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Suppose $\Sigma$ is minimal hypersurface in $R^n$, consider following function $f(x)=\frac{1}{2}r^2(x)$ where $r(x)$ is the distance to the origin, let $t$ be the distance to the hypersurface, so $$n=\Delta_{R^n}f=\frac{\partial^2f}{\partial^2 t}+\Delta_\Sigma f=1+\Delta_\Sigma f$$.

I understand first equality it is just usual laplacian calculation, I also know that for $k$ dimensional minimal surface, $\Delta_\Sigma (\frac{1}{2}r^2(x))=k$, so I guess he is trying to show that for hypersurface that $\Delta_\Sigma f=n-1$. But why do we have last two equalities?

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It is a local calculation. Suppose $p\in \Sigma$, and $\{e_1,\cdots,e_{n-1},N\}$ is local O.N. basis of $R^n$ such that $\{e_1,\cdots,e_{n-1}\}$ is local O.N. basis of $\Sigma$, $N$ is unit normal vector of $\Sigma$. Note that if $t$ represents distance to $\Sigma$, then $\frac{\partial}{\partial t}\big|_{p}=N_p$. Thus by definition of Laplacian we have $$\begin{eqnarray*}\Delta_{R^n}f &=&\Delta_\Sigma f+H(f)+(\bar{\nabla}^2f)(N,N)\end{eqnarray*}=\Delta_\Sigma f+g_{R^n}(N,N)=\Delta_\Sigma f+1$$ where $\bar{\nabla}$ is connection in $R^n$, $g_{R^n}$ is the standard metric of $R^n$. Here $H=0$ since it is minimal.

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