Dolbeault h zero one of the riemann sphere vanishes

A holomorphic function on the sphere is entire in the $z$ chart and bounded near infinity, since its $w$-chart expression is continuous at $0$. It is also bounded on each closed disk, so is bounded on all of $\mathbb{C}$. By [F2] it is constant, and all constants are global holomorphic sections of the trivial bundle. Thus [F3] gives $H^{0,0}(X,E)=\mathbb{C}\cdot 1$.

Write a global holomorphic differential as $f(z)dz$ on $\mathbb{C}$, with $f$ entire. In the other chart it is $a(w)dw$, where $a(w)=-w^{-2}f(1\mathrm{/}w)$ for $w\mathrm{\ne }0$ by [F1]. Holomorphy at $w=0$ bounds $a$ on a small closed disk. Consequently $\mathrm{\mid }f(z)\mathrm{\mid }\le C\mathrm{\mid }z{\mathrm{\mid }}^{-2}$ for sufficiently large $\mathrm{\mid }z\mathrm{\mid }$. The entire function $f$ is bounded on a closed disk and on its exterior, so [F2] makes it constant; the displayed decay forces that constant to vanish. Therefore $H^0(X,K)=0$.

The trivial dual bundle identifies $K\otimes E^{\ast }$ holomorphically with $K$. By [F3] and step 1.2, the degree-one harmonic space and hence $H^{0,1}(X,E)$ are zero. Every smooth $(0,1)$-form on a curve is closed because there are no $(0,2)$-forms; its zero cohomology class says exactly that it is ${\bar{\mathrm{\partial }}}_E$ of a global smooth function. This proves the stated exactness for any supplied compatible metric. Full AC is inherited through duality and the harmonic representative interfaces.