Two pairs of families of polyhedral norms versus $\ell _p$-norms: proximity and applications in optimization

Abstract

This paper studies four families of polyhedral norms parametrized by a single parameter. The first two families consist of the CVaR norm (which is equivalent to the D-norm, or the largest-$k$ norm) and its dual norm, while the second two families consist of the convex combination of the $\ell _1$- and $\ell _\infty$-norms, referred to as the deltoidal norm, and its dual norm. These families contain the $\ell _1$- and $\ell _\infty$-norms as special limiting cases. These norms can be represented using linear programming (LP) and the size of LP formulations is independent of the norm parameters. The purpose of this paper is to establish a relation of the considered LP-representable norms to the $\ell _p$-norm and to demonstrate their potential in optimization. On the basis of the ratio of the tight lower and upper bounds of the ratio of two norms, we show that in each dual pair, the primal and dual norms can equivalently well approximate the $\ell _p$- and $\ell _q$-norms, respectively, for $p,q\in [1,\infty ]$ satisfying $1/p+1/q=1$. In addition, the deltoidal norm and its dual norm are shown to have better proximity to the $\ell _p$-norm than the CVaR norm and its dual. Numerical examples demonstrate that LP solutions with optimized parameters attain better approximation of the $\ell _{2}$-norm than the $\ell _1$- and $\ell _\infty$-norms do.

Appendix: Proof of Propositions

1.1 Proof of Proposition 1

To show the lower bound of (6), consider the following maximization problem:

$$\begin{aligned} \begin{array}{|ll} \text{ maximize } &{} \Vert \varvec{x}\Vert _p^p\equiv |x_1|^p+\cdots +|x_n|^p \\ \text{ subject } \text{ to } &{} \langle \!\langle \varvec{x}\rangle \!\rangle _\alpha =1.\\ \end{array} \end{aligned}$$

By symmetry, the optimal value is equal to

$$\begin{aligned} \begin{array}{|ll} \text{ maximize } &{} x_1^p+\cdots +x_n^p \\ \text{ subject } \text{ to } &{} x_1+\cdots +x_{\lfloor \kappa \rfloor }+(\kappa -\lfloor \kappa \rfloor )x_{\lfloor \kappa \rfloor +1}=1,\\ &{} x_j\!\ge \! x_{\lfloor \kappa \rfloor +1},~j\!=\!1,\ldots ,\lfloor \kappa \rfloor ;~~x_{\lfloor \kappa \rfloor +1}\!\ge \! 0;~~x_{\lfloor \kappa \rfloor +1}\!=\!x_j,~j\!=\!\lfloor \kappa \rfloor \!+\!2,\ldots ,n. \end{array} \end{aligned}$$

Since this is a convex maximization over a polytope, an optimal solution is attained at some vertex of the polytope. The constraints consist of one equality and $\lfloor \kappa \rfloor +1$ inequalities while the dimension is $\lfloor \kappa \rfloor +1$, and therefore, at most one inequality can be unbinding. Besides, due to the first (equality) constraint, it is impossible that all inequalities hold with equalities.

• [Case 1: $x_{\lfloor \kappa \rfloor +1}>0$] The corresponding vertex satisfies $x_1=\cdots =x_{\lfloor \kappa \rfloor +1}=\frac{1}{\kappa }$, and its objective value is $\frac{n}{\kappa ^p}$.

• [Case 2: $x_{j}=x_{\lfloor \kappa \rfloor +1}=0$ for $j\ne \hat{j}$ and $x_{\hat{j}}>0$ for some $\hat{j}\in \{1,\ldots ,\lfloor \kappa \rfloor \}$] In this case, $x_{\hat{j}}=1$ and the vertices attain the objective value of $1$.

Therefore, we have $\Vert \varvec{x}\Vert _p/\langle \!\langle \varvec{x}\rangle \!\rangle _\alpha \!\le \!\max \{1,n^\frac{1}{p}\kappa ^{-1}\}$, and $\langle \!\langle \varvec{x}\rangle \!\rangle _\alpha /\Vert \varvec{x}\Vert _p\le \min \{1,\kappa n^{-\frac{1}{p}}\}$.

To show the upper bound of (6), consider the following minimization problem:

$$\begin{aligned} \begin{array}{|ll} \text{ minimize } &{} \Vert \varvec{x}\Vert _p^p:=|x_1|^p+\cdots +|x_n|^p \\ \text{ subject } \text{ to } &{} \langle \!\langle \varvec{x}\rangle \!\rangle _\alpha =1.\\ \end{array} \end{aligned}$$

By using the symmetry, the optimal value is equal to

$$\begin{aligned} \begin{array}{|ll} \text{ minimize } &{} x_1^p+\cdots +x_n^p \\ \text{ subject } \text{ to } &{} x_1+\cdots +x_{\lfloor \kappa \rfloor }+(\kappa -\lfloor \kappa \rfloor )x_{\lfloor \kappa \rfloor +1}=1,\\ &{} x_j\ge x_{\lfloor \kappa \rfloor +1},~j=1,\ldots ,\lfloor \kappa \rfloor ;~~ x_{\lfloor \kappa \rfloor +1}\ge 0;~~x_{\lfloor \kappa \rfloor +2}=\cdots =x_n=0, \end{array} \end{aligned}$$

and we can consider the reduced problem having $\lfloor \kappa \rfloor +1$ variables. A solution with $x_j=\{\lfloor \kappa \rfloor +(\kappa -\lfloor \kappa \rfloor )^{\frac{p}{p-1}}\}^{-1}$, $j=1,\ldots ,\lfloor \kappa \rfloor$, and $x_{\lfloor \kappa \rfloor +1}=(\kappa -\lfloor \kappa \rfloor )^{\frac{1}{p-1}}\{\lfloor \kappa \rfloor +(\kappa -\lfloor \kappa \rfloor )^{\frac{p}{p-1}}\}^{-1}$ (and $x_j=0,j=\lfloor \kappa \rfloor +2,\ldots ,n$) satisfies the Karush-Kuhn-Tucker (KKT) condition. Since the problem is a convex minimization, this solution is optimal and the optimal value is $\{\lfloor \kappa \rfloor +(\kappa -\lfloor \kappa \rfloor )^{\frac{p}{p-1}}\}^{1-p}$. Consequently, $\Vert \varvec{x}\Vert _p/\langle \!\langle \varvec{x}\rangle \!\rangle _\alpha \ge \{\lfloor \kappa \rfloor +(\kappa -\lfloor \kappa \rfloor )^{\frac{p}{p-1}}\}^{-\frac{p-1}{p}}$ and $\langle \!\langle \varvec{x}\rangle \!\rangle _\alpha /\Vert \varvec{x}\Vert _p\le \{\lfloor \kappa \rfloor +(\kappa -\lfloor \kappa \rfloor )^{\frac{p}{p-1}}\}^{\frac{p-1}{p}}$.$\Box$

1.2 Proof of Proposition 2

Similarly to the proof of Proposition 1, to prove the upper bound of (7), consider the following maximization problem:

$$\begin{aligned} \begin{array}{|ll} \text{ maximize } &{} \Vert \varvec{x}\Vert _p^p\equiv |x_1|^p+\cdots +|x_n|^p \\ \text{ subject } \text{ to } &{} \langle \!\langle \varvec{x}\rangle \!\rangle _\alpha ^*\equiv \max \{\Vert \varvec{x}\Vert _1/n(1-\alpha ),\Vert \varvec{x}\Vert _\infty \}\le 1.\\ \end{array} \end{aligned}$$

(39)

Note that due to the symmetry in the $n$ elements of $\varvec{x}$, the optimal value of (39) can be obtained by the following convex maximization program:

$$\begin{aligned} \begin{array}{|ll} \text{ maximize } &{} x_1^p+\cdots +x_n^p \\ \text{ subject } \text{ to } &{} x_1+\cdots +x_n = \kappa ,~1=x_1\ge x_2\ge \cdots \ge x_n\ge 0, \end{array} \end{aligned}$$

where $\kappa =n(1-\alpha )$. Obviously, an optimal solution can be obtained by setting $x_1=\cdots =x_{\lfloor \kappa \rfloor }=1$, $x_{\lfloor \kappa \rfloor +1}=\kappa -\lfloor \kappa \rfloor$, and $x_{\lfloor \kappa \rfloor +2}=\cdots =x_n=0$, and its optimal value is $\lfloor \kappa \rfloor +(\kappa -\lfloor \kappa \rfloor )^p$. Consequently, $\Vert \varvec{x}\Vert _p/\langle \!\langle \varvec{x}\rangle \!\rangle _\alpha ^*\le \{\lfloor \kappa \rfloor +(\kappa -\lfloor \kappa \rfloor )^p\}^\frac{1}{p}$.

Next, we prove the lower bound of (7). Consider

$$\begin{aligned} \begin{array}{|ll} \text{ minimize } &{} \Vert \varvec{x}\Vert _p^p \\ \text{ subject } \text{ to } &{} \langle \!\langle \varvec{x}\rangle \!\rangle _\alpha ^*\equiv \max \{\Vert \varvec{x}\Vert _1/n(1-\alpha ),\Vert \varvec{x}\Vert _\infty \}=1,\\ \end{array} \end{aligned}$$

(40)

which is equivalent to the minimum of the following two convex optimization problems:

$$\begin{aligned}&\begin{array}{|ll} \text{ minimize } &{} x_1^p+\cdots +x_n^p \\ \text{ subject } \text{ to } &{} x_1+\cdots +x_n = \kappa ,~1\ge x_1\ge x_2\ge \cdots \ge x_n\ge 0; \end{array}\end{aligned}$$

(41)

$$\begin{aligned}&\begin{array}{|ll} \text{ minimize } &{} x_1^p+\cdots +x_n^p \\ \text{ subject } \text{ to } &{} x_1+\cdots +x_n \le \kappa ,~1=x_1\ge x_2\ge \cdots \ge x_n\ge 0. \end{array} \end{aligned}$$

(42)

It is easy to see that the solution $x_1=\cdots =x_n=\kappa /n$ is a KKT solution to (41) and the optimal value is $n(\kappa /n)^p$. On the other hand, $x_1=1,x_2=\cdots =x_n=0$ is a KKT solution to (42) and the optimal value is $1$. Consequently, the optimal value of (40) is given by $\min \{\kappa n^{-1+\frac{1}{p}},1\}$, and we have $\Vert \varvec{x}\Vert _p/\langle \!\langle \varvec{x}\rangle \!\rangle ^*_{\alpha }\ge \min \{n^{\frac{1}{p}}(1-\alpha ),1\}$. $\square$

1.3 Proof of Lemma 1

We here prove the continuity of the function at $\kappa \in \mathbb {Z}$ since the continuity and differentiability at $\kappa \not \in \mathbb {Z}$ is evident.

• [Case: $\kappa \ge n^{\frac{1}{p}}$] We have $f_{n,p}(\kappa )=\{\lfloor \kappa \rfloor +(\kappa -\lfloor \kappa \rfloor )^{\frac{p}{p-1}}\}^{\frac{p-1}{p}}$ and $\lim \limits _{\varepsilon \rightarrow 0+} f_{n,p}(\kappa -\varepsilon )=$ $\lim \limits _{\varepsilon \rightarrow 0+} [\kappa -1+\{\kappa -\varepsilon -(\kappa -1)\}^{\frac{p}{p-1}}]^{\frac{p-1}{p}} =\kappa ^{\frac{p-1}{p}}.$ On the other hand, $\lim \limits _{\varepsilon \rightarrow 0+} f_{n,p}(\kappa +\varepsilon ) =\lim \limits _{\varepsilon \rightarrow 0+} \{\kappa +(\kappa -\varepsilon -\kappa )^{\frac{p}{p-1}}\}^{\frac{p-1}{p}} =\kappa ^{\frac{p-1}{p}}.$

• [Case: $\kappa \le n^{\frac{1}{p}}$] We have $f_{n,p}(\kappa )=\frac{n^{\frac{1}{p}}}{\kappa }\{\lfloor \kappa \rfloor +(\kappa -\lfloor \kappa \rfloor )^{\frac{p}{p-1}}\}^{\frac{p-1}{p}}$ and $\lim \limits _{\varepsilon \rightarrow 0+} f_{n,p}(\kappa -\varepsilon ) =$ $\lim \limits _{\varepsilon \rightarrow 0+} \frac{n^{\frac{1}{p}}[\kappa -1+\{\kappa -\varepsilon -(\kappa -1)\}^{\frac{p}{p-1}}]^{\frac{p-1}{p}}}{\kappa -\varepsilon } =\left( \frac{n}{\kappa }\right) ^{\frac{1}{p}}.$ On the other hand, $\lim \limits _{\varepsilon \rightarrow 0+} f_{n,p}(\kappa +\varepsilon ) =\lim \limits _{\varepsilon \rightarrow 0+} \frac{n^\frac{1}{p}\{\kappa +(\kappa -\varepsilon -\kappa )^{\frac{p}{p-1}}\}^{\frac{p-1}{p}}}{\kappa +\varepsilon } =\left( \frac{n}{\kappa }\right) ^{\frac{1}{p}}.$

1.4 Proof of Proposition 3

• [Case: $\kappa \ge n^{\frac{1}{p}}$] For any $\kappa \not \in \mathbb {Z}$, it is differentiable and $f'_{n,p}(\kappa )=\frac{(\kappa -C)^{\frac{1}{p-1}}}{\{C+(\kappa -C)^{\frac{p}{p-1}}\}^{\frac{1}{p}}}$. Since $f_{n,p}(\kappa )$ is continuous at any $\kappa \in \mathbb {Z}$, it is increasing for $\kappa \ge n^{\frac{1}{p}}$.

• [Case: $\kappa \le n^{\frac{1}{p}}$] For any $\kappa \not \in \mathbb {Z}$, it is differentiable and $f'_{n,p}(\kappa )=\frac{n^{\frac{1}{p}}C\{(\kappa -C)^{\frac{p}{p-1}}-(\kappa -C)\}}{\kappa ^2(\kappa -C)\{(\kappa -C)^{\frac{p}{p-1}}+C\}^{\frac{1}{p}}}<0$. Since $f_{n,p}(\kappa )$ is continuous at any $\kappa \in \mathbb {Z}$, it is decreasing for $\kappa \le n^{\frac{1}{p}}$. $\square$

1.5 Proof of Proposition 4

Note that the dual norm to $(\!(\varvec{x})\!)_\lambda$ is written by an LP:

$$\begin{aligned} \begin{array}{r|ll} (\!(\varvec{x})\!)^*_\lambda \equiv &{}\underset{\varvec{y}}{\text{ maximize }}&{}|\varvec{x}|^\top \varvec{y}\\ &{}\text{ subject } \text{ to } &{} \{\lambda \varvec{I}_n+(1-\lambda )\varvec{1}_{n}\varvec{1}_{n}^\top \}\varvec{y}\le \varvec{1}_n,~\varvec{y}\ge \varvec{0}, \end{array} \end{aligned}$$

(43)

where $|\varvec{x}|:=(|x_1|,\ldots ,|x_n|)^\top$. It is easy to see that any extreme point of the feasible region of the above LP is described as a solution to a system of $n$ equalities of the form

$$\begin{aligned} \left\{ \begin{array}{l} \{\lambda \varvec{I}_{|S|}+(1-\lambda )\varvec{1}_{|S|}\varvec{1}_{|S|}^\top \}\varvec{y}_{S}=\varvec{1}_{|S|},\\ y_{i}=0,~i\not \in S,\\ \end{array} \right. \end{aligned}$$

where $\varvec{y}_{S}$ denotes the vector $(y_i)_{i\in S}$ with $S\subset \{1,\ldots ,n\}$. With the Sherman-Morrison-Woodbury formula, the explicit expression of the solution is given by $\varvec{y}_S=\frac{1}{|S|-(|S|-1)\lambda }\varvec{1}_{|S|}$ and $y_i=0,i\not \in S$, and its objective value is $\frac{1}{|S|-(|S|-1)\lambda }\sum _{i\in S}|x_i|.$ Note that this solution is feasible to (43). It is easy to see that given $|S|$, the maximum objective value is $\frac{1}{|S|-(|S|-1)\lambda }\sum _{i=1}^{|S|}|x_{(i)}|.$ Consequently, the optimal value is given by the formula which is to be proved.$\square$

1.6 Proof of Proposition 5

In order to obtain the lower bound for the deltoidal norm, we consider the following optimization problem.

$$\begin{aligned} \begin{array}{|lll} \text{ maximize } &{} \Vert \varvec{x}\Vert _p^p \\ \text{ subject } \text{ to } &{} (\!(\varvec{x})\!)_\lambda \le 1.\\ \end{array} \end{aligned}$$

Taking into account the symmetry of $n$ variables, it is sufficient to solve the following optimization problem:

$$\begin{aligned} \begin{array}{|lll} \text{ maximize } &{} x_1^p+\cdots +x_n^p \\ \text{ subject } \text{ to } &{} (1-\lambda )(x_1+\cdots +x_n)+\lambda x_1=1, ~~x_1\ge x_2\ge \cdots \ge x_n\ge 0.\\ \end{array} \end{aligned}$$

(44)

Since this is a convex maximization over a polytope, we can find an optimal solution at an extreme point of the feasible region. Noting that the feasible region of (44) consists of the one equality and $n$ inequalities, no less than $n$ inequalities hold with equality at any extreme point. It is not hard to see that the extreme solution is defined by $x_1=\cdots =x_k=\frac{1}{(1-\lambda )k+\lambda }>0,~x_{k+1}=\cdots =x_n=0,$ for some $k\in \{1,2,\ldots ,n\}$, and the objective value is $f_p(k):=\frac{k}{\{(1-\lambda )k+\lambda \}^p}.$ The derivative of $f_p$ is $f_p'(k)=\frac{\{(1-\lambda )k+\lambda \}^{p-1}\{(1-p)(1-\lambda )k+\lambda \}}{\{(1-\lambda )k+\lambda \}^{2p}},$ which is increasing for $k<\frac{\lambda }{(p-1)(1-\lambda )}=:\hat{k}$, whereas decreasing for $k>\hat{k}$.

• [Case: $\lambda \in [0,1-\frac{1}{p})$] Note that this condition is equivalent to $\lfloor \hat{k}\rfloor +1\le 1$ (or $\hat{k}<1$). In this case, the maximum of $f_p$ over $k\in \{1,\ldots ,n\}$ is attained at $k=1$ and its value is 1.

• [Case: $\lambda \in [1-\frac{1}{p},\frac{(p-1)n}{(p-1)n+1})$] This condition is equivalent to $1<\lfloor \hat{k}\rfloor +1\le n$ (or $1\le \hat{k}<n$).

  The maximum is attained at $k=\lfloor \hat{k}\rfloor$ or $k=\lfloor \hat{k}\rfloor +1$, and the objective value is

  $$\begin{aligned} \max \left\{ \frac{\lfloor \hat{k}\rfloor }{\{(1-\lambda )\lfloor \hat{k}\rfloor +\lambda \}^p}, \frac{\lfloor \hat{k}\rfloor +1}{\{(1-\lambda )(\lfloor \hat{k}\rfloor +1)+\lambda \}^p} \right\} . \end{aligned}$$

• [Case: $\lambda \in [\frac{(p-1)n}{(p-1)n+1},1]$] This condition is equivalent to $n\le \lfloor \hat{k}\rfloor$ (or $\hat{k}\ge n$). The maximum is attained at $k=n$ and its objective is $\frac{n}{\{(1-\lambda )n+\lambda \}^p}.$

By taking the $p$-th root of the result above and taking the reciprocal, we obtain the lower bound, $g_{n,p}(\lambda )$, of $(\!(\varvec{x})\!)_\lambda /\Vert \varvec{x}\Vert _p$.

In order to obtain the upper bound, we consider the following optimization problem:

$$\begin{aligned} \begin{array}{|lll} \text{ minimize } &{} \Vert \varvec{x}\Vert _p^p \\ \text{ subject } \text{ to } &{} (\!(\varvec{x})\!)_\lambda =1.\\ \end{array} \end{aligned}$$

Taking into account the symmetry of $n$ variables, it is sufficient to solve the following optimization problem:

$$\begin{aligned} \begin{array}{|lll} \text{ minimize } &{} x_1^p+\cdots +x_n^p \\ \text{ subject } \text{ to } &{} (1-\lambda )(x_1+\cdots +x_n)+\lambda x_1=1, ~ x_1\ge x_2\ge \cdots \ge x_n\ge 0.\\ \end{array} \end{aligned}$$

(45)

It is easy to see that the solution $x_1=\frac{1}{(n-1)(1-\lambda )^2+1},~~x_i=\frac{1-\lambda }{(n-1)(1-\lambda )^2+1},i=2,\ldots ,n,$ satisfies the KKT condition. Since (45) is a convex program, this is an optimal solution, and the objective value is $\frac{(n-1)(1-\lambda )^p+1}{\{(n-1)(1-\lambda )^2+1\}^p}.$ Therefore, we get $\Vert \varvec{x}\Vert _p/(\!(\varvec{x})\!)_\lambda \ge \{(n-1)(1-\lambda )^p+1\}^{\frac{1}{p}}/\{(n-1)(1-\lambda )^2+1\}.$ By taking the reciprocal to the result above, we obtain the upper bound of $(\!(\varvec{x})\!)_\lambda /\Vert \varvec{x}\Vert _p$.

Next, we prove the bounds of the dual norm. Similarly to the proof of the dual CVaR norm case, we consider the maximization of $\Vert \varvec{x}\Vert _{p}$ over a constraint $(\!(\varvec{x})\!)^*\le 1$ in order to show the upper bound.

$$\begin{aligned} \begin{array}{|lll} \text{ maximize } &{} \Vert \varvec{x}\Vert _p^p \\ \text{ subject } \text{ to } &{} (\!(\varvec{x})\!)^*_\lambda \le 1.\\ \end{array} \end{aligned}$$

Based on the explicit formula (17) of $(\!(\varvec{x})\!)^*_\lambda$, this is equivalent to

$$\begin{aligned} \begin{array}{|lll} \text{ maximize } &{} x_1^p+\cdots +x_n^p \\ \text{ subject } \text{ to } &{} \max \left\{ x_{1},\frac{x_1+x_2}{2-\lambda },\ldots ,\frac{x_1+\cdots +x_n}{n-(n-1)\lambda }\right\} =1,~~ x_1\ge x_2\ge \cdots \ge x_n\ge 0,\\ \end{array} \end{aligned}$$

which is reduced to a convex maximization over a polytope

$$\begin{aligned} \begin{array}{|lll} \text{ maximize } &{} x_1^p+\cdots +x_n^p \\ \text{ subject } \text{ to } &{} x_1+\cdots +x_j\le j-(j-1)\lambda ,~j=1,\ldots ,n;~~x_1\ge x_2\ge \cdots \ge x_n\ge 0.\\ \end{array} \end{aligned}$$

Let us observe that any optimal solution satisfies the first $n$ inequality constraints with equalities. (Suppose on the contrary that there exists the smallest $k$ such that $x_1+\cdots +x_k<k-(k-1)\lambda$ at an optimal solution $\varvec{x}$. Noting that $x_1=1,x_2=\cdots =x_{k-1}=1-\lambda$ holds, this assumption implies that $x_{k}<1-\lambda$ and the objective value can increase by setting $x_k=1-\lambda$.) Therefore, the solution of the system of the $n$ equalities, $x_1=1,x_2=\cdots =x_{n}=1-\lambda$, is the unique solution, and the optimal value is then given by $1+(n-1)(1-\lambda )^p$. Consequently, $\Vert \varvec{x}\Vert _p/(\!(\varvec{x})\!)\le \{1+(n-1)(1-\lambda )^p\}^\frac{1}{p}$.

In order to obtain the lower bound for the dual deltoidal norm, we consider the following optimization problem.

$$\begin{aligned} \begin{array}{|lll} \text{ minimize } &{} \Vert \varvec{x}\Vert _p^p \\ \text{ subject } \text{ to } &{} (\!(\varvec{x})\!)^*_\lambda =1.\\ \end{array} \end{aligned}$$

(46)

Based on the explicit formula (17) of $(\!(\varvec{x})\!)^*_\lambda$, this is equivalent to

$$\begin{aligned} \begin{array}{|lll} \text{ minimize } &{} x_1^p+\cdots +x_n^p \\ \text{ subject } \text{ to } &{} \max \left\{ x_{1},\frac{x_1+x_2}{2-\lambda },\ldots ,\frac{x_1+\cdots +x_n}{n-(n-1)\lambda }\right\} =1,~~ x_1\ge x_2\ge \cdots \ge x_n\ge 0,\\ \end{array} \end{aligned}$$

whose optimal value is equal to $\min \{F_p(1),\ldots ,F_p(n)\}$ where $F_p(k)$ can be computed by the following LP:

$$\begin{aligned} \begin{array}{r|ll} F_p(k):=&{} \text{ minimize } &{} x_1^p+\cdots +x_n^p\\ &{}\text{ subject } \text{ to } &{} x_{1}+\cdots +x_{k}=k-(k-1)\lambda ,~x_1\ge x_2\ge \cdots \ge x_n\ge 0,\\ &{} &{} x_{1}+\cdots +x_{k'}\le k'-(k'-1)\lambda ,~\text{ for } k'\in \{1,\ldots ,n\}{\setminus }\{k\}.\\ \end{array} \quad \quad \end{aligned}$$

(47)

In place of (47), consider the following (relaxed) LP for some $k\in \{1,\ldots ,n\}$:

$$\begin{aligned} \begin{array}{r|ll} \hat{F}(k):=&{} \text{ minimize } &{} x_1^p+\cdots +x_n^p\\ &{}\text{ subject } \text{ to } &{} x_{1}+\cdots +x_{k}=k-(k-1)\lambda ,~ x_1\ge x_2\ge \cdots \ge x_n\ge 0.\\ \end{array} \quad \end{aligned}$$

(48)

It is easy to observe that the solution such that $x_1=\cdots =x_k=\frac{k-(k-1)\lambda }{k}$ and $x_{k+1}=\cdots =x_n=0$ satisfies the KKT condition. Since (48) is a convex program, such a KKT solution is optimal to that. Also, note that this KKT solution satisfies the inequalities $x_{1}+\cdots +x_{k'}\le k'-(k'-1)\lambda$ for any $k'\ne k$, which implies that this KKT solution is also optimal to (47), i.e., $F(k)=\hat{F}(k)$. Therefore, the optimal value of (46) can be obtained by $\min \{\hat{F}(1),\ldots ,\hat{F}(n)\}$, where $\hat{F}_p(k)=\frac{\{(1-\lambda )k+\lambda \}^p}{k^{p-1}}.$ Since the first order derivative of $\hat{F}_p(k)$ is given by $\hat{F}_p'(k)=\frac{\{(k-1)(1-\lambda )+1\}^{p-1}}{k^p}\times \{(1-\lambda )k+\lambda (1-p)\},$ we see that $\hat{F}_p(k)$ is increasing (decreasing) if $k>\frac{\lambda (p-1)}{1-\lambda }$ ($k<\frac{\lambda (p-1)}{1-\lambda }$) for $\lambda \in (0,1)$ and $p>1$, and therefore, attained its minimum at $\tilde{k}:=\frac{\lambda (p-1)}{1-\lambda }$.

• [Case: $\lambda \in [0,\frac{1}{p})$] Note that this condition is equivalent to $\lfloor \tilde{k}\rfloor +1\le 1$ (or $\tilde{k}<1$). The minimum of $F_p$ over $k\in \{1,\ldots ,n\}$ is attained at $k=1$ and its value is 1.

• [Case: $\lambda \in [\frac{1}{p},\frac{n}{n+p-1})$] This condition is equivalent to $1<\lfloor \tilde{k}\rfloor +1\le n$ (or $1\le \tilde{k}<n$). The maximum is attained at $k=\lfloor \tilde{k}\rfloor$ or $k=\lfloor \tilde{k}\rfloor +1$, and the objective value is then

  $$\begin{aligned} \min \{F(\lfloor \tilde{k}\rfloor ),F(\lfloor \tilde{k}\rfloor \!+\!1)\}\!=\! \min \left\{ \! \frac{\{(1-\lambda )\lfloor \tilde{k}\rfloor +\lambda \}^p}{\lfloor \tilde{k}\rfloor ^{p-1}}, \frac{\{(1-\lambda )(\lfloor \tilde{k}\rfloor +1)+\lambda \}^p}{(\lfloor \tilde{k}\rfloor +1)^{p-1}} \!\right\} \!. \end{aligned}$$

• [Case: $\lambda \in [\frac{n}{n+p-1},1]$] This condition is equivalent to $n\le \lfloor \tilde{k}\rfloor$ (or $\tilde{k}\ge n$). The maximum is attained at $k=n$ and its objective is $\frac{\{(1-\lambda )n+\lambda \}^p}{n^{p-1}}.$

By taking the $p$-th root, we finalize the proof of the lower bound.

1.7 Proof of Lemma 2

Note that on the interval $(\hat{\lambda }_{k-1},\hat{\lambda }_{k})$, the function $h_{n,p}(\lambda )$ is explicitly written by

$$\begin{aligned} h_{n,p,k}(\lambda ):=\frac{k^\frac{1}{p}\left\{ (n-1)(1-\lambda )^\frac{p}{p-1}+1\right\} }{(k-1)(1-\lambda )+1}, \end{aligned}$$

and its first derivative is

$$\begin{aligned} h_{n,p,k}'(\lambda ) =\frac{k^\frac{1}{p}\left\{ k-1-(n-1)(1-\lambda )^\frac{1}{p-1}\right\} }{\left\{ (n-1)(1-\lambda )^\frac{p}{p-1}+1\right\} ^\frac{1}{p}\{(k-1)(1-\lambda )+1\}^2}. \end{aligned}$$

Noting that the sign of $h_{n,p,k}'(\lambda )$ depends only on $k-1-(n-1)(1-\lambda )^\frac{1}{p-1}$, we see that for each $k\in \{1,\ldots ,n-1\}$, $h_{n,p,k}(\lambda )$ uniquely attains the minimum at $1-\left( \frac{k-1}{n-1}\right) ^{p-1}$ over $[0,1]$. Besides, we have

$$\begin{aligned}&\displaystyle \lim _{\lambda \rightarrow \hat{\lambda }_k+0}h_{n,p}'(\lambda )-\lim _{\lambda \rightarrow \hat{\lambda }_k-0}h_{n,p}'(\lambda ) =\displaystyle \lim _{\lambda \rightarrow \hat{\lambda }_k}h_{n,p,k+1}'(\lambda )-h_{n,p,k}'(\lambda ) \\&\quad =\displaystyle \frac{\left\{ k\cdot k^\frac{1}{p}-(k-1)(k+1)^\frac{1}{p}\right\} ^3\left[ \left\{ \frac{(k+1)^\frac{1}{p}-k^\frac{1}{p}}{k\cdot k^\frac{1}{p}-(k-1)(k+1)^\frac{1}{p}}\right\} ^\frac{p}{p-1}(n-1)+1\right] ^\frac{p-1}{p}}{k^\frac{1}{p}(k+1)^\frac{1}{p}} >0.\\ \end{aligned}$$

From these facts, we see that $h_{n,p}(\lambda )$ is quasiconvex and has the unique minimum. $\square$