Prove that if $T \in L(V,W)$ then $ \|T\| = \inf \{M \in \R : \, \|Tv\| \le M\|v\| \textrm{ for all } v \in V \}.$
Answer
\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 $\alpha= \|T\|$. $\Box$
Erdos
- answered
- 1544 views
- $10.00
Related Questions
- A function satifying $|f(x)-f(y)|\leq |x-y|^2$ must be constanct.
- A Real Analysis question on convergence of functions
- real analysis
- Let $f:U\subset\mathbb{R} ^3\rightarrow \mathbb{R} ^2$ given by $f(x,y,z)=(sin(x+z)+log(yz^2) ; e^{x+z} +yz)$ where $U = { (x, y, z) ∈ R^3| y, z > 0 }.$ Questions Inside.
- Mathematical modeling
- real analysis
- [Real Analysis] Show that the set $A$ is uncountable. Use this result to show that ${\displaystyle\mathbb {R}}$ is uncountable.
- Interior of union of two sets with empty interior
