Definition [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 \begin{eqnarray*} \| Tx \|_Y &\le& C\| x \|_X, \end{eqnarray*}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 $$\| T \| := \mathrm{sup} \left\{ \left. \frac{\| Tx \|_Y}{\| x \|_X} \right| x \in X \setminus \{0\} \right\}.$$ Thus the operator norm is the smallest constant $C \in \mathbf{R}$ such that \begin{eqnarray*} \| Tx \|_Y &\le& C \| x \|_X. \end{eqnarray*} 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 \begin{eqnarray*} \| T \| &=& \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\}\\ &=& \mathrm{sup} \{ \| Ty \|_Y | y \in X, \|y\| = 1 \}. \end{eqnarray*} 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. 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.

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.

Bibliography

1

E. Kreyszig, Introductory Functional Analysis With Applications, John Wiley & Sons, 1978.

2

G. Bachman, L. Narici, Functional analysis, Academic Press, 1966.