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Math Help - vector norms and matrix norms

  1. #1
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    vector norms and matrix norms

    Let x be a vector. How do you show that \left\| x^{*}  \right\| _{p} = \left\| x \right\| _{q}
    where \frac{1}{p} + \frac{1}{q} = 1 ?


    By using this definition of \left\| x^{*}  \right\| _{p} = max_{ \left\| y  \right\| _{p} =1} \left\| x^{*} y \right\| _{p}

    and Holder's inequality, I am able to prove that

    \left\| x^{*}  \right\| _{p}  \leq \left\| x \right\| _{q}

    But how do you show the other side of the inequality?
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  2. #2
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    Quote Originally Posted by math8 View Post
    Let x be a vector. How do you show that \left\| x^{*}  \right\| _{p} = \left\| x \right\| _{q}
    where \frac{1}{p} + \frac{1}{q} = 1 ?


    By using this definition of \left\| x^{*}  \right\| _{p} = max_{ \left\| y  \right\| _{p} =1} \left\| x^{*} y \right\| _{p}

    and Holder's inequality, I am able to prove that

    \left\| x^{*}  \right\| _{p}  \leq \left\| x \right\| _{q}

    But how do you show the other side of the inequality?
    If \tfrac{1}{p} + \tfrac{1}{q} = 1 then \tfrac qp + 1 = q. For convenience, suppose that the coordinates of x are all positive, so that we can drop the absolute value symbols in the formula \|x\|_q^q = \textstyle\sum |x_j|^q. Then \|x\|_q^q = \textstyle\sum x_j^{(q/p)+1} = \sum x_j^{(q/p)+1} = \sum x_j^{q/p}x_j = x^*(y), where y_j = x_j^{q/p}. But  \textstyle\sum y_j^p = \sum x_j^q = \|x\|_q^q. Thus \|y\|_p = \|x\|_q^{q/p} = \|x\|_q^{q-1} and x^*(y)=\|x\|_q^q. Now let z = y/\|x\|_q^{q-1}. Then \|z\|_p = 1 and x^*(z) = \|x\|_q

    That shows that x^* (which is a vector with the same components as x, but with the norm given as a functional on \ell^p) has norm at least \|x\|_q (which is what you wanted).

    It remains to deal with the fact that the components of x may not all be positive. If the scalars are real then you just have to ensure that, for each j, y_j has the same sign as x_j. (Then all the terms in the sum  \textstyle\sum y_jx_j will be positive.) If the scalars are complex then you have to multiply each y_j by an appropriate complex number of absolute value 1 to ensure the same result.
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    Thank you very much. Your answer is very clear and helpful
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    I was thinking about this problem again, and since I am able to prove \left\| x^{*} \right\|_{p} \leq \left\| x \right\|_{q} where \frac {1}{p} + \frac {1}{q} =1, by switching the roles of p and q, can we say that it follows that \left\| x^{*} \right\|_{q} \leq \left\| x \right\|_{p} and therefore that \left\| x^{*} \right\|_{q} = \left\| x \right\|_{p} follows from the fact that \left\| x \right\|_{p} = \left\| x^{*} \right\|_{p} and that   \left\| x \right\|_{q} =    \left\| x^{*} \right\|_{q} ?
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  5. #5
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    Quote Originally Posted by math8 View Post
    I was thinking about this problem again, and since I am able to prove \left\| x^{*} \right\|_{p} \leq \left\| x \right\|_{q} where \frac {1}{p} + \frac {1}{q} =1, by switching the roles of p and q, can we say that it follows that \left\| x^{*} \right\|_{q} \leq \left\| x \right\|_{p} and therefore that \left\| x^{*} \right\|_{q} = \left\| x \right\|_{p} follows from the fact that \left\| x \right\|_{p} = \left\| x^{*} \right\|_{p} and that   \left\| x \right\|_{q} =    \left\| x^{*} \right\|_{q} ?
    That is a cunning suggestion, and I had to think hard to see why it (unfortunately) doesn't work. In fact, if you have a sequence x \in l^p, then it defines a linear functional x^* on l^q, and \|x\|_p = \|x^*\|_q, where the right-hand side refers to the norm of the linear functional on l^q defined by x. But this x need not be an element of l^q, so it does not make sense to refer to the q-norm of x. Also, x does not necessarily define a bounded linear functional on the space l^p, so it does not make sense to refer to the p-norm of x^*.

    In other words, if x\in l^p, then you can only refer to the norm of x if you are thinking of x as an element of the space l^p (in which case you must use the p-norm), and you can only refer to the norm of x^* if you are thinking of x as defining an element of a dual space (a linear functional), in which case you must use the q-norm.
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    Thanks I got it now for the vectors. But what about for a matrix A \in C^{m \times n}, if    \left\| A^{*} \right\|_{p} \leq  \left\| A \right\|_{q} , can we say that by using A^{*} instead of A and by switching the roles of p and q, that it follows that    \left\| A \right\|_{q} \leq  \left\| A^{*} \right\|_{p}  and hence    \left\| A^{*} \right\|_{p} =  \left\| A \right\|_{q} ?
    Last edited by math8; May 13th 2010 at 02:56 PM.
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