# convex functions on Grassmannian manifolds and Lawson-Osserman problem

Y. L. Xin and Ling Yang  Institute of Mathematics, Fudan University, Shanghai
200433, China and Key Laboratory of Mathematics for Nonlinear Sciences (Fudan
University), Ministry of Education<EMAIL_ADDRESS>
###### Abstract.

We derive estimates of the Hessian of two smooth functions defined on
Grassmannian manifold. Based on it, we can derive curvature estimates for
minimal submanifolds in Euclidean space via Gauss map as [24]. In this way,
the result for Bernstein type theorem done by Jost and the first author could
be improved.

###### 1991 Mathematics Subject Classification:

49Q05, 53A07, 53A10.

The research was partially supported by NSFC and SFECC

##  1\. Introduction

The celebrated theorem of Bernstein [2] says that the only entire minimal
graphs in Euclidean 3-space are planes. Its higher dimensional generalization
was finally proved by J. Simons [19], which says that an entire minimal graph
has to be planar for dimension$\leq 7$, while Bombieri- De Giorgi-Giusti [3]
shortly afterwards produced a counterexample to such an assertion in dimension
8 and higher.

Schoen-Simon-Yau [18] gave us a direct proof for Bernstein type theorems for
$n\leq 5$ dimensional minimal graphs with the aid of curvature estimates for
stable minimal hypersurfaces.

There is a weak version of Bernstein type theorem in arbitrary dimension. It
was J. Moser [15] who proved that the entire solution $f$ to the minimal
surface equation is affine linear, provided $|\nabla f|$ is uniformly bounded.
Afterward Ecker-Huisken [9] obtained curvature estimates by a geometric
approach, as a corollary Moser’s result had been improved for the controlled
growth of $|\nabla f|$.

Moser’s theorem had been generalized to certain higher codimensional cases by
Chern-Osserman [6] for dimension $2$ and Babosa, Fischer-Colbrie for dimension
$3$ [1], [10]. But the counterexample constructed by Lawson-Osserman [13]
prevents us going further. They also raised in the same paper a question for
finding the ”best” constant possible in the theorem.

In contrast, the first author with J. Jost [12] proved the following Bernstein
type theorem without the restriction of dimension and codimension, which is an
improvement of the work done by Hildebrandt-Jost-Widman [11].

######  Theorem 1.1.

Let $z^{\alpha}=f^{\alpha}(x^{1},\cdots,x^{n}),\ \alpha=1,\cdots,m$, be smooth
functions defined everywhere in ${\tenmsb R}^{n}$. Suppose their graph
$M=(x,f(x))$ is a submanifold with parallel mean curvature in ${\tenmsb
R}^{n+m}$. Suppose that there exists a number $\beta_{0}$ with

(1.1) $\beta_{0}<\left\\{\begin{array}[]{cl}2&\mbox{when }m\geq 2,\\\
\infty&\mbox{when }m=1;\end{array}\right.$

such that

(1.2)
$\Delta_{f}=\left[\text{det}\left(\delta_{ij}+\sum_{\alpha}\frac{\partial
f^{\alpha}}{\partial x^{i}}\frac{\partial f^{\alpha}}{\partial
x^{j}}\right)\right]^{\frac{1}{2}}\leq\beta_{0}.$

Then $f^{1},\cdots,f^{m}$ has to be affine linear representing an affine
$n$-plane.

The key point of the proof is to find a geodesic convex set

$B_{JX}(P_{0})=\big{\\{}P\in\mathbb{G}_{n,m}:\mbox{ sum of any two Jordan
angles between }P\mbox{ and }P_{0}<\frac{\pi}{2}\big{\\}}$

in a geodesic polar coordinate of the Grassmannian manifold, where $P_{0}$
denotes a fixed $n-$plane. It is larger than the largest geodesic convex ball
of radius $\frac{\sqrt{2}}{4}\pi$ in $\mathbb{G}_{n,m}$. The geometric meaning
of the condition of the above result is that the image under the Gauss map of
$M$ lies in a closed subset $S\subset B_{JX}(P_{0})$.

Recently, the authors [24] studied complete minimal submanifolds whose Gauss
image lies in an open geodesic ball of radius $\frac{\sqrt{2}}{4}\pi$. They
carried out the Schoen-Simon-Yau type curvature estimates and the Ecker-
Huisken type curvature estimates, and on the basis, the corresponding
Bernstein type theorems with dimension limitation or growth assumption could
be derived.

It is natural to study the situation when $\beta_{0}$ in the condition (1.1)
and (1.2) of the Theorem 1.1 approach to $2$. The present paper will devote to
this problem. We shall follow the main idea of our previous paper [24]. But,
we view now the Grassmannian manifolds as submanifolds in Euclidean space via
Plücker imbedding. The auxiliary functions are constructed from this
viewpoint. As shown before, $B_{JX}(P_{0})$ is defined in a coordinate
neighborhood ${\tenmsb U}$ of the Grassmannian $\mathbb{G}_{n,m}$. We
introduce two functions $v$ and $u$ in ${\tenmsb U}$. Via the Gauss map we can
obtain useful functions on our minimal $n-$submanifold $M$ in ${\tenmsb
R}^{m+n}$ with $m\geq 2$. Then, we can carry out the Schoen-Simon-Yau type
curvature estimates and the Ecker-Huisken type curvature estimates, which
enable us to get the corresponding Bernstein type theorems and other
geometrical conclusions.

In Section 2, we give some facts of a Grassmannian manifold
$\mathbb{G}_{n,m}$, which can be isometric imbedding into a Euclidean space.
There is the height function for a submanifold in Euclidean space. Such a
height function is called $w-$function on $\mathbb{G}_{n,m}$. Then we have an
open domain of ${\tenmsb U}\subset\mathbb{G}_{n,m}$, where the $w-$function is
positive. Every point in ${\tenmsb U}$ has a one-to-one correspondence to an
$n\times m$ matrix. We describe canonical metric and the corresponding
connection on ${\tenmsb U}$ with respect to the coordinate. On the basis, the
Hessian of an arbitrary smooth function could be calculated.

In Section 3 we define $v=\frac{1}{w}$ on ${\tenmsb U}$. We also define
another function $u$ on ${\tenmsb U}$. In the section, we shall show $v$ and
$u$ are convex on $B_{JX}(P_{0})$ and give estimates of the Hessian of them.
The estimates are quite delicate. We use the radial compensation technique to
accurate the estimates.

In Section 4, we define four auxiliary functions, $\tilde{h}_{1}$,
$\tilde{h}_{2}$, $\tilde{h}_{3}$ and $\tilde{h}_{4}$. They are defined on the
minimal submanifolds of ${\tenmsb R}^{n+m}$ whose Gauss image is confined, and
they are expressed in term of $v$ and $u$. We also estimate the Laplacian of
them, which is useful for the next sections.

Later in Section 5, not only we give the Schoen-Simon-Yau type curvature
estimates with the aid of $\tilde{h}_{1}$ and $\tilde{h}_{3}$, but also we
obtain the Ecker-Huisken type curvature estimates with the aid of
$\tilde{h}_{2}$ and $\tilde{h}_{4}$. Our method is completely similar to the
previous paper [24], so we only describe the outline of process. From the
estimates several geometrical conclusions follow, including the following
Bernstein type theorems.

######  Theorem 1.2.

Let $M=(x,f(x))$ be an $n$-dimensional entire minimal graph given by $m$
functions $f^{\alpha}(x^{1},\cdots,x^{n})$ with $m\geq 2,n\leq 4$. If

$\Delta_{f}=\left[\text{det}\left(\delta_{ij}+\sum_{\alpha}\frac{\partial
f^{\alpha}}{\partial x^{i}}\frac{\partial f^{\alpha}}{\partial
x^{j}}\right)\right]^{\frac{1}{2}}<2,$

then $f^{\alpha}$ has to be affine linear functions representing an affine
$n$-plane.

######  Theorem 1.3.

Let $M=(x,f(x))$ be an $n$-dimensional entire minimal graph given by $m$
functions $f^{\alpha}(x^{1},\cdots,x^{n})$ with $m\geq 2$. If

$\Delta_{f}=\left[\text{det}\left(\delta_{ij}+\sum_{\alpha}\frac{\partial
f^{\alpha}}{\partial x^{i}}\frac{\partial f^{\alpha}}{\partial
x^{j}}\right)\right]^{\frac{1}{2}}<2,$

and

(1.3) $\left(2-\Delta_{f}\right)^{-1}=o(R^{\frac{4}{3}}),$

where $R^{2}=|x|^{2}+|f|^{2}$. Then $f^{\alpha}$ has to be affine linear
functions and hence $M$ has to be an affine linear subspace.

Those are what shall be done in Section 6. It is worthy to note that Theorems
1.2-1.3 still hold true when $M$ is a submanifold with parallel mean
curvature. Dong generalized Chern’s result [5] [4] to higher codimension,
which states that a graphic submanifold $M=(x,f(x))$ with parallel mean
curvature has to be minimal if the slope of $f$ is uniformly bounded. Hence
our results improve Theorem 1.1.

It is natural to ask what is the relations between the results here and that
of the previous paper [24]. Since the $v-$function varies in
$\left(\sec^{p}\left(\frac{\pi}{2\sqrt{2p}}\right),\sec\left(\frac{\sqrt{2}}{4}\pi\right)\right)$
on the open geodesic ball of radius $\frac{\sqrt{2}}{4}$ in
$\mathbb{G}_{n,m}$, where $p=\min(n,m)$, the results of the present article do
not generalize those in the previous one. Both results are complementary.

##  2\. Preliminaries on the Grassmannian manifold $\mathbb{G}_{n,m}$

Let $\mathbb{R}^{n+m}$ be an $n+m$-dimensional Euclidean space. All oriented
$n$-subspaces constitute the Grassmannian manifolds $\mathbb{G}_{n,m}$, which
is an irreducible symmetric space of compact type.

Fix $P_{0}\in\mathbb{G}_{n,m}$ in the sequel, which is spanned by a unit
$n-$vector $\varepsilon_{1}\wedge\cdots\wedge\varepsilon_{n}$. For any
$P\in\mathbb{G}_{n,m}$, spanned by a $n-$vector $e_{1}\wedge\cdots\wedge
e_{n}$, we define an important function on $\mathbb{G}_{n,m}$

$w\mathop{=}\limits^{def.}\left<P,P_{0}\right>=\left<e_{1}\wedge\cdots\wedge
e_{n},\varepsilon_{1}\wedge\cdots\wedge\varepsilon_{n}\right>=\det W,$

where $W=(\left<e_{i},\varepsilon_{j}\right>).$ It is well known that

$W^{T}W=O^{T}\Lambda O,$

where $O$ is an orthogonal matrix and

$\Lambda=\begin{pmatrix}\mu_{1}^{2}&&0\cr&\ddots&\cr
0&&\mu_{p}^{2}\end{pmatrix},\qquad p=\min(m,n),$

where each $0\leq\mu_{i}^{2}\leq 1.$ The Jordan angles between $P$ and $P_{0}$
are defined by

$\theta_{i}=\arccos(\mu_{i}).$

Denote

${\tenmsb U}=\\{P\in\mathbb{G}_{n,m}:w(P)>0\\},$

let $\\{\varepsilon_{n+\alpha}\\}$ be $m$-vectors such that
$\\{\varepsilon_{i},\varepsilon_{n+\alpha}\\}$ form an orthornormal basis of
$\mathbb{R}^{m+n}$. Then we can span arbitrary $P\in{\tenmsb U}$ by $n$
vectors $f_{i}$:

$f_{i}=\varepsilon_{i}+z_{i\alpha}\varepsilon_{n+\alpha},$

where $Z=(z_{i\alpha})$ are the local coordinate of $P$ in ${\tenmsb U}$. Here
and in the sequel we use the summation convention and agree the range of
indices:

$1\leq i,j,k,l\leq n;\qquad 1\leq\alpha,\beta,\gamma,\delta\leq m.$

The canonical metric on $\mathbb{G}_{n,m}$ in the local coordinate can be
described as (see [22] Ch. VII)

(2.1)
$g=\mbox{tr}\big{(}(I_{n}+ZZ^{T})^{-1}dZ(I_{m}+Z^{T}Z)^{-1}dZ^{T}\big{)}.$

Let $P\in{\tenmsb U}$ determined by an $n\times m$ matrix
$Z_{0}=\big{(}\lambda_{\alpha}\delta_{i\alpha}\big{)}$, where
$\lambda_{\alpha}=\tan\theta_{\alpha}$ and $\theta_{1},\cdots,\theta_{m}$ be
the Jordan angles between $P$ and $P_{0}$. (Here and in the sequel we assume
$n\geq m$ without loss of generality; for it is similar for $n<m$.) Let
$X,Y,W$ denote arbitrary $n\times m$ matrices. Then (2.1) tells us

(2.2) $\displaystyle\begin{aligned} \langle
X,Y\rangle_{P}&=\mbox{tr}\big{(}(I_{n}+Z_{0}Z_{0}^{T})^{-1}X(I_{m}+Z_{0}^{T}Z_{0})^{-1}Y^{T}\big{)}\\\
&=\sum_{i,\alpha}(1+\lambda_{i}^{2})^{-1}(1+\lambda_{\alpha}^{2})^{-1}X_{i\alpha}Y_{i\alpha}.\end{aligned}$

(Note that if $m+1\leq i\leq n$, $\lambda_{i}=0$.) Furthermore, from

$\displaystyle\begin{aligned}
&\big{(}I_{n}+(Z_{0}+tW)(Z_{0}+tW)^{T}\big{)}^{-1}X\big{(}I_{m}+(Z_{0}+tW)^{T}(Z_{0}+tW)\big{)}^{-1}Y^{T}\\\
&=\big{(}I_{n}+Z_{0}Z_{0}^{T}+t(WZ_{0}^{T}+Z_{0}W^{T})+O(t^{2})\big{)}^{-1}X\\\
&\qquad\big{(}I_{m}+Z_{0}^{T}Z_{0}+t(W^{T}Z_{0}+Z_{0}^{T}W)+O(t^{2})\big{)}^{-1}Y^{T}\\\
&=\big{(}I_{n}+t(I_{n}+Z_{0}Z_{0}^{T})^{-1}(WZ_{0}^{T}+Z_{0}W^{T})+O(t^{2})\big{)}^{-1}(I_{n}+Z_{0}Z_{0}^{T})^{-1}X\\\
&\qquad\big{(}I_{m}+t(I_{m}+Z_{0}^{T}Z_{0})^{-1}(W^{T}Z_{0}+Z_{0}^{T}W)+O(t^{2})\big{)}^{-1}(I_{m}+Z_{0}^{T}Z_{0})^{-1}Y^{T}\\\
&=\big{(}I_{n}-t(I_{n}+Z_{0}Z_{0}^{T})^{-1}(WZ_{0}^{T}+Z_{0}W^{T})+O(t^{2})\big{)}(I_{n}+Z_{0}Z_{0}^{T})^{-1}X\\\
&\qquad\big{(}I_{m}-t(I_{m}+Z_{0}^{T}Z_{0})^{-1}(W^{T}Z_{0}+Z_{0}^{T}W)+O(t^{2})\big{)}(I_{m}+Z_{0}^{T}Z_{0})^{-1}Y^{T}\\\
&=(I_{n}+Z_{0}Z_{0}^{T})^{-1}X(I_{m}+Z_{0}^{T}Z_{0})^{-1}Y^{T}\\\
&\quad-t\big{[}(I_{n}+Z_{0}Z_{0}^{T})^{-1}(WZ_{0}^{T}+Z_{0}W^{T})(I_{n}+Z_{0}Z_{0}^{T})^{-1}X(I_{m}+Z_{0}^{T}Z_{0})^{-1}Y^{T}\\\
&\qquad+(I_{n}+Z_{0}Z_{0}^{T})^{-1}X(I_{m}+Z_{0}^{T}Z_{0})^{-1}(W^{T}Z_{0}+Z_{0}^{T}W)(I_{m}+Z_{0}^{T}Z_{0})^{-1}Y^{T}\big{]}+O(t^{2}),\end{aligned}$

we have

(2.3) $\displaystyle\begin{aligned} W\langle
X,Y\rangle_{P}=&-\mbox{tr}\big{[}(I_{n}+Z_{0}Z_{0}^{T})^{-1}(WZ_{0}^{T}+Z_{0}W^{T})\\\
&\qquad\qquad(I_{n}+Z_{0}Z_{0}^{T})^{-1}X(I_{m}+Z_{0}^{T}Z_{0})^{-1}Y^{T}\\\
&\qquad+(I_{n}+Z_{0}Z_{0}^{T})^{-1}X(I_{m}+Z_{0}^{T}Z_{0})^{-1}\\\
&\qquad\qquad(W^{T}Z_{0}+Z_{0}^{T}W)(I_{m}+Z_{0}^{T}Z_{0})^{-1}Y^{T}\big{]}.\end{aligned}$

We let $E_{i\alpha}$ be the matrix with 1 in the intersection of row $i$ and
column $\alpha$ and 0 otherwise. Denote $g_{i\alpha,j\beta}=\langle
E_{i\alpha},E_{j\beta}\rangle$ and let $\big{(}g^{i\alpha,j\beta}\big{)}$ be
the inverse matrix of $\big{(}g_{i\alpha,j\beta}\big{)}$. Denote by $\nabla$
the Levi-Civita connection with respect to the canonical matric on
$\mathbb{G}_{n,m}$, and by

$\nabla_{E_{i\alpha}}E_{j\beta}=\Gamma_{i\alpha,j\beta}^{k\gamma}E_{k\gamma}.$

Then from (2.2),

(2.4) $\displaystyle
g_{i\alpha,j\beta}(P)=(1+\lambda_{i}^{2})^{-1}(1+\lambda_{\alpha}^{2})^{-1}\delta_{\alpha\beta}\delta_{ij}$

and obviously

(2.5)
$g^{i\alpha,j\beta}(P)=(1+\lambda_{i}^{2})(1+\lambda_{\alpha}^{2})\delta_{\alpha\beta}\delta_{ij}.$

Moreover, a direct calculation from (2.3) and (2.5) shows

(2.6) $\displaystyle\begin{aligned}
\Gamma_{i\alpha,j\beta}^{k\gamma}&=\frac{1}{2}g^{k\gamma,l\delta}\big{(}-E_{l\delta}\langle
E_{i\alpha},E_{j\beta}\rangle+E_{i\alpha}\langle
E_{j\beta},E_{l\delta}\rangle+E_{j\beta}\langle
E_{l\delta},E_{i\alpha}\rangle\big{)}\\\
&=-\lambda_{\alpha}(1+\lambda_{\alpha}^{2})^{-1}\delta_{\alpha
j}\delta_{\beta\gamma}\delta_{ik}-\lambda_{\beta}(1+\lambda_{\beta}^{2})^{-1}\delta_{\beta
i}\delta_{\alpha\gamma}\delta_{jk}.\end{aligned}$

From (2.5), we see that

(2.7)
$(1+\lambda_{i}^{2})^{\frac{1}{2}}(1+\lambda_{\alpha}^{2})^{\frac{1}{2}}E_{i\alpha}\
(1\leq i\leq n,1\leq\alpha\leq m)$

form an orthonormal basis of $T_{P}\mathbb{G}_{n,m}$. Denote its dual basis in
$T_{P}^{*}\mathbb{G}_{n,m}$ by

(2.8) $\omega_{i\alpha}\ (1\leq i\leq n,1\leq\alpha\leq m),$

then

(2.9) $g=\sum_{i,\alpha}\omega_{i\alpha}^{2}$

at $P$.

##  3\. Hessian estimates of two smooths functions on $\mathbb{G}_{n,m}$

On ${\tenmsb U}$, $w>0$, then we can define

(3.1) $v=w^{-1}\qquad\mbox{ on }{\tenmsb U}.$

For arbitrary $Q\in{\tenmsb U}$ determined by an $n\times m$ matrix $Z$, it is
easily seen that

(3.2)
$v(Q)=\big{[}\det(I_{n}+ZZ^{T})\big{]}^{\frac{1}{2}}=\prod_{\alpha=1}^{m}\sec\theta_{\alpha}.$

where $\theta_{1},\cdots,\theta_{m}$ denotes the Jordan angles between $Q$ and
$P_{0}$.

Now we calculate the Hessian of $v$ at $P$ whose corresponding matrix is
$Z_{0}$. At first, by noting that for any $n\times n$ orthogonal matrix $U$
and $m\times m$ orthogonal matrix $V$, $Z\mapsto UZV$ induces an isometry of
${\tenmsb U}$ which keeps $v$ invariant, we can assume
$Z_{0}=(\lambda_{\alpha}\delta_{i\alpha})$ without loss of generality, where
$\lambda_{\alpha}=\tan\theta_{\alpha}$ and $\theta_{1},\cdots,\theta_{m}$
denotes the Jordan angles between $P$ and $P_{0}$. We also need a Lemma as
follows.

######  Lemma 3.1.

Let $M$ be a manifold, $A$ be a smooth nonsingular $n\times n$ matrix-valued
function on $M$, $X,Y$ be local tangent fields, then

(3.3) $\nabla_{X}\log\det A=\text{tr}(\nabla_{X}A\cdot A^{-1})$

and

(3.4) $\nabla_{Y}\nabla_{X}\log\det A=\mbox{tr}(\nabla_{Y}\nabla_{X}A\cdot
A^{-1})-\mbox{tr}(\nabla_{X}A\cdot A^{-1}\cdot\nabla_{Y}A\cdot A^{-1}).$

###### Proof.

Assume that $e_{1},\cdots,e_{n}$ is a standard basis in ${\tenmsb R}^{n}$,
then

$\det A\ e_{1}\wedge\cdots\wedge e_{n}=Ae_{1}\wedge\cdots\wedge Ae_{n}.$

Hence

$\displaystyle\begin{aligned} \nabla_{X}\det A\ e_{1}\wedge\cdots\wedge
e_{n}=&\sum_{i}Ae_{1}\wedge\cdots\wedge Ae_{i-1}\wedge\nabla_{X}Ae_{i}\wedge
Ae_{i+1}\wedge\cdots\wedge Ae_{n}\\\ =&\sum_{i}Ae_{1}\wedge\cdots\wedge
Ae_{i-1}\wedge(\nabla_{X}A\cdot A^{-1})Ae_{i}\wedge Ae_{i+1}\wedge\cdots\wedge
Ae_{n}\\\ =&\mbox{tr}(\nabla_{X}A\cdot A^{-1})Ae_{1}\wedge\cdots\wedge
Ae_{n}\\\ =&\mbox{tr}(\nabla_{X}A\cdot A^{-1})\det A\ e_{1}\wedge\cdots\wedge
e_{n}.\end{aligned}$

Thereby (3.3) immediately follows.

(3.4) follows from (3.3) and

$A\cdot\nabla_{Y}A^{-1}+\nabla_{Y}A\cdot A^{-1}=\nabla_{Y}(AA^{-1})=0.$

∎

Now we let $M={\tenmsb U}$, $A(Z)=I_{n}+ZZ^{T}$, then $\log
v=\frac{1}{2}\log\det A$. A direct calculation shows

$\nabla_{X}A=XZ^{T}+ZX^{T},\qquad\nabla_{Y}\nabla_{X}A=XY^{T}+YX^{T}.$

Hence we compute from Lemma 3.1 that at $P$

$\displaystyle\begin{aligned} \nabla_{X}\log
v&=\frac{1}{2}\mbox{tr}\left((XZ_{0}^{T}+Z_{0}X^{T})(I_{n}+Z_{0}Z_{0}^{T})^{-1}\right)\\\
&=\sum_{\alpha}\lambda_{\alpha}(1+\lambda_{\alpha}^{2})^{-1}X_{\alpha\alpha},\end{aligned}$
$\displaystyle\nabla_{X}\nabla_{Y}\log v$ $\displaystyle=$
$\displaystyle\frac{1}{2}\mbox{tr}\left((XY^{T}+YX^{T})(I_{n}+Z_{0}Z_{0}^{T})^{-1}\right)$
$\displaystyle-\frac{1}{2}\mbox{tr}\left((XZ_{0}^{T}+Z_{0}X^{T})(I_{n}+Z_{0}Z_{0}^{T})^{-1}(YZ_{0}^{T}+Z_{0}Y^{T})(I_{n}+Z_{0}Z_{0}^{T})^{-1}\right)$
$\displaystyle=$
$\displaystyle\sum_{i,\alpha}(1+\lambda_{i}^{2})^{-1}X_{i\alpha}Y_{i\alpha}$
$\displaystyle\qquad-\frac{1}{2}\sum_{i,j}(XZ_{0}^{T}+Z_{0}X^{T})_{ij}(1+\lambda_{j}^{2})^{-1}(YZ_{0}^{T}+Z_{0}Y^{T})_{ji}(1+\lambda_{i}^{2})^{-1}$
$\displaystyle=$ $\displaystyle\sum_{m+1\leq i\leq
n,\alpha}X_{i\alpha}Y_{i\alpha}+\frac{1}{2}\sum_{\alpha,\beta}(1+\lambda_{\alpha}^{2})^{-1}X_{\alpha\beta}Y_{\alpha\beta}+\frac{1}{2}\sum_{\alpha,\beta}(1+\lambda_{\beta}^{2})^{-1}X_{\beta\alpha}Y_{\beta\alpha}$
$\displaystyle\qquad-\frac{1}{2}\sum_{\alpha,\beta}(\lambda_{\beta}X_{\alpha\beta}+\lambda_{\alpha}X_{\beta\alpha})(1+\lambda_{\beta}^{2})^{-1}(\lambda_{\alpha}Y_{\beta\alpha}+\lambda_{\beta}Y_{\alpha\beta})(1+\lambda_{\alpha}^{2})^{-1}$
$\displaystyle\qquad-\sum_{m+1\leq i\leq
n,\alpha}\lambda_{\alpha}^{2}(1+\lambda_{\alpha}^{2})^{-1}X_{i\alpha}Y_{i\alpha}$
$\displaystyle=$ $\displaystyle\sum_{m+1\leq i\leq
n,\alpha}(1+\lambda_{\alpha}^{2})^{-1}X_{i\alpha}Y_{i\alpha}+\sum_{\alpha,\beta}(1+\lambda_{\alpha}^{2})^{-1}(1+\lambda_{\beta}^{2})^{-1}X_{\alpha\beta}Y_{\alpha\beta}$
$\displaystyle\qquad-\sum_{\alpha,\beta}\lambda_{\alpha}\lambda_{\beta}(1+\lambda_{\alpha}^{2})^{-1}(1+\lambda_{\beta}^{2})^{-1}X_{\alpha\beta}Y_{\beta\alpha}.$

Furthermore,

$\displaystyle\begin{aligned} \nabla_{X}v=&v\nabla_{X}\log
v=\left(\sum_{\alpha}\lambda_{\alpha}(1+\lambda_{\alpha}^{2})^{-1}X_{\alpha\alpha}\right)v,\\\
\nabla_{X}\nabla_{Y}v=&v(\nabla_{X}\nabla_{Y}\log v+\nabla_{X}\log
v\cdot\nabla_{Y}\log v)\\\ =&\Big{(}\sum_{m+1\leq i\leq
n,\alpha}(1+\lambda_{\alpha}^{2})^{-1}X_{i\alpha}Y_{i\alpha}+\sum_{\alpha,\beta}(1+\lambda_{\alpha}^{2})^{-1}(1+\lambda_{\beta}^{2})^{-1}X_{\alpha\beta}Y_{\alpha\beta}\\\
&\qquad+\sum_{\alpha,\beta}\lambda_{\alpha}\lambda_{\beta}(1+\lambda_{\alpha}^{2})^{-1}(1+\lambda_{\beta}^{2})^{-1}(X_{\alpha\alpha}Y_{\beta\beta}-X_{\alpha\beta}Y_{\beta\alpha})\Big{)}v\\\
=&\Big{(}\sum_{i,\beta}(1+\lambda_{i}^{2})^{-1}(1+\lambda_{\beta}^{2})^{-1}X_{i\beta}Y_{i\beta}\\\
&\qquad+\sum_{\alpha,\beta}\lambda_{\alpha}\lambda_{\beta}(1+\lambda_{\alpha}^{2})^{-1}(1+\lambda_{\beta}^{2})^{-1}(X_{\alpha\alpha}Y_{\beta\beta}-X_{\alpha\beta}Y_{\beta\alpha})\Big{)}v.\end{aligned}$

In particular,

(3.5)
$\nabla_{E_{i\alpha}}v(P)=\lambda_{\alpha}(1+\lambda_{\alpha}^{2})^{-1}v\delta_{i\alpha}$

and

(3.6)
$\nabla_{E_{i\alpha}}\nabla_{E_{j\beta}}v(P)=\left\\{\begin{array}[]{cl}(1+\lambda_{i}^{2})^{-1}(1+\lambda_{\alpha}^{2})^{-1}v&i=j,\alpha=\beta;\\\
-\lambda_{\alpha}\lambda_{\beta}(1+\lambda_{\alpha}^{2})^{-1}(1+\lambda_{\beta}^{2})^{-1}v&i=\beta,j=\alpha,\alpha\neq\beta;\\\
\lambda_{\alpha}\lambda_{\beta}(1+\lambda_{\alpha}^{2})^{-1}(1+\lambda_{\beta}^{2})^{-1}v&i=\alpha,j=\beta,\alpha\neq\beta;\\\
0&\mbox{otherwise.}\end{array}\right.$

Then, from (2.6), (3.5) and (3.6) we obtain

(3.7) $\displaystyle\begin{aligned}
\mbox{Hess}(v)(E_{i\alpha},E_{j\beta})(P)=&\nabla_{E_{i\alpha}}\nabla_{E_{j\beta}}v-(\nabla_{E_{i\alpha}}E_{j\beta})v\\\
=&\nabla_{E_{i\alpha}}\nabla_{E_{j\beta}}v-\Gamma_{i\alpha,j\beta}^{k\gamma}\nabla_{E_{k\gamma}}v\\\
=&\left\\{\begin{array}[]{cl}(1+\lambda_{i}^{2})^{-1}(1+\lambda_{\alpha}^{2})^{-1}v&i=j,\alpha=\beta,i\neq\alpha;\\\
(1+2\lambda_{\alpha}^{2})(1+\lambda_{\alpha}^{2})^{-2}v&i=j=\alpha=\beta;\\\
\lambda_{\alpha}\lambda_{\beta}(1+\lambda_{\alpha}^{2})^{-1}(1+\lambda_{\beta}^{2})^{-1}v&i=\beta,j=\alpha,\alpha\neq\beta;\\\
\lambda_{\alpha}\lambda_{\beta}(1+\lambda_{\alpha}^{2})^{-1}(1+\lambda_{\beta}^{2})^{-1}v&i=\alpha,j=\beta,\alpha\neq\beta;\\\
0&\mbox{otherwise.}\end{array}\right.\end{aligned}$

In other words

(3.8) $\displaystyle\begin{aligned} \mbox{Hess}(v)_{P}&=\sum_{i\neq\alpha}v\
\omega_{i\alpha}^{2}+\sum_{\alpha}(1+2\lambda_{\alpha}^{2})v\
\omega_{\alpha\alpha}^{2}+\sum_{\alpha\neq\beta}\lambda_{\alpha}\lambda_{\beta}v(\omega_{\alpha\alpha}\otimes\omega_{\beta\beta}+\omega_{\alpha\beta}\otimes\omega_{\beta\alpha})\\\
&=\sum_{m+1\leq i\leq n,\alpha}v\
\omega_{i\alpha}^{2}+\sum_{\alpha}(1+2\lambda_{\alpha}^{2})v\
\omega_{\alpha\alpha}^{2}+\sum_{\alpha\neq\beta}\lambda_{\alpha}\lambda_{\beta}v\
\omega_{\alpha\alpha}\otimes\omega_{\beta\beta}\\\
&\qquad\qquad+\sum_{\alpha<\beta}\Big{[}(1+\lambda_{\alpha}\lambda_{\beta})v\Big{(}\frac{\sqrt{2}}{2}(\omega_{\alpha\beta}+\omega_{\beta\alpha})\Big{)}^{2}\\\
&\hskip
144.54pt+(1-\lambda_{\alpha}\lambda_{\beta})v\Big{(}\frac{\sqrt{2}}{2}(\omega_{\alpha\beta}-\omega_{\beta\alpha})\Big{)}^{2}\Big{]}.\end{aligned}$

(3.8) could be simplified further. Please note (3.5), which also tells us

(3.9) $dv=\sum_{\alpha}\lambda_{\alpha}v\ \omega_{\alpha\alpha};$

then

(3.10) $dv\otimes dv=\sum_{\alpha}\lambda_{\alpha}^{2}v^{2}\
\omega_{\alpha\alpha}^{2}+\sum_{\alpha\neq\beta}\lambda_{\alpha}\lambda_{\beta}v^{2}\
\omega_{\alpha\alpha}\otimes\omega_{\beta\beta}.$

Substituting (3.10) into (3.8) yields

(3.11) $\displaystyle\begin{aligned} \mbox{Hess}(v)_{P}&=\sum_{m+1\leq i\leq
n,\alpha}v\ \omega_{i\alpha}^{2}+\sum_{\alpha}(1+\lambda_{\alpha}^{2})v\
\omega_{\alpha\alpha}^{2}+v^{-1}\ dv\otimes dv\\\
&\qquad+\sum_{\alpha<\beta}\Big{[}(1+\lambda_{\alpha}\lambda_{\beta})v\Big{(}\frac{\sqrt{2}}{2}(\omega_{\alpha\beta}+\omega_{\beta\alpha})\Big{)}^{2}\\\
&\hskip
72.26999pt+(1-\lambda_{\alpha}\lambda_{\beta})v\Big{(}\frac{\sqrt{2}}{2}(\omega_{\alpha\beta}-\omega_{\beta\alpha})\Big{)}^{2}\Big{]}.\end{aligned}$

Note that $\lambda_{\alpha}\geq 0$ and
$1-\lambda_{\alpha}\lambda_{\beta}=1-\tan\theta_{\alpha}\tan\theta_{\beta}=\frac{\cos(\theta_{\alpha}+\theta_{\beta})}{\cos\theta_{\alpha}\cos\theta_{\beta}}$;
which implies that $\mbox{Hess}(v)_{P}$ is positive definite if and only if
$\theta_{\alpha}+\theta_{\beta}<\frac{\pi}{2}$ for arbitrary
$\alpha\neq\beta$, i.e., $P\in B_{JX}(P_{0})$.

By (3.2), $v=\prod_{\alpha}(1+\lambda_{\alpha}^{2})^{\frac{1}{2}}$, then

$\lambda_{\alpha}\lambda_{\beta}\leq\big{[}(1+\lambda_{\alpha}^{2})(1+\lambda_{\beta}^{2})\big{]}^{\frac{1}{2}}-1\leq
v-1,$

the equality holds if and only if $\lambda_{\alpha}=\lambda_{\beta}$ and
$\lambda_{\gamma}=0$ for each $\gamma\neq\alpha,\beta$. Hence, we have
$1-\lambda_{\alpha}\lambda_{\beta}\geq 2-v$. Finally we arrive at an estimate

(3.12) $\mbox{Hess}(v)\geq v(2-v)g+v^{-1}dv\otimes dv.$

Now we introduce another smooth function on ${\tenmsb U}$. For any
$Q\in{\tenmsb U}$,

(3.13) $u\mathop{=}\limits^{def.}\sum_{\alpha}\tan\theta_{\alpha}^{2}.$

where $\theta_{1},\cdots,\theta_{m}$ denotes the Jordan angles between $Q$ and
$P_{0}$. Denote by $Z$ the coordinate of $Q$, then it is easily seen that

(3.14) $u(Q)=\mbox{tr}(ZZ^{T}).$

We can calculate the Hessian of $u$ at $P\in{\tenmsb U}$ whose corresponding
matrix is $Z_{0}$ in the same way. Similar to above, we can assume
$Z_{0}=\big{(}\lambda_{a}\delta_{i\alpha}\big{)}$, where
$\lambda_{\alpha}=\tan\theta_{\alpha}$ and $\theta_{1},\cdots,\theta_{m}$ are
the Jordan angles between $P$ and $P_{0}$.

Obviously

(3.15) $\displaystyle\begin{aligned}
\nabla_{X}u=&\mbox{tr}(XZ^{T})+\mbox{tr}(ZX^{T}),\\\
\nabla_{X}\nabla_{Y}u=&\mbox{tr}(XY^{T})+\mbox{tr}(YX^{T}).\end{aligned}$

Then, at $P$

(3.16) $\displaystyle\begin{aligned}
\mbox{Hess}(u)(E_{i\alpha},E_{j\beta})=&\nabla_{E_{i\alpha}}\nabla_{E_{j\beta}}u-(\nabla_{E_{i\alpha}}E_{j\beta})u\\\
=&\nabla_{E_{i\alpha}}\nabla_{E_{j\beta}}u-\Gamma_{i\alpha,j\beta}^{k\gamma}\nabla_{E_{k\gamma}}u\\\
=&2\delta_{ij}\delta_{\alpha\beta}+\big{(}\lambda_{\alpha}(1+\lambda_{\alpha}^{2})^{-1}\delta_{\alpha
j}\delta_{\beta\gamma}\delta_{ik}+\lambda_{\beta}(1+\lambda_{\beta}^{2})^{-1}\delta_{\beta
i}\delta_{\alpha\gamma}\delta_{jk}\big{)}\\\ &\cdot
2\lambda_{\gamma}\delta_{k\gamma}\\\
=&2\delta_{ij}\delta_{\alpha\beta}+2\lambda_{\alpha}\lambda_{\beta}\big{[}(1+\lambda_{\alpha}^{2})^{-1}+(1+\lambda_{\beta}^{2})^{-1}\big{]}\delta_{\alpha
j}\delta_{\beta i}\\\
=&\left\\{\begin{array}[]{cl}2&i=j,\alpha=\beta,i\neq\alpha;\\\
2+4\lambda_{\alpha}^{2}(1+\lambda_{\alpha}^{2})^{-1}&i=j=\alpha=\beta;\\\
2\lambda_{\alpha}\lambda_{\beta}\big{[}(1+\lambda_{\alpha}^{2})^{-1}+(1+\lambda_{\beta}^{2})^{-1}\big{]}&i=\beta,j=\alpha,\alpha\neq\beta.\end{array}\right.\end{aligned}$

In other words

(3.17) $\displaystyle\begin{aligned}
\mbox{Hess}(u)_{P}=&\sum_{i\neq\alpha}2(1+\lambda_{i}^{2})(1+\lambda_{\alpha}^{2})\omega_{i\alpha}^{2}+\sum_{\alpha}(2+6\lambda_{\alpha}^{2})(1+\lambda_{\alpha}^{2})\omega_{\alpha\alpha}^{2}\\\
&\qquad+\sum_{\alpha\neq\beta}2\lambda_{\alpha}\lambda_{\beta}(2+\lambda_{\alpha}^{2}+\lambda_{\beta}^{2})\omega_{\alpha\beta}\otimes\omega_{\beta\alpha}\\\
=&\sum_{m+1\leq i\leq
n,\alpha}2(1+\lambda_{i}^{2})(1+\lambda_{\alpha}^{2})\omega_{i\alpha}^{2}+\sum_{\alpha}(2+6\lambda_{\alpha}^{2})(1+\lambda_{\alpha}^{2})\omega_{\alpha\alpha}^{2}\\\
&+2\big{[}(1+\lambda_{\alpha}^{2})(1+\lambda_{\beta}^{2})+\lambda_{\alpha}\lambda_{\beta}(2+\lambda_{\alpha}^{2}+\lambda_{\beta}^{2})\big{]}\left[\frac{\sqrt{2}}{2}(\omega_{\alpha\beta}+\omega_{\beta\alpha})\right]^{2}\\\
&+2\big{[}(1+\lambda_{\alpha}^{2})(1+\lambda_{\beta}^{2})-\lambda_{\alpha}\lambda_{\beta}(2+\lambda_{\alpha}^{2}+\lambda_{\beta}^{2})\big{]}\left[\frac{\sqrt{2}}{2}(\omega_{\alpha\beta}-\omega_{\beta\alpha})\right]^{2}\end{aligned}$

By computing,

(3.18)
$2\big{[}(1+\lambda_{\alpha}^{2})(1+\lambda_{\beta}^{2})-\lambda_{\alpha}\lambda_{\beta}(2+\lambda_{\alpha}^{2}+\lambda_{\beta}^{2})\big{]}=2(1-\lambda_{\alpha}\lambda_{\beta})(\lambda_{\alpha}^{2}+\lambda_{\beta}^{2}-\lambda_{\alpha}\lambda_{\beta}+1).$

It is positive if and only if
$1-\lambda_{\alpha}\lambda_{\beta}=1-\tan\theta_{\alpha}\tan\theta_{\beta}=\frac{\cos(\theta_{\alpha}+\theta_{\beta})}{\cos\theta_{\alpha}\cos\theta_{\beta}}\geq
0$, i.e., $\theta_{\alpha}+\theta_{\beta}<\frac{\pi}{2}$. Hence
$\mbox{Hess}(u)_{P}$ is positive definite if and only $P\in B_{JX}(P_{0})$.

Moreover, the right side of (3.18) can be estimated by

$\displaystyle\begin{aligned}
2(1-\lambda_{\alpha}\lambda_{\beta})(\lambda_{\alpha}^{2}+\lambda_{\beta}^{2}-\lambda_{\alpha}\lambda_{\beta}+1)&\geq
2\big{(}1-\frac{\lambda_{\alpha}^{2}+\lambda_{\beta}^{2}}{2}\big{)}\big{(}\frac{\lambda_{\alpha}^{2}+\lambda_{\beta}^{2}}{2}+1\big{)}\\\
&=2\big{(}1-\frac{(\lambda_{\alpha}^{2}+\lambda_{\beta}^{2})^{2}}{4}\big{)}\\\
&\geq 2(1-\frac{1}{4}u^{2})=2-\frac{1}{2}u^{2}.\end{aligned}$

(Here we used the fact
$u=\sum_{\alpha}\tan^{2}\theta_{\alpha}=\sum_{\alpha}\lambda_{\alpha}^{2}$.)
By combining it with (3.17) and (2.9), we arrive that

(3.19) $\mbox{Hess}(u)\geq\left(2-\frac{1}{2}u^{2}\right)g.$

For later applications the estimates (3.12) and (3.19) are not accurate
enough. Using the radial compensation technique we could refine those
estimates which are based on the following lemmas.

######  Lemma 3.2.

Let $V$ be a real linear space, $h$ be a nonnegative definite quadratic form
on $V$ and $\omega\in V^{*}$. $V=V_{1}\oplus V_{2}$, $h$ is positive definite
on $V_{1}$, $h(V_{1},V_{2})=0$ and $\omega(V_{2})=0$. Denote by $\omega^{*}$
the unique vector in $V_{1}$ such that for any $z\in V_{1}$,

$\omega(z)=h(\omega^{*},z).$

Then we have

(3.20) $h\geq\omega(\omega^{*})^{-1}\omega\otimes\omega.$

###### Proof.

For arbitrary $y\in V$, there exist $\lambda\in\mathbb{R}$, $z_{1}\in V_{1}$
and $z_{2}\in V_{2}$, such that $y=\lambda\omega^{*}+z_{1}+z_{2}$ and
$h(\omega^{*},z_{1})=0$. Then

$h(y,y)=\lambda^{2}h(\omega^{*},\omega^{*})+h(z_{1},z_{1})+h(z_{2},z_{2})\geq\lambda^{2}h(\omega^{*},\omega^{*})=\lambda^{2}\omega(\omega^{*})$

and

$\omega(\omega^{*})^{-1}\omega\otimes\omega(y,y)=\omega(\omega^{*})^{-1}\omega(y)^{2}=\lambda^{2}\omega(\omega^{*}).$

Hence (3.20) holds.

∎

######  Lemma 3.3.

Let $\Omega$ be a compact and convex subset of $\mathbb{R}^{k}$, such that for
every $\sigma\in\mathscr{S}(k)$ and $x=(x^{1},\cdots,x^{k})\in\Omega$,

(3.21) $T_{\sigma}(x)=(x^{\sigma(1)},\cdots,x^{\sigma(k)})\in\Omega;$

where $\mathscr{S}(k)$ denotes the permutation group of $\\{1,\cdots,k\\}$. If
$f:\Omega\rightarrow{\tenmsb R}$ is a symmetric $C^{2}$ function, and
$(D^{2}f)$ is nonpositive definite everywhere in $\Omega$, then there exists
$x_{0}=(x_{0}^{1},\cdots,x_{0}^{k})\in\Omega$, such that
$x_{0}^{1}=x_{0}^{2}=\cdots=x_{0}^{k}$ and

(3.22) $f(x_{0})=\sup_{\Omega}f.$

###### Proof.

By the compactness of $\Omega$, there exists
$x=(x^{1},\cdots,x^{k})\in\Omega$, such that $f(x)=\sup_{\Omega}f$.
Furthermore we have

$f\big{(}T_{\sigma}(x)\big{)}=f(x)=\sup_{\Omega}f\qquad\sigma\in\mathscr{S}(k)$

from the fact that $f$ is symmetric. Denote by $C_{\sigma}(x)$ the convex
closure of $\\{T_{\sigma}(x):\sigma\in\mathscr{S}(k)\\}$, then
$C_{\sigma}(x)\subset\Omega$ and
$f(y)\geq\sup_{\sigma\in\mathscr{S}(k)}f\big{(}T_{\sigma}(x)\big{)}=\sup_{\Omega}f$
for arbitrary $y\in C_{\sigma}(x)$, since $(D^{2}f)\leq 0$; which implies

$f\big{|}_{C_{\sigma}(x)}\equiv\sup_{\Omega}f.$

Denote $x_{0}^{1}=\cdots=x_{0}^{k}=\frac{1}{k}\sum_{i=1}^{k}x^{i}$, then

$x_{0}=(x_{0}^{1},\cdots,x_{0}^{k})=\frac{1}{k}\sum_{s=1}^{k}(x^{s},x^{s+1},\cdots,x^{k},x^{1},x^{2},\cdots,x^{s-1})\in
C_{\sigma}(x);$

From which (3.22) follows.

∎

By (3.12),

(3.23) $h\mathop{=}\limits^{def.}\mbox{Hess}(v)-v(2-v)g-v^{-1}dv\otimes dv$

is nonnegative definite on $T_{P}\mathbb{G}_{n,m}$. Denote

(3.24) $V_{1}=\bigoplus_{\alpha}E_{\alpha\alpha},\qquad
V_{2}=\bigoplus_{i\neq\alpha}E_{i\alpha};$

then $T_{P}\mathbb{G}_{n,m}=V_{1}\oplus V_{2}$, and (3.11), (2.9), (3.9) tell
us

$h(V_{1},V_{2})=0,\ dv(V_{2})=0$

and

(3.25) $h|_{V_{1}}=\sum_{\alpha}(v-1+\lambda_{\alpha}^{2})v\
\omega_{\alpha\alpha}^{2}.$

is positive definite. Denote by $\tilde{\nabla}v$ the unique element in
$V_{1}$ such that for any $X\in V_{1}$,

$h(\tilde{\nabla}v,X)=dv(X).$

From (3.25) and (3.9), it is not difficult to obtain

$\tilde{\nabla}v=\sum_{\alpha}\frac{\lambda_{\alpha}(1+\lambda_{\alpha}^{2})}{v-1+\lambda_{\alpha}^{2}}E_{\alpha\alpha}$

and

(3.26)
$dv(\tilde{\nabla}v)=\sum_{\alpha}\frac{\lambda_{\alpha}^{2}}{v-1+\lambda_{\alpha}^{2}}v.$

Then Lemma 3.2 and (3.23) tell us

(3.27) $\mbox{Hess}(v)\geq
v(2-v)g+\Big{[}1+\big{(}\sum_{\alpha}\frac{\lambda_{\alpha}^{2}}{v-1+\lambda_{\alpha}^{2}}\big{)}^{-1}\Big{]}v^{-1}dv\otimes
dv.$

It is necessary to estimate the upper bound of
$\sum_{\alpha}\frac{\lambda_{\alpha}^{2}}{v-1+\lambda_{\alpha}^{2}}$. Denote

(3.28) $\nu_{\alpha}=\log(1+\lambda_{\alpha}^{2}),$

then $\lambda_{\alpha}^{2}=-1+e^{\nu_{\alpha}}$; since
$v=\prod_{\alpha}(1+\lambda_{\alpha}^{2})^{\frac{1}{2}}$,

$\log
v=\frac{1}{2}\sum_{\alpha}\log(1+\lambda_{\alpha}^{2})=\frac{1}{2}\sum_{\alpha}\nu_{\alpha}$

and

$\sum_{\alpha}\frac{\lambda_{\alpha}^{2}}{v-1+\lambda_{\alpha}^{2}}=\sum_{\alpha}\frac{-1+e^{\nu_{\alpha}}}{v-2+e^{\nu_{\alpha}}}.$

Now we define

(3.29) $\Omega=\big{\\{}(\nu_{1},\cdots,\nu_{m})\in{\tenmsb
R}^{m}:\nu_{\alpha}\geq 0,\sum_{\alpha}\nu_{\alpha}=2\log v\big{\\}},$

and $f:\Omega\rightarrow{\tenmsb R}$ by

$(\nu_{1},\cdots,\nu_{m})\mapsto\sum_{\alpha}\frac{-1+e^{\nu_{\alpha}}}{v-2+e^{\nu_{\alpha}}}.$

Then obviously $\Omega$ is compact and convex, $T_{\sigma}(\Omega)=\Omega$ for
every $\sigma\in\mathscr{S}(m)$ (cf. Lemma 3.3), $f$ is a symmetric function
and a direct calculation shows

$\frac{\partial^{2}f}{\partial\nu_{\alpha}\partial\nu_{\beta}}=\frac{(v-1)e^{\nu_{\alpha}}(v-2-e^{\nu_{\alpha}})}{(v-2+e^{\nu_{\alpha}})^{3}}\delta_{\alpha\beta};$

i.e.,

$(D^{2}f)\leq 0\qquad\mbox{when }v\in(1,2].$

Then from Lemma 3.3,

$\sup_{\Omega}f=f\big{(}\frac{2\log v}{m},\cdots,\frac{2\log
v}{m}\big{)}=\frac{m(-1+v^{\frac{2}{m}})}{v-2+v^{\frac{2}{m}}};$

which is an upper bound of
$\sum_{\alpha}\frac{\lambda_{\alpha}^{2}}{v-1+\lambda_{\alpha}^{2}}$.
Substituting it into (3.27) gives

$\mbox{Hess}(v)\geq
v(2-v)g+\Big{(}\frac{v-1}{mv(v^{\frac{2}{m}}-1)}+\frac{m+1}{mv}\Big{)}dv\otimes
dv.$

In summary, we have the following Proposition.

######  Proposition 3.1.

$v$ is a convex function on $B_{JX}(P_{0})\subset{\tenmsb
U}\subset\mathbb{G}_{n,m}$, and

(3.30) $\mbox{Hess}(v)\geq
v(2-v)g+\Big{(}\frac{v-1}{pv(v^{\frac{2}{p}}-1)}+\frac{p+1}{pv}\Big{)}dv\otimes
dv$

on $\\{P\in{\tenmsb U}:v(P)\leq 2\\}$, where $p=min(n,m)$.

Similarly, we consider

(3.31)
$\tilde{h}\mathop{=}\limits^{def.}\mbox{Hess}(u)-(2-\frac{1}{2}u^{2})g;$

which is nonnegative definite on $T_{P}\mathbb{G}_{n,m}$. The definition of
$V_{1}$ and $V_{2}$ is similar to above. It is easily seen from (3.17) and
(2.9) that

$\tilde{h}(V_{1},V_{2})=0$

and

(3.32)
$\tilde{h}|_{V_{1}}=\sum_{\alpha}(8\lambda_{\alpha}^{2}+6\lambda_{\alpha}^{4}+\frac{1}{2}u^{2})\omega_{\alpha\alpha}^{2}$

is positive definite. By (3.15),

(3.33)
$du=\sum_{\alpha}2\lambda_{\alpha}(1+\lambda_{\alpha}^{2})\omega_{\alpha\alpha},$

then

$du(V_{2})=0.$

Hence Lemma 3.2 can be applied for us to obtain

(3.34)
$\mbox{Hess}(u)\geq(2-\frac{1}{2}u^{2})g+\big{(}du(\tilde{\nabla}u)\big{)}^{-1}du\otimes
du.$

where $\tilde{\nabla}u$ denotes the unique element in $V_{1}$ such that for
arbitrary $X\in V_{1}$,

$\tilde{h}(\tilde{\nabla}u,X)=du(X).$

From (3.32) and (3.33), we can derive

$\tilde{\nabla}u=\sum_{\alpha}\frac{2\lambda_{\alpha}(1+\lambda_{\alpha}^{2})^{2}}{8\lambda_{\alpha}^{2}+6\lambda_{\alpha}^{4}+\frac{1}{2}u^{2}}E_{\alpha\alpha},$

and hence

(3.35)
$du(\tilde{\nabla}u)=\sum_{\alpha}\frac{2\lambda_{\alpha}^{2}(1+\lambda_{\alpha}^{2})^{2}}{3\lambda_{\alpha}^{4}+4\lambda_{\alpha}^{2}+\frac{1}{4}u^{2}}.$

(3.34) tells us it is necessary for us to estimate the upper bound of the
right side of (3.35).

Define $\Omega=\\{(\nu_{1},\cdots,\nu_{m})\in{\tenmsb
R}^{m}:\sum_{\alpha}\nu_{\alpha}=u\\}$ and $f:\Omega\rightarrow{\tenmsb R}$

$(\nu_{1},\cdots,\nu_{m})\mapsto\sum_{\alpha}\frac{2\nu_{\alpha}(1+\nu_{\alpha})^{2}}{3\nu_{\alpha}^{2}+4\nu_{\alpha}+C}\qquad\mbox{
where }C=\frac{1}{4}u^{2}.$

Then it is easy to see that $\sup f$ is an upper bound of
$du(\tilde{\nabla}u)$, since
$u=\sum_{\alpha}\tan^{2}\theta_{\alpha}=\sum_{\alpha}\lambda_{\alpha}^{2}.$

Obviously $\Omega$ is compact and convex, $T_{\sigma}(\Omega)=\Omega$ for
every $\sigma\in\mathscr{S}(m)$, $f$ is a symmetric function and a direct
calculation shows

$\frac{\partial^{2}f}{\partial\nu_{\alpha}\partial\nu_{\beta}}=\frac{-4\big{[}(3C-1)\nu_{\alpha}^{3}+6C\nu_{\alpha}^{2}+(9C-3C^{2})\nu_{\alpha}+4C-2C^{2}\big{]}}{(3\nu_{\alpha}^{2}+4\nu_{\alpha}+C)^{3}}\delta_{\alpha\beta}.$

To show $(D^{2}f)\leq 0$ when $u\in(0,2]$, it is sufficient to prove
$F:[0,u]\rightarrow{\tenmsb R}$

$t\mapsto(3C-1)t^{3}+6Ct^{2}+(9C-3C^{2})t+4C-2C^{2}$

is a nonnegative function, where $C=\frac{u^{2}}{4}\in(0,1]$. If $F$ attains
its minimum at $t_{0}\in(0,u)$, then

(3.36) $\displaystyle
0=F^{\prime}(t_{0})=3(3C-1)t_{0}^{2}+12Ct_{0}+9C-3C^{2},$ (3.37)
$\displaystyle 0\leq F^{\prime\prime}(t_{0})=6(3C-1)t_{0}+12C.$

On the other hand, when $3C-1\geq 0$, we have $F^{\prime}(t_{0})\geq
9C-3C^{2}>0$, which causes a contradiction; when $3C-1<0$, from (3.37),
$t_{0}\leq\frac{2C}{1-3C}$, then $F^{\prime}(t_{0})\geq
F^{\prime}(0)=9C-3C^{2}>0$, which also causes a contradiction. Therefore

$\min_{[0,u]}F=\min\big{\\{}F(0),F(u)\big{\\}}.$

In conjunction with

$\displaystyle F(0)$ $\displaystyle=$ $\displaystyle 4C-2C^{2}>0$
$\displaystyle F(u)$ $\displaystyle=$
$\displaystyle(3C-1)u^{3}+6Cu^{2}+(9C-3C^{2})u+4C-2C^{2}$ $\displaystyle=$
$\displaystyle\frac{9}{16}u^{5}+\frac{11}{8}u^{4}+\frac{5}{4}u^{3}+u^{2}>0,$

$F$ is a nonnegative function. Thereby applying Lemma 3.3 we have

(3.38) $du(\tilde{\nabla}u)\leq\sup
f=f(\frac{u}{m},\cdots,\frac{u}{m})=\frac{2(u+m)^{2}}{(3+\frac{1}{4}m^{2})u+4m}.$

Substituting (3.38) into (3.34) gives

$\mbox{Hess}(u)\geq\left(2-\frac{1}{2}u^{2}\right)g+\frac{(3+\frac{1}{4}m^{2})u+4m}{2(u+m)^{2}}du\otimes
du.$

We rewrite the conclusion as the following Proposition.

######  Proposition 3.2.

$u$ is a convex function on $B_{JX}(P_{0})\subset{\tenmsb
U}\subset\mathbb{G}_{n,m}$ and

(3.39)
$\mbox{Hess}(u)\geq\left(2-\frac{1}{2}u^{2}\right)g+\frac{(3+\frac{1}{4}p^{2})u+4p}{2(u+p)^{2}}du\otimes
du$

on $\\{P\in{\tenmsb U}:u(P)\leq 2\\}$, where $p=min(n,m)$.

##  4\. The construction of auxiliary functions

Let

(4.1) $h_{1}=v^{-k}(2-v)^{k},$

where $k>0$ to be chosen, then

$\displaystyle\begin{aligned}
h^{\prime}_{1}=&-kv^{-k-1}(2-v)^{k}-kv^{-k}(2-v)^{k-1}\\\
=&-2kv^{-k-1}(2-v)^{k-1},\\\
h^{\prime\prime}_{1}=&2k(k+1)v^{-k-2}(2-v)^{k-1}+2k(k-1)v^{-k-1}(2-v)^{k-2}\\\
=&4kv^{-k-2}(2-v)^{k-2}(k+1-v).\end{aligned}$

Here ′ denotes derivative with respect to $v$. Hence, from (3.30)

(4.2) $\displaystyle\begin{aligned}
&\mbox{Hess}(h_{1})=-2kv^{-k-1}(2-v)^{k-1}\mbox{Hess}(v)+4kv^{-k-2}(2-v)^{k-2}(k+1-v)dv\otimes
dv\\\ &\hskip 43.36243pt\leq-2kv^{-k}(2-v)^{k}g-\\\
&\quad-2kv^{-k-2}(2-v)^{k-2}\Big{[}\frac{(v-1)(2-v)}{p(v^{\frac{2}{p}}-1)}+\frac{p+1}{p}(2-v)-2(k+1-v)\Big{]}dv\otimes
dv.\end{aligned}$

Please note that $\frac{v-1}{v^{\frac{2}{p}}-1}$ is an increasing function on
$[1,2]$: it is easily seen when $p$ is even, since

$\frac{v-1}{v^{\frac{2}{p}}-1}=1+v^{\frac{2}{p}}+v^{\frac{4}{p}}+\cdots+v^{1-\frac{2}{p}};$

otherwise, when $p$ is odd,

$\frac{v-1}{v^{\frac{2}{p}}-1}=\frac{v^{1-\frac{1}{p}}-1}{v^{\frac{2}{p}}-1}+\frac{v-v^{1-\frac{1}{p}}}{v^{\frac{2}{p}}-1}=1+v^{\frac{2}{p}}+v^{\frac{4}{p}}+\cdots+v^{1-\frac{3}{p}}+\frac{v^{1-\frac{1}{p}}}{v^{\frac{1}{p}}+1}$

it follows from

$\Big{(}\frac{v^{1-\frac{1}{p}}}{v^{\frac{1}{p}}+1}\Big{)}^{\prime}=\frac{1-\frac{2}{p}+(1-\frac{1}{p})v^{-\frac{1}{p}}}{(v^{\frac{1}{p}}+1)^{2}}\geq
0.$

Hence

$\frac{v-1}{v^{\frac{2}{p}}-1}\geq\frac{p}{2},$

and moreover

$\displaystyle\begin{aligned}
\frac{(v-1)(2-v)}{p(v^{\frac{2}{p}}-1)}&+\frac{p+1}{p}(2-v)-2(k+1-v)\\\
\geq&\big{(}\frac{1}{2}+\frac{p+1}{p}\big{)}(2-v)-2(k+1-v)\\\
=&\big{(}\frac{1}{2}-\frac{1}{p}\big{)}v+\big{(}3+\frac{2}{p}\big{)}-2(k+1)\\\
\geq&\frac{3}{2}+\frac{1}{p}-2k.\end{aligned}$

Now we take

(4.3) $k=\frac{3}{4}+\frac{1}{2p},$

then $\frac{(v-1)(2-v)}{p(v^{\frac{2}{p}}-1)}+\frac{p+1}{p}(2-v)-2(k+1-v)\geq
0$ and then (4.2) becomes

(4.4) $\mbox{Hess}(h_{1})\leq-2kh_{1}\
g=-\left(\frac{3}{2}+\frac{1}{p}\right)h_{1}\ g.$

Denote

(4.5) $h_{2}=h_{1}^{-\frac{6p}{3p+2}}=v^{\frac{3}{2}}(2-v)^{-\frac{3}{2}},$

then

(4.6) $\displaystyle\begin{aligned}
\mbox{Hess}(h_{2})=&-\frac{6p}{3p+2}h_{1}^{-\frac{6p}{3p+2}-1}\mbox{Hess}(h_{1})+\frac{6p}{3p+2}\big{(}\frac{6p}{3p+2}+1\big{)}h_{1}^{-\frac{6p}{3p+2}-2}dh_{1}\otimes
dh_{1}\\\ \geq&3h_{1}^{-\frac{6p}{3p+2}}\
g+\big{(}\frac{3}{2}+\frac{1}{3p}\big{)}h_{1}^{\frac{6p}{3p+2}}dh_{2}\otimes
dh_{2}\\\ =&3h_{2}\
g+\big{(}\frac{3}{2}+\frac{1}{3p}\big{)}h_{2}^{-1}dh_{2}\otimes
dh_{2}.\end{aligned}$

Let

(4.7) $h_{3}=(u+\alpha)^{-1}(2-u),$

where $\alpha>0$ to be chosen. A direct calculation shows

$\displaystyle\begin{aligned}
h^{\prime}_{3}=&-(u+\alpha)^{-2}(2-u)-(u+\alpha)^{-1}\\\
=&-(2+\alpha)(u+\alpha)^{-2},\\\
h^{\prime\prime}_{3}=&2(2+\alpha)(u+\alpha)^{-3}.\end{aligned}$

Here ′ denotes derivative with respect to $u$. Combining with (3.39), we have

(4.8) $\displaystyle\begin{aligned}
\mbox{Hess}(h_{3})=&-(2+\alpha)(u+\alpha)^{-2}\mbox{Hess}(u)+2(2+\alpha)(u+\alpha)^{-3}du\otimes
du\\\ \leq&-\frac{(2+\alpha)(u+2)}{2(u+\alpha)}h_{3}\ g\\\
&-(2+\alpha)(u+\alpha)^{-3}\Big{[}\frac{(u+\alpha)\big{(}(3+\frac{1}{4}p^{2})u+4p\big{)}}{2(u+p)^{2}}-2\Big{]}du\otimes
du.\end{aligned}$

Choose

(4.9) $\alpha=p,$

then

$\displaystyle\frac{(u+\alpha)\big{(}(3+\frac{1}{4}p^{2})u+4p\big{)}}{2(u+p)^{2}}-2=\frac{\big{(}3+\frac{1}{4}p^{2}\big{)}u+4p}{2(u+p)}-2\geq
2-2\geq 0,$

and

$\frac{(2+\alpha)(u+2)}{2(u+\alpha)}\geq\frac{2+p}{p}=1+\frac{2}{p}.$

Thereby (4.8) becomes

(4.10) $\mbox{Hess}(h_{3})\leq-\big{(}1+\frac{2}{p}\big{)}h_{3}\ g.$

Denote

(4.11)
$h_{4}=h_{3}^{-\frac{3p}{p+2}}=(u+p)^{\frac{3p}{p+2}}(2-u)^{-\frac{3p}{p+2}},$

then

(4.12) $\displaystyle\begin{aligned}
\mbox{Hess}(h_{4})=&-\frac{3p}{p+2}h_{3}^{-\frac{3p}{p+2}-1}\mbox{Hess}(h_{3})+\frac{3p}{p+2}\big{(}\frac{3p}{p+2}+1\big{)}h_{3}^{-\frac{3p}{p+2}-2}dh_{3}\otimes
dh_{3}\\\ \geq&3h_{3}^{-\frac{3p}{p+2}}\
g+\big{(}\frac{4}{3}+\frac{2}{3p}\big{)}h_{3}^{\frac{3p}{p+2}}dh_{4}\otimes
dh_{4}\\\ =&3h_{4}\
g+\big{(}\frac{4}{3}+\frac{2}{3p}\big{)}h_{4}^{-1}dh_{4}\otimes
dh_{4}.\end{aligned}$

Let $M$ be an $n$-dimensional submanifold in ${\tenmsb R}^{n+m}$ with $m\geq
2.$ The Gauss map $\gamma:M\to\mathbb{G}_{n,m}$ is defined by

$\gamma(x)=T_{x}M\in\mathbb{G}_{n,m}$

via the parallel translation in $\mathbb{R}^{m+n}$ for arbitrary $x\in M$. The
energy density of the Gauss map (see [21] Chap.3, §3.1) is

$e(\gamma)=\frac{1}{2}\left<\gamma_{*}e_{i},\gamma_{*}e_{i}\right>=\frac{1}{2}|B|^{2}.$

Ruh-Vilms proved that the mean curvature vector of $M$ is parallel if and only
if its Gauss map is a harmonic map [17].

If the Gauss image of $M$ is contained in $\\{P\in{\tenmsb
U}\subset\mathbb{G}_{n,m}:v(P)<2\\}$, then the composition function
$\tilde{h}_{1}=h_{1}\circ\gamma$ of $h_{1}$ with the Gauss map $\gamma$
defines a function on $M$. Using composition formula, we have

(4.13) $\displaystyle\Delta\tilde{h}_{1}$
$\displaystyle=\mbox{Hess}(h_{1})(\gamma_{*}e_{i},\gamma_{*}e_{i})+dh_{1}(\tau(\gamma))$
$\displaystyle\leq-\big{(}\frac{3}{2}+\frac{1}{p}\big{)}|B|^{2}\tilde{h}_{1},$

where $\tau(\gamma)$ is the tension field of the Gauss map, which is zero,
provided $M$ has parallel mean curvature by the Ruh-Vilms theorem mentioned
above. Similarly, for $\tilde{h}_{2}=h_{2}\circ\gamma$ defined on $M$, we have

(4.14) $\displaystyle\Delta\tilde{h}_{2}$
$\displaystyle=\mbox{Hess}(h_{2})(\gamma_{*}e_{i},\gamma_{*}e_{i})+dh_{2}(\tau(\gamma))$
$\displaystyle\geq 3\
\tilde{h}_{2}|B|^{2}+\big{(}\frac{3}{2}+\frac{1}{3p}\big{)}\tilde{h}_{2}^{-1}|\nabla\tilde{h}_{2}|^{2}.$

If the Gauss image of $M$ is contained in $\\{P\in{\tenmsb
U}\subset\mathbb{G}_{n,m}:u(P)<2\\}$, we can defined composition function
$\tilde{h}_{3}=h_{3}\circ\gamma$ and $\tilde{h}_{4}=h_{4}\circ\gamma$ on $M$.
Again using composition formula, we obtain

(4.15)
$\Delta\tilde{h}_{3}\leq-\Big{(}1+\frac{2}{p}\Big{)}|B|^{2}\tilde{h}_{3}$

and

(4.16) $\Delta\tilde{h}_{4}\geq 3\
\tilde{h}_{4}|B|^{2}+\Big{(}\frac{4}{3}+\frac{2}{3p}\Big{)}\tilde{h}_{4}^{-1}|\nabla\tilde{h}_{4}|^{2}.$

With the aid of $\tilde{h}_{1}$ and $\tilde{h}_{3}$, we immediately have the
following lemma.

######  Lemma 4.1.

Let $M$ be an $n$-dimensional minimal submanifold of $\mathbb{R}^{n+m}$ ($M$
needs not be complete), if the Gauss image of $M$ is contained in
$\\{P\in{\tenmsb U}\subset\mathbb{G}_{n,m}:v(P)<2\\}$ (or respectively,
$\\{P\in{\tenmsb U}\subset\mathbb{G}_{n,m}:u(P)<2\\}$), then we have

(4.17) $\displaystyle\int_{M}|\nabla\phi|^{2}$
$\displaystyle*1\geq\big{(}\frac{3}{2}+\frac{1}{p}\big{)}\int_{M}|B|^{2}\phi^{2}*1$
$\displaystyle\Big{(}\mbox{or respectively, }$
$\displaystyle\int_{M}|\nabla\phi|^{2}*1\geq\big{(}1+\frac{2}{p}\big{)}\int_{M}|B|^{2}\phi^{2}*1\Big{)}$

for any function $\phi$ with compact support $D\subset M$.

######  Remark 4.1.

For a stable minimal hypersurface there is the stability inequality, which is
one of main ingredient for Schoen-Simon-Yau’s curvature esimates for stable
minimal hypersurfaces. For minimal submanifolds with the Gauss image
restriction we have stronger inequality as shown in (4.17). Our proof is
similar to [23] and [24], so we omit the detail of it.

##  5\. Curvature estimates

We are now in a position to carry out the curvature estimates of Schoen-Simon-
Yau type.

Let $M$ be an $n$-dimensional minimal submanifold in $\mathbb{R}^{n+m}$.
Assume that the estimate

(5.1) $\int_{M}|\nabla\phi|^{2}*1\geq\lambda\int_{M}|B|^{2}\phi^{2}*1$

holds for arbitrary function $\phi$ with compact support $D\subset M$, where
$\lambda$ is a positive constant.

Replacing $\phi$ by $|B|^{1+q}\phi$ in (5.1) gives

(5.2) $\displaystyle\int_{M}|B|^{4+2q}$
$\displaystyle\phi^{2}*1\leq\lambda^{-1}\int_{M}\big{|}\nabla(|B|^{1+q}\phi)\big{|}^{2}*1$
$\displaystyle=\lambda^{-1}(1+q)^{2}\int_{M}|B|^{2q}\big{|}\nabla|B|\big{|}^{2}\phi^{2}*1+\lambda^{-1}\int_{M}|B|^{2+2q}|\nabla\phi|^{2}*1$
$\displaystyle+2\lambda^{-1}(1+q)\int_{M}|B|^{1+2q}\nabla|B|\cdot\phi\nabla\phi*1.$

Using Bochner technique, the estimate done in [14][7], and the Kato-type
inequality derived in [24], we obtain

(5.3) $\Delta|B|^{2}\geq
2\big{(}1+\frac{2}{mn}\big{)}\big{|}\nabla|B|\big{|}^{2}-3|B|^{4}.$

(For the detail, see [24] Section 2.) It is equivalent to

(5.4)
$\frac{2}{mn}\big{|}\nabla|B|\big{|}^{2}\leq|B|\Delta|B|+\frac{3}{2}|B|^{4}.$

Multiplying $|B|^{2q}\phi^{2}$ with both sides of (5.4) and integrating by
parts, we have

(5.5) $\displaystyle\frac{2}{mn}$
$\displaystyle\int_{M}|B|^{2q}\big{|}\nabla|B|\big{|}^{2}\phi^{2}*1$
$\displaystyle\qquad\leq-(1+2q)\int_{M}|B|^{2q}\big{|}\nabla|B|\big{|}^{2}\phi^{2}*1$
$\displaystyle\qquad\qquad-2\int_{M}|B|^{1+2q}\nabla|B|\cdot\phi\nabla\phi*1+\frac{3}{2}\int_{M}|B|^{4+2q}\phi^{2}*1.$

By multiplying $\frac{3}{2}$ with both sides of (5.2) and then adding up both
sides of it and (5.5), we have

(5.6)
$\displaystyle\big{(}\frac{2}{mn}+1+2q-\frac{3}{2}\lambda^{-1}(1+q)^{2}\big{)}\int_{M}|B|^{2q}\big{|}\nabla|B|\big{|}^{2}\phi^{2}*1$
$\displaystyle\qquad\leq\frac{3}{2}\lambda^{-1}\int_{M}|B|^{2+2q}|\nabla\phi|^{2}*1+\big{(}3\lambda^{-1}(1+q)-2\big{)}\int_{M}|B|^{1+2q}\nabla|B|\cdot\phi\nabla\phi*1.$

By using Young’s inequality, (5.6) becomes

(5.7)
$\displaystyle\big{(}\frac{2}{mn}+1+2q-\frac{3}{2}\lambda^{-1}(1+q)^{2}-\varepsilon\big{)}\int_{M}|B|^{2q}\big{|}\nabla|B|\big{|}^{2}\phi^{2}*1$
$\displaystyle\hskip 72.26999pt\leq
C_{1}(\varepsilon,\lambda,q)\int_{M}|B|^{2+2q}|\nabla\phi|^{2}*1.$

If

(5.8) $\lambda>\frac{3}{2}\big{(}1-\frac{2}{mn}\big{)},$

then

$\frac{2}{mn}+1+2q-\frac{3}{2}\lambda^{-1}(1+q)^{2}>0$

whenever

(5.9)
$q\in\Big{[}0,-1+\frac{2}{3}\lambda+\frac{1}{3}\sqrt{4\lambda^{2}-6\big{(}1-\frac{2}{mn}\big{)}\lambda}\
\Big{)}.$

Thus we can choose $\varepsilon$ sufficiently small, such that

(5.10) $\int_{M}|B|^{2q}\big{|}\nabla|B|\big{|}^{2}\phi^{2}*1\leq
C_{2}\int_{M}|B|^{2+2q}|\nabla\phi|^{2}*1$

where $C_{2}$ only depends on $n$, $m$, $\lambda$ and $q$.

Combining with (5.2) and (5.10), we can derive

(5.11) $\int_{M}|B|^{4+2q}\phi^{2}*1\leq
C_{3}(n,m,\lambda,q)\int_{M}|B|^{2+2q}|\nabla\phi|^{2}*1$

by again using Young’s inequality.

By replacing $\phi$ by $\phi^{2+q}$ in (5.11) and then using Hölder
inequality, we have

(5.12) $\displaystyle\int_{M}|B|^{4+2q}\phi^{4+2q}*1\leq
C\int_{M}|\nabla\phi|^{4+2q}*1.$

where $C$ is a constant only depending on $n$, $m$, $\lambda$ and $q$.

Similarly, replacing $\phi$ by $\phi^{1+q}$ in (5.11) and then again using
Hölder inequality yields

(5.13) $\displaystyle\int_{M}|B|^{4+2q}\phi^{2+2q}*1\leq
C^{\prime}\int_{M}|B|^{2}|\nabla\phi|^{2+2q}*1.$

where $C^{\prime}$ is a constant only depending on $n$, $m$, $\lambda$ and
$q$.

Let $r$ be a function on $M$ with $|\nabla r|\leq 1$. For any $R\in[0,R_{0}]$,
where $R_{0}=\sup_{M}r$, suppose

$M_{R}=\\{x\in M,\quad r\leq R\\}$

is compact.

(5.12) and Lemma 4.1 enable us to prove the following results by taking
$\phi\in C_{c}^{\infty}(M_{R})$ to be the standard cut-off function such that
$\phi\equiv 1$ in $M_{\theta R}$ and $|\nabla\phi|\leq
C(1-\theta)^{-1}R^{-1}$.

######  Theorem 5.1.

Let $M$ be an $n$-dimensional minimal submanifolds of $\mathbb{R}^{n+m}$. If
the Gauss image of $M_{R}$ is contained in $\\{P\in{\tenmsb
U}\subset\mathbb{G}_{n,m}:v(P)<2\\}$, then we have the estimate

(5.14) $\big{\|}|B|\big{\|}_{L^{s}(M_{\theta R})}\leq
C(n,m,s)(1-\theta)^{-1}R^{-1}\text{Vol}(M_{R})^{\frac{1}{s}}$

for arbitrary $\theta\in(0,1)$ and

$s\in\left[4,4+\frac{4}{3p}+\frac{2}{3}\sqrt{\big{(}3+\frac{2}{p}\big{)}\big{(}\frac{6}{mn}+\frac{2}{p}\big{)}}\
\right).$

If $p\leq 4$, and the Gauss image of $M_{R}$ is contained in $\\{P\in{\tenmsb
U}\subset\mathbb{G}_{n,m}:u(P)<2\\}$, then (5.14) still holds for arbitrary
$\theta\in(0,1)$ and

$s\in\left[4,2+\frac{4}{3}+\frac{8}{3p}+\frac{2}{3}\sqrt{\big{(}1+\frac{2}{p}\big{)}\big{(}\frac{12}{mn}+\frac{8}{p}-2\big{)}}\
\right).$

We can also fulfil the curvature estimates of Ecker-Huisken type.

Assume that $h$ is a positive function on $M$ satisfying the following
estimate

(5.15) $\Delta h\geq 3h\ g+c_{0}h^{-1}dh\otimes dh,$

where

$c_{0}>\frac{3}{2}-\frac{1}{mn}$

is a positive constant.

We compute from (5.15) and (5.3):

$\displaystyle\begin{aligned} &\Delta\big{(}|B|^{2s}h^{q}\big{)}\\\
\geq&3(q-s)|B|^{2s+2}h^{q}+2s(2s-1+\frac{2}{mn})|B|^{2s-2}\big{|}\nabla|B|\big{|}^{2}h^{q}+q(q+c_{0}-1)|B|^{2s}h^{q-2}|\nabla
h|^{2}\\\ &+4sq|B|^{2s-1}\nabla|B|\cdot h^{q-1}\nabla h.\end{aligned}$

By Young’s inequality, when $2s(2s-1+\frac{2}{mn})\cdot
q(q+c_{0}-1)\geq(2sq)^{2}$, i.e.,

(5.16) $q\geq
s\geq\frac{1}{2}-\frac{1}{mn}+\frac{1}{c_{0}-1}\big{(}\frac{1}{2}-\frac{1}{mn}\big{)}q,$

the inequality

(5.17) $\Delta\big{(}|B|^{2s}h^{q}\big{)}\geq 3(q-s)|B|^{2s+2}h^{q}$

holds. Especially,

(5.18)
$\Delta\big{(}|B|^{s-1}h^{\frac{s}{2}}\big{)}\geq\frac{3}{2}|B|^{s+1}h^{\frac{s}{2}}$

whenever

(5.19)
$s\geq\frac{2-\frac{2}{mn}}{1-\frac{1}{c_{0}-1}(\frac{1}{2}-\frac{1}{mn})}.$

Let $\eta$ be a smooth function with compact support. Integrating by parts in
conjunction with Young’s inequality lead to

(5.20) $\displaystyle\begin{aligned}
\int_{M}|B|^{2s}h^{s}\eta^{2s}*1\leq&\frac{2}{3}\int_{M}|B|^{s-1}h^{\frac{s}{2}}\eta^{2s}\Delta\big{(}|B|^{s-1}h^{\frac{s}{2}}\big{)}*1\\\
=&-\frac{2}{3}\int_{M}\Big{|}\nabla\big{(}|B|^{s-1}h^{\frac{s}{2}}\big{)}\Big{|}^{2}\eta^{2s}*1\\\
&-\frac{2}{3}\int_{M}|B|^{s-1}h^{\frac{s}{2}}\cdot
2s\eta^{2s-1}\nabla\eta\cdot\nabla\big{(}|B|^{s-1}h^{\frac{s}{2}}\big{)}*1\\\
\leq&\frac{2}{3}s^{2}\int_{M}|B|^{2s-2}h^{s}\eta^{2s-2}|\nabla\eta|^{2}*1.\end{aligned}$

By Hölder inequality,

(5.21)
$\int_{M}|B|^{2s-2}h^{s}\eta^{2s-2}|\nabla\eta|^{2}*1\leq\Big{(}\int_{M}|B|^{2s}h^{s}\eta^{2s}*1\Big{)}^{\frac{s-1}{s}}\Big{(}\int_{M}h^{s}|\nabla\eta|^{2s}*1\Big{)}^{\frac{1}{s}}.$

Substituting (5.21) into (5.20), we finally arrive at

(5.22)
$\Big{(}\int_{M}|B|^{2s}h^{s}\eta^{2s}*1\Big{)}^{\frac{1}{s}}\leq\frac{2}{3}s^{2}\Big{(}\int_{M}h^{s}|\nabla\eta|^{2s}*1\Big{)}^{\frac{1}{s}}.$

Take $\eta\in C_{c}^{\infty}(M_{R})$ to be the standard cut-off function such
that $\eta\equiv 1$ in $M_{\theta R}$ and $|\nabla\eta|\leq
C(1-\theta)^{-1}R^{-1}$; then from (5.22) we have the following estimate.

######  Theorem 5.2.

Let $M$ be an $n$-dimensional minimal submanifolds of $\mathbb{R}^{n+m}$. If
there exists a positive function $h$ on $M$ satisfying (5.15), then there
exists $C_{1}=C_{1}(n,m,c_{0}),$ such that

(5.23) $\big{\|}|B|^{2}h\big{\|}_{L^{s}(M_{\theta R})}\leq
C_{2}(s)(1-\theta)^{-2}R^{-2}\big{\|}h\big{\|}_{L^{s}(M_{R})}$

whenever $s\geq C_{1}$ and $\theta\in(0,1)$.

By (4.14) and (4.16), if the Gauss image of $M$ is contained in
$\\{P\in{\tenmsb U}\subset\mathbb{G}_{n,m}:v(P)<2\\}$, or $p\leq 4$ and the
Gauss image of $M$ is contained in $\\{P\in{\tenmsb
U}\subset\mathbb{G}_{n,m}:u(P)<2\\}$, there exists a positive function on $M$,
which is $\tilde{h}_{2}$ or respectively $\tilde{h}_{4}$, satisfying (5.15).
Hence the estimate (5.23) holds for both cases.

Furthermore, the mean value inequality for any subharmonic function on minimal
submanifolds in $\mathbb{R}^{m+n}$ (ref. [8], [16]) can be applied to yield an
estimate of the upper bound of $|B|^{2}$. We write the results as the
following theorem without detail of proof, for it is similar to [24]. Please
note that $B_{R}(x)\subset\mathbb{R}^{m+n}$ denotes a ball of radius $R$
centered at $x\in M$ and its restriction on $M$ is denoted by

$D_{R}(x)=B_{R}(x)\cap M.$

######  Theorem 5.3.

Let $x\in M$, $R>0$ such that the image of $D_{R}(x)$ under the Gauss map lies
in $\\{P\in{\tenmsb U}\subset\mathbb{G}_{n,m}:v(P)<2\\}$. Then, there exists
$C_{1}=C_{1}(n,m)$, such that

(5.24) $|B|^{2s}(x)\leq
C(n,s)R^{-(n+2s)}(\sup_{D_{R}(x)}\tilde{h}_{2})^{s}\text{Vol}(D_{R}(x)),$

for arbitrary $s\geq C_{1}$.

If $p\leq 4$, the image of $D_{R}(x)$ under the Gauss map lies in
$\\{P\in{\tenmsb U}\subset\mathbb{G}_{n,m}:u(P)<2\\}$, then there exists
$C_{2}=C_{2}(n,m)$ such that

(5.25) $|B|^{2s}(x)\leq
C(n,s)R^{-(n+2s)}(\sup_{D_{R}(x)}\tilde{h}_{4})^{s}\text{Vol}(D_{R}(x)),$

holds for any $s\geq C_{2}$.

##  6\. Bernstein type theorems and related results

If $M$ is a submanifold in $\mathbb{R}^{n+m}$, then the function $w$ defined
on $\mathbb{G}_{n,m}$ (see Section 2) and the Gauss map $\gamma$ could be
composed, yielding a smooth function on $M$, which is also denoted by $w$. By
studying the properties of $w$-function, we can obtain:

######  Proposition 6.1.

[24] Let $M$ be a complete submanifold in $\mathbb{R}^{n+m}$. If the
$w-$function is bounded below by a positive constant $w_{0}$. Then $M$ is an
entire graph with Euclidean volume growth. Precisely,

(6.1) $\text{Vol}(D_{R}(x))\leq\frac{1}{w_{0}}C(n)R^{n}.$

Now we let $M$ be a complete minimal submanifold in ${\tenmsb R}^{n+m}$ whose
Gauss image lies in $\\{P\in{\tenmsb U}\subset\mathbb{G}_{n,m}:v(P)<2\\}$.
Then $w=v^{-1}>\frac{1}{2}$ on $M$ and Proposition 6.1 tells us $M$ is an
entire graph. Precisely, the immersion $F:M\to\mathbb{R}^{m+n}$ is realized by
a graph $(x,f(x))$ with

$f:\mathbb{R}^{n}\to\mathbb{R}^{m}.$

At each point in $M$ its image $n$-plane $P$ under the Gauss map is spanned by

$f_{i}=\varepsilon_{i}+\frac{\partial f^{\alpha}}{\partial
x^{i}}\varepsilon_{\alpha}.$

Hence the local coordinate of $P$ in ${\tenmsb U}$ is

$Z=\Big{(}\frac{\partial f^{\alpha}}{\partial x^{i}}\Big{)}.$

By (3.2),

$v(P)=\left[\text{det}\left(\delta_{ij}+\sum_{\alpha}\frac{\partial
f^{\alpha}}{\partial x^{i}}\frac{\partial f^{\alpha}}{\partial
x^{j}}\right)\right]^{\frac{1}{2}}.$

Hence

(6.2) $\left[\text{det}\left(\delta_{ij}+\sum_{\alpha}\frac{\partial
f^{\alpha}}{\partial x^{i}}\frac{\partial f^{\alpha}}{\partial
x^{j}}\right)\right]^{\frac{1}{2}}<2$

at each $x\in{\tenmsb R}^{n}$. Conversely, if $M=(x,f(x))$ is a minimal graph
given by $f:{\tenmsb R}^{n}\rightarrow{\tenmsb R}^{m}$ which satisfy (6.2),
then the Gauss image of $M$ lies in $\\{P\in{\tenmsb
U}\subset\mathbb{G}_{n,m}:v(P)<2\\}$.

Let $P\in{\tenmsb U}$ such that $u(P)=\sum_{\alpha}\tan^{2}\theta_{\alpha}<2$,
then $\cos^{2}\theta_{\alpha}=(1+\tan^{2}\theta_{\alpha})^{-1}>\frac{1}{3}$
and

$w(P)=\prod_{\alpha}\cos\theta_{\alpha}>3^{-\frac{p}{2}}.$

Hence Proposition 6.1 could be applied when $M$ is a complete minimal
submanifold in ${\tenmsb R}^{n+m}$ whose Gauss image lies in $\\{P\in{\tenmsb
U}\subset\mathbb{G}_{n,m}:v(P)<2\\}$; which is hence a minimal graph given by
$f:{\tenmsb R}^{n}\rightarrow{\tenmsb R}^{m}$. Thereby (3.14) shows

(6.3) $\sum_{i,\alpha}\left(\frac{\partial f^{\alpha}}{\partial
x^{i}}\right)^{2}<2.$

And vice versa.

Theorem 5.1 and Proposition 6.1 give us the following Bernstein-type theorem.

######  Theorem 6.1.

Let $M=(x,f(x))$ be an $n$-dimensional entire minimal graph given by $m$
functions $f^{\alpha}(x^{1},\cdots,x^{n})$ with $m\geq 2,n\leq 4$. If

$\Delta_{f}=\left[\text{det}\left(\delta_{ij}+\sum_{\alpha}\frac{\partial
f^{\alpha}}{\partial x^{i}}\frac{\partial f^{\alpha}}{\partial
x^{j}}\right)\right]^{\frac{1}{2}}<2$

or

$\Lambda_{f}=\sum_{i,\alpha}\left(\frac{\partial f^{\alpha}}{\partial
x^{i}}\right)^{2}<2,$

then $f^{\alpha}$ has to be affine linear functions representing an affine
$n$-plane.

###### Proof.

If $\Delta_{f}<2$, then the Gauss image of $M$ is contained in
$\\{P\in{\tenmsb U}\subset\mathbb{G}_{n,m}:v(P)<2\\}$. We choose

$s=4+\frac{4}{3p}>4.$

Fix $x\in M$ and let $r$ be the Euclidean distance function from $x$ and
$M_{R}=D_{R}(x)$. Hence, letting $R\to+\infty$ in (5.14) yields

$\big{\|}|B|\big{\|}_{L^{s}(M)}=0.$

i.e., $|B|=0$. $M$ has to be an affine linear subspace.

For the case $\Lambda_{f}<2$, the proof is similar. ∎

Theorem 5.3 and Proposition 6.1 yield Bernstein type results as follows.

######  Theorem 6.2.

Let $M=(x,f(x))$ be an $n$-dimensional entire minimal graph given by $m$
functions $f^{\alpha}(x^{1},\cdots,x^{n})$ with $m\geq 2$. If

$\Delta_{f}=\left[\text{det}\left(\delta_{ij}+\sum_{\alpha}\frac{\partial
f^{\alpha}}{\partial x^{i}}\frac{\partial f^{\alpha}}{\partial
x^{j}}\right)\right]^{\frac{1}{2}}<2,$

and

(6.4) $\left(2-\Delta_{f}\right)^{-1}=o(R^{\frac{4}{3}}),$

where $R^{2}=|x|^{2}+|f|^{2}$. Then $f^{\alpha}$ has to be affine linear
functions and hence $M$ has to be an affine linear subspace.

######  Theorem 6.3.

Let $M=(x,f(x))$ be an $n$-dimensional entire minimal graph given by $m$
functions $f^{\alpha}(x^{1},\cdots,x^{n})$ with $p=\min\\{n,m\\}\leq 4$. If

$\Lambda_{f}=\sum_{i,\alpha}\left(\frac{\partial f^{\alpha}}{\partial
x^{i}}\right)^{2}<2,$

and

(6.5) $\left(2-\Lambda_{f}\right)^{-1}=o(R^{\frac{2(p+2)}{3p}})$

where $R^{2}=|x|^{2}+|f|^{2}$. Then $f^{\alpha}$ has to be affine linear
functions and hence $M$ has to be an affine linear subspace.

###### Proof.

Here we only give the proof of Theorem 6.2, for the proof of Theorem 6.3 is
similar.

From (4.5), it is easily seen that

$h_{2}\leq C(2-v)^{-\frac{3}{2}},$

where $C$ is a positive constant. Thus, for any point $q\in M$, by Theorem 5.3
and Proposition 6.1, we have

$|B|^{2s}(q)\leq C(n,s)R^{-2s}\left(2-v\circ\gamma\right)^{-\frac{3}{2}s}$

Letting $R\to+\infty$ in the above inequality forces $|B(q)|=0$.

∎

######  Remark 6.1.

If $n=2$ or $3$, the conclusion of Theorem 6.1-6.3 could be inferred from the
work done by Chern-Osserman [6], Babosa [1] and Fischer-Colbrie [10].

From (5.13) it is easy to obtain the following result.

######  Theorem 6.4.

Let $M=(x,f(x))$ be an $n$-dimensional entire minimal graph given by $m$
functions $f^{\alpha}(x^{1},\cdots,x^{n})$. Assume $M$ has finite total
curvature. If $\Delta_{f}<2$, or $p\leq 4$ and $\Lambda_{f}<2$, then $M$ has
to be an affine linear subspace.

There are other applications of the strong stability inequalities (4.17),
besides its key role in S-S-Y’s estimates. We state following results, whose
detail proof can be found in the previous paper of the first author [23].

######  Theorem 6.5.

Let $M=(x,f(x))$ be an $n$-dimensional entire minimal graph given by $m$
functions $f^{\alpha}(x^{1},\cdots,x^{n})$. If $\Delta_{f}<2$ or
$\Lambda_{f}<2$, then any $L^{2}$-harmonic $1$-form vanishes.

######  Theorem 6.6.

Let $M$ be one as in Theorem 6.5, $N$ be a manifold with non-positive
sectional curvature. Then any harmonic map $f:M\to N$ with finite energy has
to be constant.

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