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Math Help - Verify norm inequalities

  1. #1
    Member Mollier's Avatar
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    Verify norm inequalities

    Problem:
    verify the following inequalities:

    (a)  ||x||_\infty \leq ||x||_2

    (b)  ||x||_2 \leq \sqrt{m}||x||_\infty

    (c)  ||A||_\infty \leq \sqrt{n}||A||_2

    (d)  ||A||_2 \leq \sqrt{m}||A||_\infty

    ------------------------------------------------------------------------
    Pathetic attempt:

    (a)  (\max_{1\leq i \leq m}|x_i|)^2 \leq (x^2_1+x^2_2+...+x^2_m)
    So, if the maximum absolute entry in the vector x is at index position, say i=2, then ||x||_2 is at least that big..

    (b)
     \frac{\sqrt{x^*x}}{\sqrt{m}} \leq ||x||_\infty = \max_{1 \leq i \leq m}|x_i|
    The left side looks like the quadratic mean. Let's just say that the quadratic mean is always less than or equal to its largest absolute entry

    (c)
    Ugh, this is where I really get confused..
     (||A||_\infty)^2 \leq n\lambda_{max}
    Where \lambda_{max} is the larges eigenvalue of A^*A...

    (d)
    No idea..
    -----------------------------------------------------------------------

    These norms are giving me an abnormally hard time.
    I've done quite some reading on them, using several sources, but I just can't wrap my head around them...
    Any suggestions are greatly appreciated! Thanks.
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  2. #2
    Lord of certain Rings
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    Quote Originally Posted by Mollier View Post
    Problem:
    verify the following inequalities:

    (a)  ||x||_\infty \leq ||x||_2

    (b)  ||x||_2 \leq \sqrt{m}||x||_\infty

    (c)  ||A||_\infty \leq \sqrt{n}||A||_2

    (d)  ||A||_2 \leq \sqrt{m}||A||_\infty

    ------------------------------------------------------------------------
    Pathetic attempt:

    (a)  (\max_{1\leq i \leq m}|x_i|)^2 \leq (x^2_1+x^2_2+...+x^2_m)
    So, if the maximum absolute entry in the vector x is at index position, say i=2, then ||x||_2 is at least that big..
    Yes. Since \max_{1\leq i \leq m}|x_i| is one of the terms in the summation x^2_1+x^2_2+...+x^2_m and the rest are positive, we have the inequality.

    Quote Originally Posted by Mollier View Post
    (b)
     \frac{\sqrt{x^*x}}{\sqrt{m}} \leq ||x||_\infty = \max_{1 \leq i \leq m}|x_i|
    The left side looks like the quadratic mean. Let's just say that the quadratic mean is always less than or equal to its largest absolute entry
    This inequality is proved by observing \forall j, \, \max_{1\leq i \leq m}|x_i| \geq |x_j|. So by adding the inequalities for all values of j, (\max_{1\leq i \leq m}|x_i|)^2 \geq |x_j|^2 \implies m(\max_{1\leq i \leq m}|x_i|)^2  \geq \sum_{j=1}^{j=m}|x_j|^2. Thus we have the inequality.
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