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Length-Norm inequality to root-n times Length (using 2|uv| <= |u|^2 + |v|^2)

  1. Oct 3, 2014 #1
    I've been reading "The Qualitative Theory of Ordinary Differential Equations, An Introduction" and am now stuck on an inequality I am supposed to be able to prove. I am pretty sure the inequality comes from linear algebra, I remember seeing something about it in my intro class but I let a friend borrow my book. I have been unable to find an explanation of the 2|uv| inequality, which may be all I need.

    Definitions where bold indicates vectors.
    [tex]\text{Euclidean Length: }||\textbf{y}|| = \left(\Sigma_{i=1}^n |y_i|^2\right)^{1/2}, \text{Norm: } |\textbf{y}| = \Sigma_{i=1}^n|y_i| [/tex]

    [tex]
    \text{To Prove:} \\
    \text{If } \textbf{y} \in E_n, \text{ show that} \\
    ||\textbf{y}|| \leq |\textbf{y}| \leq \sqrt{n}||\textbf{y}|| \\
    \text{Hint, use } 2|uv| \leq |u|^2 + |v|^2 \text{ and show that } ||\textbf{y}||^2 \leq |\textbf{y}|^2 \leq n||\textbf{y}||^2[/tex]

    I was thinking that generalizing the [itex]2|uv|[/itex] inequality to

    [tex]n|u_1 u_2 \cdots u_n| \leq |u_1|^2 + |u_2|^2 + \cdots + |u_n|^2[/tex]

    may lead me in a helpful direction, but couldn't see where to go with it, or where that inequality came from to generalize it.

    That may not even be the right way to be thinking about the proof, but any direction you could give me in this direction would be greatly appreciated. Thank you!
    -Sravoff
     
  2. jcsd
  3. Oct 4, 2014 #2

    Fredrik

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    Hint for the 2|uv| thing: What is ##(|u|-|v|)^2##?
     
  4. Oct 10, 2014 #3
    For the first inequality, square both sides of ##\Vert x\Vert\leq \vert x\vert##.

    For the second, square both sides of ##\vert x\vert\leq \sqrt{n}\Vert x\Vert## and use the inequality from the hint on the cross terms that appear in ##\vert x\vert^2##.
     
  5. Oct 15, 2014 #4
    Hey Fredrik, thanks for the hint that worked great! I can see the 2|uv| identity now, but that didn't get me closer to a generalized form. I played with a couple of other forms that didn't wind up looking too promising.

    Hey platetheduke, is here an justification for where those come from in the first place? If I start with just the definitions above, how do I claim that ||x|| <= |x|?
     
  6. Oct 16, 2014 #5

    Fredrik

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    The inequalities ##\|x\|\leq|x| \leq\sqrt{n}\|x\|## are equivalent to ##\|x\|^2\leq |x|^2\leq n\|x\|^2##. The latter pair of inequalities can be obtained in a fairly straightforward way starting with the definition of ##|x|##. The result ##2|u||v|\leq|u|^2+|v|^2## is used in the proof of the second inequality, but not in the proof of the first.

    Would you like to give it a try and show us where you get stuck?

    Both proofs start like this:
    $$|x|^2=\left(\sum_{i=1}^n|x_i|\right)^2 =$$
     
    Last edited: Oct 16, 2014
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