# bounded operator

1. 1.

Suppose $X$ and $Y$ are normed vector spaces  with norms $\|\cdot\|_{X}$ and $\|\cdot\|_{Y}$. 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

 $\displaystyle\|Tx\|_{Y}$ $\displaystyle\leq$ $\displaystyle C\|x\|_{X},$
2. 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

 $\|T\|:=\mathrm{sup}\left\{\left.\frac{\|Tx\|_{Y}}{\|x\|_{X}}\right|x\in X% \setminus\{0\}\right\}.$

Thus the operator norm is the smallest constant $C\in\mathbf{R}$ such that

 $\displaystyle\|Tx\|_{Y}$ $\displaystyle\leq$ $\displaystyle C\|x\|_{X}.$

Now for any $x\in X\setminus\{0\}$, if we let $y=x/\|x\|$, then linearity implies that

 $\|Ty\|_{Y}=\left\|T\left(\frac{x}{\|x\|_{X}}\right)\right\|_{Y}=\frac{\|Tx\|_{% Y}}{\|x\|_{X}}$

and thus it easily follows that

 $\displaystyle\|T\|$ $\displaystyle=$ $\displaystyle\mathrm{sup}\left\{\left.\frac{\|Tx\|_{Y}}{\|x\|_{X}}\right|x\in X% \setminus\{0\}\right\}=\mathrm{sup}\left\{\|Ty\|_{Y}\left|x\in X\setminus\{0\}% ,y=\frac{x}{\|x\|}\right.\right\}$ $\displaystyle=$ $\displaystyle\mathrm{sup}\{\|Ty\|_{Y}|y\in X,\|y\|=1\}.$

In the special case when $X=\{\mathbf{0}\}$ is the zero vector space, any linear map $T:X\to Y$ is the zero map since $T(\mathbf{0})=T(0\mathbf{0})=0T(\mathbf{0})=0$. In this case, we define $\|T\|:=0$.

3. 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. 1.

The defined norm for mappings is a norm

2. 2.

Examples: identity operator, zero operator: see .

3. 3.

Give alternative expressions for norm of $T$.

4. 4.

Discuss boundedness and continuity

[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. 1.
2. 2.
3. 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
• 2 G. Bachman, L. Narici, Functional analysis, Academic Press, 1966.
Title bounded operator BoundedOperator 2013-03-22 14:02:17 2013-03-22 14:02:17 bwebste (988) bwebste (988) 19 bwebste (988) Definition msc 46B99 VectorNorm