Does the set of "bound factor" of linear operator always have an minimum?

Question

$\|T(u)\|\le M\|u\|$

T is a bounded operator between normed linear spaces, then is the set of all the "bound factor"(what is the proper name for this set?) M always have an minimum and why?

Answer 1

Any set contained in $[0,\infty)$ has a finite infimum. What is more useful is the fact that the constants $M$ here have a mimimum. If $\|Tu \| \leq M_n \|u\|$ for all $u$ and $n$ and if $M_n \to M$ then $\|Tu \| \leq M \|u\|$ for all $u$. Hence the infimum is attained.

Details: Let $A=\{M \geq 0: \|Tu\|\leq M \|u\| \forall u\}$. Let $m=\inf A$. Then there exists a sequence $(M_n) \subset A$ such that $M_n \to m$. Since $\|Tu\|\leq M_n \|u\|$ for all $u$ and $n$ we can let $n \to \infty$ to get $\|Tu\| \leq m \|u\|$ for all $u$. Hence $m \in A$ and $m$ is the minimum of $A$.