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?
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.
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.