so the given statement is this:
$\frac{1}{\sqrt{n}} \cdot \left \| \bar{x} \right \| \leq \left \| \bar{x} \right \|_{\infty }$ with $\bar{x} \in \mathbb{R}^{n}$
where $\left \| \bar{x} \right \|$ is the euclidean norm and $\left \| \bar{x} \right \|_{\infty }$ is the L infinity norm.
I have already proved that:
$\left \| \bar{x} \right \|_{\infty }\leq\left \| \bar{x} \right \|$
But i am not sure how i can use this to prove the shown statement. Where can i derive the square root of n from?
Best Answer
Let $\vec{x} = (x_1,\dots,x_n)$. Then $\|\vec{x}\|_\infty = \sup_i x_i$. Thus: $$ \|\vec{x}\| = \sqrt{\sum_{i=1}^n x_i^2} \leq \sqrt{\sum_{i=1}^n \left(\sup_i x_i\right)^2} = \sqrt{\sum_{i=1}^n \|\vec{x}\|_\infty^2} = \|\vec{x}\|_\infty\sqrt{\sum_{i=1}^n 1} = \sqrt{n}\|\vec{x}\|_\infty $$