Sharp Inequalities for the Hardy–Littlewood Maximal Operator on Finite Directed Graphs

Abstract

In this paper, we introduce and study the Hardy–Littlewood maximal operator $M_{\overrightarrow{G}}$ on a finite directed graph $\overrightarrow{G}$. We obtain some optimal constants for the ${\ell }^p$ norm of $M_{\overrightarrow{G}}$ by introducing two classes of directed graphs.

1. Introduction

The best constants for the Hardy–Littlewood maximal inequalities have always been a challenging topic of research. In 1997, Grafakos and Montgomery-Smith [1] first obtained the sharp $L^p\left(\mathbb{R}\right)(1<p<\infty )$ norm for the one-dimensional uncentered Hardy–Littlewood maximal operator. Since then, the best constants for Hardy–Littlewood maximal operator have been studied extensively. See [2] for the sharp $L^p\left(\mathbb{R}\right)(1<p<\infty )$ norm of the one-dimensional centered Hardy–Littlewood maximal operator as well as [3,4,5,6,7,8,9] for the optimal constants on the weak $(1,1)$ norm of the centered Hardy–Littlewood maximal operator. Recently, Soria and Tradacete [10] studied the sharp ${\ell }^p$-norm for the Hardy-Littlewoood maximal operators on finite connected graphs. It should be pointed out that geometric structure of a graph plays an important role in studying maximal operators on graphs. Given the significance of this operator, it is an interesting and natural question to ask what happens when we consider the directed graphs. It is the purpose of this paper to investigate the optimal constants for the ${\ell }^p$ norm of the Hardy–Littlewood maximal operator in directed graph setting.

Let us now recall some known notations, definitions and backgrounds. Let $G=(V,E)$ be an undirected combinatorial graph with the set of vertices V and the set of edges E. Two vertices $x,y\in V$ are called neighbors if they are connected by an edge $x\sim y\in E$. For a $v\in V$, we use the notation $N_G\left(v\right)$ to denote the set of neighbors of v. We say that G is finite if $\left|V\right|<\infty$. Here the notation $\left|A\right|$ represents the cardinality of A for each subset $A\subset V$. The graph G is called connected if for any distinct $x,y\in V$, there is a finite sequence of vertices ${\left\{x_i\right\}}_{i=0}^k$, $k\in \mathbb{N}$, such that $x=x_0\sim x_1\sim \cdots \sim x_k=y$. Let $d_G$ be the metric induced by the edges in E, i.e., given $u,v\in V$, the distance $d_G(u,v)$ is the number of edges in the shortest path connecting u and v. Let $B_G(v,r)$ be the ball centered at v, with radius r on the graph G, i.e.,

$$B_G(v,r)=\{u\in V:d_G(u,v)\le r\}.$$

For example, $B_G(v,r)=\left\{v\right\}$ if $0\le r<1$ and $B_G(v,r)=\left\{v\right\}\cup N_G\left(v\right)$ if $1\le r<2$. For a function $f:V\to \mathbb{R}$, the Hardy–Littlewood maximal operator $M_G$ on G is defined as

$$M_{G}f\left(v\right)=\underset{r\ge 0}{sup}\frac{1}{|B_G(v,r)|}\sum _{w\in B_G(v,r)}\left|f\left(w\right)\right|.$$

If G has $n(n\ge 2)$ vertices, the maximal operator $M_G$ can be rewritten by

$$M_{G}f\left(v\right)=\underset{k=0,\dots ,n-1}{max}\frac{1}{|B_G(v,k)|}\sum _{w\in B_G(v,k)}\left|f\left(w\right)\right|.$$

Over the last several years the Hardy–Littlewood maximal operator on graphs has been studied by many authors (see [10,11,12,13,14,15,16]). The Hardy-Littlewoood maximal operator on graphs was first introduced and studied by Korányi and Picardello [15] who used the above operator to explore the boundary behavior of eigenfunctions of the Laplace operator on trees. Subsequently, Cowling, Meda and Setti [12] studied the Hardy-Littlewoood maximal operator on homogeneous trees. Later, some weighted norm inequalities for the Hardy-Littlewoood maximal operators on infinite graphs were investigated by Badr and Martell [11]. Recently, Soria and Tradacete [10] studied the best constants for the ${\ell }^p$-norm of the Hardy-Littlewoood maximal operators on finite connected graphs. Later, Soria and Tradacete [16] investigated some different geometric properties on infinite graphs, related to the weak-type boundedness of the Hardy–Littlewood maximal operator on infinite connected graphs. One can consult [13,14] for the variation properties of the Hardy–Littlewood maximal operator on finite connected graphs.

We now introduce the ${\ell }^p$ spaces on graphs.

Definition 1

(${\ell }^p\left(V\right)$ space).Let $G=(V,E)$ be a graph with the set of vertices V and the set of edges E. For $0<p\le \infty$, let ${\ell }^p\left(V\right)$ be the set of all functions $f:V\to \mathbb{R}$ satisfying ${\parallel f\parallel }_{{\ell }^p\left(V\right)}<\infty$, where ${\parallel f\parallel }_{{\ell }^p\left(V\right)}=({\sum }_{v\in V}{\left|f\left(v\right)\right|}^p{)}^{1/p}$ for all $0<p<\infty$ and ${\parallel f\parallel }_{{\ell }^{\infty }\left(V\right)}={sup}_{v\in V}\left|f\left(v\right)\right|$.

By Hölder’s inequality, we have

$${\parallel f\parallel }_{{\ell }^q\left(V\right)}\le {\parallel f\parallel }_{{\ell }^p\left(V\right)}\le {\left|V\right|}^{1/p-1/q}{\parallel f\parallel }_{{\ell }^q\left(V\right)},\mathrm{for}0<p<q\le \infty .$$

(1)

On the other hand, it is easy to see that $\left|f\left(v\right)\right|\le M_{G}f\left(v\right)\le {\parallel f\parallel }_{{\ell }^{\infty }\left(V\right)}$ for all $v\in V$. This together with (1) yields that

$${\parallel f\parallel }_{{\ell }^p\left(V\right)}\le \parallel M_G{f\parallel }_{{\ell }^p\left(V\right)}\le {\left|V\right|}^{1/p}{\parallel f\parallel }_{{\ell }^{\infty }\left(V\right)}\le {\left|V\right|}^{1/p}{\parallel f\parallel }_{{\ell }^p\left(V\right)},\mathrm{for}0<p\le \infty .$$

(2)

Therefore, the ${\ell }^p$-boundedness for $M_G$ is trivial. Moreover, it follows from (2) that

$$1\le \underset{{\parallel f\parallel }_{{\ell }^p\left(V\right)}\ne 0}{sup}\frac{\parallel M_G{f\parallel }_{{\ell }^p\left(V\right)}}{{\parallel f\parallel }_{{\ell }^p\left(V\right)}}\le {\left|V\right|}^{1/p},\mathrm{for}0<p\le \infty .$$

In [10], among other things, Soria and Tradacete studied the sharp constants

$$\parallel M_G{\parallel }_p:=\underset{{\parallel f\parallel }_{{\ell }^p\left(V\right)}\ne 0}{sup}\frac{\parallel M_G{f\parallel }_{{\ell }^p\left(V\right)}}{{\parallel f\parallel }_{{\ell }^p\left(V\right)}},\mathrm{for}0<p\le \infty .$$

Precisely, they established the following result.

Theorem 1

([10]). Let $n\ge 2$.

(i) 

Let $0<p\le 1$. Then, for any graph G with n vertices, we have

$${\left(1+\frac{n-1}{n^p}\right)}^{1/p}\le {\parallel M_G\parallel }_p\le {\left(1+\frac{n-1}{2^p}\right)}^{1/p}.$$

Moreover,

(a) 

$\parallel M_G{\parallel }_p={(1+\frac{n-1}{n^p})}^{1/p}$ if and only if $G=K_n$. Here $K_n$ denotes the complete graph with n vertices, i.e., $|N_{K_n}\left(v\right)|=n-1$ for any $v\in V$.

(b) 

$\parallel M_G{\parallel }_p={(1+\frac{n-1}{2^p})}^{1/p}$ if and only if G is isomorphic to $S_n$. Here $S_n$ denotes the star graph of n vertices, i.e., there exists a unique $v\in V$ such that $|N_{S_n}\left(v\right)|=n-1$ and $|N_{S_n}\left(w\right)|=1$ for every $w\in V\setminus \left\{v\right\}$.

(ii) 

Let $1<p<\infty$, then

$${\left(1+\frac{n-1}{n^p}\right)}^{1/p}\le {\parallel M_{K_n}\parallel }_p\le {\left(1+\frac{n-1}{n}\right)}^{/p},$$

and

$${\left(1+\frac{n-1}{2^p}\right)}^{1/p}\le {\parallel M_{S_n}\parallel }_p\le {\left(\frac{n+5}{2}\right)}^{1/p}.$$

The main motivation of this paper is to extend Theorem A to the directed graph setting. Let $\overrightarrow{G}=(V,E)$ be a finite graph with the set of vertices V and the set of edges E. Given an edge $u\sim v\in E$, if $u\to v$, we say that v (resp., u) is a right (resp., left) neighbor of u (resp., v). Then we write $u\sim v=u\to v$. For $v\in V$, we denote by $N_{\overrightarrow{G},+}\left(v\right)$ (resp., $N_{\overrightarrow{G},-}\left(v\right)$) the set of right (resp., left) neighbors of v. We say that the graph $\overrightarrow{G}$ is a directed graph if every edge in E has only a unique direction and $N_{\overrightarrow{G},+}\left(v\right)\cup N_{\overrightarrow{G},-}\left(v\right)\ne \varnothing$ for all $v\in V$. The directed graph $\overrightarrow{G}$ is called connected if for any distinct $x,y\in V$, there is a finite sequence of vertices ${\left\{x_i\right\}}_{i=0}^k$, $k\in \mathbb{N}$, such that $x=x_0\to x_1\to \cdots \to x_k=y$.

In what follows, we always assume that the graph $\overrightarrow{G}=(V,E)$ with the set of vertices V and the set of edges E. Let $B_{\overrightarrow{G}}(v,r)$ be the ball centered at v, with radius r on the graph $\overrightarrow{G}$, equipped with the metric $d_{\overrightarrow{G}}$ induced by the edges in E, i.e., given $u,v\in V$, the distance $d_{\overrightarrow{G}}(u,v)$ is the number of edges in a shortest path connecting from u to v, and

$$B_{\overrightarrow{G}}(v,r)=\{u\in V:d_{\overrightarrow{G}}(v,u)\le r\}.$$

For example, $B_{\overrightarrow{G}}(v,r)=\left\{v\right\}$ if $0\le r<1$ and $B_{\overrightarrow{G}}(v,r)=\left\{v\right\}\cup N_{\overrightarrow{G},+}\left(v\right)$ if $1\le r<2$. For a function $f:V\to \mathbb{R}$, we consider the Hardy–Littlewood maximal operator on $\overrightarrow{G}$

$$M_{\overrightarrow{G}}f\left(v\right)=\underset{r\ge 0}{sup}\frac{1}{|B_{\overrightarrow{G}}(v,r)|}\sum _{w\in B_{\overrightarrow{G}}(v,r)}\left|f\left(w\right)\right|.$$

Naturally, when $\left|V\right|=n$, the maximal operator $M_{\overrightarrow{G}}$ can be redefined in the way that

$$M_{\overrightarrow{G}}f\left(v\right)=\underset{k=0,\dots ,n-1}{max}\frac{1}{|B_{\overrightarrow{G}}(v,k)|}\sum _{w\in B_{\overrightarrow{G}}(v,k)}\left|f\left(w\right)\right|.$$

There are some remarks as follows:

Remark 1.

• This type of operator $M_{\overrightarrow{G}}$ has its roots in the ergodic theory in infinite directed graph setting. More precisely, let $\overrightarrow{G_1}=(V_1,E_1)$, where $V_1=\mathbb{Z}$ and $E_1=\{i\to i+1:i\in \mathbb{Z}\}$. Then $M_{\overrightarrow{G_1}}$ is the usual one-dimensional one-sided discrete Hardy–Littlewood maximal operator $M_1$, i.e.,

  $$M_{1}f\left(n\right)=\underset{r\in \{0,1,\dots ,\}}{sup}\frac{1}{r+1}\sum _{k=0}^r\left|f(n+k)\right|,n\in \mathbb{Z}.$$
  This type of maximal operator $M_1$ first arose in Dunford and Schwartz’s work [17] and was studied by Calderón [18].

(ii) 

It was pointed out in [10] that the complete graph $K_n$ whose maximal operator $M_{K_n}$ is the smallest in the pointwise ordering among all graphs with $n\ge 2$, but there is no graph G whose maximal operator $M_G$ is the largest in the pointwise ordering among all graphs with $n\ge 2$. In Section 2, we point out that there is no directed graph $\overrightarrow{G}$ whose maximal operator $M_{\overrightarrow{G}}$ is the smallest or largest in the pointwise ordering among all graphs with $n\ge 2$ vertices, which is different from $M_G$.

(iii) 

It should be pointed out that as with $M_G$, the maximal operator $M_{\overrightarrow{G}}$ completely determines the graph $\overrightarrow{G}$(see Proposition 2).

It is not difficult to see that

$$\left|f\left(v\right)\right|\le M_{\overrightarrow{G}}f\left(v\right)\le {\parallel f\parallel }_{{\ell }^{\infty }\left(V\right)},\mathrm{for}\mathrm{all}v\in V,$$

which together with (1) leads to $\parallel M_{\overrightarrow{G}}{\parallel }_{\infty }=1$ and

$$1\le \parallel M_{\overrightarrow{G}}{\parallel }_p\le {\left|V\right|}^{1/p},\mathrm{for}0<p<\infty .$$

(3)

Based on (3) and Theorem A, finding the sharp ${\ell }^p$-norm of $M_{\overrightarrow{G}}$ is a certainly interesting issue, which is the main motivation of this work. In Section 3 we shall introduce the outward star graph $\overrightarrow{S_{O,n}}$ and the inward star graph $\overrightarrow{S_{I,n}}$ and prove that $\overrightarrow{S_{O,n}}$ and $\overrightarrow{S_{I,n}}$ are the extremal directed graphs attaining, which completely determine the lower and upper estimates of the ${\ell }^p$-norm for $M_{\overrightarrow{G}}$ in the case $0<p\le 1$, respectively (see Theorem 2). We also claim that the ${\ell }^p$-norm of $M_{\overrightarrow{G}}$ cannot determine the graph $\overrightarrow{G}$ (see Proposition 3). In Section 4, we consider the ${\ell }^p$-norm for $M_{\overrightarrow{G}}$ in the case $1<p<\infty$. Actually, the case $1<p<\infty$ is more complicated than the case $0<p\le 1$, even in the finite undirected graph setting. However, some positive results are discussed. In particular, some sharp estimates of restricted type are given in Section 4.

2. General Properties for ${\mathit{M}}_{\overrightarrow{\mathit{G}}}$

It was pointed out in [10] that there exists a smallest operator $M_{K_n}$, in the pointwise ordering, among all $M_G$, with G a graph of n vertices. However, there is no directed graph $\overrightarrow{G}$ whose maximal operator ${\mathrm{M}}_{\overrightarrow{G}}$ is the smallest or largest in the pointwise ordering among all graphs with $n\ge 2$ vertices, which can be seen by the following result.

Proposition 1.

Let $\overrightarrow{G}=(V,E)$ be a directed connected graph with $n\ge 2$ vertices. Then

(i) 

There exist $j\in V$, a function $f:V\to \mathbb{R}$ and another directed graph ${\overrightarrow{G}}_1=(V,E_1)$ with $E_1\ne E$ such that $M_{\overrightarrow{G}}f\left(j\right)>M_{{\overrightarrow{G}}_1}f\left(j\right)$;

(ii) 

There exist $j\in V$, a function $f:V\to \mathbb{R}$ and another directed graph ${\overrightarrow{G}}_2=(V,E_2)$ with $E_2\ne E$ such that $M_{\overrightarrow{G}}f\left(j\right)<M_{{\overrightarrow{G}}_2}f\left(j\right)$.

Proof. 

At first, we prove (i). When $n=2$, let $\overrightarrow{G}=(V,E)$ with $V=\{u,v\}$ and $E=\{u\to v\}$. Let us consider the function $f:V\to \mathbb{R}$ with $f\left(u\right)=1$ and $f\left(v\right)=3$, and ${\overrightarrow{G}}_1=(V,E_1)$ with $E_1=\{v\to u\}$. It is clear that $M_{\overrightarrow{G}}f\left(u\right)=2$, $M_{{\overrightarrow{G}}_1}f\left(u\right)=1$. This gives $M_{\overrightarrow{G}}f\left(j\right)>M_{{\overrightarrow{G}}_1}f\left(j\right)$ by taking $j=u$. When $n\ge 3$. There exists a vertex $u\in V$ such that $N_{\overrightarrow{G},+}\left(u\right)\ne \varnothing$. Let us consider the function $f:V\to \mathbb{R}$ with $f\left(u\right)=1$ and $f\left(v\right)=2$ for all $v\in N_{\overrightarrow{G},+}\left(u\right)$, and ${\overrightarrow{G}}_1=(V,E_1)$ be a directed graph with $N_{{\overrightarrow{G}}_1,+}\left(u\right)=\varnothing$. It is clear that $M_{\overrightarrow{G}}f\left(u\right)\ge \frac{2|N_{\overrightarrow{G},+}\left(u\right)|+1}{|N_{\overrightarrow{G},+}\left(u\right)|+1}>1$ and $M_{{\overrightarrow{G}}_1}f\left(u\right)=1$. This gives $M_{\overrightarrow{G}}f\left(j\right)>M_{{\overrightarrow{G}}_1}f\left(j\right)$ by taking $j=u$.

Now we prove (ii). When $n=2$, let $\overrightarrow{G}=(V,E)$ with $V=\{u,v\}$ and $E=\{u\to v\}$. Let us consider the function $f:V\to \mathbb{R}$ with $f\left(v\right)=1$ and $f\left(u\right)=3$, and ${\overrightarrow{G}}_2=(V,E_2)$ with $E_2=\{v\to u\}$. It is clear that $M_{\overrightarrow{G}}f\left(v\right)=1$, $M_{{\overrightarrow{G}}_2}f\left(v\right)=2$. This gives $M_{\overrightarrow{G}}f\left(j\right)<M_{{\overrightarrow{G}}_2}f\left(j\right)$ by taking $j=v$. When $n\ge 3$. There exists a vertex $u\in V$ such that $|N_{\overrightarrow{G},+}\left(u\right)|\ge 2$. Let $w\in V\setminus \left\{u\right\}$. Let us consider the function $f:V\to \mathbb{R}$ with $f\left(w\right)=2$ and $f\left(v\right)=1$ for all $v\in V\setminus \left\{w\right\}$, and ${\overrightarrow{G}}_2=(V,E_2)$ be a directed graph with $N_{{\overrightarrow{G}}_2,+}\left(u\right)=\left\{w\right\}$ and $N_{{\overrightarrow{G}}_2,+}\left(w\right)=\varnothing$. It is clear that $M_{\overrightarrow{G}}f\left(u\right)\le \frac{|N_{\overrightarrow{G},+}\left(u\right)|+2}{|N_{\overrightarrow{G},+}\left(u\right)|+1}<\frac{3}{2}$ and $M_{{\overrightarrow{G}}_2}f\left(u\right)=\frac{3}{2}$. This gives the claim (ii) by letting $j=u$. □

Using the arguments similar to those used to derive the proof of Theorem 2.4 in [10], one can get the following properties for $M_{\overrightarrow{G}}$, which tells us that the operator ${\mathrm{M}}_{\overrightarrow{G}}$ completely determines the graph $\overrightarrow{G}$.

Proposition 2.

Let $\overrightarrow{G_1}=(V,E_1)$ and $\overrightarrow{G_2}=(V,E_2)$ be two directed graphs with $V=\{1,2,\dots ,n\}$. For $j\in V$, the function ${\delta }_j$ denotes the Kronecker delta function

$${\delta }_j\left(i\right)=\left\{\begin{array}{cc}{\displaystyle 1,}\hfill & i=j;\hfill \\ 0,\hfill & i\ne j.\hfill \end{array}\right.$$

Then the following are equivalent:

(i) 

$E_1=E_2$;

(ii) 

$\overrightarrow{G_1}=\overrightarrow{G_2}$;

(iii) 

For every $f:\{1,\dots ,n\}\to \mathbb{R}$, it holds that $M_{\overrightarrow{G_1}}f=M_{\overrightarrow{G_2}}f$;

(iv) 

For every $k\in \{1,\dots ,n\}$, it holds that $M_{\overrightarrow{G_1}}{\delta }_k=M_{\overrightarrow{G_2}}{\delta }_k$.

3. Optimal Estimates for $\parallel {\mathit{M}}_{\overrightarrow{\mathit{G}}}{\parallel }_{\mathit{p}}$ with $\mathbf{0}<\mathit{p}\le \mathbf{1}$

In this section, we shall present some optimal estimates for $\parallel M_{\overrightarrow{G}}{\parallel }_p$ with $0<p\le 1$. To state the main results, the following lemma is needed.

Lemma 1.

([10]). Let $n\ge 2$ and $V=\{1,\dots ,n\}$. Assume that $T:{\ell }^p\left(V\right)\to {\ell }^p\left(V\right)$ is a sublinear operator, with $0<p\le 1$. Then

$${\parallel T\parallel }_p=\underset{k\in V}{max}{\parallel T{\delta }_k\parallel }_{{\ell }^p\left(V\right)}.$$

Before stating our main results, let us introduce two classes of directed graphs. Let $\overrightarrow{S_{I,n}}$ be the inward star graph with n vertices, i.e., there exists a unique $u\in V$ such that $N_{\overrightarrow{S_{I,n}},+}\left(u\right)=\varnothing$ and $N_{\overrightarrow{S_{I,n}},+}\left(v\right)=\left\{u\right\}$ for all $v\in V\setminus \left\{u\right\}$. Let $\overrightarrow{S_{O,n}}$ be the outward star graph with n vertices, i.e., there exists a unique $u\in V$ such that $N_{\overrightarrow{S_{O,n}},+}\left(u\right)=V\setminus \left\{u\right\}$ and $N_{\overrightarrow{S_{O,n}},+}\left(v\right)=\varnothing$ for all $v\in V\setminus \left\{u\right\}$. Recall that two graphs $\overrightarrow{G_1}=(V,E_1),\overrightarrow{G_2}=(V,E_2)$ are said to be isomorphic if there is a permutation of the vertices $\pi :V\to V$ such that $u\to v\in E_1$ if and only if $\pi \left(u\right)\to \pi \left(v\right)\in E_2$. In this case, we can write $\overrightarrow{G_1}\sim \overrightarrow{G_2}$. Noting that if $\overrightarrow{G_1}\sim \overrightarrow{G_2}$, then $M_{\overrightarrow{G_1}}f\left(v\right)=M_{\overrightarrow{G_2}}f\left(\pi \left(v\right)\right)$ and $\parallel M_{\overrightarrow{G_1}}{\parallel }_p={\parallel M_{\overrightarrow{G_2}}\parallel }_p$ for all $0<p\le \infty$. However, the converse is not true (see Proposition 3).

Theorem 2.

Let $\overrightarrow{G}=(V,E)$ be a directed graph with $n\ge 2$ vertices. Then for $0<p\le 1$, the following optimal estimates hold:

$${\left(1+\frac{1}{n^p}\right)}^{1/p}\le {\parallel M_{\overrightarrow{G}}\parallel }_p\le {\left(1+\frac{n-1}{2^p}\right)}^{1/p}.$$

(4)

Moreover,

(i) 

$\parallel M_{\overrightarrow{G}}{\parallel }_p={(1+\frac{n-1}{2^p})}^{1/p}$ if and only if $\overrightarrow{G}\sim \overrightarrow{S_{I,n}}$;

(ii) 

$\parallel M_{\overrightarrow{G}}{\parallel }_p={(1+\frac{1}{n^p})}^{1/p}$ if and only if $\overrightarrow{G}\sim \overrightarrow{S_{O,n}}$.

Proof. 

At first, we shall prove (4). Without loss of generality, we may assume that $V=\{1,\dots ,n\}$. Invoking Lemma 4, one has

$$\parallel M_{\overrightarrow{G}}{\parallel }_p^p=\underset{k\in V}{max}\sum _{j=1}^n{\left(M_{\overrightarrow{G}}{\delta }_k\left(j\right)\right)}^p.$$

(5)

Fix $k\in [1,n]$, it is clear that $M_{\overrightarrow{G}}{\delta }_k\left(k\right)=1$ and

$$M_{\overrightarrow{G}}{\delta }_k\left(j\right)=\underset{r\in \{1,\dots ,n\}}{max}\frac{1}{|B_{\overrightarrow{G}}(j,r)|}\sum _{u\in B_{\overrightarrow{G}}(j,r)}{\delta }_k\left(u\right)=\frac{{\chi }_{d_{\overrightarrow{G}}(j,k)\ne 0}}{|B_{\overrightarrow{G}}(j,d_{\overrightarrow{G}}(j,k))|}$$

(6)

for all $j\in V\setminus \left\{k\right\}$. Since $|B_{\overrightarrow{G}}(j,d_{\overrightarrow{G}}(j,k))|\ge 2$ when $d_{\overrightarrow{G}}(j,k)\ge 1$, then $M_{\overrightarrow{G}}{\delta }_k\left(j\right)\le \frac{1}{2}$ for all $j\in V\setminus \left\{k\right\}$ This together with (5) implies that $\parallel M_{\overrightarrow{G}}{\parallel }_p\le {(1+\frac{n-1}{2^p})}^{1/p}$. Therefore, to prove (4), it is enough to show that

$$\parallel M_{\overrightarrow{G}}{\parallel }_p\ge {\left(1+\frac{1}{n^p}\right)}^{1/p}.$$

(7)

Fix $k\in V$. If $d_{\overrightarrow{G}}(j,k)=0$ for all $j\in V\setminus \left\{k\right\}$, then $M_{\overrightarrow{G}}{\delta }_k\left(j\right)=0$ for all $j\in V\setminus \left\{k\right\}$ and ${\sum }_{j=1}^n{\left(M_{\overrightarrow{G}}{\delta }_k\left(j\right)\right)}^p=1$. If there exists $j_0\in V\setminus \left\{k\right\}$ such that $d_{\overrightarrow{G}}(j_0,k)\ge 1$, then we get from (6) that $M_{\overrightarrow{G}}{\delta }_k\left(j_0\right)\ge \frac{1}{n}$ since $|B_{\overrightarrow{G}}(j_0,d_G(j_0,k))|\le n$. Then we have ${\sum }_{j=1}^n{\left(M_{\overrightarrow{G}}{\delta }_k\left(j\right)\right)}^p\ge 1+\frac{1}{n^p}$. Therefore, inequality (7) holds.

Next we prove part (i). Without loss of generality, we may assume that $\overrightarrow{S_{I,n}}=(V,E)$, where $V=\{1,\dots ,n\}$ and $E=\{2\to 1,\dots ,n\to 1\}$. It is not difficult to see that $\parallel {\delta }_k{\parallel }_{{\ell }^p\left(V\right)}=1$ for all $k\in V$ and $M_{\overrightarrow{S_{I,n}}}{\delta }_1\left(1\right)=1$, $M_{\overrightarrow{S_{I,n}}}{\delta }_1\left(i\right)=\frac{1}{2}$ for all $i=2,\dots ,n$. Moreover, $M_{\overrightarrow{S_{I,n}}}{\delta }_k\left(k\right)=1$, $M_{\overrightarrow{S_{I,n}}}{\delta }_k\left(i\right)=0$ for all $i\in V\setminus \left\{k\right\}$ and $k\in \{2,\dots ,n\}$. Hence, $\parallel M_{\overrightarrow{S_{I,n}}}{\delta }_1{\parallel }_{{\ell }^p\left(V\right)}={(1+\frac{n-1}{2^p})}^{1/p}$ and $\parallel M_{\overrightarrow{S_{I,n}}}{\delta }_k{\parallel }_{{\ell }^p\left(V\right)}=1$ for all $k\in V\setminus \left\{1\right\}$. Invoking Lemma 1, we have

$$\parallel M_{\overrightarrow{S_{I,n}}}{\parallel }_p={\left(1+\frac{n-1}{2^p}\right)}^{1/p},\mathrm{if}0<p\le 1.$$

Assume that $\parallel M_{\overrightarrow{G}}{\parallel }_p={(1+\frac{n-1}{2^p})}^{1/p}$. By Lemma 1, we have

$$\underset{k\in V}{max}\sum _{j=1}^n{\left(M_{\overrightarrow{G}}{\delta }_k\left(j\right)\right)}^p=1+\frac{n-1}{2^p}.$$

We may assume without loss of generality that

$$\sum _{j=1}^n{\left(M_{\overrightarrow{G}}{\delta }_1\left(j\right)\right)}^p=\underset{k\in V}{max}\sum _{j=1}^n{\left(M_{\overrightarrow{G}}{\delta }_k\left(j\right)\right)}^p.$$

This implies that

$$\sum _{j=1}^n{\left(M_{\overrightarrow{G}}{\delta }_1\left(j\right)\right)}^p=1+\frac{n-1}{2^p}.$$

(8)

Noting that $M_{\overrightarrow{G}}{\delta }_1\left(1\right)=1$ and $M_{\overrightarrow{G}}{\delta }_1\left(j\right)\le \frac{1}{2}$ for all $j\in \{2,\dots ,n\}$. This together with (8) yields that $M_{\overrightarrow{G}}{\delta }_1\left(j\right)=\frac{1}{2}$ for all $j\in \{2,\dots ,n\}$, which is equivalent to that $N_{\overrightarrow{G},+}\left(j\right)=\left\{1\right\}$ for all $j\in \{2,\dots ,n\}$. This leads to $\overrightarrow{G}\sim \overrightarrow{S_{I,n}}$ and finishes the proof of part (i).

It remains to prove part (ii). Without loss of generality, we may assume that $\overrightarrow{S_{O,n}}=(V,E)$, where $V=\{1,\dots ,n\}$ and $E=\{1\to 2,\dots ,1\to n\}$. Clearly, $M_{\overrightarrow{S_{O,n}}}{\delta }_1\left(1\right)=1$ and $M_{\overrightarrow{S_{O,n}}}{\delta }_1\left(i\right)=0$ for all $i\in \{2,\dots ,n\}$. For $k\in \{2,\dots ,n\}$, we have that $M_{\overrightarrow{S_{O,n}}}{\delta }_k\left(k\right)=1$, $M_{\overrightarrow{S_{O,n}}}{\delta }_k\left(1\right)=\frac{1}{n}$ and $M_{\overrightarrow{S_{O,n}}}{\delta }_k\left(i\right)=0$ for all $i\in V\setminus \{1,k\}$. Clearly, $\parallel {\delta }_k{\parallel }_{{\ell }^p\left(V\right)}=1$ for all $k\in V$ and $\parallel M_{\overrightarrow{S_{O,n}}}{\delta }_1{\parallel }_{{\ell }^p\left(V\right)}=1$, $\parallel M_{\overrightarrow{S_{O,n}}}{\delta }_k{\parallel }_{{\ell }^p\left(V\right)}={(1+n^{-p})}^{1/p}$ for all $k\in \{2,\dots ,n\}$. Invoking Lemma 1, we have

$$\parallel M_{\overrightarrow{S_{O,n}}}{\parallel }_p={(1+n^{-p})}^{1/p},\mathrm{if}0<p\le 1.$$

Assume that $\parallel M_{\overrightarrow{G}}{\parallel }_p={(1+\frac{1}{n^p})}^{1/p}$. We get by Lemma 1 that

$$\underset{k\in V}{max}\sum _{j=1}^n{\left(M_{\overrightarrow{G}}{\delta }_k\left(j\right)\right)}^p=1+\frac{1}{n^p}.$$

We may assume without loss of generality that

$$\sum _{j=1}^n{\left(M_{\overrightarrow{G}}{\delta }_1\left(j\right)\right)}^p=\underset{k\in V}{max}\sum _{j=1}^n{\left(M_{\overrightarrow{G}}{\delta }_k\left(j\right)\right)}^p.$$

It follows that

$$\sum _{j=1}^n{\left(M_{\overrightarrow{G}}{\delta }_1\left(j\right)\right)}^p=1+\frac{1}{n^p}.$$

(9)

Noting that $M_{\overrightarrow{G}}{\delta }_1\left(1\right)=1$. Moreover, if $M_{\overrightarrow{G}}{\delta }_1\left(j\right)\ne 0$ for some $j\in \{2,\dots ,n\}$, then $M_{\overrightarrow{G}}{\delta }_1\left(j\right)\ge \frac{1}{n}$. Therefore, from (9) we see that there exists $j_0\in \{2,\dots ,n\}$ such that $M_{\overrightarrow{G}}{\delta }_1\left(j_0\right)=\frac{1}{n}$ and $M_{\overrightarrow{G}}{\delta }_1\left(j\right)=0$ for all $j\in V\setminus \{1,j_0\}$. Assume that there exist $i_1,i_2\in V\setminus \left\{j_0\right\}$ such that $i_1\to i_2\in E$. In this case we have $M_{\overrightarrow{G}}{\delta }_{i_2}\left(i_2\right)=1$, $M_{\overrightarrow{G}}{\delta }_{i_2}\left(i_1\right)=\frac{1}{2}$ and $M_{\overrightarrow{G}}{\delta }_{i_2}\left(j_0\right)=\frac{1}{n}$. Consequently,

$$\sum _{j=1}^n{\left(M_{\overrightarrow{G}}{\delta }_{i_2}\left(j\right)\right)}^p\ge 1+\frac{1}{2^p}+\frac{1}{n^p}>1+\frac{1}{n^p},$$

which is a contradiction. Hence, we have $N_{\overrightarrow{G},+}\left(j_0\right)=V\setminus \left\{j_0\right\}$ and $N_{\overrightarrow{G},+}\left(j\right)=\varnothing$ for $j\ne j_0$. So $\overrightarrow{G}\sim \overrightarrow{S_{O,n}}$. This completes the proof of part (ii). □

It should be pointed out that parts (i) and (ii) in Theorem 2 show that the ${\ell }^p$-norm of $M_{\overrightarrow{G}}$ can determine the property of graph $\overrightarrow{G}$. However, the following proposition tells us that the ${\ell }^p$-norm of $M_{\overrightarrow{G}}$ cannot determine the concrete graph $\overrightarrow{G}$ generally.

Proposition 3.

Let $0<p\le 1$. There exist two graphs $\overrightarrow{G_1}=(V,E_1)$ and $\overrightarrow{G_2}=(V,E_2)$ with $E_1\ne E_2$ such that $\parallel M_{\overrightarrow{G_1}}{\parallel }_p={\parallel M_{\overrightarrow{G_2}}\parallel }_p$.

Proof. 

Let $\overrightarrow{G_1}=\overrightarrow{P_n}=(V,E_1)$, where $V=\{1,\dots ,n\}$ and $E_1=\{n\to n-1,n-1\to n-2,\dots ,2\to 1\}$. Given $k\in V$, one can easily check that

$$M_{\overrightarrow{P_n}}{\delta }_k\left(i\right)=\left\{\begin{array}{cc}0,\hfill & i=1,\dots ,k-1;\hfill \\ \frac{1}{i-k+1},\hfill & i=k,\dots ,n.\hfill \end{array}\right.$$

Then we get by Lemma 1 that

$$\parallel M_{\overrightarrow{P_n}}{\parallel }_p={\left(\sum _{i=1}^n{i}^{-p}\right)}^{1/p},\mathrm{if}0<p\le 1.$$

Let $\overrightarrow{G_2}=\overrightarrow{C_n}=(V,E_2)$, where $V=\{1,\dots ,n\}$ and $E_2=\{1\to 2,2\to 3,\dots ,n-1\to n,n\to 1\}$. It is clear that $M_{\overrightarrow{C_n}}{\delta }_1\left(1\right)=1$, $M_{\overrightarrow{C_n}}{\delta }_1\left(i\right)=\frac{1}{n+2-i}$ for all $i=2,\dots ,n$. Then we have

$$\parallel M_{\overrightarrow{C_n}}{\delta }_1{\parallel }_{{\ell }^p\left(V\right)}^p=1+\sum _{i=2}^n\frac{1}{{(n+2-i)}^p}.$$

Invoking Lemma 1, we have

$$\parallel M_{\overrightarrow{C_n}}{\parallel }_p={\left(\sum _{i=1}^n{i}^{-p}\right)}^{1/p},\mathrm{if}0<p\le 1.$$

Observing that $\parallel M_{\overrightarrow{P_n}}{\parallel }_p={\parallel M_{\overrightarrow{C_n}}\parallel }_p={\left({\sum }_{i=1}^n{i}^{-p}\right)}^{1/p}$. This proves Proposition 3. □

4. Optimal Estimates for $\parallel {\mathit{M}}_{\overrightarrow{\mathit{G}}}{\parallel }_{\mathit{p}}$ with $\mathbf{1}<\mathit{p}<\infty$

This section is devoted to presenting some positive results for the $\parallel M_{\overrightarrow{G}}{\parallel }_p$ with $1<p<\infty$. Before formulating the main results, let us give the following observation, which is useful in our proof.

Lemma 2.

Let ${(X,\parallel ·\parallel }_X{),(Y,\parallel ·\parallel }_Y)$ be two normed spaces and let $T:X\to Y$ be a sublinear operator, with $0<p<\infty$. Then the following is valid:

$${\parallel T\parallel }_{X\to Y}=\underset{{f:\parallel f\parallel }_X\ne 0}{sup}\frac{{\parallel Tf\parallel }_Y}{{\parallel f\parallel }_X}=\underset{{f:\parallel f\parallel }_X=1}{sup}{\parallel Tf\parallel }_Y.$$

At first, we present the ${\ell }^p$-norm for ${\mathrm{M}}_{\overrightarrow{S_{I,n}}}$ with $1<p<\infty$.

Theorem 3.

Let $n\ge 2$.

(i) 

If $1<p<\infty$, then

$${\left(1+\frac{n-1}{2^p}\right)}^{1/p}\le {\parallel M_{\overrightarrow{S_{I,n}}}\parallel }_p<{\left(1+\frac{n-1}{2}\right)}^{1/p}.$$

(ii) 

If $p=2$, then

$$\parallel M_{\overrightarrow{S_{I,n}}}{\parallel }_2=\underset{k\in [2,n]}{max}{\left(-\frac{(3x_0^2-2x_0-1)(k-1)}{4(n-k+1+(k-1)x_0^2)}+1\right)}^{1/2},$$

where

$$x_0:=\frac{2(k-1)-3n+\sqrt{9n^2-8n(k-1)}}{2(k-1)}.$$

Proof. 

At first, we shall prove part (i). Without loss of generality, we may assume that $\overrightarrow{S_{I,n}}=(V,E)$, where $V=\{1,\dots ,n\}$ and $E=\{2\to 1,\dots ,n\to 1\}$. Clearly, $\parallel {\delta }_1{\parallel }_{{\ell }^p\left(V\right)}=1$ and $M_{\overrightarrow{S_{I,n}}}{\delta }_1\left(1\right)=1$, $M_{\overrightarrow{S_{I,n}}}{\delta }_1\left(i\right)=\frac{1}{2}$ for all $i=2,\dots ,n$. It follows that $\parallel M_{\overrightarrow{S_{I,n}}}{\delta }_1{\parallel }_{{\ell }^p\left(V\right)}={(1+\frac{n-1}{2^p})}^{1/p}$, which gives

$$\parallel M_{\overrightarrow{S_{I,n}}}{\parallel }_p\ge {\left(1+\frac{n-1}{2^p}\right)}^{1/p},\mathrm{for}1<p<\infty .$$

We now prove

$$\parallel M_{\overrightarrow{S_{I,n}}}{\parallel }_p<{\left(1+\frac{n-1}{2}\right)}^{1/p},\mathrm{for}1<p<\infty .$$

(10)

Given a function $f={\sum }_{i=1}^n{a}_i{\delta }_i$, we write

$$M_{\overrightarrow{S_{I,n}}}f\left(i\right)=\left\{\begin{array}{cc}{\displaystyle |a_1|,}\hfill & \mathrm{for}i=1;\hfill \\ {\displaystyle max\left\{|a_i|,\frac{1}{2}\left(\right|a_i|+|a_1\left|\right)\right\},}\hfill & \mathrm{for}i=2,\dots ,n.\hfill \end{array}\right.$$

Invoking Lemma 2, one has

$$\begin{array}{cc}\hfill & {\displaystyle \parallel M_{\overrightarrow{S_{I,n}}}{\parallel }_p^p=max\left\{|a_1{|}^p+\sum _{i=2}^{n}max\left\{|a_i{|}^p,{\left(\frac{1}{2}\left(\right|a_i|+|a_1\left|\right)\right)}^p\right\}:\sum _{i=1}^n{\left|a_i\right|}^p=1\right\}}\hfill \\ \hfill & {\displaystyle =max\{a_1^p+\sum _{i=2}^{n}max\left\{a_i^p,{\left(\frac{1}{2}(a_i+a_1)\right)}^p\right\}:}\hfill \\ \hfill & {\displaystyle \sum _{i=1}^n{a}_i^p=1,a_i\ge 0,i=1,\dots ,n\}.}\hfill \end{array}$$

For a given sequence ${\left\{a_i\right\}}_{i=1}^n$ with ${\sum }_{i=1}^n{a}_i^p=1$ and all $a_i\ge 0(i=1,\dots ,n)$. We set

$$N_1:=\{j\in \{2,\dots ,n\}:a_j\ge a_1\},N_2:=\{j\in \{2,\dots ,n\}:a_j<a_1\}.$$

By the Jensen’s inequality we have ${\left(\frac{a_i+a_1}{2}\right)}^p<\frac{1}{2}(a_i^p+a_1^p)$ for all $i\in N_2$ since $1<p<\infty$. Therefore, we have

$$\begin{array}{cc}\hfill & {\displaystyle a_1^p+\sum _{i=2}^{n}max\left\{a_i^p,{\left(\frac{1}{2}(a_i+a_1)\right)}^p\right\}}\hfill \\ \hfill & {\displaystyle =a_1^p+\sum _{i\in N_1}{a}_i^p+\sum _{i\in N_2}{\left(\frac{a_i+a_1}{2}\right)}^p}\hfill \\ \hfill & {\displaystyle =1+\sum _{i\in N_2}\left({\left(\frac{a_i+a_1}{2}\right)}^p-a_i^p\right)}\hfill \\ \hfill & {\displaystyle <1+\frac{1}{2}\sum _{i\in N_2}(a_1^p-a_i^p)<1+\frac{n-1}{2}.}\hfill \end{array}$$

This proves (10).

Next, we prove part (ii). Let $f={\sum }_{j=1}^n{a}_j{\delta }_j$ with each $a_j\ge 0$. If $a_j\ge a_1$ for all $j=2,\dots ,n$, then ${\mathrm{M}}_{\overrightarrow{S_{I,n}}}f\left(i\right)=f\left(i\right)$ for all $i=1,\dots ,n$. It follows that $\parallel {\mathrm{M}}_{\overrightarrow{S_{I,n}}}{f\parallel }_{{\ell }^2\left(V\right)}\le {\parallel f\parallel }_{{\ell }^2\left(V\right)}$. Otherwise, there exists $j_0\in \{2,\dots ,n\}$ such that $a_{j_0}<a_1$. Without loss of generality we may assume that

$$a_2\le \dots \le a_k\le a_1\le a_{k+1}\le \dots \le a_n.$$

Then we have

$$\begin{array}{cc}\hfill & {\displaystyle \parallel M_{\overrightarrow{S_{I,n}}}{f\parallel }_{{\ell }^2\left(V\right)}^2=a_1^2+\sum _{i=2}^{n}max\left\{a_i^2,{\left(\frac{1}{2}(a_i+a_1)\right)}^2\right\}}\hfill \\ \hfill & {\displaystyle =a_1^2+\sum _{j=2}^k{\left(\frac{1}{2}(a_j+a_1)\right)}^2+\sum _{j=k+1}^n{a}_j^2}\hfill \\ \hfill & {\displaystyle =\left(\frac{k-1}{4}+1\right)a_1^2+\frac{1}{4}\sum _{j=2}^k{a}_j^2+\frac{1}{2}\sum _{j=2}^k{a}_1{a}_j+\sum _{j=k+1}^n{a}_j^2.}\hfill \end{array}$$

(11)

Applying the AM-GM inequality, one finds

$$2\sum _{j=2}^k{a}_1{a}_j\le \sum _{j=2}^k\left(x{a}_1^2+\frac{1}{x}{a}_j^2\right),$$

for all $x\in (0,\infty )$, where the above equality is attained if and only if $a_j=x{a}_1$ for all $j=2,\dots ,k$. This together with (11) implies that

$$\begin{array}{cc}\hfill & \parallel M_{\overrightarrow{S_{I,n}}}{f\parallel }_{{\ell }^2\left(V\right)}^2\hfill \\ \hfill & {\displaystyle \le \left(\frac{k-1}{4}+1\right)a_1^2+\frac{1}{4}\sum _{j=2}^k{a}_j^2+\frac{1}{4}\sum _{j=2}^k\left(x{a}_1^2+\frac{1}{x}{a}_j^2\right)+\sum _{j=k+1}^n{a}_j^2}\hfill \\ \hfill & {\displaystyle =\left(\frac{k-1}{4}+1+\frac{(k-1)x}{4}\right)a_1^2+\left(\frac{1}{4}+\frac{1}{4x}\right)\sum _{j=2}^k{a}_j^2+\sum _{j=k+1}^n{a}_j^2}\hfill \\ \hfill & {\displaystyle =\left(\frac{k-1}{4}{x}^2+\frac{k-1}{2}x+\frac{k+3}{4}\right)a_1^2+\sum _{j=k+1}^n{a}_j^2}\hfill \end{array}$$

(12)

for all $x\in (0,\infty )$, were the first equality in (12) is attained if and only if $a_j=x{a}_1$ for all $j=2,\dots ,k$. Assume that $a_j=x{a}_1$ for all $j=2,\dots ,k$ and some $x\in (0,\infty )$, then

$$\sum _{j=k+1}^n{a}_j^2={\parallel f\parallel }_{{\ell }^2\left(V\right)}^2-((k-1)x^2+1)a_1^2.$$

(13)

Combining (13) and (12) implies that

$$\begin{array}{cc}\hfill & {\displaystyle \parallel M_{\overrightarrow{S_{I,n}}}{f\parallel }_{{\ell }^2\left(V\right)}^2=\left(-\frac{3(k-1)}{4}{x}^2+\frac{k-1}{2}x+\frac{k-1}{4}\right)a_1^2+{\parallel f\parallel }_{{\ell }^2\left(V\right)}^2}\hfill \\ \hfill & {\displaystyle =-\frac{(k-1)}{4}(3x^2-2x-1)a_1^2+{\parallel f\parallel }_{{\ell }^2\left(V\right)}^2}\hfill \end{array}$$

(14)

for all $x\in (0,\infty )$. Here $a_j=x{a}_1$ for all $j=2,\dots ,k$ and some $x\in (0,\infty )$. Let us consider two cases:

(i)

If $x\ge 1$, then $3x^2-2x-1\ge 0$. We get from (14) that

$$\parallel M_{\overrightarrow{S_{I,n}}}{f\parallel }_{{\ell }^2\left(V\right)}\le {\parallel f\parallel }_{{\ell }^2\left(V\right)}.$$

(ii)

If $0<x<1$, then $3x^2-2x-1<0$. Observing that

$${\parallel f\parallel }_{{\ell }^2\left(V\right)}^2=a_1^2+\sum _{j=2}^k{a}_j^2+\sum _{j=k+1}^n{a}_j^2\ge \left(n-k+1+(k-1)x^2\right)a_1^2,$$

is equivalent to

$$a_1^2\le \frac{1}{n-k+1+(k-1)x^2}{\parallel f\parallel }_{{\ell }^2\left(V\right)}^2.$$

(15)

Here the equality in (15) is attained if and only if $a_j=x{a}_1$ for all $j=2,\dots ,k$ and $a_j=a_1$ for all $j=k+1,\dots ,n$. In light of (14) and (15) we would have

$${\displaystyle \parallel M_{\overrightarrow{S_{I,n}}}{f\parallel }_{{\ell }^2\left(V\right)}^2\le \left(-\frac{(k-1)}{4}\frac{3x^2-2x-1}{n-k+1+(k-1)x^2}+1\right){\parallel f\parallel }_{{\ell }^2\left(V\right)}^2.}$$

(16)

Here the equality in (16) is attained if and only if $a_j=x{a}_1$ for all $j=2,\dots ,k$ and $a_j=a_1$ for all $j=k+1,\dots ,n$. Let $g\left(x\right)=\frac{3x^2-2x-1}{n-k+1+(k-1)x^2}$, $x\in (0,1)$. One can easily check that there exists a unique $x_0\in (0,\infty )$ such that $g\left(x_0\right)=\underset{x\in (0,+\infty )}{min}g\left(x\right)$, where

$$x_0:=x_0(n,k)=\frac{2(k-1)-3n+\sqrt{9n^2-8n(k-1)}}{2(k-1)}\in (0,1).$$

For convenience, let

$$h(x,n,k)=-\frac{(k-1)}{4}\frac{3x^2-2x-1}{n-k+1+(k-1)x^2}+1.$$

Then we have

$${\displaystyle \parallel M_{\overrightarrow{S_{I,n}}}{f\parallel }_{{\ell }^2\left(V\right)}^2\le \underset{k\in [2,n]}{max}h(x_0,n,k){\parallel f\parallel }_{{\ell }^2\left(V\right)}^2.}$$

(17)

In particular, for fixed $k\in [2,n]$, let $f={\sum }_{j=1}^n{a}_j{\delta }_j$, where $a_j=x_0$ for all $j\in \{2,\dots ,k\}$ and $a_j=1$ for all $j\in \{k+1,\dots ,n\}$. One can easily check that

$$\frac{\parallel M_{\overrightarrow{S_{I,n}}}{f\parallel }_{{\ell }^2\left(V\right)}^2}{{\parallel f\parallel }_{{\ell }^2\left(V\right)}^2}=h(x_0,n,k).$$

This together with (17) yields the conclusion of part (ii). □

As applications of Theorem 3, we get

Corollary 1. 

(i) 

If $n=2$, then $\parallel M_{\overrightarrow{S_{I,2}}}{\parallel }_2={\left(\frac{\sqrt{5}+3}{2}\right)}^{1/2}$. Moreover,

$$\parallel M_{\overrightarrow{S_{I,2}}}{f\parallel }_{{\ell }^2\left(V\right)}={\left(\frac{\sqrt{5}+3}{2}\right)}^{1/2}{\parallel f\parallel }_{{\ell }^2\left(V\right)}$$

if and only if $f\left(2\right)=(\sqrt{5}-2)f\left(1\right)\ne 0$.

(ii) 

If $n=3$, then $\parallel M_{\overrightarrow{S_{I,3}}}{\parallel }_2={\left(\frac{\sqrt{33}+7}{8}\right)}^{1/2}$. Moreover,

$$\parallel M_{\overrightarrow{S_{I,3}}}{f\parallel }_{{\ell }^2\left(V\right)}={\left(\frac{\sqrt{33}+7}{8}\right)}^{1/2}{\parallel f\parallel }_{{\ell }^2\left(V\right)}$$

if and only if $f\left(2\right)=f\left(3\right)=\frac{\sqrt{33}-5}{4}f\left(1\right)\ne 0$.

(iii) 

If $n=4$, then $\parallel M_{\overrightarrow{S_{I,4}}}{\parallel }_2={\left(\frac{2+\sqrt{3}}{2}\right)}^{1/2}$. Moreover,

$$\parallel M_{\overrightarrow{S_{I,4}}}{f\parallel }_{{\ell }^2\left(V\right)}={\left(\frac{2+\sqrt{3}}{2}\right)}^{1/2}{\parallel f\parallel }_{{\ell }^2\left(V\right)}$$

if and only if $f\left(2\right)=f\left(3\right)=f\left(4\right)=(\frac{2}{3}\sqrt{3}-1)f\left(1\right)\ne 0$.

Proof. 

Let $x_0$ and $h(x,n,k)$ be given as in the proof of Theorem 3. When $n=2$ and $k=2$, we have $x_0=\sqrt{5}-2$ and $h(x_0,2,2)={\left(\frac{\sqrt{5}+3}{2}\right)}^{1/2}$.

When $n=3$, if $k=2$, then $x_0=\frac{\sqrt{57}-7}{2}$ and $h(x_0,3,2)=\frac{\sqrt{57}+11}{16}$. If $k=3$, then $x_0=\frac{\sqrt{33}-5}{4}$ and $h(x_0,3,2)=\frac{\sqrt{33}+7}{8}$. It is clear that $\frac{\sqrt{33}+7}{8}>\frac{\sqrt{57}+11}{16}$. Therefore, applying Theorem 3 we get

$$\parallel M_{\overrightarrow{S_{I,3}}}{f\parallel }_{{\ell }^2\left(V\right)}\le {\left(\frac{\sqrt{33}+7}{8}\right)}^{1/2}{\parallel f\parallel }_{{\ell }^2\left(V\right)}.$$

(18)

We note that the equality in (18) is attained if and only if $f\left(2\right)=f\left(3\right)=\frac{\sqrt{33}-5}{4}f\left(1\right)\ne 0$. Actually, let $\alpha \ne 0$ and $f:V\to \mathbb{R}$ be defined by $f\left(1\right)=\alpha$ and $f\left(2\right)=f\left(3\right)=\frac{\sqrt{33}-5}{4}\alpha$. One can easily check that $\parallel M_{\overrightarrow{S_{I,3}}}{f\parallel }_{{\ell }^2\left(V\right)}^2=\frac{33-\sqrt{33}}{16}$ and ${\parallel f\parallel }_{{\ell }^2\left(V\right)}^2=\frac{33-5\sqrt{33}}{4}$. Therefore,

$$\frac{\parallel M_{\overrightarrow{S_{I,3}}}{f\parallel }_{{\ell }^2\left(V\right)}^2}{{\parallel f\parallel }_{{\ell }^2\left(V\right)}^2}=\frac{\sqrt{33}+7}{8},$$

which together with (18) yields that

$$\parallel M_{\overrightarrow{S_{I,3}}}{\parallel }_2={\left(\frac{\sqrt{33}+7}{8}\right)}^{1/2}.$$

When $n=4$. If $k=2$, then $x_0=2\sqrt{7}-5$ and $h(x_0,4,2)=\frac{4+\sqrt{7}}{6}$. If $k=3$, then $x_0=\sqrt{5}-2$ and $h(x_0,4,3)=\frac{\sqrt{5}+3}{4}$. If $k=4$, then $x_0=\frac{2}{3}\sqrt{3}-1$ and $h(x_0,4,4)=\frac{\sqrt{3}+2}{2}$. Therefore, we get by Theorem 3 that

$$\parallel M_{\overrightarrow{S_{I,4}}}{f\parallel }_{{\ell }^2\left(V\right)}\le {\left(\frac{\sqrt{3}+2}{2}\right)}^{1/2}{\parallel f\parallel }_{{\ell }^2\left(V\right)}.$$

(19)

It should be pointed out that the equality in (19) is attained if and only if $f\left(2\right)=f\left(3\right)=f\left(4\right)=(\frac{2}{3}\sqrt{3}-1)f\left(1\right)\ne 0$. This proves part (ii) and completes the proof. □

The following result presents the estimates for $\parallel M_{\overrightarrow{S_{O,n}}}{\parallel }_p$ with $1<p<\infty$.

Theorem 4. 

Let $n\ge 3$.

(i) 

If $1<p<\infty$, then

$${(1+n^{-p})}^{1/p}\le {\parallel M_{\overrightarrow{S_{O,n}}}\parallel }_p<{(1+n^{-1})}^{1/p}.$$

(ii) 

If $p=2$, then

$$\parallel M_{\overrightarrow{S_{O,n}}}{\parallel }_2={\left(1+\frac{1}{2n}+\underset{k\in [1,n-1]}{max}\left(-\frac{1}{2k}+\frac{\sqrt{(n+3k)(n-k)}}{2nk}\right)\right)}^{1/2}.$$

Proof. 

At first, we shall prove part (i). We may assume without loss of generality that $\overrightarrow{S_{O,n}}=(V,E)$, where $V=\{1,\dots ,n\}$ and $E=\{1\to 2,\dots ,1\to n\}$. It is obvious that $M_{\overrightarrow{S_{O,n}}}{\delta }_2\left(2\right)=1$, $M_{\overrightarrow{S_{O,n}}}{\delta }_2\left(1\right)=\frac{1}{n}$ and $M_{\overrightarrow{S_{O,n}}}{\delta }_2\left(i\right)=0$ for all $i\in V\setminus \{1,2\}$. Moreover, $\parallel {\delta }_2{\parallel }_{{\ell }^p\left(V\right)}=1$, $\parallel M_{\overrightarrow{S_{O,n}}}{\delta }_2{\parallel }_{{\ell }^p\left(V\right)}={(1+n^{-p})}^{1/p}$. Hence,

$$\parallel M_{\overrightarrow{S_{O,n}}}{\parallel }_p\ge {(1+n^{-p})}^{1/p},\mathrm{for}1<p<\infty .$$

We now prove

$$\parallel M_{\overrightarrow{S_{O,n}}}{\parallel }_p<{(1+n^{-1})}^{1/p},\mathrm{for}1<p<\infty .$$

(20)

Fix $f={\sum }_{i=1}^n{a}_i{\delta }_i$ with ${\sum }_{i=1}^n{\left|a_i\right|}^p=1$, we can write

$$M_{\overrightarrow{S_{O,n}}}f\left(i\right)=\left\{\begin{array}{cc}{\displaystyle max\left\{|a_1|,\frac{1}{n}\sum _{j=1}^n\left|a_j\right|\right\},}\hfill & i=1;\hfill \\ {\displaystyle |a_i|,}\hfill & i=2,\dots ,n.\hfill \end{array}\right.$$

Then we have

$$\parallel M_{\overrightarrow{S_{O,n}}}{f\parallel }_{{\ell }^p\left(V\right)}^p=\sum _{i=2}^n{\left|a_i\right|}^p+max\left\{|a_1{|}^p,{\left(\frac{1}{n}\sum _{j=1}^n\left|a_j\right|\right)}^p\right\}.$$

Therefore, to prove (20), it suffices to show that

$$\sum _{i=2}^n{\left|a_i\right|}^p+max\left\{|a_1{|}^p,{\left(\frac{1}{n}\sum _{j=1}^n\left|a_j\right|\right)}^p\right\}<1+\frac{1}{n}.$$

(21)

for any sequence ${\left\{a_i\right\}}_{i=1}^n$ with ${\sum }_{i=1}^n{\left|a_i\right|}^p=1$.

Given a sequence ${\left\{a_i\right\}}_{i=1}^n$ with ${\sum }_{i=1}^n{\left|a_i\right|}^p=1$, we consider two cases:

(a)

If $|a_1|\ge \frac{1}{n}{\sum }_{j=1}^n\left|a_j\right|$. Then

$$\sum _{i=2}^n{\left|a_i\right|}^p+max\left\{|a_1{|}^p,{\left(\frac{1}{n}\sum _{j=1}^n\left|a_j\right|\right)}^p\right\}=1.$$

This proves (21) in this case.

(b)

If $|a_1|<\frac{1}{n}{\sum }_{j=1}^n\left|a_j\right|$. By the Jensen’s inequality, we have

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{i=2}^n{\left|a_i\right|}^p+max\left\{|a_1{|}^p,{\left(\frac{1}{n}\sum _{j=1}^n\left|a_j\right|\right)}^p\right\}}\hfill \\ \hfill & {\displaystyle =\sum _{i=2}^n|a_i{|}^p+{\left(\frac{1}{n}\sum _{j=1}^n\left|a_j\right|\right)}^p<\sum _{i=2}^n|a_i{|}^p+\frac{1}{n}\sum _{j=1}^n{\left|a_j\right|}^p<1+\frac{1}{n}.}\hfill \end{array}$$

This proves (21) in this case.

Next, we prove part (ii). Let $f={\sum }_{i=1}^n{a}_i{\delta }_i$ with each $a_i\ge 0(i=1,\dots ,n)$ and ${\sum }_{j=1}^n{a}_j^2=1$. Without loss of generality we may assume that

$$a_2\le \dots \le a_k\le \frac{1}{n}\sum _{j=1}^n{a}_j\le a_{k+1}\le \dots \le a_n.$$

Assume that $a_1<\frac{1}{n}{\sum }_{j=1}^n{a}_j$. Then we have

$$\begin{array}{cc}\hfill & {\displaystyle \parallel M_{\overrightarrow{S_{O,n}}}{f\parallel }_{{\ell }^2\left(V\right)}^2=\sum _{i=2}^n{a}_i^2+{\left(\frac{1}{n}\sum _{j=1}^n{a}_j\right)}^2}\hfill \\ \hfill & {\displaystyle =\sum _{i=2}^n{a}_i^2+\frac{1}{n^2}\sum _{j=1}^n{a}_j^2+\frac{2}{n^2}(\sum _{1\le i<j\le k}{a}_i{a}_j}\hfill \\ \hfill & {\displaystyle +\sum _{k+1\le i<j\le n}{a}_i{a}_j+\sum _{\genfrac{}{}{0pt}{}{1\le i\le k}{k+1\le j\le n}}{a}_i{a}_j)}\hfill \\ \hfill & {\displaystyle \le 1+\frac{1}{n^2}-a_1^2+\frac{k-1}{n^2}\sum _{i=1}^k{a}_i^2}\hfill \\ \hfill & {\displaystyle +\frac{n-k-1}{n^2}\sum _{i=k+1}^n{a}_i^2+\frac{2}{n^2}\sum _{\genfrac{}{}{0pt}{}{1\le i\le k}{k+1\le j\le n}}{a}_i{a}_j}\hfill \\ \hfill & {\displaystyle =1+\frac{1}{n^2}+\left(\frac{k(k-1)}{n^2}-1\right){\alpha }^2+\frac{(n-k)(n-k-1)}{n^2}{\beta }^2}\hfill \\ \hfill & {\displaystyle +\frac{2(n-k)k}{n^2}\alpha \beta .}\hfill \end{array}$$

(22)

Here the equality in (22) is attained if and only if $a_i=\alpha$ for all $1\le i\le k$ and $a_j=\beta$ for all $k+1\le j\le n$. Moreover, $\alpha ,\beta$ satisfy $k{\alpha }^2+(n-k){\beta }^2=1$ and $\alpha <\beta$. Please note that the AM-GM inequality holds:

$$2\alpha \beta \le x{\alpha }^2+\frac{1}{x}{\beta }^2,$$

for all $x\in (0,\infty )$, where the above equality holds if and only if $\beta =x\alpha$. This combines with (22) leads to

$$\begin{array}{cc}\hfill & {\displaystyle \left(\frac{k(k-1)}{n^2}-1\right){\alpha }^2+\frac{(n-k)(n-k-1)}{n^2}{\beta }^2+\frac{2(n-k)k}{n^2}\alpha \beta }\hfill \\ \hfill & {\displaystyle \le \left(\frac{k(k-1)}{n^2}-1+\frac{k(n-k)x}{n^2}\right){\alpha }^2}\hfill \\ \hfill & {\displaystyle +\left(\frac{(n-k)(n-k-1)}{n^2}+\frac{k(n-k)}{n^{2}x}\right){\beta }^2}\hfill \\ \hfill & {\displaystyle =\left(\frac{k-1}{n^2}-\frac{1}{k}+\frac{(n-k)x}{n^2}\right)k{\alpha }^2+\left(\frac{n-k-1}{n^2}+\frac{k}{n^{2}x}\right)(n-k){\beta }^2}\hfill \end{array}$$

(23)

for all $x\in (0,\infty )$. Here the equality in (23) is attained if and only if $\beta =x\alpha$. There exists a unique $x_0\in (0,\infty )$ such that

$$\frac{k-1}{n^2}-\frac{1}{k}+\frac{(n-k)x_0}{n^2}=\frac{n-k-1}{n^2}+\frac{k}{n^2{x}_0},$$

(24)

where

$$x_0:=\frac{(n+2k)(n-k)+n\sqrt{(n+3k)(n-k)}}{2k(n-k)}.$$

It follows from (22)–(24) that

$$\begin{array}{cc}\hfill & \parallel M_{\overrightarrow{S_{O,n}}}{f\parallel }_{{\ell }^2\left(V\right)}^2\hfill \\ \hfill & {\displaystyle \le \underset{k\in [1,n-1]}{max}\frac{k}{n^2}+1-\frac{1}{k}+\frac{(n+2k)(n-k)+n\sqrt{(n+3k)(n-k)}}{2n^{2}k}}\hfill \\ \hfill & {\displaystyle =1+\frac{1}{2n}+\underset{k\in [1,n-1]}{max}\left(-\frac{1}{2k}+\frac{\sqrt{(n+3k)(n-k)}}{2nk}\right).}\hfill \end{array}$$

(25)

Here the equality in (25) is attained if and only if $a_i=\alpha >0$ for all $1\le i\le k$ and $a_j=\beta >0$ for all $k+1\le j\le n$. Moreover, $\beta =x_0\alpha$ and $\alpha <\beta$. This proves part (ii). □

As applications of Theorem 4, we get

Corollary 2. 

(i)

If $n=3$, then $\parallel M_{\overrightarrow{S_{O,3}}}{\parallel }_2={\left(\frac{2+\sqrt{3}}{3}\right)}^{1/2}$. Moreover,

$$\parallel M_{\overrightarrow{S_{O,3}}}{f\parallel }_{{\ell }^2\left(V\right)}={\left(\frac{2+\sqrt{3}}{3}\right)}^{1/2}{\parallel f\parallel }_{{\ell }^2\left(V\right)}$$

if and only if $f\left(2\right)=f\left(3\right)=\frac{5+3\sqrt{3}}{2}f\left(1\right)\ne 0$.

(ii) 

If $n=4$, then $\parallel M_{\overrightarrow{S_{O,4}}}{\parallel }_2={\left(\frac{5+\sqrt{21}}{8}\right)}^{1/2}$. Moreover,

$$\parallel M_{\overrightarrow{S_{O,4}}}{f\parallel }_{{\ell }^2\left(V\right)}={\left(\frac{5+\sqrt{21}}{8}\right)}^{1/2}{\parallel f\parallel }_{{\ell }^2\left(V\right)}$$

if and only if $f\left(2\right)=f\left(3\right)=f\left(4\right)=(3+\frac{2\sqrt{21}}{3})f\left(1\right)\ne 0$.

Proof. 

At first, we prove part (i). For convenience, we set

$$h(n,k)=1+\frac{1}{2n}-\frac{1}{2k}+\frac{\sqrt{(n+3k)(n-k)}}{2nk}.$$

Let $x_0$ be given as in (24). When $n=3$. If $k=1$, then $x_0=\frac{5+3\sqrt{3}}{2}$ and $h(3,1)=\frac{2+\sqrt{3}}{3}$. If $k=2$, then $x_0=4$ and $h(3,1)=\frac{7}{6}$. Hence, we get by Theorem 4 that

$$\parallel M_{\overrightarrow{S_{O,3}}}{f\parallel }_{{\ell }^2\left(V\right)}\le {\left(\frac{2+\sqrt{3}}{3}\right)}^{1/2}{\parallel f\parallel }_{{\ell }^2\left(V\right)},$$

(26)

where the equality in (26) is attained if and only if $f\left(2\right)=f\left(3\right)=\frac{5+3\sqrt{3}}{2}f\left(1\right)\ne 0$.

We now prove part (ii). When $n=4$. If $k=1$, then $x_0=3+\frac{2\sqrt{21}}{3}$ and $h(4,1)=\frac{5+\sqrt{21}}{8}$. If $k=2$, then $x_0=2+\sqrt{5}$ and $h(4,2)=\frac{7+\sqrt{5}}{8}$. If $k=3$, then $x_0=\frac{5+2\sqrt{13}}{3}$ and $h(4,3)=\frac{23+\sqrt{13}}{24}$. Invoking Theorem 4, we get

$$\parallel M_{\overrightarrow{S_{O,4}}}{f\parallel }_{{\ell }^2\left(V\right)}\le {\left(\frac{5+\sqrt{21}}{8}\right)}^{1/2}{\parallel f\parallel }_{{\ell }^2\left(V\right)},$$

(27)

where the equality in (27) is attained if and only if $f\left(j\right)=(3+\frac{2\sqrt{21}}{3})f\left(1\right)\ne 0$ for all $j=2,3,4$. This proves part (ii). □

Remark 2.

Some challenging questions are to find the sharp constants for $\parallel M_{\overrightarrow{G}}{\parallel }_p$ with $p\in (1,\infty )$ and $p\ne 2$.

To obtain some sharp constants for $\parallel M_{\overrightarrow{G}}{\parallel }_p$ in the range $1<p<\infty$, we consider the following restricted-type estimate:

$$\parallel M_{\overrightarrow{G}}{\parallel }_{p,\mathrm{rest}}=\underset{A\subset V}{max}\frac{\parallel M_{\overrightarrow{G}}\left({\chi }_A\right){\parallel }_{{\ell }^p\left(V\right)}}{\parallel {\chi }_A{\parallel }_{{\ell }^p\left(V\right)}},$$

where $\overrightarrow{G}=(V,E)$. From the definition we see that

$$\parallel M_{\overrightarrow{G}}{\parallel }_{p,\mathrm{rest}}\le {\parallel M_{\overrightarrow{G}}\parallel }_p.$$

(28)

We have the following estimate for $\parallel M_{\overrightarrow{S_{I,n}}}{\parallel }_{p,\mathrm{rest}}$.

Theorem 5.

Let $0<p<\infty$ and $n\ge 2$. Then

$$\parallel M_{\overrightarrow{S_{I,n}}}{\parallel }_{p,\mathrm{rest}}=\left\{\begin{array}{cc}{\displaystyle 2^{1/p},}\hfill & {log}_2(n-1)\le p<\infty ;\hfill \\ {\displaystyle {\left(1+\frac{n-1}{2^p}\right)}^{1/p},}\hfill & 0<p<{log}_2(n-1).\hfill \end{array}\right.$$

Proof. 

Without loss of generality, we may assume that $\overrightarrow{S_{I,n}}=(V,E)$, where $V=\{1,\dots ,n\}$ and $E=\{2\to 1,\dots ,n\to 1\}$. For $A\subset V$ with $\left|A\right|=k<n$. Please note that $\parallel {\chi }_A{\parallel }_{{\ell }^p\left(V\right)}=k^{1/p}$.

We consider two cases:

Case 1: $1\in A$.

We have

$$M_{\overrightarrow{S_{I,n}}}{\chi }_A\left(i\right)=\left\{\begin{array}{cc}{\displaystyle 1,}\hfill & i\in A;\hfill \\ {\displaystyle \frac{1}{2},}\hfill & i\notin A.\hfill \end{array}\right.$$

Therefore,

$$\parallel M_{\overrightarrow{S_{I,n}}}{\chi }_A{\parallel }_{{\ell }^p\left(V\right)}={\left(\sum _{j=1}^n{\left(M_{\overrightarrow{S_{I,n}}}{\chi }_A\left(j\right)\right)}^p\right)}^{1/p}={\left(k+\frac{n-k}{2^p}\right)}^{1/p}.$$

It follows that

$$\frac{\parallel M_{\overrightarrow{S_{I,n}}}{\chi }_A{\parallel }_{{\ell }^p\left(V\right)}}{\parallel {\chi }_A{\parallel }_{{\ell }^p\left(V\right)}}={\left(1+\frac{1}{2^p}\left(\frac{n}{k}-1\right)\right)}^{1/p}.$$

Case 2: $1\notin A$.

We have

$$M_{\overrightarrow{S_{I,n}}}{\chi }_A\left(i\right)=\left\{\begin{array}{cc}{\displaystyle 1,}\hfill & i\in A\cup \left\{1\right\};\hfill \\ {\displaystyle 0,}\hfill & i\notin A,i\ne 1.\hfill \end{array}\right.$$

Therefore,

$$\parallel M_{\overrightarrow{S_{I,n}}}{\chi }_A{\parallel }_{{\ell }^p\left(V\right)}={\left(\sum _{j=1}^n{\left(M_{\overrightarrow{S_{I,n}}}{\chi }_A\left(j\right)\right)}^p\right)}^{1/p}={(k+1)}^{1/p}.$$

It follows that

$$\frac{\parallel M_{\overrightarrow{S_{I,n}}}{\chi }_A{\parallel }_{{\ell }^p\left(V\right)}}{\parallel {\chi }_A{\parallel }_{{\ell }^p\left(V\right)}}={\left(1+\frac{1}{k}\right)}^{1/p}.$$

Hence, we have

$$\begin{array}{cc}\hfill & {\displaystyle \parallel M_{\overrightarrow{S_{I,n}}}{\parallel }_{p,\mathrm{rest}}=max\left\{{\left(1+\frac{n-1}{2^p}\right)}^{1/p},2^{1/p}\right\}}\hfill \\ \hfill & =\left\{\begin{array}{cc}{\displaystyle 2^{1/p},}\hfill & {log}_2(n-1)\le p<\infty ;\hfill \\ {\displaystyle {\left(1+\frac{n-1}{2^p}\right)}^{1/p},}\hfill & 0<p<{log}_2(n-1).\hfill \end{array}\right.\hfill \end{array}$$

This yields the conclusion of Theorem 5. □

Remark 3.

Let $\overrightarrow{S_{I,n}}=(V,E)$ with $V=\{1,\dots ,n\}$ and $E=\{2\to 1,\dots ,n\to 1\}$ and $A\subset V$. Then

(i) When ${log}_2(n-1)\le p<\infty$, then $\parallel M_{\overrightarrow{S_{I,n}}}{\chi }_A{\parallel }_{{\ell }^p\left(V\right)}=2^{1/p}{\parallel {\chi }_A\parallel }_{{\ell }^p\left(V\right)}$ if and only if $A=\left\{i\right\}$ for some $i\in \{2,\dots ,n\}$.

(ii) When $0<p<{log}_2(n-1)$, then $\parallel M_{\overrightarrow{S_{I,n}}}{\chi }_A{\parallel }_{{\ell }^p\left(V\right)}={(1+\frac{n-1}{2^p})}^{1/p}{\parallel {\chi }_A\parallel }_{{\ell }^p\left(V\right)}$ if and only if $A=\left\{1\right\}$.

Theorem 6.

Let $1<p\le \infty$. Then

$$\parallel M_{\overrightarrow{S_{O,n}}}{\parallel }_{p,\mathrm{rest}}=\left\{\begin{array}{cc}{\displaystyle {\left(1+\frac{{(n-1)}^{p-1}}{n^p}\right)}^{1/p},}\hfill & 1<p<\infty ;\hfill \\ {\displaystyle {\left(1+\frac{1}{n^p}\right)}^{1/p},}\hfill & 0<p\le 1.\hfill \end{array}\right.$$

Proof. 

Without loss of generality, we may assume that $\overrightarrow{S_{O,n}}=(V,E)$, where $V=\{1,\dots ,n\}$ and $E=\{1\to 2,\dots ,1\to n\}$. For $A\subset V$ with $\left|A\right|=k<n$. It is clear that $\parallel {\chi }_A{\parallel }_{{\ell }^p\left(V\right)}=k^{1/p}$. We consider two cases:

Case 1: $1\in A$.

We have

$$M_{\overrightarrow{S_{O,n}}}{\chi }_A\left(i\right)=\left\{\begin{array}{cc}{\displaystyle 1,}\hfill & i\in A;\hfill \\ {\displaystyle 0,}\hfill & i\notin A.\hfill \end{array}\right.$$

Therefore,

$$\parallel M_{\overrightarrow{S_{O,n}}}{\chi }_A{\parallel }_{{\ell }^p\left(V\right)}={\left(\sum _{j=1}^n{\left(M_{\overrightarrow{S_{O,n}}}{\chi }_A\left(j\right)\right)}^p\right)}^{1/p}=k^{1/p}.$$

It follows that

$$\frac{\parallel M_{\overrightarrow{S_{O,n}}}{\chi }_A{\parallel }_{{\ell }^p\left(V\right)}}{\parallel {\chi }_A{\parallel }_{{\ell }^p\left(V\right)}}=1.$$

Case 2: $1\notin A$.

We have

$$M_{\overrightarrow{S_{O,n}}}{\chi }_A\left(i\right)=\left\{\begin{array}{cc}{\displaystyle 1,}\hfill & i\in A;\hfill \\ {\displaystyle \frac{k}{n},}\hfill & i\notin A,i=1;\hfill \\ {\displaystyle 0,}\hfill & i\notin A,i\ne 1.\hfill \end{array}\right.$$

Therefore,

$$\parallel M_{\overrightarrow{S_{O,n}}}{\chi }_A{\parallel }_{{\ell }^p\left(V\right)}={\left(\sum _{j=1}^n{\left(M_{\overrightarrow{S_{I,n}}}{\chi }_A\left(j\right)\right)}^p\right)}^{1/p}={\left(k+\frac{k^p}{n^p}\right)}^{1/p}.$$

It follows that

$$\frac{\parallel M_{\overrightarrow{S_{O,n}}}{\chi }_A{\parallel }_{{\ell }^p\left(V\right)}}{\parallel {\chi }_A{\parallel }_{{\ell }^p\left(V\right)}}={\left(1+\frac{k^{p-1}}{n^p}\right)}^{1/p}.$$

Hence, we have

$${\displaystyle \parallel M_{\overrightarrow{S_{O,n}}}{\parallel }_{p,\mathrm{rest}}=\underset{1\le k\le n-1}{max}{\left(1+\frac{k^{p-1}}{n^p}\right)}^{1/p}=\left\{\begin{array}{cc}{\displaystyle {\left(1+\frac{{(n-1)}^{p-1}}{n^p}\right)}^{1/p},}\hfill & 1<p<\infty ;\hfill \\ {\displaystyle {\left(1+\frac{1}{n^p}\right)}^{1/p},}\hfill & 0<p\le 1.\hfill \end{array}\right.}$$

This proves Theorem 6. □

Remark 4.

Let $\overrightarrow{S_{O,n}}=(V,E)$ with $V=\{1,\dots ,n\}$ and $E=\{1\to 2,\dots ,1\to n\}$ and $A\subset V$. Then

(i) 

When $1<p<\infty$, then $\parallel M_{\overrightarrow{S_{O,n}}}{\chi }_A{\parallel }_{{\ell }^p\left(V\right)}={\left(1+\frac{{(n-1)}^{p-1}}{n^p}\right)}^{1/p}{\parallel {\chi }_A\parallel }_{{\ell }^p\left(V\right)}$ if and only if $A=\{2,\dots ,n\}$.

(ii) 

When $0<p\le 1$, then $\parallel M_{\overrightarrow{S_{O,n}}}{\chi }_A{\parallel }_{{\ell }^p\left(V\right)}={\left(1+\frac{1}{n^p}\right)}^{1/p}{\parallel {\chi }_A\parallel }_{{\ell }^p\left(V\right)}$ if and only if $A=\left\{i\right\}$ for some $i\in \{2,\dots ,n\}$.