Bounded Linear Operator

Bounded Linear Operators

$H_1,H_2$ be two normed vector spaces, a linear operator $T:H_1\to H_2$ is bounded iff $\exists M>0$ such that $\Vert Tx\Vert \le M\Vert x\Vert ,\forall x\in H_1$. Then the operator norm is defined as follow: $\Vert T\Vert =\inf \{M>0,:\Vert Tx\Vert \le M\Vert x\Vert \forall x\in H_1\}={\sup }_{\Vert x\Vert =1}\Vert Tx\Vert$ $={\sup }_{x\in H_1,x\ne 0}\frac{\Vert Tx\Vert }{\Vert x\Vert }={\sup }_{\Vert x\Vert =\Vert y\Vert =1}|⟨Tx,y⟩|$

Remark

• $T$ is bounded if and only if $T$ is continuous.
• We denote the set of bounded linear maps from $H_1$ to $H_2$ as $B(H_1,H_2)$

Theorem

For $X,Y$ normed vector spaces over the same fields, $(B(X,Y),\Vert \cdot {\Vert }_{X\to Y})$ with $\Vert T{\Vert }_{X\to Y}={\sup }_{\Vert x{\Vert }_{X=1,x\in X}}\Vert Tx{\Vert }_Y$ is a Banach space as long is $Y$ is complete.

Proof Let $\{T_n\}$ be a Cauchy sequence in $B(X,Y)$, i.e. $\forall ϵ>0$, $\exists N\in \mathbb{N}$ such that $\Vert T_n-T_m{\Vert }_{X\to Y}<ϵ$, $\forall m,n\ge N$. Then $\forall x\in X$, $\{T_{n}x\}$ is Cauchy in $Y$ as $\Vert T_{n}x-T_{m}x\Vert \le \Vert T_n-T_m{\Vert }_{X\to Y}\Vert x\Vert <ϵ\Vert x\Vert ,\forall m,n\ge N.$ $Y$ complete then we can define $Tx={\lim }_{n\to \infty }{T}_{n}x$. $\Vert Tx\Vert =\Vert {\lim }_{n\to \infty }{T}_{n}x\Vert ={\lim }_{n\to \infty }\Vert T_{n}x\Vert \le {\mathrm{limsup}}_{n\to \infty }\Vert T_n\Vert \Vert x\Vert$, so $T$ is bounded. $T$ is linear as limit preserve linearity. $\Vert T_n-T{\Vert }_{X\to Y}={\sup }_{x\in Y,\Vert x\Vert =1}\Vert T_{n}x-Tx\Vert ={\sup }_{x\in Y,\Vert x\Vert =1}{\lim }_{m\to \infty }\Vert T_{n}x-T_{m}x\Vert <ϵ$ So $\Vert T_n-T{\Vert }_{X\to Y}\to 0$. We have $(B(X,Y),\Vert \cdot {\Vert }_{X\to Y})$ complete. $\square$