Talk:Norm on Bounded Linear Transformation is Submultiplicative

Why are the operator norms not specified?
I think the following is a mistake.
 * $\norm \cdot$ is submultiplicative.

I guess what is meant is, if $X=Y=Z$, then:
 * $\norm \cdot _{\map B {X,X} }$ is submultiplicative.

Isn't it? Dropping the associated space causes such confusion. --Usagiop (talk) 11:06, 19 November 2022 (UTC)


 * We have established that for arbitrary $X$, $Y$ and $X$ exhibit the property they do.


 * The conclusion that is made is that as a consequence, the norm, as defined on a bounded linear transformation according to the multiple definitions on its definition page, has been demonstrated generay to exhibit the property of submultiplicativity, therefore this norm can as a consequence be classified as a submultiplicative norm.


 * That's my take from having watched all this take place at a distance. It's just part of the long tedious path towards demonstrating that these norms that have been created do in fact behave in a way consistent with such norms to be expected to behave, and so we can apply norm theoretical results to bounded linear transformations and hence allowing their treatment on a more abstract level.


 * I could be completely wrong. --prime mover (talk) 11:19, 19 November 2022 (UTC)