- #1
ELESSAR TELKONT
- 39
- 0
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?
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?
