I am trying to understand why $\dim H^0(\mathbb{P}^1,nK)=0$ for $n\geq 1$ and $K=-2\infty$, a canonical divisor on $\mathbb{P}^1$.
By Riemann-Roch, we have that $h^0(nK)=h^0(K-nK)+\deg(nK)+1$, which isn't obviously helpful since $\deg(K-nK)=-2+2n$ so we can't conclude that $H^0(K-nK)=0$.
So instead, looking directly at $H^0(\mathbb{P}^1,nK)$, we have $H^0(\mathbb{P}^1,nK)=\{f\in K(\mathbb{P}^1)\mid \operatorname{div}(f)\geq 2n\infty\}$. But I'm not sure why this should just be the constant functions. For instance, if we let $K(\mathbb{P}^1)=K(t)$, then $1/t^{2n}$ is a non-constant function in this space. So am I missing something?