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 $Z$, this operation exhibits the property demonstrated.


 * 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 generally 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)


 * Yes this was just an oversight on my part. $\norm \cdot$ somehow denoted three different norms. Caliburn (talk) 12:12, 19 November 2022 (UTC)


 * Oh yeah that doesn't mean the conclusion doesn't really make sense. Maybe say "if $X = Y = Z$ then this shows that $\norm \cdot_{\map B {X, X} }$ is submultiplicative. It's not a huge deal. Caliburn (talk) 12:13, 19 November 2022 (UTC)


 * Yes, or there also could be a more general definition of submultiplicativity of norms. --Usagiop (talk) 14:22, 19 November 2022 (UTC)


 * It should be noted that the term submultiplicative generally refers to a multiplication or scalar multiplication and compostition is neither for general $X$, $Y$ and $Z$.


 * If $Y = Z$ it is a scalar multiplication and if $X = Y = Z$ it is a multiplication. --Leigh.Samphier (talk) 22:39, 19 November 2022 (UTC)