I was reading a proof that used the following result


Let $C$ be a hyperelliptic of genus $\ge 3$ and $\tau \colon C \to C$ the hyperelliptic involution. If $D$ is an effective divisor of degree $g-1$ such that $h^0(D)>1$ then $D = x + \tau(x) + D'$ where $D'$ is an effective divisor.


My question is, how is this result proved? It seems equivalent to showing that $|D|$ contains the unique $g^1\_2$ and this made me think of Clifford's theorem but this didn't lead to much. For $g = 3$ the result holds because then $|D| = g^1\_2$. But already for $g = 4$ I'm stuck. I tried playing around with the Riemann-Roch theorem but didn't get far.



Special divisors on hyperelliptic curves
Suppose that $D=x\_1+x\_2+\cdots+x\_{g-1}$. We may assume that $\tau(x\_i)\neq x\_j$
for all $i\neq j$. Now, assume that $D'=y\_1+\cdots+y\_{g-1}$ is an element of
$|K-D|$. Then $x\_1+x\_2+\cdots+x\_{g-1}+y\_1+\cdots+y\_{g-1}$ is an element of $|K|$
but we know that any such element is of the form
$z\_1+\tau(z\_1)+\cdots+z\_{g-1}+\tau(z\_{g-1})$. After possibly renumbering the $z\_i$ (as
well as possibly replacing $z\_i$ by $\tau(z\_i)$) we may assume $x\_1=z\_1$ and
then $x\_2=z\_2$ and so on. This gives $D'=\tau(D)$ which implies
$h^0(K-D)\leq 1$. However, R-R and gives $h^0(D)=h^0(K-D)\leq1$ which
contradicts assumptions.


