In this post, we proved the following result (appeared in a paper by Y.Y. Li published in *J. Eur. Math. Soc.* (2004))

**Lemma 1**. For and , let be a function defined on and valued in satisfying

Then is constant or .

Later, we considered the equality case in this post and proved the following result:

**Lemma 2**. Let , and . Suppose that for every there exists such that

Then for some , and

In this post, we consider the third result which can be stated as follows:

**Lemma 3**. For and , let be a function defined on and valued in satisfying

Then , restricted to , is constant or .

In the statement of Lemma 3 above, by we mean those points such that the last coordinate . Then by we mean the boundary of . Hence, we may identify .

In addition, to make the difference possible for each and , we may identify by a new point sitting in . Hence, by definition, we simply set

where, and from now on, we denote for each point . We are now in a position to prove Lemma 3, see a paper by Dou and Zhu.

*Proof of Lemma 3*. For each we set with such that as . Then we choose in such a way that

Note that this is possible since we can always choose such that . We also set

Then we immediately obtain

Therefore, by our assumption

Since

and

we have

Our lemma follows since are arbitrary.

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