bounded operatorBoundedOperator2003-10-15 01:30:322007-08-07 22:47:14Definition%fancy typeface/symbols
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%% \include{argawarga}{\bf Definition} \cite{kreyszig}
\begin{enumerate}
\item 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 {\bf bounded operator}.
\item Let $X$ and $Y$ be as above, and let $T : X \to Y$ be a bounded
operator. Then the {\bf norm} of $T$ is defined as the real number
$$\| T \| := \mathrm{sup} \left\{ \left. \frac{\| Tx \|_Y}{\| x \|_X} \right| x \in X \setminus \{0\} \right\}.$$ %this is a hack, if you know how to do it correctly, please change this
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 $\|T \| := 0$.
\item 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.
\end{enumerate}
\subsubsection{TO DO:}
\begin{enumerate}
\item The defined norm for mappings is a norm
\item Examples: identity operator, zero operator: see \cite{kreyszig}.
\item Give alternative expressions for norm of $T$. \item Discuss boundedness and continuity
\end{enumerate}
{\bf Theorem} \cite{kreyszig, bachman} 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.
{\bf Theorem} Suppose $T : X \to Y$ is a
linear map between normed vector spaces $X$ and $Y$. The following are equivalent:
\begin{enumerate}
\item $T$ is continuous in some point $x_{0} \in X$
\item $T$ is uniformly continuous in $X$
\item $T$ is bounded
\end{enumerate}
{\bf Lemma} Any bounded operator with a finite dimensional kernel and cokernel has a closed image.
{\bf Proof} By Banach's isomorphism theorem.
\begin{thebibliography}{9}
\bibitem{kreyszig} E. Kreyszig,
\emph{Introductory Functional Analysis With Applications},
John Wiley \& Sons, 1978.
\bibitem{bachman} G. Bachman, L. Narici, \emph{Functional analysis}, Academic Press, 1966.
\end{thebibliography}