Is there any heterogeneous impact of mandatory disclosure on corporate social responsibility dimensions? Evidence from a quasi-natural experiment in China

1.   Introduction

In this paper, we study a conformal Hessian quotient inequality:

where $\frac{S_{k}}{S_{l}}\left(A^{g}u\right): = \frac{S_{k}}{S_{l}}\left(\lambda\left(A^{g}u\right)\right)$, $0\leq l<k\leq n$, $k\in\mathbb{N^+}$, $l\in\mathbb{N}$, $u$ is the unknown function, $\alpha$ is a non-negative constant. Note that in Euclidean space $\mathbb{R}^n$, the conformal Hessian matrix $A^gu$ in (1) has the form

where $a(x)$ and $b(x)$ denote the functions of $x$, $Du$ and $D^2u$ denote the gradient vector and Hessian matrix of $u$ respectively, $I$ is the $n\times n$ identity matrix. In (1), the operator $S_{k}$ denotes the Hessian operator or the $k$-th order elementary symmetric polynomial given by

where $k = 1, \ldots, n$, $i_1, \ldots, i_s\in{\left\{1, \ldots, n\right\}}$ and $\lambda = \left(\lambda_1, \ldots, \lambda_n\right)$ denotes the eigenvalues of the matrix $A^gu$. $S_{k}(A^{g}u): = S_{k}(\lambda(A^{g}u))$ denotes the sum of the $k\times k$ principle minors of the matrix $A^{g}u$. As usual, we define $S_0(\lambda)\equiv1$, see [17]. Then, naturally we can define the quotient operator in (1) to be

When $l = 0$, the leading term $S_{k}(A^{g}u)$ in the inequality (1) is related to the $k$-Yamabe problem, see [15,18]. Specially, when $k = 1$, it is related to the well-known Yamabe problem, see [1,2,14,16,19]. According to Caffarelli-Nirenberg-Spruck (see [4]), we say that $u$ is a $k$-admissible function of (2) if

We give the following nonexistence result of positive $k$-admissible solution of the inequality (1).

Theorem 1.1. The inequality (1) with $A^gu$ in the form (2) has no entire positive $k$-admissible solution in $\mathbb{R}^n$ for either $\inf\limits_{x\in\mathbb{R}^n}(na(x)-b(x))\geq0, \ \alpha>0$ or $\inf\limits_{x\in\mathbb{R}^n}(na(x)-b(x))>0, \ \alpha\geq0$.

Remark 1. Note that the functions $a(x)$ and $b(x)$ should satisfy the conditions in Theorem 1.1. When $a(x)$ is positive, $b(x)$ can be positive or negative. When $a(x)$ is negative, $b(x)$ can only be negative.

The above theorem is a nonexistence result of positive solutions to a single Hessian quotient inequality. Naturally, we can consider the system of Hessian quotient inequalities. We then study the following system of Hessian quotient inequalities:

where $\alpha\geq0$, $\beta\geq0$, $0\leq l<k\leq n$, $0\leq l'<k'\leq n$, $k'\in\mathbb{N^+}$, $l'\in\mathbb{N}$, $A^gu$ is defined in (2), and

If a $k$-admissible $u$ and a $k'$-admissible $v$ satisfy (6), we call $(u, v)$ the $(k, k')$-admissible solution pair. In some situation, we need to assume that

holds for some constant $p\geq1$.

We formulate the nonexistence result of positive admissible solution of the coupled inequalities (6).

Theorem 1.2. Assume that the system (6) of Hessian quotient inequalities satisfies (2) and (7). Assume also that either $\inf\limits_{x\in\mathbb{R}^n}(na(x)-b(x))\geq0, \ \alpha>0, \ \beta>0$, (8), or $\inf\limits_{x\in\mathbb{R}^n}(na(x)-b(x))>0, \ \alpha\geq0, \ \beta\geq0$ hold. Then the system (6) has no entire positive $(k, k')$-admissible solution pair $(u, v)$ in $\mathbb{R}^n$.

We recall some related studies on the entire solutions of Hessian equations and Hessian inequalities. The classification of the nonnegative entire solutions of the equation

had been deduced for $1\leq\alpha<\frac{n+2}{n-2}$ in [5] and $\alpha = \frac{n+2}{n-2}$ in [3], respectively. The similar classification results are extended in [8] to admissible positive solutions of the conformal $k$-Hessian equations

where $g = u^{-2}dx^2$ is a locally conformally flat matrix in $\mathcal{M}$, $A^gu$ is given by $A^gu = g^{-1}A^{g_0}u$, $g_0$ is a given metric on $\mathcal{M}$. In [11], the same classification result for special case of $\mathbb{R}^n$ ($n = 2k+1$) is also obtained by suitable choices of the text functions and the argument of integration. Using the method as in [5] and [11], a nonexistence result of the Hessian inequality

is proved in [12] for $2k<n$ and $\alpha \in (-\infty, nk/(n-2k)]$. The conformal $k$-Hessian inequality

with $A^gu = u\left(D^2u\right)-\frac{1}{2}|Du|^2I$ is considered in [13], where the nonexistence results for $2k<n$ and $\alpha\in[k, \infty)$ is formulated. The conformal $k$-Hessian inequality $S_k\left(A^gu\right)\geq u^\alpha$ with $A^gu = D^2u-\left(\frac{1}{2}|Du|^2I-Du\otimes Du\right)$ is considered in [6]. Note that $A^gu$ in (2) has a similar structure to that in [6], but our $A^gu$ here is more general because of the existence of $a(x)$ and $b(x)$.

The organization of this paper is as follows. In Section 2, we first formulate a nonexistence result, Lemma 2.1, for an inequality involving the Laplace operator. Then we give its proof by appropriate choice of text functions and the method of integration by parts. The proof is divided into two cases according to $\alpha$ and the coefficient of the $|Du|^2$, where Schwarz's inequality and Young's inequality are properly used. In section 3, we establish the relations between Hessian quotient operator and the Laplace operator by using Maclaurin's inequality. Based on this relationship and Lemma 2.1, we prove the nonexistence result in Theorem 3.2. The proof of Theorem 1.1 is completed by using Taylor's expansion and Theorem 3.2. At the end, we give the proof of nonexistence result of the coupled inequalities (6), Theorem 1.2. The proof is divided into two cases, one of which needs to be proved under condition (8), and the other is a direct consequence of Theorem 1.1.

2.   A preliminary result - semilinear case

Before proving Theorem 1.1, we first introduce a nonexistence result for an inequality involving the Laplace operator in this section, which is a preliminary result for proving Theorem 1.1. Since there is a quadratic term $|Du|^2$ in the inequality, the preliminary nonexistence result is nontrivial and different from [7] and [10].

Lemma 2.1. The inequality

has no entire positive solution for either $\inf\limits_{x\in\mathbb{R}^n}m(x)\geq0, \ \alpha>1$ or $\inf\limits_{x\in\mathbb{R}^n}m(x)>0, \ \alpha\geq0$, where $m(x)$ denotes the function of $x$.

When $m(x)\equiv 0$, the inequality (13) becomes $\Delta u\geq u^{\alpha}$, whose existence and nonexistence of entire positive solutions are related to the Keller-Osserman condition, see [7], [10]. The Keller-Osserman condition of $\Delta u = f(u)$ is to check whether $\int^\infty\left(\int_0^\tau f(s)ds\right)^{-\frac{1}{2}}d\tau$ equals to $+\infty$ or not, where the lower limit is omitted in the integral to admit any positive constant. By taking $f(u) = u^\alpha$, we have

Therefore, the equation $\Delta u = u^\alpha$ has an entire positive solution when $0<\alpha \le 1$, but has no entire positive solution when $\alpha>1$, which corresponds to our Lemma 2.1 in the case of $\inf\limits_{x\in\mathbb{R}^n}m(x)\geq0$.

However, when $\inf\limits_{x\in\mathbb{R}^n}m(x)>0$, the nonlinear term $m(x)|Du|^2$ comes into play so that the nonexistence of entire positive solutions can hold in a wider range of $\alpha$, namely $\alpha \in [0, \infty)$. Comparing with the classical results related to the Keller-Osserman condition, this is an interesting and different point. We will prove Lemma 2.1 by multiplying proper test functions to the inequality (13) and integrating by parts.

When $k = 1$ and $l = 0$ in (1), the operator in (1) is just $\Delta u-m(x)|Du|^2$ with $m(x) = na(x)-b(x)$, which is the same as the operator in (13). In Section 3, we will prove the main results, Theorems 1.1 and 1.2, based on the preliminary result in Lemma 2.1.

Proof of Lemma 2.1. It is enough to prove nonexistence of entire positive solution for either $\inf\limits_{x\in\mathbb{R}^n}m(x)\geq0, \ \alpha>1$ or $\inf\limits_{x\in\mathbb{R}^n}m(x)>0, \ 0\leq\alpha\leq1$. In fact, the situation when $\inf\limits_{x\in\mathbb{R}^n}m(x)>0, \ \alpha>1$ is already covered by the former case.

Suppose that the inequality (13) has an entire positive solution $u$. We will deduce the contradiction.

Multiplying both sides of (13) by $u^\delta\zeta^\theta$ and integrating over $\mathbb{R}^n$, we have

where $\delta, \theta$ are constants to be determined, $\zeta\in C^2$ is a cut-off function satisfying:

where $B_R$ denotes a ball in $\mathbb{R}^n$ centered at $0$ with radius $R$, and $C$ is a positive constant. In order to deal with the last term of (15), we use the integration by parts to get

where $U: = {\rm supp}|D\zeta| = \left\{x\in\mathbb{R}^n:R<|x|<2R\right\}$.

Inserting (18) into (15), we have

We next split the proof into the following two cases of $\alpha$ and $m(x)$:

$({\it i})$ $\inf\limits_{x\in\mathbb{R}^n}m(x) = a_0>0$, $0\leq\alpha\leq 1$,

$({\it ii})$ $\inf\limits_{x\in\mathbb{R}^n}m(x) = a_0\geq0$, $\alpha>1$.

In case $(i)$, we further consider the two subcases:

$({\it a})$ $\inf\limits_{x\in\mathbb{R}^n}m(x) = a_0>0$, $\alpha = 0$,

$({\it b})$ $\inf\limits_{x\in\mathbb{R}^n}m(x) = a_0>0$, $0<\alpha\leq 1$.

In case $(i)(a)$, we fix the constants $\delta$ and $\theta$ such that $\delta = 0,$ $\theta>n.$ The last term in (19) becomes

where Schwarz's inequality is used to obtain the first inequality, Young's inequality $ab\leq{\varepsilon_2a^p}/p+{C_{\varepsilon_2}b^q}/q$ with the exponent pair $(p, q) = (\theta/(\theta-2), \theta/2)$ is used to obtain the second inequality. We emphasize that $\varepsilon_1$ and $\varepsilon_2$ are positive constants to be determined, $C_{\varepsilon_1}$ and $C_{\varepsilon_2}$ denote some positive constants depending on $\varepsilon_1$ and $\varepsilon_2$ respectively.

Inserting (20) into (19), we have

By selecting appropriate $\varepsilon_1$ and $\varepsilon_2$ such that $\varepsilon_1\leq\frac{a_0}{\theta}$, $\varepsilon_2<\frac{1}{C_{\varepsilon_1}(\theta-2)}$, and using (17), we get from (21) that

where $w_n$ denotes the volume of the unit ball in $\mathbb{R}^n$. Letting $R\rightarrow \infty$ in (22), since $\theta>n$, we have

Since $\zeta\in C^2$ is a cut-off function satisfying (16)-(17), we get a contradiction in case $(i)(a)$.

In case $(i)(b)$, we fix the constants $\delta$ and $\theta$ such that

For the last term in (19), we have

where Schwarz's inequality is used to obtain the first inequality, Young's inequality $ab\leq{\varepsilon_4a^p}/p+{C_{\varepsilon_4}b^q}/q$ with the exponent pair $(p, q) = (2/(2-\alpha), 2/\alpha)$ is used to obtain the second inequality, $\varepsilon_3$ and $\varepsilon_4$ are positive constants to be determined, $C_{\varepsilon_3}$ and $C_{\varepsilon_4}$ denote some positive constants depending on $\varepsilon_3$ and $\varepsilon_4$ respectively.

Inserting (25) into (19), we have

By selecting appropriate $\varepsilon_3$ and $\varepsilon_4$ such that $\varepsilon_3\leq {\rm min}\left\{\frac{2\delta}{\theta C_{\varepsilon_4}\alpha}, 1\right\}$, $\varepsilon_4\leq\frac{2a_0}{\theta(2-\alpha)}$, we can discard the first two terms on the right hand side of (26). Hence, (26) becomes

where Young's inequality $ab\leq\varepsilon_5a^p/p+C_{\varepsilon_5}b^q/q$ with the exponent pair $(p, q) = \left(\frac{\alpha+\delta}{\alpha/2+\delta}, \frac{2\alpha+2\delta}{\alpha}\right)$ is used to obtain the last inequality, $\varepsilon_5$ is a positive constant to be determined, and $C_{\varepsilon_5}$ denote some positive constants depending on $\varepsilon_5$.

By choosing $\varepsilon_5<\frac{2\alpha+2\delta}{\theta C_{\varepsilon_3}(\alpha+2\delta)}$ and using (17), we get from (27) that

where $w_n$ denotes the volume of the unit ball in $\mathbb{R}^n$. By (24), the exponent $n-4(1+\delta/\alpha)$ of $R$ is negative. Letting $R\rightarrow \infty$ in (28), we can obtain

Since $\zeta\in C^2$ is a cut-off function satisfying (16)-(17) and $u$ is positive, we get a contradiction in case $(i)(b)$.

In case $(ii)$, we fix the constants $\delta$ and $\theta$ such that $\delta>{\rm max}\left\{{(n-2)}\alpha-n+1, 0\right\}$, $\theta>\frac{2(\alpha+\delta)}{\alpha-1}.$ Hence, we always have $\delta>0$ and $\theta>2$. For the last term in (18), using Schwarz's inequality, we have

where $\varepsilon_6$ is any positive constant and $C_{\varepsilon_6}$ denotes some positive constant depending on $\varepsilon_6$. Inserting (30) into (19), we have

Hence, by taking $\varepsilon_6\leq\delta/\theta$ in (31), we have

Then $\varepsilon_6$ is now fixed. Applying Young's inequality to the last term in (32), we have

where $p, q$ are positive constants satisfying

$\varepsilon_7$ is a positive constant, and $C_{\varepsilon_7}$ denotes some positive constant depending on $\varepsilon_7$.

Setting $s = \frac{\alpha+\delta}{\delta+1}>1$, we get

Inserting $p, q, s$ and $t$ into (33), we have

Taking $\varepsilon_7<\frac{\alpha+\delta}{\theta C_{\varepsilon_6}(\delta+1)}$, and using (17), we get from (34) that

where $w_n$ denotes the volume of the unit ball in $\mathbb{R}^n$. Now the constant $\varepsilon_7$ is fixed. Recalling that $\delta>{\rm max}\{{(n-2)}\alpha-n+1, 0\}$. It is easy to check that $n-\frac{2(\alpha+\delta)}{\alpha-1}<0$. Letting $R\rightarrow \infty$ in (35), we can obtain

Since $\zeta\in C^2$ is a cut-off function satisfying (16)-(17) and $u$ is positive, we get a contradiction in case $(ii)$.

We have completed the proof of Lemma 2.1.

3.   Proofs of the main results

In this section, we prove the nonexistence result of entire positive solutions of single conformal Hessian quotient inequality (1). Then we will consider the nonexistence result of entire positive solutions of the system (6) of Hessian quotient inequalities.

3.1.   Single conformal Hessian quotient inequality

In order to prove the Theorem 1.1, we first establish the relationship between Hessian quotient operator and the Laplace operator by using Maclaurin's inequality of Lemma 15.13 in [9].

Lemma 3.1. For $\lambda\in\Gamma^k$, and $n\geq k>l\geq0$, $r>s\geq0$, $k\geq r$, $l\geq s$, we have

and the inequality holds if and only if $\lambda_1 = \lambda_2 = \cdots = \lambda_n>0$.

The inequality (37) is a consequence of Newton's inequality, see [9].

Taking $r = 1, \ s = 0$ in (37), we can obtain the following relationship between Hessian quotient operator and Laplace operator,

which is crucial in proving the nonexistence of (1). Since

when $\lambda(A^gu)\in\Gamma^k$, we have from (38) that

Since $A^gu$ has the form (2), it can be readily calculated that

Hence, combining (39)-(40) with Lemma 2.1, we get the following theorem.

Theorem 3.2. The inequality

with $A^gu$ in the form (2) has no entire positive $k$-admissible solution in $\mathbb{R}^n$ for either $\inf\limits_{x\in\mathbb{R}^n}(na(x)-b(x))\geq 0$, $\alpha>k-l$ or $\inf\limits_{x\in\mathbb{R}^n}(na(x)-b(x))>0$, $\alpha\ge 0$.

Proof. We replace $\alpha$ with $\frac{\alpha}{k-l}$ in (13) and take $m(x) = na(x)-b(x)$. Then we repeat the proof of the Lemma 2.1. Suppose that the inequality (41) has an entire positive solution $u$. We can get that

has no entire positive $k$-admissible solution in $\mathbb{R}^n$ for either $\inf\limits_{x\in\mathbb{R}^n}(na(x)-b(x))\geq 0$, $\alpha>k-l$ or $\inf\limits_{x\in\mathbb{R}^n}(na(x)-b(x))>0$, $\alpha\ge 0$. Combining (39), (40) and (42), we get that

which implies that the inequality (41) has no entire positive $k$-admissible solution in $\mathbb{R}^n$ for either $\inf\limits_{x\in\mathbb{R}^n}(na(x)-b(x))\geq0$, $\alpha>k-l$ or $\inf\limits_{x\in\mathbb{R}^n}(na(x)-b(x))>0$, $\alpha\ge 0$.

Remark 2. In Theorem 3.2, when $\inf\limits_{x\in\mathbb{R}^n}(na(x)-b(x))\geq0$, the nonexistence of entire positive $k$-admissible solution is proved for $\alpha > k-l$. According to the analysis below (2.1), it is reasonable to guess that the corresponding Keller-Osserman type condition for the Hessian quotient equation $\frac{S_k}{S_l}(D^2u) = f(u)$ may has the following form

This kind of conditions for Hessian quotient equations will be discussed independently in a sequel.

Next, we need to use the Taylor's expansion and pick the right value of $\alpha$ to prove Theorem 1.1.

Proof of Theorem 1.1. Suppose that the inequality (1) has entire positive solutions. We will deduce the contradiction.

Taking $k = 1$ and $l = 0$, we have

where (39) and (1) are used to obtain the first and second inequalities, respectively, Taylor's expansion for some $\theta\in\left(0, \frac{\alpha}{k-l}u\right)$ is used in the last equality, the last inequality holds since $\frac{e^\theta}{(n+1)!}\left(\frac{\alpha}{k-l}\right)^{n+1}u^{n+1}$ is positive. We next split the proof into the following two cases of $\alpha$ and $n$:

$({\it i})$ $\inf\limits_{x\in\mathbb{R}^n}(na(x)-b(x))\geq0, \alpha>0$,

$({\it ii})$ $\inf\limits_{x\in\mathbb{R}^n}(na(x)-b(x))>0, \alpha\geq0$.

In case (i), we get from (44) that

From case (ii) in Lemma 2.1, $\Delta u-(na(x)-b(x))|Du|^2\geq c u^2$ (with a positive constant $c$) has no entire positive solution. So we get a contradiction in case (i).

In case (ii), we get from (44) that

From case (i) in Lemma 2.1, $\Delta u-(na(x)-b(x))|Du|^2\geq 1$ has no entire positive solution. So we get a contradiction in case (ii).

We have completed the proof of Theorem 1.1.

3.2.   System of conformal Hessian quotient inequalities

In this section, we consider the nonexistence result of positive admissible solution pair of the system of conformal Hessian quotient inequalities (6).

Proof of Theorem 1.2. Suppose that the system (6) have entire positive $(k, k')$-admissible solution pair $(u, v)$ in $\mathbb{R}^n$. We will deduce the contradiction. We next split the proof into the following two cases:

$({\it i})$ $\inf\limits_{x\in\mathbb{R}^n}(na(x)-b(x))\geq0$, $\alpha>0$, $\beta>0$ and (8) hold;

$({\it ii})$ $\inf\limits_{x\in\mathbb{R}^n}(na(x)-b(x))>0$, $\alpha\geq0$ and $\beta\geq0$ hold.

In case (i), Using Taylor's expansion, we have that

where $0<\theta<v$. Hence, we have

for $\alpha>0$. In the same way, we have

for $\beta>0$. So we can derive from (6) to get

Using Maclaurin's inequality (39), the system (50) can be converted to the following system of inequalities:

Since we only consider the situation that $\frac{\alpha}{k-l} = \frac{\beta}{k'-l'} = p\geq1$ in this case, by adding the two inequalities in (51), we get

Setting $w = u+v$, we get

where Cauchy's inequality and Jensen's inequality are used. By taking $m(x) = \frac{1}{2}(na(x)-b(x))$ in (13), there is only a coefficient difference between (53) and (13). Hence, the nonexistence of (53) can be proved in the same way as Section $2$, which contracts with the existence of the solution pair $(u, v)$.

In case (ii), since $\alpha\geq0$ and $u, v>0$, we get from (6) that

It contradicts with the conclusion of Theorem 1.1 under the assumption $\inf\limits_{x\in\mathbb{R}^n}(na(x)-b(x))>0, \alpha\geq0$.

We have completed the proof of Theorem 1.2.

Remark 3. It is interesting to ask whether condition (8) can be removed or not in Theorem 1.2.

Acknowledgments

The authors would like to thank the anonymous referee for his/her useful suggestions, especially for the comments about Theorem 1.2 in an earlier version of the paper.