3
$\begingroup$

Let $x$ be real vector with $\|x\|_1=x_1+\ldots +x_{2n}$.

How to bound from above $(x_1+\ldots+x_n)(x_{n+1}+\ldots+x_{2n})$ by $l_2$ norm of the vector $x$.

Of course, using $\|x\|\leq\sqrt {2n}\|x\|_2$ I can bound $ (x_1+\ldots+x_n)(x_{n+1}+\ldots+x_{2n})\leq\|x\|^2_1\leq 2n\|x\|_2^2 $

But I would like to get an upper bount not greater then $1/2\|x\|_2^2$. Is it possible to get such a bound?

  • 0
    Are you sure $x_1+\ldots+x_{2n}$ is a norm? Shouln't be $|x_1|+\ldots+|x_{2n}|$?2012-08-02

1 Answers 1

4

$(x_1 + \ldots + x_n)(x_{n+1} + \ldots x_{2n}) = \sum_{i=1}^n \sum_{j=n+1}^{2n} x_i x_j \le \sum_{i=1}^n \sum_{j=n+1}^{2n} \frac{x_i^2 + x_j^2}{2} = \frac{n}{2} \|x\|_2^2 $ and this is an equality when all $x_i$ are equal.