hyperelliptic curve, then $\bar{f}=f(x,-y-h_1(x))$.
Let's say the curve is in imaginary model with a single point at infinity $P_\infty$, and let's take $f,g$ in the Riemann-Roch space $\mathcal L(mP_\infty)$ for some $m>2g+1$ (so that $y\in L(mP_\infty)$).
Let $f,g \in L$ such that $v_{P_\infty}(f)=v_{P_\infty}(g)=-m$ and such that $N(f)=N(g)$. This says that the principal divisors are of the form $(f)=P_1+\dotsb+P_m-mP_\infty$ and $(g)=\epsilon_1P_1+\dotsb+\epsilon_mP_m -mP_\infty$, with $\epsilon_i=\pm1$, and abusing the notation $-P_i=\iota(P_i)$.
Up to one case where one $P_i$ is a two torsion element in the Jacobian then I think I can prove that this implies that $g=f$ or $g=\bar{f}$ up to a constant. However, I find the proof not very elegant and not really intuititve (it relies on elementary algebraic consideration on univariate polynomials).
My question is: is this a known result, or maybe a special case of a more general result ? I can elaborate on the proof if needed.
Thanks in advance