Let $a=(a_1, \ldots, a_n)$ be a vector in $R^n$.
I am wondering for which vectors the following would be true: $ \|a\|_2^2\geq c \sum_{i\ne j,i,j}a_ia_j, \quad i,j=1, \ldots, n. $ Here $c>0$ is an absolute consrant.
Thank you.
Let $a=(a_1, \ldots, a_n)$ be a vector in $R^n$.
I am wondering for which vectors the following would be true: $ \|a\|_2^2\geq c \sum_{i\ne j,i,j}a_ia_j, \quad i,j=1, \ldots, n. $ Here $c>0$ is an absolute consrant.
Thank you.
Note that $\sum_{i\neq j} x_i x_j = \sum_i x_i \sum_j x_j -\sum_i x_i^2 = (x_1+\cdots+x_n)^2 -\|x\|^2 = |\langle e , x \rangle |^2 - \|x\|^2$, with $ e = (1,...1)^T$. Hence for given $c>0$, you are looking for $x$ that satisfy $c (|\langle e , x \rangle |^2-\|x\|^2) \leq \|x\|^2$, or equivalently, $x$ satisfying $|\langle e , x \rangle | \leq \sqrt{1+\frac{1}{c}} \|x\|$ Clearly $x=0$ will always satisfy this inequality, so suppose $x\neq 0$. Note that $\|e\| = \sqrt{n}$. Then the inequality can be written as $|\langle \frac{e}{\sqrt{n}} , \frac{x}{\|x\|} \rangle | \leq K = \frac{1}{\sqrt{n}} \sqrt{1+\frac{1}{c}}$ If $K\geq 1$, then all $x$ satisfy this inequality (Cauchy-Schwarz), otherwise it consists of all $x$ whose angle $\theta$ with the vector $e$ satisfies $|\cos \theta| \leq K$.