port from mathematics-physics notes
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# Compactness
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> *Definition 1*: a metric space $X$ is **compact** if every sequence in $X$ has a convergent subsequence. A subset $M$ of $X$ is compact if every sequence in $M$ has a convergent subsequence whose limit is an element of $M$.
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A general property of compact sets is expressed in the following proposition.
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> *Proposition 1*: a compact subset $M$ of a metric space $(X,d)$ is closed and bounded.
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??? note "*Proof*:"
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Will be added later.
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The converse of this proposition is generally false.
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??? note "*Proof*:"
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Will be added later.
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However, for a finite dimensional normed space we have the following proposition.
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> *Proposition 2*: in a finite dimensional normed space $(X, \|\cdot\|)$ a subset $M \subset X$ is compact if and only if $M$ is closed and bounded.
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??? note "*Proof*:"
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Will be added later.
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A source of interesting results is the following lemma.
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> *Lemma 1*: let $Y$ and $Z$ be subspaces of a normed space $(X, \|\cdot\|)$, suppose that $Y$ is closed and that $Y$ is a strict subset of $Z$. Then for every $\alpha \in (0,1)$ there exists a $z \in Z$, such that
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>
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> 1. $\|z\| = 1$,
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> 2. $\forall y \in Y: \|z - y\| \geq \alpha$.
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??? note "*Proof*:"
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Will be added later.
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Lemma 1 gives the following remarkable proposition.
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> *Proposition 3*: if a normed space $(X, \|\cdot\|)$ has the property that the closed unit ball $M = \{x \in X | \|x\| \leq 1\}$ is compact, then $X$ is finite dimensional.
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??? note "*Proof*:"
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Will be added later.
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Compact sets have several basic properties similar to those of finite sets and not shared by non-compact sets. Such as the following.
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> *Proposition 4*: let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces and let $T: X \to Y$ be a continuous mapping. Let $M$ be a compact subset of $(X,d_X)$, then $T(M)$ is a compact subset of $(Y,d_Y)$.
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??? note "*Proof*:"
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Will be added later.
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From this proposition we conclude that the following property carries over to metric spaces.
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> *Corollary 1*: let $M \subset X$ be a compact subset of a metric space $(X,d)$ over a field $F$, a continuous mapping $T: M \to F$ attains a maximum and minimum value.
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??? note "*Proof*:"
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Will be added later.
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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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# Normed spaces
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> *Definition 1*: a vector space $X$ is a **normed space** if a norm $\| \cdot \|: X \to F$ is defined on $X$, satisfying
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>
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> 1. $\forall x \in X: \|x\| \geq 0$,
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> 2. $\|x\| = 0 \iff x = 0$,
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> 3. $\forall x \in X, \alpha \in F: \|\alpha x\| = |\alpha| \|x\|$,
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> 4. $\forall x, y \in X: \|x + y\| \leq \|x\| + \|y\|$.
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Also called a *normed vector space* or *normed linear space*.
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> *Proposition 1*: a norm on a vector space $X$ defines a metric $d$ on $X$ given by
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>
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> $$
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> d(x,y) = \|x - y\|,
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> $$
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>
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> for all $x, y \in X$ and is called a **metric induced by the norm**.
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??? note "*Proof*:"
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Will be added later.
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Furthermore, there is a category of normed spaces with interesting properties which is given in the following definition.
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> *Definition 2*: a **Banach space** is a complete normed space with its metric induced by the norm.
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If we define the norm $\| \cdot \|$ of the Euclidean vector space $\mathbb{R}^n$ by
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$$
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\|x\| = \sqrt{\sum_{j=1}^n |x(j)|^2},
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$$
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for all $x \in \mathbb{R}^n$, then it yields the metric
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$$
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d(x,y) = \|x - y\| = \sqrt{\sum_{j=1}^n |x(j) - y(j)|^2},
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$$
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for all $x, y \in \mathbb{R}^n$ which imposes completeness. Therefore $(\mathbb{R}^n, \|\cdot\|)$ is a Banach space.
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This adaptation also works for $C$, $l^p$ and $l^\infty$, obviously. Obtaining that $\mathbb{R}^n$, $C$, $l^p$ and $l^\infty$ are all Banach spaces.
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> *Lemma 1*: a metric $d$ induced by a norm on a normed space $(X, \|\cdot\|)$ satisfies
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>
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> 1. $\forall x, y \in X, \alpha \in F: d(x + \alpha, y + \alpha) = d(x,y)$,
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> 2. $\forall x, y \in X, \alpha \in F: d(\alpha x, \alpha y) = |\alpha| d(x,y)$.
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??? note "*Proof*:"
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We have
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$$
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d(x + \alpha, y + \alpha) = \|x + \alpha - (y + \alpha)\| = \|x - y\| = d(x,y),
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$$
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and
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$$
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d(\alpha x, \alpha y) = \|\alpha x - \alpha y\| = |\alpha| \|x - y\| = |\alpha| d(x,y).
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$$
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By definition, a subspace $M$ of a normed space $X$ is a subspace of $X$ with its norm induced by the norm on $X$.
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> *Definition 3*: let $M$ be a subspace of a normed space $X$, if $M$ is closed then $M$ is a **closed subspace** of $X$.
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By definition, a subspace $M$ of a Banach space $X$ is a subspace of $X$ as a normed space. Hence, we do not require $M$ to be complete.
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> *Theorem 1*: a subspace $M$ of a Banach space $X$ is complete if and only if $M$ is a closed subspace of $X$.
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??? note "*Proof*:"
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Will be added later.
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Convergence in normed spaces follows from the definition of convergence in metric spaces and the fact that the metric is induced by the norm.
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## Convergent series
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> *Definition 4*: let $(x_k)_{k \in \mathbb{N}}$ be a sequence in a normed space $(X, \|\cdot\|)$. We define the sequence of partial sums $(s_n)_{n \in \mathbb{N}}$ by
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>
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> $$
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> s_n = \sum_{k=1}^n x_k,
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> $$
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>
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> if $s_n$ converges to $s \in X$, then
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>
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> $$
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> \lim_{n \to \infty} \sum_{k=1}^n x_k,
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> $$
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>
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> is convergent, and $s$ is the sum of the series, writing
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>
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> $$
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> s = \lim_{n \to \infty} \sum_{k=1}^n x_k = \sum_{k=1}^\infty x_k = \lim_{n \to \infty } s_n.
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> $$
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>
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> If the series
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>
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> $$
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> \sum_{k=1}^\infty \|x_k\|,
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> $$
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>
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> is convergent in $F$, then the series is **absolutely convergent**.
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From the notion of absolute convergence the following theorem may be posed.
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> *Theorem 2*: absolute convergence of a series implies convergence if and only if $(X, \|\cdot\|)$ is complete.
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??? note "*Proof*:"
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Will be added later.
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## Schauder basis
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> *Definition 5*: let $(X, \|\cdot\|)$ be a normed space and let $(e_k)_{k \in \mathbb{N}}$ be a sequence of vectors in $X$, such that for every $x \in X$ there exists a unique sequence of scalars $(\alpha_k)_{k \in \mathbb{N}}$ such that
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>
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> $$
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> \lim_{n \to \infty} \|x - \sum_{k=1}^n \alpha_k e_k\| = 0,
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> $$
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>
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> then $(e_k)_{k \in \mathbb{N}}$ is a **Schauder basis* of $(X, \|\cdot\|)$.
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The expansion of a $x \in X$ with respect to a Schauder basis $(e_k)_{k \in \mathbb{N}}$ is given by
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$$
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x = \sum_{k=1}^\infty \alpha_k e_k.
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$$
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> *Lemma 2*: if a normed space has a Schauder basis then it is seperable.
|
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|
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??? note "*Proof*:"
|
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|
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Will be added later.
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|
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## Completion
|
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|
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> *Theorem 3*: for every normed space $(X, \|\cdot\|_X)$ there exists a Banach space $(Y, \|\cdot\|_Y)$ that contains a subspace $W$ that satisfies the following conditions
|
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>
|
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> 1. $W$ is a normed space isometric with $X$.
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> 2. $W$ is dense in $Y$.
|
||||
|
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??? note "*Proof*:"
|
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|
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Will be added later.
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|
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The Banach space $(Y, \|\cdot\|_Y)$ is unique up to isometry.
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|
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## Finite dimension
|
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|
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> *Lemma 3*: let $\{x_k\}_{k=1}^n$ with $n \in \mathbb{N}$ be a linearly independent set of vectors in a normed space $(X, \|\cdot\|)$, then there exists a $c > 0$ such that
|
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>
|
||||
> $$
|
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> \Big\| \sum_{k=1}^n \alpha_k x_k \Big\| \geq c \sum_{k=1}^n |\alpha_k|,
|
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> $$
|
||||
>
|
||||
> for all $\{\alpha_k\}_{k=1}^n \in F$.
|
||||
|
||||
??? note "*Proof*:"
|
||||
|
||||
Will be added later.
|
||||
|
||||
As a first application of this lemma, let us prove the following.
|
||||
|
||||
> *Theorem 4*: every finite-dimensional subspace $M$ of a normed space $(X, \|\cdot\|)$ is complete.
|
||||
|
||||
??? note "*Proof*:"
|
||||
|
||||
Will be added later.
|
||||
|
||||
In particular, every finite dimensional normed space is complete.
|
||||
|
||||
> *Proposition 2*: every finite-dimensional subspace $M$ of a normed space $(X, \|\cdot\|)$ is a closed subspace of $X$.
|
||||
|
||||
??? note "*Proof*:"
|
||||
|
||||
Will be added later.
|
||||
|
||||
Another interesting property of finite-dimensional vector space $X$ is that all norms on $X$ lead to the same topology for $X$. That is, the open subsets of $X$ are the same, regardless of the particular choice of a norm on $X$. The details are as follows.
|
||||
|
||||
> *Definition 6*: a norm $\|\cdot\|_1$ on a vector space $X$ is **equivalent** to a norm $\|\cdot\|_2$ on $X$ if there exists $a,b>0$ such that
|
||||
>
|
||||
> $$
|
||||
> \forall x \in X: a \|x\|_1 \leq \|x\|_2 \leq b \|x\|_1.
|
||||
> $$
|
||||
|
||||
This concept is motivated by the following proposition.
|
||||
|
||||
> *Proposition 3*: equivalent norms on $X$ define the same topology for $X$.
|
||||
|
||||
??? note "*Proof*:"
|
||||
|
||||
Will be added later.
|
||||
|
||||
Using lemma 3 we may now prove the following theorem.
|
||||
|
||||
> *Theorem 5*: on a finite dimensional vector space $X$ any norm $\|\cdot\|_1$ is equivalent to any other norm $\|\cdot\|_2$.
|
||||
|
||||
??? note "*Proof*:"
|
||||
|
||||
Will be added later.
|
||||
|
||||
This theorem is of considerable importance. For instance, it implies that convergence or divergence of a sequence in a finite dimensional vector space does not depend on the particular choice of a norm on that space.
|
||||
|
|
@ -0,0 +1,100 @@
|
|||
# Vector spaces
|
||||
|
||||
> *Definition 1*: a **vector space** $X$ over a **scalar field** $F$ is a non-empty set, on which two algebraic operations are defined; vector addition and scalar multiplication. Such that
|
||||
>
|
||||
> 1. $(X, +)$ is a commutative group with neutral element 0.
|
||||
> 2. the scalar multiplication satisfies $\forall x, y \in X$ and $\lambda, \mu \in F$
|
||||
> * $\lambda (x + y) = \lambda x + \lambda y$,
|
||||
> * $(\lambda + \mu) x = \lambda x + \mu x$,
|
||||
> * $\lambda (\mu x) = (\lambda \mu) x$,
|
||||
> * $1 x = x$.
|
||||
|
||||
When $F = \mathbb{R}$ we have a real vector space while when $F = \mathbb{C}$ we have a complex vector space.
|
||||
|
||||
We have that the metric spaces $\mathbb{R}^n$, $C$, $l^p$ and $l^\infty$ are also vector spaces.
|
||||
|
||||
??? note "*Proof*:"
|
||||
|
||||
I am too lazy to add this trivial proof. Maybe some time in the future, if I do not forget.
|
||||
|
||||
> *Definition 2*: a **subspace** of a vector space $X$ is a non-empty subset $M$ of $X$, such that $\forall x, y \in M$ and $\lambda, \mu \in F$:
|
||||
>
|
||||
> $$
|
||||
> \lambda x + \mu y \in M,
|
||||
> $$
|
||||
>
|
||||
> with $M$ itself a vector space.
|
||||
|
||||
A special subspace $M$ of a vector space $X$ is the *improper subspace* $M = X$. Every other subspace of $X$ is a *proper subspace*.
|
||||
|
||||
## Linear combinations
|
||||
|
||||
> *Definition 3*: a **linear combination** of the vectors $\{x_i\}_{i=1}^n$ with $n \in \mathbb{N}$ is vector of the form
|
||||
>
|
||||
> $$
|
||||
> \alpha_1 x_1 + \dots + \alpha_n x_n = \sum_{i=1}^n \alpha_i x_i,
|
||||
> $$
|
||||
>
|
||||
> with $\{\alpha_i\}_{i=1}^n \in F$.
|
||||
|
||||
The set of all linear combinations of a set of vectors is defined as follows.
|
||||
|
||||
> *Definition 4*: the **span** of a subset $M \subset X$ of a vector space $X$, denoted by $\mathrm{span}(M)$, is the set of all linear combinations of vectors from $M$.
|
||||
|
||||
It follows that $\mathrm{span}(M)$ is a subspace of $X$.
|
||||
|
||||
## Linear independence
|
||||
|
||||
> *Definition 5*: a finite subset of vectors $M = \{x_i\}_{i=1}^n$ is **linearly independent** if
|
||||
>
|
||||
> $$
|
||||
> \sum_{i=1}^n \alpha_i x_i = 0 \implies \forall i \in \{1, \dots, n\}: \alpha_i = 0.
|
||||
> $$
|
||||
|
||||
The converse may also be defined.
|
||||
|
||||
> *Definition 6*: a finite subset of vectors $M = \{x_i\}_{i=1}^n$ is **linearly dependent** if $\exists \{\alpha_i\}_{i=1}^n \in F$ not all zero such that
|
||||
>
|
||||
> $$
|
||||
> \sum_{i=1}^n \alpha_i x_i = 0.
|
||||
> $$
|
||||
|
||||
The notions of linear dependence and independence may also be extended to infinite subsets.
|
||||
|
||||
> *Definition 7*: a subset $M$ of a vector space $X$ is **linearly independent** if every non-empty finite subset of $M$ is linearly independent.
|
||||
|
||||
While the converse in this case is defined by the contradiction.
|
||||
|
||||
> *Definition 8*: a subset $M$ of a vector space $X$ is **linearly dependent** if $M$ is not linearly independent.
|
||||
|
||||
## Dimension and basis
|
||||
|
||||
> *Definition 9*: a vector space $X$ is **finite dimensional** if there exists a $n \in \mathbb{N}$, such that $X$ contains a set of $n$ linearly independent vectors, while every set of $n+1$ vectors in $X$ is linearly dependent. In this case $n$ is the dimension of $X$, denoted by $\dim X = n$.
|
||||
|
||||
By definition $X = \{0\}$ is finite dimensional and $\dim X = 0$.
|
||||
|
||||
> *Definition 10*: if a vector space $X$ is not finite dimensional then $X$ is **infinite dimensional**.
|
||||
|
||||
The following definition of a basis is both relevant to finite and infinite dimensional vector spaces.
|
||||
|
||||
> *Definition 11*: a **basis** $B$ of a vector space $X$ is a linearly independent subset of $X$, that spans $X$.
|
||||
|
||||
Such a set $B$ is also called a *Hamel basis* of $X$.
|
||||
|
||||
> *Theorem 1*: every vector space $X$ has a Hamel basis.
|
||||
|
||||
??? note "*Proof*:"
|
||||
|
||||
Read it again, a proof is not necessary.
|
||||
|
||||
> *Theorem 2*: let $X$ be a vector space with $\dim X = n \in \mathbb{N}$. Then any proper subspace $M \subset X$ has dimension less than $n$.
|
||||
|
||||
??? note "*Proof*:"
|
||||
|
||||
If $n = 0$, then $X = \{0\}$ and $X$ has no proper subspace.
|
||||
|
||||
If $\dim M = 0$, then $M = \{0\}$ and $X \neq M \implies \dim X \geq 1$.
|
||||
|
||||
If $\dim M = n$ then $M$ would have a basis of $n$ elements, which would also be a basis for $X$ since $\dim X = n$, so that $X = M$.
|
||||
|
||||
This shows that any linearly independent set of vectors in $M$ must have fewer than $n$ elements and $\dim M < n$.
|
||||
Loading…
Add table
Add a link
Reference in a new issue