Lecture 08 - Linear Operators

#math

Let $X, Y$ be Banach spaces and

T : D (T) \subset X \overset{l i n e a r}{\to} Y

$D (T)$ is called the domain of $T,$ and is a subspace of $X .$
$R (T)$ is called the range of T and is a subspace of $Y .$
If $\overset{―}{D (T)} = X$ then $T$ is said to be densely defined.

Definition

The norm of T is defined to be:

| | T | | = \sup_{x \in D (T)} \frac{| | T x | |_{Y}}{| | x | |_{X}}

Equivalently:

| | T | | = \sup_{| | x | |_{X} \leq 1} | | T x | |_{Y}

| | T | | = \sup_{| | x | |_{X} = 1} | | T x | |_{Y}

If $| | T | |$ is finite then $T$ is said to be bounded on its domain.

Proposition

$| | T x | | \leq | | T | | | | x | |$

Theorem

Let $T : D (T) \subset X \to Y$ be linear. TFAE:

$T$ is bounded on $D (T)$
$T$ is continuous at every point on $D (T)$
T is continuous at some points of $D (T)$
$T$ is continuous at 0

Proposition

If $T$ is bounded on $D (T)$ , then it has a unique norm preserving extension to $\overset{―}{D (T)}$ , i.e. , $\exists! S : \overset{―}{D (T)} \to Y$ such that $S x = T x$ on $D (T)$ and $| | S | | = | | T | |$

We denote $B (X \to Y) : {T : X \to Y : T is bounded on X}$ , and if $Y = C$ (or whatever scalar field) we state say that it is the dual space of $X .$