Let x be a vector. How do you show that

where ?

By using this definition of

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

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

Printable View

- April 5th 2010, 05:42 PMmath8vector norms and matrix norms
Let x be a vector. How do you show that

where ?

By using this definition of

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

But how do you show the other side of the inequality? - April 6th 2010, 01:25 AMOpalg
If then . For convenience, suppose that the coordinates of x are all positive, so that we can drop the absolute value symbols in the formula . Then , where . But . Thus and . Now let . Then and

That shows that (which is a vector with the same components as x, but with the norm given as a functional on ) has norm at least (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, has the same sign as . (Then all the terms in the sum will be positive.) If the scalars are complex then you have to multiply each by an appropriate complex number of absolute value 1 to ensure the same result. - April 6th 2010, 05:56 PMmath8
Thank you very much. Your answer is very clear and helpful :)

- May 11th 2010, 01:29 PMmath8
I was thinking about this problem again, and since I am able to prove where , by switching the roles of p and q, can we say that it follows that and therefore that follows from the fact that and that ?

- May 12th 2010, 12:20 PMOpalg
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 , then it defines a linear functional on , and , where the right-hand side refers to the norm of the linear functional on defined by x. But this x need not be an element of , 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 , so it does not make sense to refer to the p-norm of .

In other words, if , then you can only refer to the norm of x if you are thinking of x as an element of the space (in which case you must use the p-norm), and you can only refer to the norm of 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, 03:42 PMmath8
Thanks I got it now for the vectors. But what about for a matrix , can we say that by using instead of and by switching the roles of p and q, that it follows that and hence ?