Linear operators

Definition 1: a linear operator \(T\) is a linear mapping such that

1. the domain \(\mathscr{D}(T)\) of \(T\) is a vector space and the range \(\mathscr{R}(T)\) of \(T\) is contained in a vector space over the same field as \(\mathscr{D}(T)\).
2. \(\forall x, y \in \mathscr{D}(T): T(x + y) = Tx + Ty\).
3. \(\forall x \in \mathscr{D}(T), \alpha \in F: T(\alpha x) = \alpha Tx\).

Observe the notation; we \(Tx\) and \(T(x)\) are equivalent, most of the time.

Definition 2: let \(\mathscr{N}(T)\) be the null space of \(T\) defined as

\[ \mathscr{N}(T) = \{x \in \mathscr{D}(T) \;|\; Tx = 0\}. \]

We have the following properties.

Proposition 1: let \(T\) be a linear operator, then

1. \(\mathscr{R}(T)\) is a vector space,
2. \(\mathscr{N}(T)\) is a vector space,
3. if \(\dim \mathscr{D}(T) = n \in \mathbb{N}\) then \(\dim \mathscr{R}(T) \leq n\).

Proof:

Will be added later.

An immediate consequence of statement 3 is that linear operators preserve linear dependence.

Proposition 2: let \(Y\) be a vector space, a linear operator \(T: \mathscr{D}(T) \to Y\) is injective if

\[ \forall x_1, x_2 \in \mathscr{D}(T): Tx_1 = Tx_2 \implies x_1 = x_2. \]

Proof:

Will be added later.

Injectivity of \(T\) is equivalent to \(\mathscr{N}(T) = \{0\}\).

Proof:

Will be added later.

Theorem 1: if a linear operator \(T: \mathscr{D}(T) \to \mathscr{R}(T)\) is injective there exists a mapping \(T^{-1}: \mathscr{R}(T) \to \mathscr{D}(T)\) such that

\[ y = Tx \iff T^{-1} y = x, \]

for all \(x \in \mathscr{D}(T)\), denoted as the inverse operator.

Proof:

Will be added later.

Proposition 3: let \(T: \mathscr{D}(T) \to \mathscr{R}(T)\) be an injective linear operator, if \(\mathscr{D}(T)\) is finite-dimensional, then

\[ \dim \mathscr{D}(T) = \dim \mathscr{R}(T). \]

Proof:

Will be added later.

Lemma 1: let \(X,Y\) and \(Z\) be vector spaces and let \(T: X \to Y\) and \(S: Y \to Z\) be injective linear operators, then \((ST)^{-1}: Z \to X\) exists and

\[ (ST)^{-1} = T^{-1} S^{-1}. \]

Proof:

Will be added later.

We finish this subsection with a definition of the space of linear operators.

Definition 3: let \(\mathscr{L}(X,Y)\) denote the set of linear operators mapping from a vector space \(X\) to a vector space \(Y\).

From this definition the following theorem follows.

Theorem 2: let \(X\) and \(Y\) be vectors spaces, the set of linear operators \(\mathscr{L}(X,Y)\) is a vector space.

Proof:

Will be added later.

Therefore, we may also call \(\mathscr{L}(X,Y)\) the space of linear operators.

Bounded linear operators

Definition 4: let \((X, \|\cdot\|_X)\) and \((Y,\|\cdot\|_Y)\) be normed spaces over a field \(F\) and let \(T: \mathscr{D}(T) \to Y\) be a linear operator with \(\mathscr{D}(T) \subset X\). Then \(T\) is a bounded linear operator if

\[ \exists c \in F \forall x \in \mathscr{D}(T): \|Tx\|_Y \leq c \|x\|_X. \]

In this case we may also define the set of all bounded linear operators.

Definition 5: let \(\mathscr{B}(X,Y)\) denote the set of bounded linear operators mapping from a vector space \(X\) to a vector space \(Y\).

We have the following theorem.

Theorem 3: let \(X\) and \(Y\) be vectors spaces, the set of bounded linear operators \(\mathscr{B}(X,Y)\) is a subspace of \(\mathscr{L}(X,Y)\).

Proof:

Will be added later.

Likewise, we may call \(\mathscr{B}(X,Y)\) the space of bounded linear operators.

The smallest possible \(c\) such that the statement in definition 4 still holds is denoted as the norm of \(T\) in the following definition.

Definition 5: the norm of a bounded linear operator \(T \in \mathscr{B}(X,Y)\) is defined by

\[ \|T\|_{\mathscr{B}} = \sup_{x \in \mathscr{D}(T) \backslash \{0\}} \frac{\|Tx\|_Y}{\|x\|_X}, \]

with \(X\) and \(Y\) vector spaces.

The operator norm makes \(\mathscr{B}\) into a normed space.

Lemma 2: let \(X\) and \(Y\) be normed spaces, the norm of a bounded linear operator \(T \in \mathscr{B}(X,Y)\) may be given by

\[ \|T\|_\mathscr{B} = \sup_{\substack{x \in \mathscr{D}(T) \\ \|x\|_X = 1}} \|Tx\|_Y, \]

and the norm of a bounded linear operator is a norm.

Proof:

Will be added later.

Note that the second statement in lemma 2 is non trivial, as the norm of a bounded linear operator is only introduced by a definition.

Proposition 4: if \((X, \|\cdot\|)\) is a finite-dimensional normed space, then every linear operator on \(X\) is bounded.

Proof:

Will be added later.

By linearity of the linear operators we have the following.

Theorem 4: let \(X\) and \(Y\) be normed spaces and let \(T: \mathscr{D}(T) \to Y\) be a linear operator with \(\mathscr{D}(T) \subset X\). Then the following statements are equivalent

1. \(T\) is bounded,
2. \(T\) is continuous in \(\mathscr{D}(T)\),
3. \(T\) is continuous in a point in \(\mathscr{D}(T)\).

Proof:

Will be added later.

Corollary 1: let \(T \in \mathscr{B}\) and let \((x_n)_{n \in \mathbb{N}}\) be a sequence in \(\mathscr{D}(T)\), then we have that

1. \(x_n \to x \in \mathscr{D}(T) \implies Tx_n \to Tx\) as \(n \to \infty\),
2. \(\mathscr{N}(T)\) is closed.

Proof:

Will be added later.

Furthermore, bounded linear operators have the property that

\[ \|T_1 T_2\| \leq \|T_1\| \|T_2\|, \]

for \(T_1, T_2 \in \mathscr{B}\).

Proof:

Will be added later.

Theorem 5: if \(X\) is a normed space and \(Y\) is a Banach space, then \(\mathscr{B}(X,Y)\) is a Banach space.

Proof:

Will be added later.

Definition 6: let \(T_1, T_2 \in \mathscr{L}\) be linear operators, \(T_1\) and \(T_2\) are equal if and only if

1. \(\mathscr{D}(T_1) = \mathscr{D}(T_2)\),
2. \(\forall x \in \mathscr{D}(T_1) : T_1x = T_2x\).

Restriction and extension

Definition 7: the restriction of a linear operator \(T \in \mathscr{L}\) to a subspace \(A \subset \mathscr{D}(T)\), denoted by \(T|_A: A \to \mathscr{R}(T)\) is defined by

\[ T|_A x = Tx, \]

for all \(x \in A\).

Furthermore.

Definition 8: the extension of a linear operator \(T \in \mathscr{L}\) to a vector space \(M\) is an operator denoted by \(\tilde T: M \to \mathscr{R}(T)\) such that

\[ \tilde T|_{\mathscr{D}(T)} = T. \]

Which implies that \(\tilde T x = Tx\; \forall x \in \mathscr{D}(T)\). Hence, \(T\) is the resriction of \(\tilde T\).

Theorem 6: let \(X\) be a normed space and let \(Y\) be Banach space. Let \(T \in \mathscr{B}(M,Y)\) with \(A \subset X\), then there exists an extension \(\tilde T: \overline M \to Y\), with \(\tilde T\) a bounded linear operator and \(\| \tilde T \| = \|T\|\).

Proof:

Will be added later.