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Thread: Norm versus inner product

  1. #1
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    Norm versus inner product

    Hey,

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

    $\displaystyle
    \Vert \bf{v} \Vert = \sup \{ | \langle \bf{v},\bf{w} \rangle| \big| \bf{w}\in V, \Vert w\Vert=1\}
    $

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

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

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

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

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

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

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


    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 "$\displaystyle \bf{v}\in \mathcal{H}$. Where did $\displaystyle \mathcal{H}$ come from?

    I am concerned about the whole thing. Since, in an inner product space, $\displaystyle \bf{||v||}$ is defined as $\displaystyle \sqrt{<v, v>}$, 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 $\displaystyle \bf{v} \in V$ i would like to show the following:

    $\displaystyle
    \Vert \bf{v} \Vert = \sup \{ | \langle \bf{v},\bf{w} \rangle| \big| \bf{w}\in V, \Vert w\Vert=1\}
    $

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

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

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

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

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

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

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