vector norms and matrix norms

• Apr 5th 2010, 04:42 PM
math8
vector norms and matrix norms
Let x be a vector. How do you show that $\displaystyle \left\| x^{*} \right\| _{p} = \left\| x \right\| _{q}$
where $\displaystyle \frac{1}{p} + \frac{1}{q} = 1$ ?

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

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

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

But how do you show the other side of the inequality?
• Apr 6th 2010, 12:25 AM
Opalg
Quote:

Originally Posted by math8
Let x be a vector. How do you show that $\displaystyle \left\| x^{*} \right\| _{p} = \left\| x \right\| _{q}$
where $\displaystyle \frac{1}{p} + \frac{1}{q} = 1$ ?

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

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

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

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

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

That shows that $\displaystyle x^*$ (which is a vector with the same components as x, but with the norm given as a functional on $\displaystyle \ell^p$) has norm at least $\displaystyle \|x\|_q$ (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, $\displaystyle y_j$ has the same sign as $\displaystyle x_j$. (Then all the terms in the sum $\displaystyle \textstyle\sum y_jx_j$ will be positive.) If the scalars are complex then you have to multiply each $\displaystyle y_j$ by an appropriate complex number of absolute value 1 to ensure the same result.
• Apr 6th 2010, 04:56 PM
math8
• May 11th 2010, 12:29 PM
math8
I was thinking about this problem again, and since I am able to prove $\displaystyle \left\| x^{*} \right\|_{p} \leq \left\| x \right\|_{q}$ where $\displaystyle \frac {1}{p} + \frac {1}{q} =1$, by switching the roles of p and q, can we say that it follows that $\displaystyle \left\| x^{*} \right\|_{q} \leq \left\| x \right\|_{p}$ and therefore that $\displaystyle \left\| x^{*} \right\|_{q} = \left\| x \right\|_{p}$ follows from the fact that $\displaystyle \left\| x \right\|_{p} = \left\| x^{*} \right\|_{p}$ and that $\displaystyle \left\| x \right\|_{q} = \left\| x^{*} \right\|_{q}$ ?
• May 12th 2010, 11:20 AM
Opalg
Quote:

Originally Posted by math8
I was thinking about this problem again, and since I am able to prove $\displaystyle \left\| x^{*} \right\|_{p} \leq \left\| x \right\|_{q}$ where $\displaystyle \frac {1}{p} + \frac {1}{q} =1$, by switching the roles of p and q, can we say that it follows that $\displaystyle \left\| x^{*} \right\|_{q} \leq \left\| x \right\|_{p}$ and therefore that $\displaystyle \left\| x^{*} \right\|_{q} = \left\| x \right\|_{p}$ follows from the fact that $\displaystyle \left\| x \right\|_{p} = \left\| x^{*} \right\|_{p}$ and that $\displaystyle \left\| x \right\|_{q} = \left\| x^{*} \right\|_{q}$ ?

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 $\displaystyle x \in l^p$, then it defines a linear functional $\displaystyle x^*$ on $\displaystyle l^q$, and $\displaystyle \|x\|_p = \|x^*\|_q$, where the right-hand side refers to the norm of the linear functional on $\displaystyle l^q$ defined by x. But this x need not be an element of $\displaystyle l^q$, 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 $\displaystyle l^p$, so it does not make sense to refer to the p-norm of $\displaystyle x^*$.

In other words, if $\displaystyle x\in l^p$, then you can only refer to the norm of x if you are thinking of x as an element of the space $\displaystyle l^p$ (in which case you must use the p-norm), and you can only refer to the norm of $\displaystyle x^*$ 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.
• May 13th 2010, 02:42 PM
math8
Thanks I got it now for the vectors. But what about for a matrix $\displaystyle A \in C^{m \times n}, if$ $\displaystyle \left\| A^{*} \right\|_{p} \leq \left\| A \right\|_{q}$, can we say that by using $\displaystyle A^{*}$ instead of $\displaystyle A$ and by switching the roles of p and q, that it follows that $\displaystyle \left\| A \right\|_{q} \leq \left\| A^{*} \right\|_{p}$ and hence $\displaystyle \left\| A^{*} \right\|_{p} = \left\| A \right\|_{q}$?