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Math Help - Norms in R^m

  1. #1
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    Norms in R^m

    Define ||x||=\sqrt{(x1)^2 +...+(xm)^2} in R^m, and ||x||_{p} = \sqrt[p]{|x1|^{p} +...+|x|^{p}}. I need to show that for every m and p there are scalars c and C such that, for every x in R^m, c||x|| \leq ||x||_{p} \leq C||x||.

    My only thought is to try to find the limit of C||x||-||x||_{p} as each coordinate goes to infinity. If it converges the I think I'm done, no? But I don't see how to do it.
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  2. #2
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    However, being in the first part of an analysis course, the notion of a limit is not yet defined so this is probably not allowed.
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    Let me write this a little better (now that I can do it on my computer rather than an iPhone).

    Define ||\text{\bf x}|| = \sqrt{(x_{1})^{2} + ... + (x_{m})^{2}} for \text{\bf x} \in \mathbb{R}^{m}, and ||\text{\bf x}||_{p} = \sqrt[p]{|x_{1}|^{p} + ... + |x_{m}|^{p}}. Then prove what was said above: for any m, p \geq 1, there are scalars c, C > 0 such that for any \text{\bf x} \in \mathbb{R}^{m} we have c||\text{\bf x}|| \leq ||\text{\bf x}||_{p} \leq C||\text{\bf x}||.
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  4. #4
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    Quote Originally Posted by ragnar View Post
    Define ||\text{\bf x}|| = \sqrt{(x_{1})^{2} + ... + (x_{m})^{2}} for \text{\bf x} \in \mathbb{R}^{m}, and ||\text{\bf x}||_{p} = \sqrt[p]{|x_{1}|^{p} + ... + |x_{m}|^{p}}. Then prove what was said above: for any m, p \geq 1, there are scalars c, C > 0 such that for any \text{\bf x} \in \mathbb{R}^{m} we have c||\text{\bf x}|| \leq ||\text{\bf x}||_{p} \leq C||\text{\bf x}||.
    Hint: Let \|{\bf x}\|_\infty = \max\{|x_1|,|x_2|,\ldots,|x_m|\}. Show that \|{\bf x}\|_\infty\leqslant\|{\bf x}\|_p\leqslant\|{\bf x}\|_1\leqslant m\|{\bf x}\|_\infty. Notice also that \|{\bf x}\|=\|{\bf x}\|_2.
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    Okay, I see how to prove the inequality of your hint. What I don't see is how this contributes to the assignment. I also see that ||\text{\bf x}||_{\infty} \leq ||\text{\bf x}||_{2} \leq ||\text{\bf x}||_{1}, but I'm still stuck.
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    I think I may have part of it. If I multiply ||\text{\bf x}|| by \sqrt{m} then for whichever term is the max, I'll have m many of it, thus meaning that the stuff under the radical will be closer to being demonstrably larger than ||\text{\bf x}||_{p}.
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  7. #7
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    Quote Originally Posted by ragnar View Post
    Okay, I see how to prove the inequality of your hint. What I don't see is how this contributes to the assignment. I also see that ||\text{\bf x}||_{\infty} \leq ||\text{\bf x}||_{2} \leq ||\text{\bf x}||_{1}, but I'm still stuck.
    Once you have proved those inequalities, you're practically there:

    \|{\bf x}\|_p \leqslant m\|{\bf x}\|_\infty \leqslant m\|{\bf x}\|_2,

    \|{\bf x}\|_2\leqslant m\|{\bf x}\|_\infty \leqslant m\|{\bf x}\|_p and hence m^{-1}\|{\bf x}\|_2\leqslant \|{\bf x}\|_p.

    Thus m^{-1}\|{\bf x}\| \leqslant \|{\bf x}\|_p \leqslant m\|{\bf x}\|.
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  8. #8
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    Ah, got it... But WOW I'm amazed I was expected to have figured this all out without even the hint you gave.

    Thank you very much for your help! That was very good.
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