Shannon Type Inequalities for Kapur’s Entropy

Abstract

In the paper, by methods of the theory of majorization, the authors establish the Schur m-convexity and Shannon type inequalities for Kapur’s entropy.

1. Introduction and Main Results

Let $p=(p_1,p_2,\dots ,p_n)$ be a probability vector, that is, $p_i\ge 0$ for $1\le i\le n$ and ${\sum }_{i=1}^n{p}_i=1$. The quantity

$$H\left(p\right)=-\sum _{i=1}^n{p}_{i}ln{p}_i,$$

where, if $p_i=0$, we put $p_{i}ln{p}_i=0$ in the sum, is known as entropy of p, or the Shannon information entropy of p.

Theorem 1

([1] (p. 101)). Let $p=(p_1,p_2,\dots ,p_n)$ be a probability vector. Then $H\left(p\right)$ is Schur-concave on ${[0,1]}^n$. In particular, we have

$$H(1,0,\dots ,0)\le H\left(p\right)\le H\left(p_0\right)=lnn,$$

where $p_0=\left(\frac{1}{n},\frac{1}{n},\dots ,\frac{1}{n}\right)$.

Let p be a probability vector and $t\in (0,\infty )$. The quantity

$$H_t\left(p\right)=H_t(p_1,p_2,\dots ,p_n)=-\frac{1}{{\sum }_{i=1}^n{p}_i^t}\sum _{i=1}^n{p}_i^{t}ln{p}_i^t$$

is known [2] as Kapur’s entropy of order 1 and type $t\in (0,\infty )$. It is easy to see that

$$H_t\left(p_0\right)=-ln{A}_n\left(p_0^t\right)=tlnn,$$

where $p^t=\left(p_1^t,p_2^t,\dots ,p_n^t\right)$ and $A_n\left(p^t\right)=\frac{1}{n}{\sum }_{i=1}^n{p}_i^t$. It is easy to see that the inequality $H_1\left(p\right)\le lnn$ holds for each probability vector p. However, the inequality

$$H_t\left(p\right)\le lnn$$

(1)

does not hold for each $t\in (0,\infty )$ and each probability vector p. Stolarsky [3] shows that the inequality (1) holds for $t\ge t_0\left(n\right)$, where $t_0\left(2\right)=\frac{1}{2}$. Hereafter, Clausing [4] proved that if $n>3$ and $t=t_0\left(n\right)$, then the inequality (1) holds for each probability vector p. Thus, with respect to $t\in (0,\infty )$, the function $H_t\left(p\right)$ is not strictly Schur-convex. For more information about this topic, the reader is referred to the papers [2,4,5,6,7,8,9,10,11,12,13,14] and the closely related references therein.

In this paper, we will establish the Schur m-convexity and some Shannon type inequalities for Kapur’s entropy $H_t\left(p\right)$.

Our main results are the following two theorems.

Theorem 2.

Let $p\in {(0,\infty )}^n$, $t\in (0,\infty )$, and ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n\in [0,1]$ with ${\sum }_{i=1}^n{\lambda }_i=1$ for $n\ge 2$. Then the function $H_t\left(p\right)$ is Schur m-concave on ${(0,\infty )}^n$ for $m=t$ and

$$H_t\left(p\right)\le H_t\left(M_t(1,\lambda ;p),M_t(2,\lambda ;p),\dots ,M_t(n,\lambda ;p)\right)\le -ln{A}_n\left(p^t\right)$$

(2)

with the equality in (2) holding for $p_0=\left(\frac{1}{n},\frac{1}{n},\dots ,\frac{1}{n}\right)$, where

$$M_t(k,\lambda ;p)={\left(\sum _{i=1}^n{\lambda }_i{p}_{i+k-1}^t\right)}^{1/t},k=1,2,\dots ,n$$

(3)

with $p_{n+i}=p_i$ for $i=1,2,\dots ,n-1$.

Theorem 3.

Let $p\in {(0,1)}^n$, $t\in (0,1]$, and ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n\in [0,1]$ with ${\sum }_{i=1}^n{p}_i={\sum }_{i=1}^n{\lambda }_i=1$ for $n\ge 1$.

1.

If $0<t<\frac{1}{2}$ and $p\in [q,1)$ with $\frac{1}{q}\ge n$, then

$$lnn\le -\sum _{k=1}^n{\left[M_1(k,\lambda ;p)\right]}^{t}ln{M}_1(k,\lambda ;p).$$

2.

If $0<t<\frac{1}{2}$ and $p\in {(0,q]}^n$ with $\frac{1}{q}\le n$, or if $\frac{1}{2}\le t\le 1$ and $p\in {(0,1]}^n$, then

$$\begin{array}{c}\hfill \frac{H_t\left(p\right)}{t}\le -\frac{1}{n}\sum _{k=1}^n{\left[M_1(k,\lambda ;p)\right]}^{t}ln{M}_1(k,\lambda ;p)\le min\left\{\frac{1}{t},n^{1-t}\right\}lnn,\end{array}$$

where $q=e^{(2t-1)/t(1-t)}$ for $0<t<\frac{1}{2}$ and $M_1(k,\lambda ;p)$ is defined by (3).

2. Definitions and Lemmas

For proving our main results, we need several definitions and lemmas below.

It is well known that a function $\phi (x_1,x_2,\dots ,x_n)$ of n variables is said to be symmetric if its value is unchanged for any permutation of its n variables $x_1,x_2,\dots ,x_n$.

Definition 1

([1,15]). Let $x=(x_1,x_2,\dots ,x_n)$ and $y=(y_1,y_2,\dots ,y_n)\in {\mathbb{R}}^n$.

1.

A tuple x is said to be majorized by y (in symbols $x\prec y$) if ${\sum }_{i=1}^k{x}_{\left[i\right]}\le {\sum }_{i=1}^k{y}_{\left[i\right]}$ for $k=1,2,\dots ,n-1$ and ${\sum }_{i=1}^n{x}_i={\sum }_{i=1}^n{y}_i$, where $x_{\left[1\right]}\ge x_{\left[2\right]}\ge \dots \ge x_{\left[n\right]}$ and $y_{\left[1\right]}\ge y_{\left[2\right]}\ge \dots \ge y_{\left[n\right]}$ are rearrangements of x and y in a descending order.

2.

A set $\mathsf{\Omega }\subseteq {\mathbb{R}}^n$ is called convex if $(\lambda x_1+(1-\lambda )y_1,\lambda x_2+(1-\lambda )y_2,\dots ,\lambda x_n+(1-\lambda )y_n)\in \mathsf{\Omega }$ for each $x,y\in \mathsf{\Omega }$ and $\lambda \in [0,1]$.

3.

Let $\mathsf{\Omega }\subseteq {\mathbb{R}}^n$. A function $\phi :\mathsf{\Omega }\to \mathbb{R}$ is said to be Schur-convex on Ω if $x\prec y$ on Ω implies $\phi \left(x\right)\le$$\phi \left(y\right)$. A function φ is said to be Schur-concave on Ω if and only if $-\phi$ is Schur-convex.

Definition 2

([16]). Let $\mathsf{\Omega }\subseteq {(0,\infty )}^n$ and $\phi :\mathsf{\Omega }\to (0,\infty )$.

1.

The set Ω is said to be geometrically convex if $\left(x_1^{\lambda }{y}_1^{1-\lambda },x_2^{\lambda }{y}_2^{1-\lambda },\dots ,x_n^{\lambda }{y}_n^{1-\lambda }\right)\in \mathsf{\Omega }$ for each $x,y\in \mathsf{\Omega }$ and $\lambda \in [0,1]$.

2.

The function φ is said to be Schur-geometrically convex on Ω if the majorization $lnx=(ln{x}_1,ln{x}_2,\dots ,ln{x}_n)\prec lny=(ln{y}_1,ln{y}_2,\dots ,ln{y}_n)$ implies $\phi \left(x\right)\le \phi \left(y\right)$ for each $x,y\in \mathsf{\Omega }$.

Definition 3

([17]). Let $\mathsf{\Omega }\subseteq {(0,\infty )}^n$ and $\phi :\mathsf{\Omega }\to (0,\infty )$.

1.

The set Ω is said to be harmonically convex if $\frac{xy}{\lambda x+(1-\lambda )y}\in \mathsf{\Omega }$ for each $x,y\in \mathsf{\Omega }$ and $\lambda \in [0,1]$.

2.

The function φ is said to be Schur-harmonically convex on Ω if the majorization $\frac{1}{x}=\left(\frac{1}{x_1},\frac{1}{x_2},\dots ,\frac{1}{x_n}\right)\prec \frac{1}{y}=\left(\frac{1}{y_1},\frac{1}{y_2},\dots ,\frac{1}{y_n}\right)$ implies $\phi \left(x\right)\le \phi \left(y\right)$ for each $x,y\in \mathsf{\Omega }$.

Definition 4

([18,19,20]). Let $f_m\left(x\right)=\left\{\begin{array}{cc}\frac{x^m-1}{m},\hfill & m\ne 0;\hfill \\ lnx,\hfill & m=0.\hfill \end{array}\right.$ A function $\phi :\mathsf{\Omega }\subseteq {(0,\infty )}^n\to \mathbb{R}$ is said to be Schur m-power convex on Ω if the majorization relation $(f_m\left(x_1\right),f_m\left(x_2\right),\dots f_m\left(x_n\right))\prec (f_m\left(y_1\right),f_m\left(y_2\right),\dots f_m\left(y_n\right))$ on Ω implies that $\phi \left(x\right)\le \phi \left(y\right)$.

Remark 1.

It is not difficult to see that, when $m\ne 0$,

$$\left(\frac{x_1^m-1}{m},\frac{x_2^m-1}{m},\dots ,\frac{x_n^m-1}{m}\right)\prec \left(x_1^m-1,x_2^m-1,\dots ,x_n^m-1\right)\prec \left(x_1^m,x_2^m,\dots ,x_n^m\right).$$

When taking $m=1,0,-1$, we derive that $f_1\left(x\right)=x-1$, $f_0\left(x\right)=lnx$, and $f_{-1}\left(x\right)=1-\frac{1}{x}$ in Definition 4. Therefore, we can derive from Definition 4 the Schur-convexity, Schur-geometric convexity, and Schur-harmonic convexity, respectively.

Lemma 1

([1,15,21]). Let $\mathsf{\Omega }\subseteq {\mathbb{R}}^n$ be a symmetric convex set with nonempty interior and let $\phi :\mathsf{\Omega }\to (0,\infty )$ be a continuous symmetric function which is differentiable in the interior ${\mathsf{\Omega }}^{\circ }$. Then φ is a Schur-convex function on Ω if and only if

$$(x_1-x_2)\left[\frac{\partial \phi \left(x\right)}{\partial x_1}-\frac{\partial \phi \left(x\right)}{\partial x_2}\right]\ge 0,x\in {\mathsf{\Omega }}^{\circ }.$$

Lemma 2

([18,19,20]). Let $\mathsf{\Omega }\subseteq {(0,\infty )}^n$ be a symmetric set with nonempty interior ${\mathsf{\Omega }}^{\circ }$ and let $\phi :\mathsf{\Omega }\to \mathbb{R}$ be continuous on Ω and differentiable in ${\mathsf{\Omega }}^{\circ }$. Then φ is Schur m-power convex on Ω if and only if φ is symmetric on Ω such that

$$\frac{x_1^m-x_2^m}{m}\left[x_1^{1-m}\frac{\partial \phi \left(x\right)}{\partial x_1}-x_2^{1-m}\frac{\partial \phi \left(x\right)}{\partial x_2}\right]\ge 0,m\ne 0$$

and

$$(ln{x}_1-ln{x}_2)\left[x_1\frac{\partial \phi \left(x\right)}{\partial x_1}-x_2\frac{\partial \phi \left(x\right)}{\partial x_2}\right]\ge 0,m=0$$

for $x\in {\mathsf{\Omega }}^{\circ }$.

For more information about various convexity named after Schur, please refer to [22,23,24,25,26,27,28,29,30,31,32,33,34] and closely related references therein.

3. Shannon Type Inequalities for Kapur’s Entropy

Now we are in a position to prove our main results.

Proof of Theorem 2.

Calculating the partial derivatives of function $H_t\left(p\right)$ in $p_1$ and $p_2$ yields

$$\frac{\partial H_t\left(p\right)}{\partial p_1}=\frac{-t{p}_1^{t-1}}{{[p_1^t+p_2^t+A\left(p\right)]}^2}\left[p_1^t+p_2^t+A\left(p\right)-B\left(p\right)+(p_2^t+A\left(p\right))ln{p}_1^t-p_2^{t}ln{p}_2^t\right]$$

and

$$\frac{\partial H_t\left(p\right)}{\partial p_2}=\frac{-t{p}_2^{t-1}}{{[p_1^t+p_2^t+A\left(p\right)]}^2}\left[p_1^t+p_2^t+A\left(p\right)-B\left(p\right)+(p_1^t+A\left(p\right))ln{p}_2^t-p_1^{t}ln{p}_1^t\right],$$

where $A\left(p\right)={\sum }_{i=3}^n{p}_i^t$ and $B\left(p\right)={\sum }_{i=3}^n{p}_i^{t}ln{p}_i^t$ under the condition that any empty sum is understood as null.

Letting $t=m$ in Definition 4 and utilizing Lemma 2 give

$$\begin{array}{c}\frac{p_1^m-p_2^m}{m}\left[p_1^{1-m}\frac{\partial H_t\left(p\right)}{\partial p_1}-p_2^{1-m}\frac{\partial H_t\left(p\right)}{\partial p_2}\right]\hfill \\ =-\frac{t(p_1^m-p_2^m)}{m{(p_1^t+p_2^t+A\left(p\right))}^2}[\left(p_1^t+p_2^t+A\left(p\right)-B\left(p\right)\right)\left(p_1^{t-m}-p_2^{t-m}\right)\hfill \\ +A\left(p\right)\left(p_1^{t-m}ln{p}_1^t-p_2^{t-m}ln{p}_2^t\right)+p_1^{t-m}{p}_2^{t-m}\left(p_1^m+p_2^m\right)\left(ln{p}_1^t-ln{p}_2^t\right)]\hfill \\ =-\frac{p_1^t-p_2^t}{{(p_1^t+p_2^t+A\left(p\right))}^2}\left[A\left(p\right)\left(ln{p}_1^t-ln{p}_2^t\right)+\left(p_1^t+p_2^t\right)\left(ln{p}_1^t-ln{p}_2^t\right)\right]\hfill \\ =-\frac{1}{{(p_1^t+p_2^t+A\left(p\right))}^2}\left(p_1^t-p_2^t\right)\left(ln{p}_1^t-ln{p}_2^t\right)\sum _{i=1}^n{p}_i^t\hfill \\ \le 0.\hfill \end{array}$$

Hence, the function $H_t\left(p\right)$ is Schur m-concave on ${(0,\infty )}^n$ for $m=t$.

For ${\ell }_1,{\ell }_2,\dots ,{\ell }_k\in \{1,2,\dots ,n\}$ and $1\le k\le n$, we have

$$\sum _{j=1}^k{p}_{\left[j\right]}^t-\sum _{j=1}^k\sum _{i=1}^n{\lambda }_i{p}_{i+{\ell }_j-1}^t=\sum _{j=1}^k{p}_{\left[j\right]}^t-\sum _{i=1}^n{\lambda }_i\sum _{j=1}^k{p}_{i+{\ell }_j-1}^t\ge \sum _{j=1}^k{p}_{\left[j\right]}^t-\sum _{i=1}^n{\lambda }_i\sum _{j=1}^k{p}_{\left[j\right]}^t=0.$$

Let $y_j={\sum }_{i=1}^n{\lambda }_i{p}_{i+j-1}^t$ for $j=1,2,\dots ,n$. If there exists $1\le k_0<n$ such that ${\sum }_{j=1}^{k_0}{y}_{\left[j\right]}\ge k_0{A}_n\left(p^t\right)$ and ${\sum }_{j=1}^{k_0+1}{y}_{\left[j\right]}<(k_0+1)A_n\left(p^t\right)$, that is,

$$\sum _{i=1}^n{p}_i^t=\sum _{j=1}^{k_0+1}{y}_{\left[j\right]}+\sum _{k_0+2}^n{y}_{\left[j\right]}<(k_0+1)A_n\left(p^t\right)+(n-k_0-1)A_n\left(p^t\right)=\sum _{i=1}^n{p}_i^t,$$

Then, it follows that

$$\begin{array}{c}\hfill \left(A_n\left(p^t\right),A_n\left(p^t\right),\dots ,A_n\left(p^t\right)\right)\prec \left({\left[M_t(1,\lambda ;p)\right]}^t,{\left[M_t(2,\lambda ;p)\right]}^t,\dots ,{\left[M_t(n,\lambda ;p)\right]}^t\right)\prec \left(p_1^t,p_2^t,\dots ,p_n^t\right).\end{array}$$

Further by virtue of Definition 4, we have

$$H_t\left(p\right)\le H_t\left(M_t(1,\lambda ;p),M_t(2,\lambda ;p),\dots ,M_t(n,\lambda ;p)\right)\le -ln{A}_n\left(p^t\right).$$

The proof of Theorem 2 is thus complete. □

Proof of Theorem 3.

For all $p\in {(0,1)}^n$ and $t\in (0,1]$ with ${\sum }_{i=1}^n{p}_i=1$, by the easily-understandable inequality

$$A_n\left(p^t\right)\ge \frac{1}{n}\sum _{i=1}^n{p}_i=\frac{1}{n}$$

(4)

for $0<t\le 1$, we obtain

$$-\frac{1}{n^{1-t}}\sum _{i=1}^n{p}_i^{t}ln{p}_i^t\le H_t\left(p\right)\le -\sum _{i=1}^n{p}_i^{t}ln{p}_i^t.$$

Let $F_t\left(p\right)=-{\sum }_{i=1}^n{p}_i^{t}ln{p}_i^t$. Then

$$\frac{\partial F_t\left(p\right)}{\partial p_1}=-t{p}_1^{t-1}\left(1+ln{p}_1^t\right)\mathrm{and}\frac{\partial F_t\left(p\right)}{\partial p_2}=-t{p}_2^{t-1}\left(1+ln{p}_2^t\right).$$

By Lemma 1, we obtain

$$(p_1-p_2)\left[\frac{\partial F_t\left(p\right)}{\partial p_1}-\frac{\partial F_t\left(p\right)}{\partial p_2}\right]=t(p_1-p_2)\left[g\left(p_2\right)-g\left(p_1\right)\right],$$

where $g\left(p\right)=p^{t-1}+p^{t-1}ln{p}^t$ for $p\in (0,1]$. Then

• when $0<t<\frac{1}{2}$, the function $g\left(p\right)$ is increasing on $(0,q]$ and decreasing on $[q,1)$;
• when $\frac{1}{2}\le t\le 1$, the function $g\left(p\right)$ is increasing on $(0,1]$.

Therefore, by Lemma 1, we find that

• if $0<t<\frac{1}{2}$, the function $F_t\left(p\right)$ is Schur-convex on ${[q,1)}^n$;
• if $0<t<\frac{1}{2}$, the function $F_t\left(p\right)$ is Schur-concave on ${(0,q]}^n$;
• if $\frac{1}{2}\le t\le 1$, the function $F_t\left(p\right)$ is Schur-concave on ${(0,1]}^n$.

For ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n\in [0,1]$ with ${\sum }_{i=1}^n{\lambda }_i=1$, we have

$$\left(\frac{1}{n},\frac{1}{n},\dots ,\frac{1}{n}\right)\prec \left(\sum _{i=1}^n{\lambda }_i{p}_i,\sum _{i=1}^n{\lambda }_i{p}_{i+1},\dots ,\sum _{i=1}^n{\lambda }_i{p}_{i+n-1}\right)\prec p.$$

By Definition 1, we see that

• if $0<t<\frac{1}{2}$ and $p\in {[q,1)}^n$ with $\frac{1}{q}\ge n$, then

  $$H_t\left(p\right)\ge n^{t-1}{F}_t\left(p\right)\ge -\frac{1}{n^{1-t}}\sum _{k=1}^n{\left[M_1(k,\lambda ;p)\right]}^{t}ln{\left[M_1(k,\lambda ;p)\right]}^t\ge \frac{tlnn}{n};$$
• if $0<t<\frac{1}{2}$ and $p\in {(0,q]}^n$ with $\frac{1}{q}\le n$, then

  $$H_t\left(p\right)\le F_t\left(p\right)\le -\sum _{k=1}^n{\left[M_1(k,\lambda ;p)\right]}^{t}ln{\left[M_1(k,\lambda ;p)\right]}^t\le t{n}^{1-t}lnn;$$

  (5)
• if $\frac{1}{2}\le t\le 1$ and $p\in {(0,1)}^n$, the inequality (5) still holds.

The proof of Theorem 3 is thus complete. □

4. Remarks

Finally we list several remarks on our main results and closely related ones.

Remark 2.

Let $p\in {(0,\infty )}^n$, $t\ge 1$, ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n\in [0,1]$ with ${\sum }_{i=1}^n{\lambda }_i=1$ for $n\ge 2$. Since ${\left[A_n\left(p\right)\right]}^t\le A_n\left(p^t\right)$ for $t\ge 1$, from Theorem 2, it follows that

$$H_t\left(p\right)\le H_t(M_t(1,\lambda ;p),M_t(2,\lambda ;p),\dots ,M_t(n,\lambda ;p))\le -ln{A}_n\left(p^t\right)\le -tln{A}_n\left(p\right)$$

with equality for $p_1=p_2=\dots =p_n>0$, where $M_1(k,\lambda ;p)$ is defined by (3).

Remark 3.

Let $p\in {(0,1)}^n$, $t\in (0,\infty )$, and ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n\in [0,1]$ with ${\sum }_{i=1}^n{p}_i={\sum }_{i=1}^n{\lambda }_i=1$ and $n\ge 2$. From Remark 2 and the inequality (4), we conclude the following results.

1.

If $t\ge 1$, we have

$$H_t\left(p\right)\le H_t(M_t(1,\lambda ;p),M_t(2,\lambda ;p),\dots ,M_t(n,\lambda ;p))\le -ln{A}_n\left(p^t\right)\le tlnn$$

(6)

with equality for $p_0=\left(\frac{1}{n},\frac{1}{n},\dots ,\frac{1}{n}\right)$.

2.

If $0<t<1$, we have

$$H_t\left(p\right)\le H_t(M_t(1,\lambda ;p),M_t(2,\lambda ;p),\dots ,M_t(n,\lambda ;p))\le -ln{A}_n\left(p^t\right)\le lnn.$$

In particular, when ${\lambda }_1={\lambda }_2=\frac{1}{2}$ and ${\lambda }_3=\dots ={\lambda }_n=0$,

1.

if $t\ge 1$, then

$$H_t\left(p\right)\le H_t\left(\sqrt[t]{\frac{p_1^t+p_2^t}{2}},\sqrt[t]{\frac{p_2^t+p_3^t}{2}},\dots ,\sqrt[t]{\frac{p_n^t+p_1^t}{2}}\right)\le -ln{A}_n\left(p^t\right)\le tlnn$$

with equality for $p_0=\left(\frac{1}{n},\frac{1}{n},\dots ,\frac{1}{n}\right)$;

2.

if $0<t<1$, then

$$H_t\left(p\right)\le H_t\left(\sqrt[t]{\frac{p_1^t+p_2^t}{2}},\sqrt[t]{\frac{p_2^t+p_3^t}{2}},\dots ,\sqrt[t]{\frac{p_n^t+p_1^t}{2}}\right)\le -ln{A}_n\left(p^t\right)\le lnn.$$

Remark 4.

When $0<t<1$, the inequality (6) does not necessarily hold. For $t=0.369$, ${\lambda }_1=1$, ${\lambda }_2={\lambda }_3=0$, and $p=(0.3,0.69,0.01)$, we have

$$0.4053\dots =tlnn<H_t\left(p\right)=0.4215\dots <lnn=1.0986\dots .$$

Remark 5.

Under conditions of Theorems 2 and 3, if $0<t<\frac{1}{2}$ and $p\in [q,1)$ with $\frac{1}{q}\ge n$, then

$$\begin{array}{c}\frac{tlnn}{n}\le -\frac{1}{n}\sum _{k=1}^n{\left[M_1(k,\lambda ;p)\right]}^{t}ln{\left[M_1(k,\lambda ;p)\right]}^t\le H_t\left(p\right)\hfill \\ \le H_t(M_t(1,\lambda ;p),M_t(2,\lambda ;p),\dots ,M_t(n,\lambda ;p))\le -ln{A}_n\left(p^t\right)\le min\{1,t{n}^{1-t}\}lnn.\hfill \end{array}$$