λ1×1+⋯+λkxk=max{x1,⋯,xk}\lambda_1x_1 +\cdots+\lambda_kx_k = \max\{x_1,\cdots,x_k\} implies x1=x2=⋯=xkx_1 = x_2 = \cdots = x_k?

Is it true, and i case of it being true, how can I show that:

(1) \lambda_1x_1 +\cdots+\lambda_kx_k = \max\{x_1,\cdots,x_k\}\lambda_1x_1 +\cdots+\lambda_kx_k = \max\{x_1,\cdots,x_k\}

(2) \sum_{i=1}^{k}\lambda_i = 1\sum_{i=1}^{k}\lambda_i = 1 and \lambda_i \in (0 ,1] \lambda_i \in (0 ,1] for i = 1,\cdots, ki = 1,\cdots, k

Implies that x_1 = x_2 = \cdots = x_kx_1 = x_2 = \cdots = x_k ?

=================

2

 

Presumably also 0\leq \lambda_i\leq 10\leq \lambda_i\leq 1?
– Alex R.
2 days ago

  

 

Yes, sorry. You are right, actually \lambda_i \in (0,1]\lambda_i \in (0,1]. I have edited the question. @alex .
– InterpolationKid
2 days ago

=================

3 Answers
3

=================

Let x_1 = max\{x_1, x_2, \cdots, x_k\}x_1 = max\{x_1, x_2, \cdots, x_k\}.

Then we have that

\lambda_1x_1 + \lambda_2x_2 + \cdots \lambda_kx_k = x_1 \\
\lambda_1 + \lambda_2 + \cdots + \lambda_k = 1

\lambda_1x_1 + \lambda_2x_2 + \cdots \lambda_kx_k = x_1 \\
\lambda_1 + \lambda_2 + \cdots + \lambda_k = 1

\lambda_1x_1 + \lambda_2x_2 + \cdots \lambda_kx_k = x_1 \ (1) \\
\lambda_1x_1 + \lambda_2x_1 + \cdots + \lambda_kx_1 = x_1 \ (2).

\lambda_1x_1 + \lambda_2x_2 + \cdots \lambda_kx_k = x_1 \ (1) \\
\lambda_1x_1 + \lambda_2x_1 + \cdots + \lambda_kx_1 = x_1 \ (2).

Take (1) – (2)(1) – (2) to get that

\lambda_2 (x_2-x_1) + \lambda_3 (x_3 – x_1) + \cdots + \lambda_k (x_k – x_1) = 0.

\lambda_2 (x_2-x_1) + \lambda_3 (x_3 – x_1) + \cdots + \lambda_k (x_k – x_1) = 0.

which can be written as

\sum_{i=2}^{k} \lambda_i (x_i – x_1) = 0.

\sum_{i=2}^{k} \lambda_i (x_i – x_1) = 0.

Since \lambda_i > 0\lambda_i > 0 and x_i – x_1 \leq 0x_i – x_1 \leq 0 we have that \lambda_i (x_i – x_1) \leq 0\lambda_i (x_i – x_1) \leq 0. From where we conclude that \lambda_i (x_i – x_1) = 0\lambda_i (x_i – x_1) = 0.

Let M= \max \{x_k \;|\; k=1,2,\dots ,n\}M= \max \{x_k \;|\; k=1,2,\dots ,n\}. Then:
x_k \le Mx_k \le M
\lambda_k x_k \le \lambda_k M\lambda_k x_k \le \lambda_k M
\sum_{k=1}^{n} \lambda_k x_k \le \sum_{k=1}^{n} \lambda_k M = M\sum_{k=1}^{n} \lambda_k x_k \le \sum_{k=1}^{n} \lambda_k M = M

For the last inequality to turn into an equality, all the previous inequalities must be equalities as well (since any one strict inequality would cause the last one to be strict, too). Therefore x_k = Mx_k = M for all k=1,2, \dots,nk=1,2, \dots,n.

You must mean \lambda_i> 0,\lambda_i> 0, not \lambda_i\ne 0,\lambda_i\ne 0, otherwise it’s false.

Let M=\max (x_1,..,x_n).M=\max (x_1,..,x_n). For each ii we have (1)\quad \lambda_i x_i\leq \lambda_iM.(1)\quad \lambda_i x_i\leq \lambda_iM. Therefore (2) \quad \sum_{i=1}^n\lambda_i x_i\leq \sum_{i=1}^n\lambda_iM.(2) \quad \sum_{i=1}^n\lambda_i x_i\leq \sum_{i=1}^n\lambda_iM. But if x_i