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Assume your surface is conformally equivalent to a disc $D$ and $e^\phi$ be the conformal factor. From completeness, $\phi(x)\to\infty$ as $x\to \partial D$. Gauss curvature can be expressed as $K=-\frac12{\cdot} e^{-\phi}{\cdot}\Delta\phi$. Thus, $\Delta\phi\le 0$. The later contradicts maximum principle.

P.S. As Igor noticed, the argument has a gap: we only have that upper limit of $\phi(x)$ is $\infty$ as $x$ converge to any point on the boundary. He also give a ref with a complete proof. The argument would work if for any superharmonic function $\phi$ on $D$ there is a curve $\gamma$ from $0$ to the boundary such that $$\int\limits_\gamma e^\phi<\infty.$$ It The later is likely to be true, but I do not know enough to prove itproved by Fedja Nazarov here.

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Assume your surface is conformally equivalent to a disc $D$ and $e^\phi$ be the conformal factor. From completeness, $\phi(x)\to\infty$ as $x\to \partial D$. Gauss curvature can be expressed as $K=-\frac12{\cdot} e^{-\phi}{\cdot}\Delta\phi$. Thus, $\Delta\phi\le 0$. The later contradicts maximum principle.

P.S. As Igor noticed, the argument has a gap: we only have that upper limit of $\phi(x)$ is $\infty$ as $x$ converge to any point on the boundary. He also give a ref with a complete proof. The argument would work if for any superharmonic function $\phi$ on $D$ there is a curve $\gamma$ from $0$ to the boundary such that $$\int\limits_\gamma e^\phi<\infty.$$ It is likely to be true, but I do not know enough to prove it.

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Assume your surface is conformally equivalent to a disc $D$ and $e^\phi$ be the conformal factor. From completeness, $\phi(x)\to\infty$ as $x\to \partial D$. Gauss curvature can be expressed as $K=-\tfrac12{\cdot} K=-\frac12{\cdot} e^{-\phi}{\cdot}\Delta\phi$. Thus, $\Delta\phi\le 0$. The later contradicts maximum principle.

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