I have one final enquiry, and that is the relation,

$\displaystyle ||B|| = \sup_{||x||\leq 1}||Bx|| = \sup_{||x|| = 1}||Bx||.$

I don't exactly know what this is saying. I mean, I could interpret it as the least upper bound on the sphere, $\displaystyle ||x|| \leq 1$, and that that is the same as the least upper bound on the surface of the sphere. Thinking in that way, I can see kind of why this is the case. Is there any way of proving it formally?

Further thoughts:

The norm of a bounded linear operator is given by,

$\displaystyle \displaystyle ||A|| = \sup_{||x||\leq 1}||Ax|| = \sup_{||x|| = 1} ||Ax||.$

Now, the way I think about this is that this is the least upper bound on the closed unit sphere, and that this can be viewed as the same as the least upper bound on the surface of the unit sphere, I think. However, how can this be proved formally?

Thanks in advance.