1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook