Added the sections linear operators and linear functionals to functional analysis.
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# Linear functionals
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> *Definition 1*: a **linear functional** $f$ is a linear operator with its domain in a vector space $X$ and its range in a scalar field $F$ defined in $X$.
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The norm can be a linear functional $\|\cdot\|: X \to F$ under the condition that the norm is linear. Otherwise, it would solely be a functional.
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> *Definition 2*: a **bounded linear functional** $f$ is a bounded linear operator with its domain in a vector space $X$ and its range in a scalar field $F$ defined in $X$.
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## Dual space
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> *Definition 3*: the set of linear functionals on a vector space $X$ is defined as the **algebraic dual space** $X^*$ of $X$.
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From this definition we have the following.
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> *Theorem 1*: the algebraic dual space $X^*$ of a vector space $X$ is a vector space.
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??? note "*Proof*:"
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Will be added later.
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Furthermore, a secondary type of dual space may be defined as follows.
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> *Definition 4*: the set of bounded linear functionals on a normed space $X$ is defined as **dual space** $X'$.
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In this case, a rather interesting property of a dual space emerges.
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> *Theorem 2*: the dual space $X'$ of a normed space $(X,\|\cdot\|_X)$ is a Banach space with its norm $\|\cdot\|_{X'}$ given by
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>
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> $$
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> \|f\|_{X'} = \sup_{x \in X\backslash \{0\}} \frac{|f(x)|}{\|x\|_X} = \sup_{\substack{x \in X \\ \|x\|_X = 1}} |f(x)|,
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> $$
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>
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> for all $f \in X'$.
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??? note "*Proof*:"
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Will be added later.
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# Linear operators
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> *Definition 1*: a **linear operator** $T$ is a linear mapping such that
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>
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> 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 $\mathscr{D}(T)$.
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> 2. $\forall x, y \in \mathscr{D}(T): T(x + y) = Tx + Ty$.
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> 3. $\forall x \in \mathscr{D}(T), \alpha \in F: T(\alpha x) = \alpha Tx$.
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Observe the notation; we $Tx$ and $T(x)$ are equivalent, most of the time.
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> *Definition 2*: let $\mathscr{N}(T)$ be the **null space** of $T$ defined as
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>
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> $$
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> \mathscr{N}(T) = \{x \in \mathscr{D}(T) \;|\; Tx = 0\}.
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> $$
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We have the following properties.
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> *Proposition 1*: let $T$ be a linear operator, then
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>
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> 1. $\mathscr{R}(T)$ is a vector space,
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> 2. $\mathscr{N}(T)$ is a vector space,
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> 3. if $\dim \mathscr{D}(T) = n \in \mathbb{N}$ then $\dim \mathscr{R}(T) \leq n$.
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??? note "*Proof*:"
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Will be added later.
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An immediate consequence of statement 3 is that linear operators preserve linear dependence.
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> *Proposition 2*: let $Y$ be a vector space, a linear operator $T: \mathscr{D}(T) \to Y$ is injective if
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>
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> $$
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> \forall x_1, x_2 \in \mathscr{D}(T): Tx_1 = Tx_2 \implies x_1 = x_2.
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> $$
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??? note "*Proof*:"
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Will be added later.
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Injectivity of $T$ is equivalent to $\mathscr{N}(T) = \{0\}$.
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??? note "*Proof*:"
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Will be added later.
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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
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>
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> $$
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> y = Tx \iff T^{-1} y = x,
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> $$
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>
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> for all $x \in \mathscr{D}(T)$, denoted as the **inverse operator**.
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??? note "*Proof*:"
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Will be added later.
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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
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>
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> $$
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> \dim \mathscr{D}(T) = \dim \mathscr{R}(T).
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> $$
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??? note "*Proof*:"
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Will be added later.
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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
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>
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> $$
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> (ST)^{-1} = T^{-1} S^{-1}.
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> $$
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??? note "*Proof*:"
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Will be added later.
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We finish this subsection with a definition of the space of linear operators.
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> *Definition 3*: let $\mathscr{L}(X,Y)$ denote the set of linear operators mapping from a vector space $X$ to a vector space $Y$.
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From this definition the following theorem follows.
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> *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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??? note "*Proof*:"
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Will be added later.
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Therefore, we may also call $\mathscr{L}(X,Y)$ the space of linear operators.
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## Bounded linear operators
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> *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
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>
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> $$
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> \exists c \in F \forall x \in \mathscr{D}(T): \|Tx\|_Y \leq c \|x\|_X.
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> $$
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In this case we may also define the set of all bounded linear operators.
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> *Definition 5*: let $\mathscr{B}(X,Y)$ denote the set of bounded linear operators mapping from a vector space $X$ to a vector space $Y$.
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We have the following theorem.
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> *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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??? note "*Proof*:"
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Will be added later.
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Likewise, we may call $\mathscr{B}(X,Y)$ the space of bounded linear operators.
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The smallest possible $c$ such that the statement in definition 4 still holds is denoted as the norm of $T$ in the following definition.
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> *Definition 5*: the norm of a bounded linear operator $T \in \mathscr{B}(X,Y)$ is defined by
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>
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> $$
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> \|T\|_{\mathscr{B}} = \sup_{x \in \mathscr{D}(T) \backslash \{0\}} \frac{\|Tx\|_Y}{\|x\|_X},
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> $$
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>
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> with $X$ and $Y$ vector spaces.
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The operator norm makes $\mathscr{B}$ into a normed space.
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> *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
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>
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> $$
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> \|T\|_\mathscr{B} = \sup_{\substack{x \in \mathscr{D}(T) \\ \|x\|_X = 1}} \|Tx\|_Y,
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> $$
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>
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> and the norm of a bounded linear operator is a norm.
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??? note "*Proof*:"
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Will be added later.
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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.
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> *Proposition 4*: if $(X, \|\cdot\|)$ is a finite-dimensional normed space, then every linear operator on $X$ is bounded.
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??? note "*Proof*:"
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Will be added later.
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By linearity of the linear operators we have the following.
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> *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
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>
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> 1. $T$ is bounded,
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> 2. $T$ is continuous in $\mathscr{D}(T)$,
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> 3. $T$ is continuous in a point in $\mathscr{D}(T)$.
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??? note "*Proof*:"
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Will be added later.
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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
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>
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> 1. $x_n \to x \in \mathscr{D}(T) \implies Tx_n \to Tx$ as $n \to \infty$,
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> 2. $\mathscr{N}(T)$ is closed.
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??? note "*Proof*:"
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Will be added later.
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Furthermore, bounded linear operators have the property that
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$$
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\|T_1 T_2\| \leq \|T_1\| \|T_2\|,
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$$
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for $T_1, T_2 \in \mathscr{B}$.
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??? note "*Proof*:"
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Will be added later.
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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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??? note "*Proof*:"
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Will be added later.
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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
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>
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> 1. $\mathscr{D}(T_1) = \mathscr{D}(T_2)$,
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> 2. $\forall x \in \mathscr{D}(T_1) : T_1x = T_2x$.
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## Restriction and extension
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> *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
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>
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> $$
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> T|_A x = Tx,
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> $$
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>
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> for all $x \in A$.
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Furthermore.
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> *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
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>
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> $$
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> \tilde T|_{\mathscr{D}(T)} = T.
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> $$
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Which implies that $\tilde T x = Tx\; \forall x \in \mathscr{D}(T)$. Hence, $T$ is the resriction of $\tilde T$.
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> *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 $\| \tilde T \| = \|T\|$.
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??? note "*Proof*:"
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Will be added later.
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