If $x\in \mathbb{R}^n$ and is a unit vector, why is $\sum\limits_{j,k=1}^n |x_j||x_k| < n^2$?

Question

This is an excerpt of a larger proof:

Other pertinent information:

• $A$ is a positive definite $n \times n$ matrix
• The set $C$ is the unit sphere

I don't get the last inequality:

$\gamma \sum \limits_{j,k=1}^n |x_j||x_k| < \gamma n^2$

Why is $\sum \limits_{j,k=1}^n |x_j||x_k| < n^2$?

This looks like it might have something to do with $x$ being in $C$ but the expression in the summation is not the norm for $x$ so I don't see how that is helpful.

Answer 1

If $C$ is the unit $\ell^2$-sphere, Cauchy-Schwarz inequality yields $$\sum\limits_{j,k=1}^n |x_j||x_k|=\left(\sum \limits_{j=1}^n |x_j|\right)^2\leqslant\left(\sum_{j=1}^n1^2\right)\cdot\left(\sum \limits_{j=1}^n |x_j|^2\right)=n.$$