Maximum principle for minimal hypersurfaces The well known local version of the maximum principle for minimal hypersurfaces asserts that if two minimal hypersurfaces $ M_1 $ and $ M_2 $ of $ R^n $ has a common point $ x_0 \in M_1 \cap M_2 $ where $ M_1 $ lies (locally) on one side of $ M_2 $ then $ M_1 = M_2 $ in a neighbourod of $ x_0 $. This result can be found for example in 'A course in minimal surfaces' of Colding Minicozzi. 
Now i want to prove that if one of them is complete, for example $ M_1 $, then $ M_2 \subset M_1 $.
It is quite obvious to proceed in this way: if $ f_i:M_i \rightarrow R^n $, $ i=1,2 $, are the immersion maps, for the local version of the maximum principle there exist $ U_i \subset M_i $ such that $ f_i :U_i \rightarrow R^n $ is an embedding and $ f_1(U_1)=f_2(U_2) $. Now it is easy to check that $f_1^{-1} \circ f_2 : U_2 \rightarrow U_1 $ is an isometry. Now the most obviuos thing to do is an application of the unique continuation theorem since $ f_1 $ and $ f_2 $ are harmonic maps on the manifolds (unique continuation is suggested in Colding Minicozzi, but they do not give a proof). My problem is that i don't know how i can apply the unique continuation for elliptic operators? In fact $ f_1 $ and $ f_2 $ are defined on two hypersurfaces of $ R^n $ that a priori can be distinct.
Thank you
 A: I think that the following argument should give an answer to my own question. We prove the following theorem 
Let $ f_{1}:M_{1}\rightarrow \mathbb{R}^{n} $ and $ f_{2}:M_{2}\rightarrow \mathbb{R}^{n} $ be two minimal immersed hypersurfaces. Suppose $ M_{1} $ complete. If $ f_{1}(M_{1})\cap f_{2}(M_{2}) $ contains at least a point where $ f_2(M_2) $ lies locally on side of $ f_1(M_1) $ then $ f_{2}(M_{2})\subseteq f_{1}(M_{1}) $.
In the proof we use the following two results:
(Lemma 1)Let $ \Omega\subseteq \mathbb{R}^{n} $ be a bounded connected open set and let $ u_{1}, u_{2}:\Omega \rightarrow \mathbb{R} $ be two (smooth) solutions of the minimal equation. Suppose $ u_{1}, u_{2} \in C^{0}(\overline{\Omega}) $. Then $ v=u_{1}-u_{2} $ satisfies an equation of the form
$ \sum _{ij}a_{ij}\frac{\partial^{2}v}{\partial x_{i} \partial x_{j}}+ \sum_{i}b_{i}\frac{\partial v}{\partial x_{i}}=0 $\
Moreover it holds that:\
$\sum _{ij}a_{ij}(x)\xi_{i}\xi_{j}\geq \mu^{2}|\xi|^{2} $\
for every $ x \in \Omega $ and $ \xi \in \mathbb{R}^{n} $.
As an immediate consequence of Lemma 1 and the Maximum principle for elliptic equation we obtain:
(Lemma 2)Let $ \Omega \subseteq \mathbb{R}^{n} $ be a bounded open neighbourhood of the origin and let $ u_{1},u_{2}:\Omega\rightarrow \mathbb{R} $ be two solutions of the minimal equation such that $ u_{1}(0)=u_{2}(0) $ and $ u_{1} \leq u_{2} $ in $ \Omega $. Then $ u_{1}=u_{2} $.
(Proof of the Theorem)From Lemma 2 we easily conclude that there exist two open subsets $ V \subseteq M_{2} $ and $ U \subseteq M_{1} $  such that $ f_{1}(U)=f_{2}(V) $. Now let $ \mathscr{S} $ be the collection of open subsets of $ M_{2} $ such that for each $ V \in \mathscr{S} $ there exists an open subset $ U \subseteq M_{1} $ such that $ f_{1}(U)=f_{2}(V) $. Clearly $ \mathscr{S}\neq \varnothing $. Moreover observe that if $ f_{2}(V)=f_{1}(U) $ then $ f_{2}(\overline{V})\subseteq f_{1}(\overline{U}) $. In fact let $ q \in \partial V $ and let $ q_{n} \in V $ such that $ q_{n}\rightarrow q $. Let $ p_{n}\in U $ such that $ f_{1}(p_{n})=f_{2}(q_{n}) $. Since $ f_{2}^{-1}\circ f_{1}:U \rightarrow V $ is an isometry we have that $ \{p_{n}  \} $ is a Cauchy sequence in $ M_{1} $. Since $ M_{1} $ is complete there exist $ p \in M_{1} $ such that $ p_{n}\rightarrow p $. Therefore $ p \in \overline{U} $ and $ f_{1}(p)=f_{2}(q) $.\
Let $ V_{0} \in \mathscr{S} $ be a maximal set and let $ U_{0}\subseteq M_{1}  $ such that $ f_{1}(U_{0})=f_{2}(V_{0}) $. If $ \overline{V}_{0}=M_{2} $ then
$ f_{2}(M_{2})=f_{2}(\overline{V}_{0}) \subseteq f_{1}(\overline{U}_{0})\subseteq f_{1}(M_{1}) $
We now suppose $ \overline{V}_{0}\subsetneq M_{2} $ and we obtain a contradiction. Let $ q \in \partial V_{0} $ and let $ p \in \partial U_{0} $ such that $ f_{1}(p)=f_{2}(q)=x $. Note $ T_{p}(M_{1})=T_{q}(M_{2}) $ (as subspaces of $ T_{x}\mathbb{R}^{n} $). Let $ u_{1}, u_{2}:\Omega \rightarrow \mathbb{R} $ be the functions representing $ f_{1} $ and $ f_{2} $ in a neighbourhood of $ x $ ($ \Omega \subseteq \mathbb{R}^{n-1} $ is a bounded open subset). From Lemma 1 the difference function $ v=u_{1}-u_{2} $ is a solution of a linear elliptic equation. Moreover, since $ f_{1}(U_{0})=f_{2}(V_{0}) $, there exists $ \Omega' \subseteq \Omega $ such that $ v|_{\Omega'}=0 $. By Unique continuation $ v=0 $ in $ \Omega $. Therefore we can find $ V' \in \mathscr{S} $ such that $ V_{0} \subsetneq V' $, contradicting the maximality of $ V_{0} $.\
