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Math Help - Norm versus inner product

  1. #1
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    Norm versus inner product

    Hey,

    For any \bf{v} \in V i would like to show the following:

    <br />
\Vert \bf{v} \Vert = \sup \{ | \langle \bf{v},\bf{w} \rangle| \big| \bf{w}\in V, \Vert w\Vert=1\}<br />

    I have done the following: To show an equality one can show that the case is true for both \geq and \leq. For the \geq-case I have used the Cauchy-Schwarz inequality:

    |\langle \bf{v},\bf{w} \rangle| \leq \Vert \bf{v} \Vert \Vert \bf{w} \Vert

    Now since \Vert \bf{w}\Vert=1 we have that:

    |\langle \bf{v},\bf{w} \rangle| \leq \Vert \bf{v} \Vert

    Taking the \sup on both sides of the inequality yields:

    \sup|\langle \bf{v},\bf{w} \rangle| \leq \sup\Vert \bf{v} \Vert

    If I could somehow get rid of the \sup on the right side I would be home free. I could be tempted to reason that \sup\Vert\bf{v}\Vert=\Vert \bf{v}\Vert, but somehow that does not sound correct since \bf{v}\in \mathcal{H}.


    Comments would be greatly appreciated.
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  2. #2
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    "||v||= sup ||v||" is trivial because ||v|| is a single number amd the "sup" of a single number is that number. Also, although you have been talking about vectors in V you say " \bf{v}\in \mathcal{H}. Where did \mathcal{H} come from?

    I am concerned about the whole thing. Since, in an inner product space, \bf{||v||} is defined as \sqrt{<v, v>}, shouldn't there be square roots throughout that?
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  3. #3
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    Quote Originally Posted by surjective View Post
    Hey,

    For any \bf{v} \in V i would like to show the following:

    <br />
\Vert \bf{v} \Vert = \sup \{ | \langle \bf{v},\bf{w} \rangle| \big| \bf{w}\in V, \Vert w\Vert=1\}<br />

    I have done the following: To show an equality one can show that the case is true for both \geq and \leq. For the \geq-case I have used the Cauchy-Schwarz inequality:

    |\langle \bf{v},\bf{w} \rangle| \leq \Vert \bf{v} \Vert \Vert \bf{w} \Vert

    Now since \Vert \bf{w}\Vert=1 we have that:

    |\langle \bf{v},\bf{w} \rangle| \leq \Vert \bf{v} \Vert

    Taking the \sup on both sides of the inequality yields:

    \sup|\langle \bf{v},\bf{w} \rangle| \leq \sup\Vert \bf{v} \Vert That should say \color{red}\sup|\langle \bf{v},\bf{w} \rangle| \leq \| \bf{v} \| see HallsofIvy's comment.
    What you should now do is to take \bf{w} = \frac{\bf{v}}{\|\bf{v}\|}. Then \|\bf{w}\| = 1 and \langle \bf{v},\bf{w}\rangle = \frac{\|\bf{v}\|^2}{\|\bf{v}\|} = \|\bf{v}\|.

    (Of course, you can't divide by 0 so that argument won't deal with the case when \bf{v}=\bf{0}. So you have to deal with that case separately.)
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