# Norm proof help

1. Aug 20, 2007

### ELESSAR TELKONT

I have the next problem. I have to proof that $\left\vert x_{i}\right\vert\leq\left\vert\left\vert x\right\vert\right\vert$ $\forall x\in\mathbb{R}^{n}$ with the usual scalar product and norm.

It's obvious that $x_{i}^{2}=\left\vert x_{i}\right\vert^{2}\leq \max\left\{\left\vert x_{i}\right\vert\mid i=1,\dots, n\right\}^{2}=M^{2},\; \forall i$. Then we have $M^{2}\leq\left\vert\left\vert x\right\vert\right\vert^{2}=\sum_{i=1}^{n}x_{i}^{2}\leq nM^{2}$. Because of this

$\left\vert x_{i}\right\vert^{2}\leq M^{2}\leq\left\vert\left\vert x\right\vert\right\vert^{2}=\sum_{i=1}^{n}x_{i}^{2}\leq nM^{2}$ and making square root.

$\left\vert x_{i}\right\vert\leq M\leq\left\vert\left\vert x\right\vert\right\vert=\sqrt{\sum_{i=1}^{n}x_{i}^{2}}\leq \sqrt{n}M$

Is this proof correct? or are something missing or that lacks of justification?

2. Aug 20, 2007

### quasar987

$$||x||^2=\sum_{j=1}^n x_j^2\geq x_i^2$$ (for any i)
$$\Leftrightarrow ||x||\geq \sqrt{x_i^2}=|x_i|$$