Erratum to: The interior gradient estimate of prescribed Hessian quotient curvature equations

1 Erratum to: Manuscripta Math. (2017) https://doi.org/10.1007/s00229-016-0877-4

In this note we correct a typo in the proof of the main result of [1], that is (3.10).

(3.10) should be

$$\begin{aligned} F^{1i} u_{1i} = \frac{{\partial F}}{{\partial a_{1i} }}\frac{1}{{W^2 }}u_{1i} = \frac{{\partial F}}{{\partial a_{1i} }}a_{i1} \le 0. \end{aligned}$$

The last inequality in the above fomula holds from the concavity of $\big [\frac{\sigma _k }{\sigma _l }\big ]^{\frac{1}{k-l}}$. We can give a detailed proof as follows. Following the proof in [1], we assume $A =\{ a_{ij} \}_{n \times n}$ with $a_{ij} = a_{ji}$, and at $x_0$, we have $a_{ij} =0$ for $2 \le i \ne j \le n$. For any $i_0 \ge 2$, consider

$$\begin{aligned} g(t) = \big [\frac{\sigma _k (B+tC) }{\sigma _l (B+tC) }\big ]^{\frac{1}{k-l}}, \quad (-1 \le t \le 1), \end{aligned}$$

where $B =\{ b_{ij} \}_{n \times n}$ with $b_{1i_0} = b_{i_01} =0$ and $b_{ij} = a_{ij}$ otherwise, and $C =\{c_{ij} \}_{n \times n}$ with $c_{1i_0} = c_{i_01} = a_{i_01}$ and $c_{ij} = 0$ otherwise. Then we have $B+C=A$.

Since $\sigma _1 (B + tC) = \sigma _1 (B)$ and

$$\begin{aligned} \sigma _m (B + tC) = \sigma _m (B) - t^2 a_{1i_0 } ^2 \sigma _{m-2} (B|1i_0 ) \end{aligned}$$

for $m \ge 2$, we can get

$$\begin{aligned} \sigma _m (B + tC) =&\sigma _m (A) +(1 - t^2) a_{1i_0 } ^2 \sigma _{m-2} (B |1i_0 ) \nonumber \\ =&\sigma _m (A) +(1 - t^2) a_{1i_0 } ^2 \sigma _{m-2} (A |1i_0 ) \nonumber \\ \ge&\sigma _m (A) \end{aligned}$$

(0.1)

for $t \in [-1, 1]$ and $\sigma _m (B + tC) = \sigma _m (B - tC)$, which can also be obtained from

$$\begin{aligned} \left| {\begin{array}{*{20}c} {a_{11} } &{}\quad \cdots &{}\quad {ta_{1i_0 } } &{}\quad \cdots \\ \vdots &{}\quad \ddots &{}\quad {} &{}\quad {} \\ {ta_{i_0 1} } &{}\quad {} &{}\quad {a_{i_0 i_0 } } &{}\quad {} \\ \vdots &{}\quad {} &{}\quad {} &{}\quad \ddots \\ \end{array}} \right| _{m \times m} = \left| {\begin{array}{*{20}c} {a_{11} } &{}\quad \cdots &{}\quad { - ta_{1i_0 } } &{}\quad \cdots \\ \vdots &{}\quad \ddots &{}\quad {} &{}\quad {} \\ { - ta_{i_0 1} } &{}\quad {} &{}\quad {a_{i_0 i_0 } } &{}\quad {} \\ \vdots &{}\quad {} &{}\quad {} &{}\quad \ddots \\ \end{array}} \right| _{m \times m}. \end{aligned}$$

Hence we know the eigenvalues of $B+tC$ are in the convex cone $\Gamma _k$ for $t \in [-1, 1]$ and $g(-1)=g(1)$. From the concavity of $\big [\frac{\sigma _k }{\sigma _l }\big ]^{\frac{1}{k-l}}$ in $\Gamma _k$, we have g(t) is concave with respect to $t \in [-1, 1]$. Hence

$$\begin{aligned} 0\ge g'(1) = \frac{1}{k-l} \big [\frac{\sigma _k (A) }{\sigma _l (A) }\big ]^{\frac{1}{k-l}-1} \big [ \frac{\partial {\frac{\sigma _k (A) }{\sigma _l (A)}}}{\partial a_{1i_0}} a_{i_0 1 }+ \frac{\partial {\frac{\sigma _k (A) }{\sigma _l (A)}}}{\partial a_{i_01}} a_{1 i_0}\big ]. \end{aligned}$$

(0.2)

Hence $\frac{{\partial F}}{{\partial a_{1i_0} }}a_{i_01} \le 0$ for any $i_0 \ge 2$.