1 Introduction

For a finite set V and a positive integer r we denote by \(\left( \begin{array}{c}V \\ r \end{array}\right) \) the family of all r-subsets of V. An r-uniform graph G is a set V(G) of vertices together with a set \(E(G)\subseteq \left( \begin{array}{c} V(G)\\ r \end{array}\right) \) of edges. An r-uniform graph H is a subgraph of an r-uniform graph G if \(V(H)\subseteq V(G)\) and \(E(H)\subseteq E(G)\). H is an induced subgraph of an r-uniform graph G if \(E(H)=E(G)\bigcap \left( \begin{array}{c} V(H) \\ r \end{array}\right) \). The density of an r-uniform graph G is defined to be \(d(G)=|E(G)|/|\left( \begin{array}{c} V(G) \\ r \end{array}\right) |\). Let \(\mathscr {F}\) be a family of r-uniform graphs. We say that an r-graph G is \(\mathscr {F}\)-free if G does not contain an isomorphic copy of any member of \(\mathscr {F}\) as a subgraph. The Turán density of \(\mathscr {F}\), denoted by \(t_r(\mathscr {F})\) is the limit of the maximum density of an \(\mathscr {F}\)-free r-uniform graph of order n as \(n\rightarrow \infty \). Finding good estimates of Turán densities in hypergraphs is believed to be one of the most challenging problems in extremal set theory. A real number \(\alpha \in [0,1)\) is a jump for an integer \(r\ge 2\) if there exists a constant \(c>0\) such that any number in \((\alpha , \alpha +c]\) cannot be the Turán density of a family of r-uniform graphs. It is pointed out in [6] that it is also equivalent to the following definition.

Definition 1.1

A real number \(\alpha \in [0,1)\) is a jump for an integer \(r\ge 2\) if there exists a constant \(c>0\) such that for any \(\epsilon >0\) and any integer m, \(m\ge r\), there exists an integer \(n_0(\epsilon ,m)\) such that any r-uniform graph with \(n\ge n_0(\epsilon ,m)\) vertices and density \(\ge \alpha +\epsilon \) contains a subgraph with m vertices and density \(\ge \alpha +c\).

Erdős et al. [3, 4] showed that every \(\alpha \in [0,1)\) is a jump for 2. Erdős [2] proved that every \(\alpha \in [0,\frac{r!}{r^r})\) is a jump for \(r\ge 3\). Furthermore, Erdős proposed the well-known jumping constant conjecture: Every \(\alpha \in [0,1)\) is a jump for every integer \(r\ge 2\). Frankl and Rödl [6] disproved this conjecture by showing that

Theorem 1.2

For \(r\ge 3\), \(1-\frac{1}{l^{r-1}}\) is a non-jump for r if \(l>2r\).

Using a similar approach, more non-jumping numbers were obtained in [5, 7, 9,10,11,12] and some other papers. Recently, Baber and Talbot [1] showed that every number in \([0.2299,0.2316)\bigcup [0.2871, \frac{8}{27})\) is a jump for \(r=3\) using Razborov’s flag algebra method. Pikhurko [13] showed that the set of non-jumps for every \(r\ge 3\) has cardinality of the continuum. However, there are still a lot of unknowns on determining whether a number is a jump for \(r\ge 3\). Following the approach by Frankl and Rödl [6], we prove the following result.

Theorem 1.3

Let \(l\ge 3\) and \(r\ge 4\) be integers. Then \(1+\frac{r-1}{l^{r-1}}-\frac{r}{l^{r-2}}\) is a non-jumpfor r.

For \(r=4\) and \(r=5\), Theorem 1.3 implies the main result given in  [7, 9] respectively. We do not know whether the same result holds for \(r=3\). In fact, when \(r=3\) and \(l=3\), \(1+\frac{r-1}{l^{r-1}}-\frac{r}{l^{r-2}}={2 \over 9}\), and determining whether \({2 \over 9}\) is a jump or not for \(r=3\) is perhaps the most important question regarding this subject. Erdős offered $500 for answering this question.

2 Lagrangians and Other Tools

We first give a definition of the Lagrangian of an r-uniform graph.

Definition 2.1

For an r-uniform graph G with vertex set \(\{1,2,\ldots ,n\}\), edge set E(G) and a vector \(\vec {x}=(x_1,x_2,\ldots ,x_n)\in \mathbb {R}^n\), define

$$\begin{aligned} \lambda (G, \vec {x})=\sum _{\{i_1,i_2,\ldots ,i_r\}\in E(G)}x_{i_1}x_{i_2}\ldots ,x_{i_r}, \end{aligned}$$

where \(x_i\) is called the weight of vertex i.

Definition 2.2

Let \( S=\{\vec {x}=(x_1,x_2,\ldots ,x_n):\sum _{i=1}^n x_i=1,x_i\ge 0 ~ for i=1,2,\ldots ,n\}.\) The Lagrangian of G, denoted by \(\lambda (G)\), is defined as

$$\begin{aligned} \lambda (G)=max\{\lambda (G, \vec {x}):\vec {x}\in S\}. \end{aligned}$$

A vector \(\vec {y}\in S\) is called an optimum vector of \(\lambda (G)\) if \(\lambda (G,\vec {y})=\lambda (G).\)

Fact 2.3

Let \(G_1,G_2\) be r-uniform graphs and \(G_1\subset G_2\). Then \(\lambda (G_1)\le \lambda (G_2)\).

We call two vertices i, j of an r-uniform graph G equivalent if for all , \(f\cup \{j\}\in E(G)\) if and only if \(f\cup \{i\}\in E(G)\).

Lemma 2.4

([6]) Suppose G is an r-uniform graph on vertex set \(\{1,2,\ldots ,n\}\). If vertices \(i_1,\ldots ,i_t\) are pairwise equivalent, then there exists an optimum vector \(\vec {y}=(y_1,y_2,\ldots ,y_n)\) of \(\lambda (G)\) such that \(y_{i_1}=y_{i_2}=\cdots =y_{i_t}\).

We also introduce the blowup of an r-uniform graph which will allow us to construct r-uniform graphs with large number of vertices and densities close to \(r!\lambda (G)\).

Definition 2.5

Let G be an r-uniform graph with \(V(G)=\{1,2,\ldots ,m\}\) and \(\vec {n}=(n_1,n_2,\ldots ,n_m)\) be a positive integer vector. Define the \(\vec {n}\) blow-up of G, \(\vec {n}\otimes G\) as an m-partite r-uniform graph with vertex set \(V_1\bigcup \cdots \bigcup V_m,|V_i|=n_i,1\le i\le m\), and edge set \(E(\vec {n}\otimes G)=\{\{v_{i_1},v_{i_2},\ldots ,v_{i_r}\}: v_{i_k}\in V_{i_k}~ for~1\le k\le r,\{i_1,i_2,\ldots ,i_r\}\in E(G)\}\).

We make the following easy remark proved in [8].

Remark 2.6

Let G be an r-uniform graph with m vertices and \(\vec {y}=(y_1,y_2,\ldots , y_m)\) be an optimum vector of \(\lambda (G)\). Then for any \(\epsilon >0\), there exists an integer \(n_1(\epsilon )\), such that for any integer \(n\ge n_1(\epsilon )\),

$$\begin{aligned} d\left( (\lfloor ny_1\rfloor ,\lfloor ny_2\rfloor ,\ldots ,\lfloor ny_m\rfloor )\otimes G\right) \ge r!\lambda (G)-\epsilon . \end{aligned}$$

Let us also state a fact relating the Lagrangian of an r-uniform graph to the Lagrangian of its blow-up.

Fact 2.7

([6]) Let \(\vec {n}=(n,n,\ldots ,n),n\ge 1.\) Then for every r-uniform graph G and every integer n, \(\lambda (\vec {n}\otimes G)=\lambda (G)\) holds.

The following lemmma proved in [6] gives a necessary and sufficient condition for a number \(\alpha \) to be a jump.

Lemma 2.8

([6]) The following two properties are equivalent.

  1. (i)

     \(\alpha \) is jump for r.

  2. (ii)

     There exists some finite family \(\mathscr {F}\) of r-uniform graphs satisfying \(\lambda (F)>\frac{\alpha }{r!}\) for all \(F\in \mathscr {F}\) and \(t_r(\mathscr {F})\le \alpha .\)

We also need the following lemma from [6].

Lemma 2.9

([6]) For any \(\delta \ge 0\) and any integer \(k\ge r\), there exists \(t_0(k,\delta )\) such that for every \(t>t_0(k,\delta )\), there exists an r-uniform graph A satisfying:

  1. 1.

    \(|V(A)|=t\),

  2. 2.

    \(|E(A)|\ge \delta t^{r-1}\),

  3. 3.

    For all \(V_0\subset V(A),r\le |V_0|\le k\), we have \(|E(A)\bigcap \left( \begin{array}{c}V_0 \\ r \end{array}\right) |\le |V_0|-r+1.\)

The approach in proving Theorem 1.3 is sketched as follows: Let \(\alpha \) be a number to be proved to be a non-jump. Assuming that \(\alpha \) is a jump, we will derive a contradiction by the following steps.

Step 1. :

Construct an r-uniform graph with the Lagrangian close to but slightly smaller than \(\frac{\alpha }{r!}\), then use Lemma 2.9 to add an r-uniform graph with enough number of edges but sparse and obtain an r-uniform graph with the Lagrangian \(\ge \frac{\alpha }{r!}+\epsilon \) for some positive \(\epsilon \). Then we blow up this r-uniform graph to an r-uniform graph, say H with large enough number of vertices and density \(>\alpha +\frac{\epsilon }{2}\) (see Remark 2.6). If \(\alpha \) is a jump, by Lemma 2.8, \(t_r(\mathscr {F})\le \alpha \) for some finite family \(\mathscr {F}\) of r-uniform graphs with Lagrangians \(>\frac{\alpha }{r!}\). So H must contain some member of \(\mathscr {F}\) as a subgraph.

Step 2. :

We show that any subgraph of H with the number of vertices not greater than \(max\{|V(F)|,F\in \mathscr {F}\}\) has the Lagrangian \(\le \frac{\alpha }{r!}\) and derive a contradiction.

3 Proof of Theorem 1.3

In this section, we give a proof of Theorem 1.3. Let \(l\ge 3\) and \(r\ge 4\) be integers. Let

$$\begin{aligned} \alpha =1+\frac{r-1}{l^{r-1}}-\frac{r}{l^{r-2}}. \end{aligned}$$

Suppose that \(\alpha \) is a jump. By Lemma 2.8, there exists a finite family \(\mathscr {F}\) of r-uniform graphs satisfying:

  1. (i)

    \(\lambda (F)>\frac{\alpha }{r!}\) for all \(F\in \mathscr {F}\), and

  2. (ii)

    \(t_r(\mathscr {F})\le \alpha \).

Let t be a large enough integer determined later. Define an r-uniform hypergraph G(r, l, t) on l pairwise disjoint sets \(V_1,\ldots ,V_l\), each with order t and \(E(G(r,l,t))=\{\{v_{i_1},\ldots ,v_{i_r}\}:\{v_{i_1},\ldots ,v_{i_r}\}\in \left( \begin{array}{c} V(G(r,l,t)) \\ r \end{array}\right) \setminus \big (\bigcup ^l_{i=1}\left( \begin{array}{c} V_i \\ r \end{array}\right) \bigcup \bigcup ^l_{i=1}\bigcup ^l_{j=1,j\ne i}\left( \begin{array}{c} V_i \\ r-1 \end{array}\right) \times \left( \begin{array}{c} V_j \\ 1 \end{array}\right) \big )\}\). Note that

$$\begin{aligned} |E(G(r,l,t))|&=\left( \begin{array}{c} lt \\ r \end{array}\right) -l\left( \begin{array}{c} t \\ r \end{array}\right) -l(l-1)t\left( \begin{array}{c} t \\ r-1 \end{array}\right) \nonumber \\&=\frac{\alpha }{r!}(lt)^r-c_0(l)t^{r-1}+o(t^{r-2}), \end{aligned}$$
(3.1)

where \(c_0(l)=\frac{\left( \begin{array}{c} r \\ 2 \end{array}\right) (l^{r-1}-l)}{r!}- \frac{l(l-1)\left( \begin{array}{c} r-1 \\ 2 \end{array}\right) }{(r-1)!}>0\).

It is easy to verify that d(G(r, l, t)) is close to \(\alpha \) when t is large enough.

Take \(\vec {x}=(x_1,\ldots ,x_{lt})\), where \(x_i=\frac{1}{lt}\) for each i, \(1\le i\le lt\). Then

$$\begin{aligned} \lambda (G(r,l,t))&\ge \lambda (G(r,l,t),\vec {x}) \\&=\frac{|E(G(r,l,t))|}{(lt)^r}\\&=\frac{\alpha }{r!}-\frac{c_0(l)}{l^rt}+o\left( \frac{1}{t}\right) , \end{aligned}$$

which is close to \(\frac{\alpha }{r!}\) when t is large enough.

Set \(k_0=max_{F\in \mathscr {F}}|V(F)|\) and \(\delta _0=2c_0(l)\). Let \(t_0(k_0,\delta _0)\) be given as in Lemma 2.9. Take an integer \(t>t_0(k_0,\delta _0)\) and an r-uniform graph \(A_{k_0,\delta _0(t)}\) satisfying the conditions in Lemma 2.9 with \(V(A_{k_0,\delta _0(t)})=V_1\). The r-uniform graph H(r, l, t) is obtained by adding \(A_{k_0,\delta _0(t)}\) to the r-uniform graph G(r, l, t). Note that

$$\begin{aligned} \lambda (H(r,l,t))\ge \frac{|E(H(r,l,t))|}{(lt)^r}. \end{aligned}$$

In view of the construction of H(r, l, t) and Eq. (3.1), we have

$$\begin{aligned} \frac{|E(H(r,l,t))|}{(lt)^r}=\frac{|E(G(r,l,t))|+\delta _0t^{r-1}}{(lt)^r}\ge \frac{\alpha }{r!}+\frac{c_o(l)}{l^rt} \end{aligned}$$

for sufficiently large t. Consequently,

$$\begin{aligned} \lambda (E(H(r,l,t))\ge \frac{\alpha }{r!}+\frac{c_o(l)}{l^rt}. \end{aligned}$$

Now suppose \(\vec {y}=(y_1,y_2,\ldots ,y_{lt})\) is an optimum vector of \(\lambda (E(H(r,l,t))\). Let \(\epsilon =\frac{c_0(l)}{2l^rt}\) and \(n>n_1(\epsilon )\) as in Remark 2.6. Then the r-uniform graph \(S_n=(\lfloor ny_1\rfloor ,\ldots ,\lfloor ny_{lt}\rfloor )\otimes H(r,l,t)\) has density not less than \(\alpha +\epsilon \). Since \(t_r(\mathscr {F})\le \alpha \), some member of \(\mathscr {F}\) is a subgraph of \(S_n\) for \(n\ge n_1(\epsilon )\). For such \(F\in \mathscr {F}\), there exists a subgraph M of H(r, l, t) with \(|V(M)|\le |V(F)|\le k_0\) so that \(F\subset \vec {n}\otimes M\). By Facts 2.3 and 2.7, we have

$$\begin{aligned} \lambda (F)\le \lambda (\vec {n}\otimes M)=\lambda (M). \end{aligned}$$
(3.2)

Theorem 1.3 will follow from the following Lemma 3.1.

Lemma 3.1

Let M be any subgraph of H(r, l, t) with \(|V(M)|\le k_0\). Then

$$\begin{aligned} \lambda (M)\le \frac{\alpha }{r!} \end{aligned}$$

holds.

Applying Lemma 3.1 to (3.2), we have

$$\begin{aligned} \lambda (F)\le \frac{\alpha }{r!}, \end{aligned}$$

which contradicts the fact that \(\lambda (F)>\frac{\alpha }{r!}\) for all \(F\in \mathscr {F}.\) \(\square \)

To complete the proof of Theorem 1.3, it is sufficient to show Lemma 3.1.

3.1 Proof of Lemma 3.1

Define \(U_i=V(M)\bigcap V_i\). Let \(\vec {\xi }=(x_1,x_2,\ldots ,x_{lt})\). Let \(a_i\) be the sum of the weights in \(U_i,1\le i\le l\) respectively. Define \(M_1=(U_1,E(M)\bigcap \left( \begin{array}{c} U_1 \\ r \end{array}\right) )\). Again, by Fact 2.3, it is enough to show Lemma 3.1 for the case \(E(M_1)\ne \emptyset \). Thus we may assume \(|V(M_1)|=r-1+d\) with d a positive integer. By Lemma 2.9, \(M_1\) has at most d edges. Let \(V(M_1)=\{v_1,v_2,\ldots ,v_{r-1+d}\}\) and \(\vec {\eta }=(x_1,x_2,\ldots ,x_{r-1+d})\) be an optimum vector for \(\lambda (M_1)\) with \(x_1\ge x_2\ge \ldots \ge x_{r-1+d}\). The following Claim was proved in [6].

Claim 3.2

$$\begin{aligned} \sum _{\{v_{i_1},v_{i_2},\ldots ,v_{i_r}\}\in E(M_1)}x_{v_{i_1}}x_{v_{i_2}}\cdots x_{v_{i_r}}\le \sum _{r\le i\le r-1+d}x_1x_2\cdots x_{r-1}x_i. \end{aligned}$$

By Claim 3.2, we may assume that

$$\begin{aligned} E(M_1)=\{\{v_1,v_2,\ldots ,v_{r-1},v_i\}:r\le i\le r-1+d\}. \end{aligned}$$

Since \(v_1,v_2,\ldots ,v_{r-1}\) are equivalent, by Lemma 2.4, we may assume that \(x_1=x_2=\cdots =x_{r-1}{\mathop {=}\limits ^{\mathrm {def}}}\rho _0\), Notice that

$$\begin{aligned} \left\{ \begin{array}{l} \sum ^l_{i=1}a_i=1,\\ \alpha _i\ge 0,1\le i\le l,\\ 0\le \rho _0\le \frac{\alpha _1}{r-1}. \end{array}\right. \end{aligned}$$

Now we give an upper bound for \(\lambda (M,\vec {\xi })\). Observing that each term in \(\lambda (M,\vec {\xi })\) appears r! times in the expansion \((x_1+x_2+\cdots +x_m)^r\) but this expansion contains lots of terms not appearing in \(\lambda (M)\) as well. Since \(E(M)=\{v_1,\ldots ,v_{r-1},v_{i}:v_i\in \{v_r,\ldots ,v_{r-1+d}\}\subseteq U_1\}\bigcup \{\{v_{i_1},\ldots ,v_{i_r}\}:\{v_{i_1},\ldots ,v_{i_r}\}\in \left( \begin{array}{c} V(H(r,l,t)) \\ r \end{array}\right) {\setminus }\big (\bigcup ^l_{i=1}\left( \begin{array}{c} V_i \\ r \end{array}\right) \bigcup \bigcup ^l_{i=1}\bigcup ^l_{j=1,j\ne i}\left( \begin{array}{c} V_i \\ r-1 \end{array}\right) \times \left( \begin{array}{c} V_j \\ 1 \end{array}\right) \big )\}\), \(r!\sum _{1\le j\le d}x_1\ldots x_{r-1}x_{r-1+j}\) will be added and \(\sum ^l_{j=1}\alpha ^r_j\) and \(r\sum ^l_{i=1}\alpha ^{r-1}_i(1-\alpha _i)\) should be subtracted in this expansion. Also note that \(\{v_i,v_i,v_{i_3},\ldots , v_{i_{r-2}}, v_{s_2},v_{s_3}\}\) is not an edge in M, where \(1\le i\le r-1\), and \(\{i_3,\ldots ,i_{r-2}\}\) is an \((r-4)\)-subset of \(\{1,2,\ldots ,r-1\}-\{i\}\) and \(s_2,s_3\) (allow that \(s_2=s_3\)) are any vertices in \(\bigcup ^l_{j=2}U_j\). Since each of the corresponding terms appears at least \(\frac{r!}{4}\) times in the expansion, then \((r-1)\left( \begin{array}{c} r-2\\ r-4\end{array}\right) \frac{r!}{4}\rho ^{r-2}_0(1-\alpha _1)^2 =\frac{(r-1)(r-2)(r-3)}{2}\frac{r!}{4}\rho ^{r-2}_0(1-\alpha _1)^2 \ge (r-1)\frac{r!}{4}\rho ^{r-2}_0(1-\alpha _1)^2\) should be subtracted from the expansion. Therefore,

$$\begin{aligned} \lambda (M,\vec {\xi })&\le \frac{1}{r!}\left\{ 1-\sum ^l_{i=1}\alpha ^r_i+ r!\sum _{1\le j\le d}x_1\ldots x_{r-1}x_{r-1+j}\right. \\&\quad \left. -r\sum ^l_{i=1}\alpha ^{r-1}_i(1-\alpha _i)-(r-1)\frac{r!}{4}\rho ^{r-2}_0(1-\alpha _1)^2\right\} \\&\le \frac{1}{r!}\left\{ 1-\sum ^l_{i=1}[r-(r-1)\alpha _i]\alpha ^{r-1}_i\right. \\&\quad \left. +r!\rho ^{r-2}_0\left[ \alpha _1\rho _0-(r-1)\rho ^2_0-\frac{(r-1)}{4}(1-\alpha _1)^2\right] \right\} . \end{aligned}$$

Lemma 3.1 follows directly from the following claim.

Claim 3.3

Let

$$\begin{aligned} f(\alpha _1,\alpha _2,\ldots ,\alpha _l,\rho _0)&=1-\sum ^l_{i=1}\left[ r-(r-1)\alpha _i\right] \alpha ^{r-1}_i\\&\quad + r!\rho ^{r-2}_0\left[ \alpha _1\rho _0-(r-1)\rho ^2_0-\frac{(r-1)}{4}(1-\alpha _1)^2\right] . \end{aligned}$$

Then

$$\begin{aligned} f(\alpha _1,\alpha _2,\ldots ,\alpha _l,\rho _0)\le 1+\frac{r-1}{l^{r-1}}-\frac{r}{l^{r-2}} \end{aligned}$$

holds under the constraints

$$\begin{aligned} \left\{ \begin{array}{ll} \sum ^l_{i=1}a_i=1,\\ \alpha _i\ge 0,1\le i\le l\text{, }\\ 0\le \rho _0\le \frac{\alpha _1}{r-1}. \end{array}\right. \end{aligned}$$

Proof of Claim 3.3

We consider three cases as follows.

Case 1. \(\alpha _1=0\).

Note that \(\rho _0=0\). We have

$$\begin{aligned} f(0,\alpha _2,\ldots ,\alpha _l,0)=1-\sum ^l_{i=2}[r-(r-1)\alpha _i]\alpha ^{r-1}_i. \end{aligned}$$

Let \(g(\alpha _2,\alpha _3,\ldots ,\alpha _l)=1-\sum ^l_{i=2}[r-(r-1)\alpha _i]\alpha ^{r-1}_i\), where \(\sum ^l_{i=2}\alpha _i=1, 0\le \alpha _i\le 1, i=2,3,\ldots ,l\). Let \(L(\alpha _2,\alpha _3,\ldots ,\alpha _l,\lambda )=g(\alpha _2,\alpha _3,\ldots ,\alpha _l)+ \lambda (\sum ^l_{i=2}\alpha _i-1)\), where \(\lambda \) is a real variable. By Lagrange multiplier method, an interior optimal point must satisfy

$$\begin{aligned} \left\{ \begin{array}{ll} L_{a_i}=-r(r-1)(1-a_i)a_i^{r-2}+\lambda =0, &{}\quad i=2,3,\ldots ,l\text{; }\\ L_\lambda =\sum ^l_{i=2}\alpha _i-1=0.\\ \end{array} \right. \end{aligned}$$

Thus \(\alpha _2=\alpha _3=\cdots =\alpha _l=\frac{1}{l-1}\) is the only possible interior optimal point and \(1+\frac{r-1}{(l-1)^{r-1}}-\frac{r}{(l-1)^{r-2}}\) is the corresponding possible optimal value for g. Similarly, for the boundary points with i zeros, \(1+\frac{r-1}{(l-1-i)^{r-1}}-\frac{r}{(l-1-i)^{r-2}}\) is the only possible optimal value for g.

Recall that \(r\ge 4\). Let \(h(x)=\frac{r-1}{x^{r-1}}-\frac{r}{x^{r-2}}\), where \(x\in Z^{+}\). Then \(h^\prime (x)=\frac{-(r-1)^2+r(r-2)x}{x^{r-2}}\). If \(x\ge 2\), then \(-(r-1)^2+r(r-2)x\ge -(r-1)^2+2r(r-2)=r^2-2r-1\ge 7>0\) and \(h^\prime (x)>0\). Also note that \(h(1)< h(2)\). Thus h(x) is monotonically increasing on \(Z^{+}\). Therefore, for \(0\le i\le l-2\), \(1+\frac{r-1}{(l-1-i)^{r-1}}-\frac{r}{(l-1-i)^{r-2}}< 1+\frac{r-1}{l^{r-1}}-\frac{r}{l^{r-2}}\). It settles this case.

Case 2. \(\alpha _1=1.\)

Note that

$$\begin{aligned} f(1,0,\ldots ,0,\rho _0)&=r!\rho ^{r-1}_0[1-(r-1)\rho _0]. \end{aligned}$$

Since the geometric mean is no more than the arithmetic mean, we obtain that

$$\begin{aligned} f(1,0,\ldots ,0,\rho _0)&\le r!\left[ \frac{(r-1)\rho _0+1-(r-1)\rho _0}{r}\right] ^r=\frac{(r-1)!}{r^{r-1}}. \end{aligned}$$

Recall that \(h(l)=\frac{r-1}{l^{r-1}}-\frac{r}{l^{r-2}}\) is monotonically increasing on \(l\ge 3\). Thus

$$\begin{aligned} \frac{r-1}{l^{r-1}}-\frac{r}{l^{r-2}}&\ge \frac{r-1}{3^{r-1}}-\frac{r}{3^{r-2}} =-\frac{2r+1}{3^{r-1}},\\ 1+\frac{r-1}{l^{r-1}}-\frac{r}{l^{r-2}}-\frac{(r-1)!}{r^{r-1}}&\ge 1- \left( \frac{2r+1}{3^{r-1}}+\frac{(r-1)!}{r^{r-1}}\right) . \end{aligned}$$

Let \(h_1(r)=\frac{2r+1}{3^{r-1}}\) and \(h_2(r)=\frac{(r-1)!}{r^{r-1}}\) for \(r\ge 4\). Since \(h^\prime _1(r)=\frac{2-(2r+1)ln3}{3^{r-1}}<0\) and \(\frac{h_2(r+1)}{h_2(r)}=(\frac{r}{r+1})^r<1\), \(h_1(r)\) and \(h_2(r)\) are both monotonically decreasing on \(r\ge 4\). Thus

$$\begin{aligned} 1+\frac{r-1}{l^{r-1}}-\frac{r}{l^{r-2}}-\frac{(r-1)!}{r^{r-1}}&\ge 1- \left( \frac{2r+1}{3^{r-1}}+ \frac{(r-1)!}{r^{r-1}}\right) \\&\ge 1- \left( \frac{9}{3^3}+\frac{3!}{4^3}\right) =\frac{55}{96}. \end{aligned}$$

Therefore,

$$\begin{aligned} f(1,0,\ldots ,0,\rho _0)\le \frac{(r-1)!}{r^{r-1}}< 1+\frac{r-1}{l^{r-1}}-\frac{r}{l^{r-2}}. \end{aligned}$$

Case 3. \(0<\alpha _1<1\).

Let \(g(\alpha _1,\alpha _2,\ldots ,\alpha _l)=1-\sum ^l_{i=1}[r-(r-1)\alpha _i]\alpha ^{r-1}_i\), where \(\sum ^l_{i=1}\alpha _i=1, 0\le \alpha _i\le 1, i=1,2,\ldots ,l\). Similar to case 1, we have

$$\begin{aligned} 1-\sum ^l_{i=1}[r-(r-1)\alpha _i]\alpha ^{r-1}_i\le 1+\frac{r-1}{l^{r-1}}-\frac{r}{l^{r-2}}. \end{aligned}$$

If \(\rho _0=0\), then \(f(\alpha _1,\alpha _2,\ldots ,\alpha _l,0)= 1-\sum ^l_{i=1}[r-(r-1)\alpha _i]\alpha ^{r-1}_i\le 1+ \frac{r-1}{l^{r-1}}-\frac{r}{l^{r-2}}.\)

So we may assume that \(\rho _0>0\). Also recall that \(\rho _0\le \frac{\alpha _1}{r-1}\). We consider two subcases as follows.

Subcase 3.1. \(0<\alpha _1\le 1-\frac{1}{r}\).

Note that

$$\begin{aligned} f(\alpha _1,\alpha _2,\ldots ,\alpha _l,\rho _0)&\le 1+\frac{r-1}{l^{r-1}}-\frac{r}{l^{r-2}}\\&\quad + r!\rho ^{r-2}_0\left[ \alpha _1\rho _0-(r-1)\rho ^2_0-\frac{(r-1)}{4}(1-\alpha _1)^2\right] . \end{aligned}$$

Let \(\Delta _1(\rho _0)=r!\rho ^{r-2}_0\Delta _2(\rho _0)\), where \(\Delta _2(\rho _0)=\alpha _1\rho _0-(r-1)\rho ^2_0-\frac{(r-1)}{4}(1-\alpha _1)^2\). Then \(\Delta _2^{\prime }(\rho _0)=\alpha _1-2(r-1)\rho _0\), and \(\Delta _2^{\prime }(\rho _0)>0\) when \(0<\rho _0<\frac{\alpha _1}{2(r-1)}\) and \(\Delta _2^{\prime }(\rho _0)<0\) when \(\frac{\alpha _1}{2(r-1)}<\rho _0\le \frac{\alpha _1}{r-1}\). Thus \(\Delta _1(\rho _0)=r!\rho ^{r-2}_0\Delta _2(\rho _0)\le r!\rho ^{r-2}_0\Delta _2(\frac{\alpha _1}{2(r-1)})=\frac{r!\rho ^{r-2}_0}{4(r-1)}[\alpha ^2_1-(r-1)^2(1-\alpha _1)^2]\le 0\) since \(\alpha _1\le 1-\frac{1}{r}\). Therefore,

$$\begin{aligned} f(\alpha _1,\alpha _2,\ldots ,\alpha _l,\rho _0)&\le 1+\frac{r-1}{l^{r-1}}-\frac{r}{l^{r-2}}. \end{aligned}$$

Subcase 3.2. \(1-\frac{1}{r}\le \alpha _1<1\).

Note that

$$\begin{aligned} f(\alpha _1,\alpha _2,\ldots ,\alpha _l,\rho _0)\le 1-[r-(r-1)\alpha _1]\alpha ^{r-1}_1+ r!\rho ^{r-1}_0[\alpha _1-(r-1)\rho _0]. \end{aligned}$$

Let \(\Delta _3(\alpha _1)=1-[r-(r-1)\alpha _1]\alpha ^{r-1}_1\). Then

$$\begin{aligned} \Delta ^{\prime }_3(\alpha _1)=-r(r-1)\alpha ^{r-2}_1(1-\alpha _1)<0. \end{aligned}$$

Thus \(\Delta _3(\alpha _1)\) is monotonically decreasing on \([1-\frac{1}{r},1)\).

To prove this subcase, now we need the following useful claim.

Claim 3.4

\((2-\frac{1}{r})(1-\frac{1}{r})^{r-1}\ge \frac{2}{e}\) for \(r\ge 4\).

Proof of Claim 3.4

It is easy to verify that the claim is true for \(r=4\). Note that \((2-\frac{1}{r})(1-\frac{1}{r})^{r-1}\rightarrow \frac{2}{e}(r\rightarrow +\infty )\). Let \(N>0\) be a sufficiently large integer and \(c_1(r)=(r-1)ln(1-\frac{1}{r})+ ln (2-\frac{1}{r})\) for \(r\in [4,N]\). It is sufficient to proved that \(c_1^{\prime }(r)<0\). Note that

$$\begin{aligned} c_1^{\prime }(r)&=ln\left( 1-\frac{1}{r}\right) +(r-1)\cdot \frac{r}{r-1}\cdot \frac{1}{r^2} +\frac{r}{2r-1}\cdot \frac{1}{r^2}\\&= ln\left( 1-\frac{1}{r}\right) +\frac{1}{r}+\frac{1}{r(2r-1)}. \end{aligned}$$

Let \(c_2(r)=ln(1-\frac{1}{r})+\frac{1}{r}+\frac{1}{r(2r-1)}\) for \(r\in [4, N]\). Then \(c_2^{\prime }(r)=\frac{r}{r-1}\cdot \frac{1}{r^2}-\frac{1}{r^2}-\frac{4r-1}{(2r-1)^2r^2} =\frac{r}{(r-1)(2r-1)^2r^2}>0\). Thus \(c_2(r)\) is monotonically increasing continuous function on \(r\in [4, N]\). Clearly, \(c_2(4)<0, c_2(N)\rightarrow 0(N\rightarrow +\infty ).\) Hence \(c_1^{\prime }(r)<0\) for \(r\in [4,N]\). \(\square \)

By Claim 3.4, \(\Delta _3(\alpha _1)\le \Delta _3(1-\frac{1}{r})\le 1 -\frac{2}{e}<\frac{55}{96}\). From Case 2, we have

$$\begin{aligned} r!\rho ^{r-1}_0[\alpha _1-(r-1)\rho _0]\le & {} \frac{(r-1)!}{r^{r-1}},\\ 1+\frac{r-1}{l^{r-1}}-\frac{r}{l^{r-2}}-\frac{(r-1)!}{r^{r-1}}\ge & {} \frac{55}{96}. \end{aligned}$$

Therefore,

$$\begin{aligned} f(\alpha _1,\alpha _2,\ldots ,\alpha _l,\rho _0)&\le 1-[r-(r-1)\alpha _1]\alpha ^{r-1}_1 +\frac{(r-1)!}{r^{r-1}}\\&\le 1+\frac{r-1}{l^{r-1}}-\frac{r}{l^{r-2}}. \end{aligned}$$

\(\square \)

Remark 3.5

For \(r=5\) and \(l=2\), we can combine case 2 with subcase 3.2 in the proof of Claim 3.3, and verify that \(1+\frac{r-1}{l^{r-1}}-\frac{r}{l^{r-2}}\) is not jump for \(r=5,l\ge 2\). This result is given in [7].