vector norms and matrix norms

**math8** · April 5th 2010, 04:42 PM

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?

**Opalg** · April 6th 2010, 12:25 AM

Originally Posted by 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?

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.

**math8** · April 6th 2010, 04:56 PM

Thank you very much. Your answer is very clear and helpful

**math8** · May 11th 2010, 12:29 PM

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 $\left\| x \right\|_{q} = \left\| x^{*} \right\|_{q}$ ?

**Opalg** · May 12th 2010, 11:20 AM

Originally Posted by math8

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.

**math8** · May 13th 2010, 02:42 PM

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 $\left\| A^{*} \right\|_{p} = \left\| A \right\|_{q}$ ?

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