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How do you show the other side of the inequality?

  • Thread starter math8
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  • #1
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Let x be a vector. How do you show that [tex]\left\| x^{*} \right\| _{p} = \left\| x \right\| _{q}[/tex]
where [tex]\frac{1}{p} + \frac{1}{q} = 1[/tex]

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

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

[tex]\left\| x^{*} \right\| _{p} \leq \left\| x \right\| _{q}[/tex]

But how do you show the other side of the inequality?
 

Answers and Replies

  • #2
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I am unsure of what you're doing. Is x an element in Lp? If so, are you asking to prove that the dual space of Lp is isometrically isomorphic to Lq?
 
  • #3
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x is a column vector. The initial problem that I am trying to prove is to show that for a matrix [tex]A \in \textbf{C}^{m\times n}[/tex],

[tex]\left\| A \right\|^{2}_{2} \leq \left\| A \right\|_{p} \left\| A \right\|_{q}[/tex]
where [tex]\frac{1}{p} + \frac{1}{q} = 1[/tex]

I use the fact that for any Hermitian matrix, the 2-norm is less or equal than any matrix norm induced by a vector norm. I use the fact that the matrix A*A is Hermitian and I can show that

[tex]\left\| A \right\|^{2}_{2} = \left\|A ^{\ast}A \right\|_{2} \leq \left\| A^{*} \right\|_{q} \left\| A \right\|_{q}[/tex]

So now my claim is to show that
[tex] \left\| A^{*} \right\|_{q} = \left\| A \right\|_{p}[/tex]

So that's why I first want to prove the claim for [tex] A= x [/tex] being a mX1 matrix (column vector).
 
Last edited:

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