# Prove that if $T \in L(V,W)$ then $\|T\| = \inf \{M \in \R : \, \|Tv\| \le M\|v\| \textrm{ for all } v \in V \}.$

Prove that if $T \in L(V,W)$ then $\|T\| = \inf \{M \in \R : \, \|Tv\| \le M\|v\| \textrm{ for all } v \in V \}.$

\textbf{Solution:} By definition $\|T\|=\sup_{v \in V, v \neq 0} \frac{\|Tv\|}{\|v\|}\geq \frac{\|Tv\|}{\|v\|} \text{ for all } v \in V, v \neq 0.$ Thus, $\|Tv\| \leq \|T\|\|v\|, \ \ \forall v\in V.$ Note that the above inequality trivially holds for $v=0$ since $T(0)=0.$ Therefore $\alpha \leq \|T\|,$ where $\alpha = \inf \{M \in \R : \|Tv\|\leq M\|v\| , \forall v \in V \}$.\\ Next we show that $\alpha \geq \|T\|$. Let $\epsilon >0$. Then there exists $M<\alpha+\epsilon$ such that $\|Tv\|\leq M\|v\| \rightarrow \frac{\|Tv\|}{\|v\|}\leq M < \alpha+\epsilon, \forall v \in V, v\neq 0.$ Thus $\|T\|=\sup_{v \in V, v \neq 0} \frac{\|Tv\|}{\|v\|}\leq M <\alpha+\epsilon,$ and hence $\|T\|<\alpha+\epsilon.$ Letting $\epsilon \rightarrow 0$ we get $\|T\| \leq \alpha$, and hence $\alpha= \|T\|$.  $\Box$