notes/docs/mathematics/functional-analysis/normed-spaces/linear-operators.md

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.

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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.

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

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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.

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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).

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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}.

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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.

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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).

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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.

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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.

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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).

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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.

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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}.

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

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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\|.

??? note "Proof:"

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