[proofplan] We compare the density of $\eta$ with the density of the complete Poincare metric on each smaller disk $\Delta_R := \{z \in \mathbb{C}: |z| < R\}$. The curvature inequality turns into a differential inequality for the logarithm of the ratio between the two densities. If that ratio had an interior maximum greater than $1$, the elementary second-derivative test at the maximum would contradict the curvature inequality. Letting $R \uparrow 1$ gives the desired estimate on $\Delta$, and the holomorphic-map statement follows by applying the metric estimate to the pullback metric. [/proofplan] [step:Write the curvature inequality as a logarithmic differential inequality] Since $\eta$ is a smooth conformal Hermitian $(1,1)$-form on $\Delta$, write \begin{align*} \eta = i\lambda(z)\, dz \wedge d\bar z, \end{align*} where $\lambda:\Delta \to [0,\infty)$ is smooth. On the [open set](/page/Open%20Set) \begin{align*} U_+ := \{z \in \Delta : \lambda(z) > 0\}, \end{align*} the Gaussian curvature of $\eta$ is \begin{align*} K_\eta(z) = -\frac{2}{\lambda(z)}\frac{\partial^2}{\partial z\,\partial \bar z}\log \lambda(z). \end{align*} The hypothesis $K_\eta(z) \leq -\kappa$ therefore gives, for every $z \in U_+$, \begin{align*} \frac{\partial^2}{\partial z\,\partial \bar z}\log \lambda(z) \geq \frac{\kappa}{2}\lambda(z). \end{align*} For $0<R<1$, define the Poincare density on $\Delta_R$ by \begin{align*} \lambda_R:\Delta_R &\to (0,\infty) \\ z &\mapsto \frac{2R^2}{(R^2-|z|^2)^2}. \end{align*} Thus \begin{align*} \omega_R := i\lambda_R(z)\, dz \wedge d\bar z \end{align*} is the pullback of $\omega_\Delta$ under the biholomorphism $z \mapsto z/R$ from $\Delta_R$ to $\Delta$, and hence has Gaussian curvature $-1$. Equivalently, \begin{align*} \frac{\partial^2}{\partial z\,\partial \bar z}\log \lambda_R(z) = \frac{1}{2}\lambda_R(z). \end{align*} [/step] [step:Compare $\eta$ with the Poincare metric on a smaller disk] Fix $R \in (0,1)$. Define the continuous comparison function \begin{align*} q_R:\Delta_R &\to [0,\infty) \\ z &\mapsto \frac{\kappa\lambda(z)}{\lambda_R(z)}. \end{align*} Since $\lambda$ is smooth on $\Delta$, it is bounded on every compact subset of $\Delta$. Since $\lambda_R(z) \to \infty$ as $|z| \uparrow R$, the function $q_R$ extends continuously to $\overline{\Delta_R}$ by setting $q_R(z)=0$ for $|z|=R$. We claim that $q_R \leq 1$ on $\Delta_R$. Suppose otherwise. Then $q_R$ attains a maximum value $M>1$ at some point $z_0 \in \Delta_R$. Because $M>1$, we have $\lambda(z_0)>0$, so $z_0 \in U_+$. In a neighbourhood of $z_0$, define \begin{align*} u_R:\Delta_R \cap U_+ &\to \mathbb{R} \\ z &\mapsto \log q_R(z) = \log \kappa + \log \lambda(z)-\log \lambda_R(z). \end{align*} The point $z_0$ is a local maximum of $u_R$, so the real Hessian of $u_R$ at $z_0$ is negative semidefinite. Since \begin{align*} \frac{\partial^2}{\partial z\,\partial \bar z} = \frac{1}{4}\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\right) \end{align*} for $z=x+iy$, this gives \begin{align*} \frac{\partial^2}{\partial z\,\partial \bar z}u_R(z_0) \leq 0. \end{align*} On the other hand, using the two curvature identities above, \begin{align*} \frac{\partial^2}{\partial z\,\partial \bar z}u_R(z_0) &= \frac{\partial^2}{\partial z\,\partial \bar z}\log \lambda(z_0) - \frac{\partial^2}{\partial z\,\partial \bar z}\log \lambda_R(z_0) \\ &\geq \frac{\kappa}{2}\lambda(z_0) - \frac{1}{2}\lambda_R(z_0) \\ &= \frac{1}{2}\lambda_R(z_0)\bigl(q_R(z_0)-1\bigr) \\ &= \frac{1}{2}\lambda_R(z_0)(M-1) \\ &>0. \end{align*} This contradicts the second-derivative test. Hence $q_R \leq 1$ on $\Delta_R$, which is exactly \begin{align*} \eta \leq \frac{1}{\kappa}\omega_R \end{align*} on $\Delta_R$. [/step] [step:Let the smaller disk expand to the unit disk] Let $z \in \Delta$ be arbitrary. For every $R$ with $|z|<R<1$, the preceding step gives \begin{align*} \lambda(z) \leq \frac{1}{\kappa}\lambda_R(z) = \frac{1}{\kappa}\frac{2R^2}{(R^2-|z|^2)^2}. \end{align*} Taking the limit as $R \uparrow 1$ yields \begin{align*} \lambda(z) \leq \frac{1}{\kappa}\frac{2}{(1-|z|^2)^2}. \end{align*} Multiplying by $i\,dz\wedge d\bar z$ gives \begin{align*} \eta \leq \frac{1}{\kappa}\omega_\Delta \end{align*} at the point $z$. Since $z \in \Delta$ was arbitrary, the inequality holds on all of $\Delta$. [/step] [step:Apply the metric estimate to holomorphic pullbacks] Let \begin{align*} f:\Delta \to Y \end{align*} be holomorphic. Since $\Delta$ is one-dimensional, the pullback form $f^*\omega_Y$ is a smooth semipositive conformal Hermitian $(1,1)$-form, so there exists a smooth function $\lambda_f:\Delta \to [0,\infty)$ such that \begin{align*} f^*\omega_Y = i\lambda_f(z)\, dz \wedge d\bar z. \end{align*} At every point $z \in \Delta$ with $\lambda_f(z)>0$, the differential \begin{align*} df_z:T_z\Delta \to T_{f(z)}Y \end{align*} has nonzero image in the complex line spanned by $df_z(\partial/\partial z)$. The Gaussian curvature of the pullback metric at such a point is bounded above by the holomorphic sectional curvature of $\omega_Y$ in that complex tangent direction; this is the one-dimensional Gauss curvature comparison for holomorphic curves, where the second fundamental form contributes a nonpositive term. Since the holomorphic sectional curvature of $\omega_Y$ is at most $-\kappa$, we obtain \begin{align*} K_{f^*\omega_Y}(z) \leq -\kappa \end{align*} wherever $\lambda_f(z)>0$. Thus $f^*\omega_Y$ satisfies the hypotheses already proved for $\eta$. Applying the first part with $\eta=f^*\omega_Y$ gives \begin{align*} f^*\omega_Y \leq \frac{1}{\kappa}\omega_\Delta. \end{align*} This proves the stated Schwarz-type estimate for every holomorphic map $f:\Delta \to Y$. [/step]