bounded operator
Definition [1]

1.
Suppose $X$ and $Y$ are normed vector spaces^{} with norms $\parallel \cdot {\parallel}_{X}$ and $\parallel \cdot {\parallel}_{Y}$. Further, suppose $T$ is a linear map $T:X\to Y$. If there is a $C\in \mathbf{R}$ such that for all $x\in X$ we have
${\parallel Tx\parallel}_{Y}$ $\le $ $C{\parallel x\parallel}_{X},$ then $T$ is a bounded operator^{}.

2.
Let $X$ and $Y$ be as above, and let $T:X\to Y$ be a bounded operator. Then the norm of $T$ is defined as the real number
$$\parallel T\parallel :=\mathrm{sup}\left\{\frac{{\parallel Tx\parallel}_{Y}}{{\parallel x\parallel}_{X}}\rightx\in X\setminus \{0\}\}.$$ Thus the operator norm is the smallest constant $C\in \mathbf{R}$ such that
${\parallel Tx\parallel}_{Y}$ $\le $ $C{\parallel x\parallel}_{X}.$ Now for any $x\in X\setminus \{0\}$, if we let $y=x/\parallel x\parallel $, then linearity implies that
$${\parallel Ty\parallel}_{Y}={\parallel T\left(\frac{x}{{\parallel x\parallel}_{X}}\right)\parallel}_{Y}=\frac{{\parallel Tx\parallel}_{Y}}{{\parallel x\parallel}_{X}}$$ and thus it easily follows that
$\parallel T\parallel $ $=$ $\mathrm{sup}\left\{{\displaystyle \frac{{\parallel Tx\parallel}_{Y}}{{\parallel x\parallel}_{X}}}\rightx\in X\setminus \{0\}\}=\mathrm{sup}\left\{{\parallel Ty\parallel}_{Y}\rightx\in X\setminus \{0\},y={\displaystyle \frac{x}{\parallel x\parallel}}\}$ $=$ $\mathrm{sup}\{{\parallel Ty\parallel}_{Y}y\in X,\parallel y\parallel =1\}.$ In the special case when $X=\{\mathrm{\U0001d7ce}\}$ is the zero vector space, any linear map $T:X\to Y$ is the zero map since $T(\mathrm{\U0001d7ce})=T(\mathrm{\U0001d7ce\U0001d7ce})=0T(\mathrm{\U0001d7ce})=0$. In this case, we define $\parallel T\parallel :=0$.

3.
To avoid cumbersome notational stuff usually one can simplify the symbols like ${x}_{X}$ and ${Tx}_{Y}$ by writing only $x$, $Tx$ since there is a little danger in confusing which is space about calculating norms.
0.0.1 TO DO:

1.
The defined norm for mappings is a norm

2.
Examples: identity operator, zero operator: see [1].

3.
Give alternative expressions for norm of $T$.

4.
Discuss boundedness and continuity
Theorem^{} [1, 2] Suppose $T:X\to Y$ is a linear map between normed vector spaces $X$ and $Y$. If $X$ is finitedimensional, then $T$ is bounded^{}.
Theorem Suppose $T:X\to Y$ is a linear map between normed vector spaces $X$ and $Y$. The following are equivalent^{}:

1.
$T$ is continuous^{} in some point ${x}_{0}\in X$

2.
$T$ is uniformly continuous^{} in $X$

3.
$T$ is bounded
Lemma Any bounded operator with a finite dimensional kernel and cokernel has a closed image.
Proof By Banach’s isomorphism theorem.
References
 1 E. Kreyszig, Introductory Functional Analysis^{} With Applications, John Wiley & Sons, 1978.
 2 G. Bachman, L. Narici, Functional analysis, Academic Press, 1966.
Title  bounded operator 

Canonical name  BoundedOperator 
Date of creation  20130322 14:02:17 
Last modified on  20130322 14:02:17 
Owner  bwebste (988) 
Last modified by  bwebste (988) 
Numerical id  19 
Author  bwebste (988) 
Entry type  Definition 
Classification  msc 46B99 
Related topic  VectorNorm 