is totally umbilical.

In particular, if ${\Sigma}$ is a topological sphere then $(I,II)$ is totally umbilical.

Several proofs of this result when the ambient space is $\r^3$ or $\h^3$ have been given by Hopf [@Ho], Chern [@Ch], Hartman and Wintner [@HW], Bryant [@Br] or Alencar, do Carmo and Tribuzy [@AdCT].

Let us consider ${\Sigma}$ as a Riemann surface with the conformal structure induced by $I$. Given a local conformal parameter $z$, we can write the fundamental pair $(I,II)$ as in (\[parholomorfo\]). Hence we have $$\label{levicivita}
\nabla_{\frac{\partial\
}{\partial z}}\frac{\partial\ }{\partial z}=\frac{\lambda_z}{\lambda}\
\frac{\partial\ }{\partial z},\qquad \nabla_{\frac{\partial\ }{\partial
z}}\frac{\partial\ }{\partial {\bar{z}}}=0$$ and the shape operator $S$ becomes $$\label{endomorfismoweingarten}
S\frac{\partial\ }{\partial z}=H\,\frac{\partial\ }{\partial z}+\frac{Q}{\lambda}\
\frac{\partial\ }{\partial {\bar{z}}}.$$ Consequently, if we take $X=\frac{\partial\ }{\partial z}$ and $Y=\frac{\partial\ }{\partial {\bar{z}}}$ in the Codazzi equation (\[ecuacioncodazzi\]) we get $$\label{qzbarra}
Q_{{\bar{z}}}=\lambda\,H_z.$$ In addition, from (\[endomorfismoweingarten\]) we obtain that the extrinsic curvature is given by $$\label{modulodeq}
K=H^2-\frac{|Q|^2}{\lambda^2}.$$ Thus, differentiating the equality $W(H,K)=0$ with respect to $z$ $$\begin{aligned}
0&=&H_{z}\,W_x(H,K)+K_{z}\,W_y(H,K)\\
&=&H_{z}\,W_x(H,K)+\left(2H\,H_{z}-|Q|^2\left(\frac{1}{\lambda^2}\right)_{z}-
\frac{Q_{z}\overline{Q}+Q\,\overline{Q}_{z}}{\lambda^2}\right)\,W_y(H,K),\end{aligned}$$ and using (\[qzbarra\]) $$(W_x(H,K)+2H\,W_y(H,K))\,Q_{{\bar{z}}}=\lambda\,W_y(H,K)\,\left(|Q|^2\left(\frac{1}{\lambda^2}\right)_{z}+
\frac{Q_{z}\overline{Q}+Q\,\overline{Q}_{z}}{\lambda^2}\right).$$

Therefore, from (\[hipotesis\]), if $p\in{\Sigma}$ is an umbilical point (i.e. $Q(p)=0$ or equivalently $H^2=K$), there exists a continuous function $h$ in a neighborhood $U$ of $p$ such that $|Q_{{\bar{z}}}|\leq h\,|Q|$ on $U$.

Hence, from 