How do you show the other side of the inequality?

[math8]

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?

 

[VeeEight]

I am unsure of what you're doing. Is x an element in L^p? If so, are you asking to prove that the dual space of L^p is isometrically isomorphic to L^q?

 

[math8]

x is a column vector. The initial problem that I am trying to prove is to show that for a matrix $$A \in \textbf{C}^{m\times n}$$,

$$\left\| A \right\|^{2}_{2} \leq \left\| A \right\|_{p} \left\| A \right\|_{q}$$
where $$\frac{1}{p} + \frac{1}{q} = 1$$

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

$$\left\| A \right\|^{2}_{2} = \left\|A ^{\ast}A \right\|_{2} \leq \left\| A^{*} \right\|_{q} \left\| A \right\|_{q}$$

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

So that's why I first want to prove the claim for $$A= x$$ being a mX1 matrix (column vector).