Question about dual norms on Banach spaces

I've been exploring the properties of tuning map that minimize various
Lp distances to the JIP. Paul has suggested the name Lp-TOP for these,
which I like. The special case Linf-TOP is just called "TOP," and the
case L2-TOP is called "TE."

I'm trying to figure out what, exactly, is being minimized when one
minimizes the Lp distance to the JIP. I've read this article here:

http://en.wikipedia.org/wiki/Dual_norm

So I note that for some covector V, ||V|| = sup {|V(M)|, ||M|| <=1},
where M is an element of the dual space.

What I'm trying specifically to figure out is if this definition is
equivalent to the above: ||V|| = sup { |V(M)| / ||M|| }, where ||M||
can be unbounded, for all Lp norms.

I'd like to say "yes." But, the fact that the definition of the dual
norm bounds ||M|| to be <= 1, rather than just = 1, makes me think
that there's some strange corner case where this isn't true.

Does anyone have any insight?

-Mike

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:

> Does anyone have any insight?

It's all the same.

On Sun, Mar 25, 2012 at 9:05 PM, genewardsmith <genewardsmith@sbcglobal.net>
wrote:
>
> --- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...>
> wrote:
>
> > Does anyone have any insight?
>
> It's all the same.

I assume you're right, but what's the basic reasoning that you use to
see this? It seemed to be related to that norms are defined so that
||k*x|| = k*||x|| for any scalar k, but I couldn't figure out how to
connect the dots to prove that there wasn't some Lp case with p < 0
where things didn't work out neatly.

For instance, if there's some case where the sup V(M) ends up being
produced by an M with ||M|| < 1, then for that M, V(M)/||M|| > V(M).
(And if there's no such case, then why is the dual norm defined with
||M|| <= 1 rather than ||M|| = 1?)

-Mike

On Sun, Mar 25, 2012 at 11:23 PM, Mike Battaglia <battaglia01@gmail.com> wrote:
>
> For instance, if there's some case where the sup V(M) ends up being
> produced by an M with ||M|| < 1, then for that M, V(M)/||M|| > V(M).
> (And if there's no such case, then why is the dual norm defined with
> ||M|| <= 1 rather than ||M|| = 1?)

Never mind, I see how to prove it in general. Here's the proof:

Assume that V is an element in the dual space to some vector space,
and we want to find ||V||. Then assume that some vector M with ||M|| <
1 produces the supremum of (V(M)). Then, also, assume that k is some
scalar such that k||M|| = 1. Since k||M|| = 1, then k = 1/||M||, and
since we've said that ||M|| < 1, we know k > 1. But by the definition
of a norm, k||M|| = ||kM||, and hence V(kM) = k*V(M), so we've arrived
at a contradiction.

Thus, we know that the supremum must lie on the unit circle. Note that
any vector can be expressed in the form km, where k is a real number
and m is a vector on the unit circle. Thus, the expression V(M)/||M||
can be rewritten as V(km)/||km||, which simplifies to V(m)/||m||, with
||m|| being bounded at 1, satisfying our original definition.

There must be some extremely strange and exotic space in which the
above is not true, but it's got to have some strange properties. If a
vector with norm less than 1 can produce the supremum, and if norms
and linear functionals must both be linear with respect to scalar
multiplication, then I'm not sure how it'd work. Must be some
situation where you can't scale a vector by enough to get a norm of 1.

-Mike