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Norm proof help

  1. Aug 20, 2007 #1
    I have the next problem. I have to proof that [itex]\left\vert x_{i}\right\vert\leq\left\vert\left\vert x\right\vert\right\vert[/itex] [itex]\forall x\in\mathbb{R}^{n}[/itex] with the usual scalar product and norm.

    It's obvious that [itex]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[/itex]. Then we have [itex]M^{2}\leq\left\vert\left\vert x\right\vert\right\vert^{2}=\sum_{i=1}^{n}x_{i}^{2}\leq nM^{2}[/itex]. Because of this

    [itex]\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}[/itex] and making square root.

    [itex]\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[/itex]

    Is this proof correct? or are something missing or that lacks of justification?
     
  2. jcsd
  3. Aug 20, 2007 #2

    quasar987

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    How about simply:

    [tex]||x||^2=\sum_{j=1}^n x_j^2\geq x_i^2[/tex] (for any i)

    [tex]\Leftrightarrow ||x||\geq \sqrt{x_i^2}=|x_i|[/tex]
     
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