What the relationship between curvature and the sectional curvature of a Riemannian manifold? In the paper Curvature estimates for minimal hypersurfaces written by R.Schoen, L.Simon and S.T.Yau, in the middle of page 279, there is a inequality
$2|K_{n+1,iji}|\leq K_1-K_2$
where $K_{n+1,iji}$is the curvature of a Riemannian manifold $N$ and $K_1$ and $K_2$ are upper bound and lower bound of sectional curvature of $N$ respectively.
I don't know how this inequality is obtained.
 A: Call the elements of the local orthonormal frame $e_i$, and for brevity write $\nu$ for $e_{n+1}$. The component in question is then
\begin{equation}
K_{n+1,iji} = R(\nu, e_i, e_j, e_i),
\end{equation}
where $R$ is the rank-4 Riemann curvature tensor. We can assume that $i \neq j$ because $K_{n+1,iii} = 0$, and then the inequality is trivially true.
The trick is to define $e_\pm = \nu \pm e_j$, so that
\begin{equation}
\nu = \frac{1}{2}(e_+ + e_-), \qquad e_j = \frac{1}{2}(e_+ - e_-).
\end{equation}
Substituting this decomposition in,
\begin{align}
K_{n+1,iji}
&= \frac{1}{4}R(e_+ + e_-, e_i, e_+ - e_-, e_i) \\
&= \frac{1}{4}R(e_+, e_i, e_+, e_i)
- \frac{1}{4}R(e_-, e_i, e_-, e_i) \\
&\qquad + \frac{1}{4}R(e_-, e_i, e_+, e_i)
- \frac{1}{4}R(e_+, e_i, e_-, e_i).
\end{align}
Since $R(X,Y,Z,W) = R(Z,W,X,Y)$, the last terms cancel. The remaining two terms can be written as sectional curvatures using the definition
\begin{equation}
K(X,Y) = \frac{R(X,Y,X,Y)}{g(X,X)g(Y,Y) - g(X,Y)^2}.
\end{equation}
By direct computation one sees that $g(e_+,e_+) = g(e_-,e_-) = 2$ and $g(e_+,e_i) = g(e_-,e_i) = 0$, so
\begin{equation}
K_{n+1,iji} = \frac{1}{2}(K(e_+,e_i) - K(e_-,e_i)).
\end{equation}
Clearly this can be no larger than $(K_1 - K_2)/2$ and no smaller than $(K_2 - K_1)/2$, which proves the inequality.
