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Math Help - Infinity-norm and inner-product

  1. #1
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    Infinity-norm and inner-product

    Hey,

    Could someone assist me in the following math-problem.

    I want to show that the norm \Vert \cdot \Vert_{\infty} does not come from an inner product.
    Hint: Show that the norm \Vert \cdot \Vert_{\infty} does not satisfy the paralellogram law, e.g., for

    f(x)= \left\lbrace \begin{matrix} x \hspace{0,4cm}, x\in[0,1] \\ 2-x \hspace{0,4cm}, x\in [1,2] \end{matrix} \right.

    g(x)= \left\lbrace \begin{matrix} 1-x \hspace{0,4cm}, x\in[0,1] \\ x-1 \hspace{0,4cm}, x\in [1,2] \end{matrix} \right.

    The paralellogram law states that if V is a vector space with an inner product \langle \cdot, \cdot \rangle and associated norm \Vert \cdot \Vert, then:

    \Vert \textbf{v}+\textbf{w} \Vert^{2} + \Vert \textbf{v}-\textbf{w} \Vert^{2}= 2\left( \Vert \textbf{v} \Vert^{2} + \Vert \textbf{w} \Vert^{2} \right)


    Thanks in advance.
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  2. #2
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    Have you tried to this yourself? What problem did you have? You are told exactly what to do. What are ||f|| and ||g||? What are ||f+ g|| and ||f- g||? What is 2(||f||^2+ ||g||^2)?

    Surely you see that f+ g= 1 for all x in [0, 2] and that f- g= 2x-1 for x in [0, 1] and 3- 2x for x in [1, 2].
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  3. #3
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    Infinity norm and inner-product

    Hey,

    Yes indeed I have tried myself. The following is what I have so far:

    By definition I know that:

    \Vert h \Vert_{\infty}=\displaystyle\smash{\max_{x \in [a,b]}}\vert h(x) \vert

    It is clear that:

    \displaystyle\smash{\max_{x \in [0,2]}}\vert f(x)+g(x)\vert=1

    \displaystyle\smash{\max_{x \in [0,1]}}\vert f(x)-g(x)\vert=\displaystyle\smash{\max_{x \in [0,1]}}\vert 2x-1\vert=1

    \displaystyle\smash{\max_{x \in [1,2]}}\vert f(x)-g(x)\vert=\displaystyle\smash{\max_{x \in [1,2]}}\vert -2x+3\vert=1

    \displaystyle\smash{\max_{x \in [0,2]}}\vert f(x)\vert=1

    \displaystyle\smash{\max_{x \in [0,2]}}\vert g(x)\vert=1

    Applying the definition in the beginning we have:

    \Vert f+g \Vert_{\infty}^{2}+\Vert f-g \Vert_{\infty}^{2}=2\left( \Vert f \Vert_{\infty}^{2}+\Vert g \Vert_{\infty}^{2} \right) \Leftrightarrow

    1+1=2(1+1) \Leftrightarrow

    2=4

    We conclude that the norm \Vert \cdot \Vert_{\infty} does not satisfy the parallelogram law. My question is : "How does this show that the infinity-norm does not come from an inner-product? Is is simply because the above shows that the infinity-norm does not fulfill the parallelogram law which is one of the requirements for a vectorspace with an inner-product and associated norm?"

    Thanks.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by surjective View Post
    Hey,

    Yes indeed I have tried myself. The following is what I have so far:

    By definition I know that:

    \Vert h \Vert_{\infty}=\displaystyle\smash{\max_{x \in [a,b]}}\vert h(x) \vert

    It is clear that:

    \displaystyle\smash{\max_{x \in [0,2]}}\vert f(x)+g(x)\vert=1

    \displaystyle\smash{\max_{x \in [0,1]}}\vert f(x)-g(x)\vert=\displaystyle\smash{\max_{x \in [0,1]}}\vert 2x-1\vert=1

    \displaystyle\smash{\max_{x \in [1,2]}}\vert f(x)-g(x)\vert=\displaystyle\smash{\max_{x \in [1,2]}}\vert -2x+3\vert=1

    \displaystyle\smash{\max_{x \in [0,2]}}\vert f(x)\vert=1

    \displaystyle\smash{\max_{x \in [0,2]}}\vert g(x)\vert=1

    Applying the definition in the beginning we have:

    \Vert f+g \Vert_{\infty}^{2}+\Vert f-g \Vert_{\infty}^{2}=2\left( \Vert f \Vert_{\infty}^{2}+\Vert g \Vert_{\infty}^{2} \right) \Leftrightarrow

    1+1=2(1+1) \Leftrightarrow

    2=4

    We conclude that the norm \Vert \cdot \Vert_{\infty} does not satisfy the parallelogram law. My question is : "How does this show that the infinity-norm does not come from an inner-product? Is is simply because the above shows that the infinity-norm does not fulfill the parallelogram law which is one of the requirements for a vectorspace with an inner-product and associated norm?"

    Thanks.
    You have proved that the parallelogram law need not apply for \|\cdot\|_{\infty} but every norm satisfies this. Thus...
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  5. #5
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    Quote Originally Posted by Drexel28 View Post
    You have proved that the parallelogram law need not apply for \|\cdot\|_{\infty} but every norm satisfies this. Thus...
    I think you mean every inner product, not every norm. Since \|\cdot\|_\infty is a norm.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by putnam120 View Post
    I think you mean every inner product, not every norm. Since \|\cdot\|_\infty is a norm.
    Yes, thank you.
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