Schur convexity for the ratios of the Hamy and generalized Hamy symmetric functions

Abstract

In this paper, we present the Schur convexity and monotonicity properties for the ratios of the Hamy and generalized Hamy symmetric functions and establish some analytic inequalities. The achieved results is inspired by the paper of Hara et al. [J. Inequal. Appl. 2, 387-395, (1998)], and the methods from Guan [Math. Inequal. Appl. 9, 797-805, (2006)]. The inequalities we obtained improve the existing corresponding results and, in some sense, are optimal.

2010 Mathematics Subject Classification: Primary 05E05; Secondary 26D20.

1 Introduction

Throughout this paper, we denote ${ℝ}_{+}^n=\left\{x=\left(x_1,x_2,\dots ,x_n\right)|x_i>0,i=1,2,\dots ,n\right\}.$. For $x\in {ℝ}_{+}^n$, the Hamy symmetric function [1] is defined as

$$F_n\left(x,r\right)=F_n\left(x_1,x_2,\dots ,x_n;r\right)=\sum _{1\le i_1<i_2<\cdots <i_r\le n}{\left(\prod _{j=1}^r{x}_{i_j}\right)}^{\frac{1}{r}},$$

(1.1)

where r is an integer and 1 ≤ r ≤ n.

The generalized Hamy symmetric function was introduced by Guan [2] as follows

$$F_n^{*}\left(x,r\right)=F_n^{*}\left(x_1,x_2,\dots ,x_n;r\right)=\sum _{i_1+i_2+\cdots +i_n=r}{\left(x_1^{i_1}{x}_2^{i_2}\dots x_n^{i_n}\right)}^{\frac{1}{r}},$$

(1.2)

where r is a positive integer.

In [2], Guan proved that both F _n (x,r) and $F_n^{*}\left(x,r\right)$ are Schur concave in ${ℝ}_{+}^n$. The main of this paper is to investigate the Schur convexity for the functions $\frac{F_n\left(x,r\right)}{F_n\left(x,r-1\right)}$ and $\frac{F_n^{*}\left(x,r\right)}{F_n^{*}\left(x,r-1\right)}$ and establish some analytic inequalities by use of the theory of majorization.

For convenience of readers, we recall some definitions as follows, which can be found in many references, such as [3].

Definition 1.1. The n-tuple x is said to be majorized by the n-tuple y (in symbols x ≺ y), if

$$\sum _{i=1}^k{x}_{\left[i\right]}\le \sum _{i=1}^k{y}_{\left[i\right]},\sum _{i=1}^n{x}_{\left[i\right]}=\sum _{i=1}^n{y}_{\left[i\right]},$$

where 1 ≤ k ≤ n - 1, and x_[i]denotes the i th largest component of x.

Definition 1.2. Let E ⊆ ℝⁿbe a set. A real-valued function F : E → ℝ is said to be Schur convex on E if F(x) ≤ F(y) for each pair of n-tuples x = (x₁,..., x _n ) and y = (y₁,..., y _n ) in E, such that x ≺ y. F is said to be Schur concave if -F is Schur convex.

The theory of Schur convexity is one of the most important theories in the fields of inequalities. It can be used in combinatorial optimization [4], isoperimetric problems for polytopes [5], theory of statistical experiments [6], graphs and matrices [7], gamma functions [8], reliability and availability [9], optimal designs [10] and other related fields.

Our aim in what follows is to prove the following results.

Theorem 1.1. Let $x\in {ℝ}_{+}^n,2\le r\le n$ is an integer, then the function ${\varphi }_r\left(x\right)=\frac{F_n\left(x,r\right)}{F_n\left(x,r-1\right)}$ is Schur concave in ${ℝ}_{+}^n$ and increasing with respect to x _i (i= 1,2, ...,n).

Theorem 1.2. Let $x\in {ℝ}_{+}^n,2\le r\le n$ is an integer, then the function ${\varphi }_r^{*}\left(x\right)=\frac{F_n^{*}\left(x,r\right)}{F_n^{*}\left(x,r-1\right)}$ is Schur concave in ${ℝ}_{+}^n$ and increasing with respect to x _i (i= 1,2,... n).

Corollary 1.1. If $x_i>0,i=1,2,\dots ,n,{\sum }_{i=1}^n{x}_i=s$ and that c ≥ s, then

$$\frac{G_n\left(x\right)}{G_n\left(c-x\right)}=\frac{F_n\left(x,n\right)}{F_n\left(c-x,n\right)}\le \frac{F_n\left(x,n-1\right)}{F_n\left(c-x,n-1\right)}\le \cdots \le \frac{F_n\left(x,1\right)}{F_n\left(c-x,1\right)}=\frac{A_n\left(x\right)}{A_n\left(c-x\right)}$$

and

$$\frac{G_n\left(x\right)}{G_n\left(c+x\right)}=\frac{F_n\left(x,n\right)}{F_n\left(c+x,n\right)}\le \frac{F_n\left(x,n-1\right)}{F_n\left(c+x,n-1\right)}\le \cdots \le \frac{F_n\left(x,1\right)}{F_n\left(c+x,1\right)}=\frac{A_n\left(x\right)}{A_n\left(c+x\right)},$$

where $A_n\left(x\right)=\frac{1}{n}{\sum }_{i=1}^n{x}_i,G_n\left(x\right)={\left({\prod }_{i=1}^n{x}_i\right)}^{\frac{1}{n}}$ are the arithmetic and geo-metric means of x, respectively.

Corollary 1.2. If $x_i>0,i=1,2,\dots ,n,{\sum }_{i=1}^n{x}_i=s$ and that c ≥ s, then

$$\frac{F_n^{*}\left(x,r\right)}{F_n^{*}\left(c-x,r\right)}\le \frac{F_n^{*}\left(x,r-1\right)}{F_n^{*}\left(c-x,r-1\right)}\le \cdots \le \frac{F_n^{*}\left(x,2\right)}{F_n^{*}\left(c-x,2\right)}\le \frac{F_n^{*}\left(x,1\right)}{F_n^{*}\left(c-x,1\right)}=\frac{A_n\left(x\right)}{A_n\left(c-x\right)}$$

and

$$\frac{F_n^{*}\left(x,r\right)}{F_n^{*}\left(c+x,r\right)}\le \frac{F_n^{*}\left(x,r-1\right)}{F_n^{*}\left(c+x,r-1\right)}\le \cdots \le \frac{F_n^{*}\left(x,2\right)}{F_n^{*}\left(c+x,2\right)}\le \frac{F_n^{*}\left(x,1\right)}{F_n^{*}\left(c+x,1\right)}=\frac{A_n\left(x\right)}{A_n\left(c+x\right)}.$$

2 Lemmas

In order to establish our main results, we need several lemmas, which we present in this section.

Lemma 2.1 (see [3]). Let E ⊆ ℝⁿbe a symmetric convex set with nonempty interior intE and φ : E → ℝ be a continuous symmetric function. If φ is differentiable on intE, then φ is Schur convex (or Schur concave, respectively) on E if and only if

$$\left(x_i-x_j\right)\left(\frac{\partial \phi }{\partial x_i}-\frac{\partial \phi }{\partial x_j}\right)\ge 0\left(\mathsf{\text{or}}\le 0,\mathsf{\text{respectively}}\right)$$

for all i,j = 1,2,...,n and x = (x₁,...,x _n ) ∈ intE.

The r th elementary symmetric function (see [11]) is defined as

$$E_n\left(x,r\right)=E_n\left(x_1,x_2,\dots ,x_n;r\right)=\sum _{1\le i_1<i_2<\cdots <i_r\le n}\left(\prod _{j=1}^r{x}_{i_j}\right),$$

(2.1)

where 1 ≤ r ≤ n is a positive integer, and E _n (x, 0) = 1.

By (2.1) and simple computations, we have the following lemma.

Lemma 2.2. Let $x\in {ℝ}_{+}^n,1\le i\le n$, if

$$\overline{x_i}=\left(x_1,x_2,\dots ,x_{i-1},x_{i+1},\dots ,x_n\right).$$

Then,

$$E_n\left(x_1,x_2,\dots ,x_n;r\right)=x_i{E}_{n-1}\left(\overline{x_i},r-1\right)+E_{n-1}\left(\overline{x_i},r\right).$$

(2.2)

Lemma 2.3 (see [11]). Let $x\in {ℝ}_{+}^n,r$ is an integer and 1 ≤ r ≤ n - 1.

Then,

$${\left(E_n\left(x,r\right)\right)}^2>E_n\left(x,r-1\right)E_n\left(x,r+1\right).$$

(2.3)

Another important symmetric function is the complete symmetric function (see [3]), which is defined by

$$C_r\left(x\right)=C_r\left(x_1,x_2,\dots ,x_n\right)=\sum _{i_1+i_2+\cdots +i_n=r}{x}_1^{i_1}{x}_2^{i_2}\dots x_n^{i_n},$$

where i₁, i₂,..., i _n are non-negative integer, r ∈ {1, 2,...} and C₀(x) = 1.

Lemma 2.4 (see [12]). Let x _i > 0, i = 1, 2,..., n, and $\overline{x_i}=\left(x_1,x_2,\dots ,x_{i-1},x_{i+1},\dots ,x_n\right)$.

Then,

$$C_r\left(x\right)=x_i{C}_{r-1}\left(x\right)+C_r\left(\overline{x_i}\right).$$

Lemma 2.5 (see [13]). If $0<r<s,x\in {ℝ}_{+}^n$, then

$$C_r\left(x\right)C_{s-1}\left(x\right)>C_{r-1}{C}_s\left(x\right).$$

Lemma 2.6 (see [14]). If $x_i>0,i=1,2,\dots ,n,{\sum }_{i=1}^n{x}_i=s$ and c ≥ s, then

• (1) $\frac{c-x}{\frac{nc}{s}-1}=\left(\frac{c-x_1}{\frac{nc}{s}-1},\frac{c-x_2}{\frac{nc}{s}-1},\dots ,\frac{c-x_n}{\frac{nc}{s}-1}\right)\prec \left(x_1,x_2,\dots ,x_n\right)=x,$,

(2)$\frac{c+x}{s+nc}=\left(\frac{c+x_1}{s+nc},\frac{c+x_2}{s+nc},\dots ,\frac{c+x_n}{s+nc}\right)\prec \left(\frac{x_1}{s},\frac{x_2}{s},\dots ,\frac{x_n}{s}\right)=\frac{x}{s}$.

3 Proof of Theorems

Proof of Theorem 1.1. It is obvious that ϕ _r (x) is symmetric and has continuous partial derivatives in ${ℝ}_{+}^n$. By Lemma 2.1, we only need to prove that

$$\left(x_1-x_2\right)\left(\frac{\partial {\varphi }_r\left(x\right)}{\partial x_1}-\frac{\partial {\varphi }_r\left(x\right)}{\partial x_2}\right)\le 0.$$

(3.1)

For any fixed 2 ≤ r ≤ n, let $u_i=\sqrt[r]{x_i},i=1,2,\dots ,n$ and $u=\left(u_1,u_2,\dots ,u_n\right)\in {ℝ}_{+}^n$, we have

$${\varphi }_r\left(x\right)=\frac{F_n\left(x,r\right)}{F_n\left(x,r-1\right)}=\frac{E_n\left(u,r\right)}{E_n\left(u,r-1\right)}.$$

Differentiating ϕ _r (x) with respect to x₁ yields

$$\frac{\partial {\varphi }_r\left(x\right)}{\partial x_1}=\frac{1}{E_n^2\left(u,r-1\right)}\left[E_n\left(u,r-1\right)\frac{\partial E_n\left(u,r\right)}{\partial u_1}\frac{\partial u_1}{\partial x_1}-E_n\left(u,r\right)\frac{\partial E_n\left(u,r-1\right)}{\partial u_1}\frac{\partial u_1}{\partial x_1}\right].$$

(3.2)

Using Lemma 2.2 repeatedly, we get

$$\begin{array}{ll}\hfill E_n\left(u,r\right)& =u_1{u}_2{E}_{n-2}\left(u_3,\dots ,u_n;r-2\right)+\left(u_1+u_2\right)E_{n-2}\left(u_3,\dots ,u_n;r-1\right)\\ +E_{n-2}\left(u_3,\dots ,u_n;r\right).\end{array}$$

(3.3)

Equations (3.2) and (3.3) lead to

$$\frac{\partial {\varphi }_r\left(x\right)}{\partial x_1}=\frac{1}{r{E}_n^2\left(u,r-1\right)}\left(u_1^{1-r}{u}_{2}A+u_1^{1-r}B\right),$$

(3.4)

where

$$A=E_n\left(u,r-1\right)E_{n-2}\left(u_3,\dots ,u_n;r-2\right)-E_n\left(u,r\right)E_{n-2}\left(u_3,\dots ,u_n;r-3\right)$$

and

$$B=E_n\left(u,r-1\right)E_{n-2}\left(u_3,\dots ,u_n;r-1\right)-E_n\left(u,r\right)E_{n-2}\left(u_3,\dots ,u_n;r-2\right).$$

Similarly, we can deduce that

$$\frac{\partial {\varphi }_r\left(x\right)}{\partial x_2}=\frac{1}{r{E}_n^2\left(u,r-1\right)}\left(u_1{u}_2^{1-r}A+u_2^{1-r}B\right).$$

(3.5)

From (3.4) and (3.5), one has

$$\begin{array}{c}\left(x_1-x_2\right)\left(\frac{\partial {\varphi }_r\left(x\right)}{\partial x_1}-\frac{\partial {\varphi }_r\left(x\right)}{\partial x_2}\right)\\ =\frac{x_1-x_2}{r{E}_n^2\left(u,r-1\right)}\left[x_1^{\frac{1}{r}}{x}_2^{\frac{1}{r}}\left(x_1^{-1}-x_2^{-1}\right)A+\left(x_1^{\frac{1}{r}-1}-x_2^{\frac{1}{r}-1}\right)B\right].\end{array}$$

(3.6)

It follows from (3.3) and Lemma 2.3 that

$$\begin{array}{c}A=(u_1+u_2)[E_{n-2}^2(u_3,\dots ,u_n;r-2)-E_{n-2}(u_3,\dots ,u_n;r-1)\\ \times E_{n-2}(u_3,\dots ,u_n;r-3)]+E_{n-2}(u_3,\dots ,u_n;r-1)E_{n-2}(u_3,\dots ,u_n;r-2)\\ -E_{n-2}(u_3,\dots ,u_n;r)E_{n-2}(u_3,\dots ,u_n;r-3)\\ >0.\end{array}$$

Similarly, we can get B > 0.

It follows from the function $x^{\frac{k-r}{r}}\left(k=0,1\right)$ is decreasing in (0, +∞) that

$$\left(x_1-x_2\right)\left(x_1^{\frac{k-r}{r}}-x_2^{\frac{k-r}{r}}\right)\le 0,\left(k=0,1\right).$$

(3.7)

Therefore, inequality (3.1) follows from (3.6) and (3.7) together with A > 0 and B > 0.

Next, we prove that ${\varphi }_r\left(x\right)=\frac{F_n\left(x,r\right)}{F_n\left(x,r-1\right)}$ is increasing with respect to x _i (i= 1,2,...,n).

By the symmetry of ϕ _r (x) with respect to x _i (i = 1, 2,..., n), we only need to prove that

$$\frac{\partial {\varphi }_r\left(x\right)}{\partial x_1}\ge 0,$$

which can be derived directly from A > 0 and B > 0 together with Equation (3.4).

Proof of Theorem 1.2. It is obvious that ${\varphi }_r^{*}\left(x\right)$ is symmetric and has continuous partial derivatives in ${ℝ}_{+}^n$. By Lemma 2.1, we only need to prove that

$$\left(x_1-x_2\right)\left(\frac{\partial {\varphi }_r^{*}\left(x\right)}{\partial x_1}-\frac{\partial {\varphi }_r^{*}\left(x\right)}{\partial x_2}\right)\le 0.$$

(3.8)

For any fixed 2 ≤ r ≤ n, let $u_i=\sqrt[r]{x_i},i=1,2,\dots ,n$ and $u=\left(u_1,u_2,\dots ,u_n\right)\in {ℝ}_{+}^n$. Then,

$${\varphi }_r^{*}\left(x\right)=\frac{F_n^{*}\left(x,r\right)}{F_n^{*}\left(x,r-1\right)}=\frac{C_r\left(u\right)}{C_{r-1}\left(u\right)}.$$

(3.9)

Differentiating ${\varphi }_r^{*}\left(x\right)$ with respect to x₁, we have

$$\frac{\partial {\varphi }_r^{*}\left(x\right)}{\partial x_1}=\frac{1}{C_{r-1}^2\left(u\right)}\left[C_{r-1}\left(u\right)\frac{\partial C_r\left(u\right)}{\partial u_1}\frac{\partial u_1}{\partial x_1}-C_r\left(u\right)\frac{\partial C_{r-1}\left(u\right)}{\partial u_1}\frac{\partial u_1}{\partial x_1}\right].$$

(3.10)

It follows from Lemma 2.4 that

$$\begin{array}{ll}\hfill \frac{\partial C_r\left(u\right)}{\partial u_1}& =C_{r-1}\left(u\right)+u_1\frac{\partial C_{r-1}\left(u\right)}{\partial u_1}\\ =C_{r-1}\left(u\right)+u_1\left[C_{r-2}\left(u\right)+u_1\frac{\partial C_{r-2}\left(u\right)}{\partial u_1}\right]\\ =C_{r-1}\left(u\right)+u_1{C}_{r-2}\left(u\right)+u_1^2\frac{\partial C_{r-2}\left(u\right)}{\partial u_1}\\ =\cdots \cdots \\ =C_{r-1}\left(u\right)+u_1{C}_{r-2}\left(u\right)+u_1^2{C}_{r-3}\left(u\right)+\cdots +u_1^{r-2}{C}_1\left(u\right)+u_1^{r-1}.\end{array}$$

(3.11)

Equations (3.10) and (3.11) lead to

$$\begin{array}{c}\frac{\partial {\varphi }_r^{*}(x)}{\partial x_1}=\frac{1}{C_{r-1}^2(u)}\{[C_{r-1}^2(u)-C_r(u)C_{r-2}(u)]+u_1[C_{r-1}(u)C_{r-2}(u)\\ -C_r(u)C_{r-3}(u)]+\cdots +u_1^{r-2}[C_{r-1}(u)C_1(u)-C_r(u)C_0(u)]\\ +C_{r-1}(u)u_1^{r-1}\}\frac{1}{r}{u}_1^{1-r}.\end{array}$$

(3.12)

Similarly, we have

$$\begin{array}{c}\frac{\partial {\varphi }_r^{*}(x)}{\partial x_2}=\frac{1}{C_{r-1}^2(u)}\{[C_{r-1}^2(u)-C_r(u)C_{r-2}(u)]+u_2[C_{r-1}(u)C_{r-2}(u)\\ -C_r(u)C_{r-3}(u)]+\cdots +u_2^{r-2}[C_{r-1}(u)C_1(u)-C_r(u)C_0(u)]\\ +C_{r-1}(u)u_2^{r-1}\}\frac{1}{r}{u}_2^{1-r}.\end{array}$$

(3.13)

From (3.12) and (3.13), one has

$$\begin{array}{l}(x_1-x_2)\left(\frac{\partial {\varphi }_r^{*}(x)}{\partial x_1}-\frac{\partial {\varphi }_r^{*}(x)}{\partial x_2}\right)\\ =\frac{x_1-x_2}{r{C}_{r-1}^2(u)}\{[C_{r-1}^2(u)-C_r(u)C_{r-2}(u)]\left(x_1^{\frac{1-r}{r}}-x_2^{\frac{1-r}{r}}\right)+[C_{r-1}(u)C_{r-2}(u)\\ -C_r(u)C_{r-3}(u)]\left(x_1^{\frac{2-r}{r}}-x_2^{\frac{2-r}{r}}\right)+\cdots +[C_{r-1}(u)C_1(u)-C_r(u)C_0(u)]\\ \times \left(x_1^{\frac{(r-1)-r}{r}}-x_2^{\frac{(r-1)-r}{r}}\right)\}.\end{array}$$

(3.14)

By Lemma 2.5, we know that

$$\begin{array}{c}{C}_{r-1}^2\left(u\right)-C_r\left(u\right)C_{r-2}\left(u\right)>0,\\ C_{r-1}\left(u\right)C_{r-2}\left(u\right)-C_r\left(u\right)C_{r-3}\left(u\right)>0,\\ \cdots \cdots ,\\ C_{r-1}\left(u\right)C_1\left(u\right)-C_r\left(u\right)C_0\left(u\right)>0.\end{array}$$

(3.15)

The monotonicity of the function $x^{\frac{j-r}{r}}\left(1\le j\le r-1\right)$ in (0, +∞) leads to the conclusion that

$$\left(x_1-x_2\right)\left(x_1^{\frac{j-r}{r}}-x_2^{\frac{j-r}{r}}\right)\le 0.$$

(3.16)

Therefore, inequality (3.8) follows from (3.14)-(3.16).

Next, we prove that ${\varphi }_r^{*}\left(x\right)=\frac{F_n^{*}\left(x,r\right)}{F_n^{*}\left(x,r-1\right)}$ is increasing with respect to x _i (i= 1,2,...,n).

From (3.12) and (3.15), we clearly see that

$$\frac{\partial {\varphi }_r^{*}\left(x\right)}{\partial x_1}\ge 0.$$

(3.17)

Inequality (3.17) implies that ${\varphi }_r^{*}\left(x\right)$ is increasing with respect to x₁, then from the symmetry of ${\varphi }_r^{*}\left(x\right)$ with respect to x _i (i = 1, 2,..., n) we know that ${\varphi }_r^{*}\left(x\right)$ is increasing with respect to each x _i (i = 1, 2,..., n).

Proof of Corollary 1.1. By Theorem 1.1 and Lemma 2.6, we have ${\varphi }_r\left(\frac{c-x}{\frac{nc}{s}-1}\right)\ge {\varphi }_r\left(x\right)$ and ${\varphi }_r\left(\frac{c+x}{s+nc}\right)\ge {\varphi }_r\left(\frac{x}{s}\right)$ which imply Corollary 1.1.

Remark 1. Let $0<x_i\le \frac{1}{2},i=1,2,\dots ,n$, then

$$\frac{G_n\left(x\right)}{G_n\left(1-x\right)}\le \frac{A_n\left(x\right)}{A_n\left(1-x\right)},$$

(3.18)

where (1 - x) = (1 - x₁, 1 - x₂,... , 1 - x _n ), commonly referred to as Ky Fan inequality (see [15]), which has attracted the attention of a considerable number of mathematicians (see [16–20]).

Letting ${\sum }_{i=1}^n{x}_i\le 1$ and taking c = 1 in Corollary 1.1, we get

$$\frac{G_n\left(x\right)}{G_n\left(1-x\right)}=\frac{F_n\left(x,n\right)}{F_n\left(1-x,n\right)}\le \frac{F_n\left(x,n-1\right)}{F_n\left(1-x,n-1\right)}\le \cdots \le \frac{F_n\left(x,1\right)}{F_n\left(1-x,1\right)}=\frac{A_n\left(x\right)}{A_n\left(1-x\right)}.$$

(3.19)

It is obvious that inequality (3.19) can be called Ky Fan-type inequality.

Remark 2. Let x _i > 0, i = 1, 2,..., n, the following inequalities

$$\prod _{i=1}^n\left(x_i^{-1}-1\right)\ge {\left(n-1\right)}^n$$

and

$$\prod _{i=1}^n\left(x_i^{-1}+1\right)\ge {\left(n+1\right)}^n$$

are the well-known Weierstrass inequalities (see [11]).

Taking c = s = 1 in Corollary 1.1, one has

$$\prod _{i=1}^n\left(x_i^{-1}-1\right)\ge {\left(\frac{F_n\left(1-x,n-1\right)}{F_n\left(x,n-1\right)}\right)}^n\ge \cdots \ge {\left(\frac{F_n\left(1-x,2\right)}{F_n\left(x,2\right)}\right)}^n\ge {\left(n-1\right)}^n$$

and

$$\prod _{i=1}^n\left(x_i^{-1}+1\right)\ge {\left(\frac{F_n\left(1+x,n-1\right)}{F_n\left(x,n-1\right)}\right)}^n\ge \cdots \ge {\left(\frac{F_n\left(1+x,2\right)}{F_n\left(x,2\right)}\right)}^n\ge {\left(n+1\right)}^n.$$

It is obvious that our inequalities can be called Weierstrass-type inequalities.

Proof of Corollary 1.2. By Theorem 1.2 and Lemma 2.6, we have ${\varphi }_r^{*}\left(\frac{c-x}{\frac{nc}{s}-1}\right)\ge {\varphi }_r^{*}\left(x\right)$ and ${\varphi }_r^{*}\left(\frac{c+x}{s+nc}\right)\ge {\varphi }_r^{*}\left(\frac{x}{s}\right)$, which imply Corollary 1.2.