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 $C\in 𝐑$ 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

   $$\parallel T\parallel :=\sup \left\{\frac{{\parallel Tx\parallel }_Y}{{\parallel x\parallel }_X}\right|x\in X\setminus \{0\}\}.$$

   Thus the operator norm is the smallest constant $C\in 𝐑$ 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

   $\parallel T\parallel$ $=$ $\sup \left\{{\displaystyle \frac{{\parallel Tx\parallel }_Y}{{\parallel x\parallel }_X}}\right|x\in X\setminus \{0\}\}=\sup \left\{{\parallel Ty\parallel }_Y\right|x\in X\setminus \{0\},y={\displaystyle \frac{x}{\parallel x\parallel }}\}$

   $=$ $\sup \{{\parallel Ty\parallel }_Y|y\in X,\parallel y\parallel =1\}.$

   In the special case when $X=\{\mathrm{𝟎}\}$ is the zero vector space, any linear map $T:X\to Y$ is the zero map since $T(\mathrm{𝟎})=T(\mathrm{𝟎𝟎})=0T(\mathrm{𝟎})=0$. 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 finite-dimensional, 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.

Title	bounded operator
Canonical name	BoundedOperator
Date of creation	2013-03-22 14:02:17
Last modified on	2013-03-22 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