port from mathematics-physics notes
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# Direct sums
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> *Definition 1*: in a metric space $(X,d)$, the **distance** $\delta$ from an element $x \in X$ to a nonempty subset $M \subset X$ is defined as
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>
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> $$
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> \delta = \inf_{\tilde y \in M} d(x,\tilde y).
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> $$
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In a normed space $(X, \|\cdot\|)$ this becomes
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$$
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\delta = \inf_{\tilde y \in M} \|x - \tilde y\|.
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$$
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> *Definition 2*: let $X$ be a vector space and let $x, y \in X$, the **line segment** $l$ between the vectors $x$ and $y$ is defined as
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>
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> $$
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> l = \{z \in X \;|\; \exists \alpha \in [0,1]: z = \alpha x + (1 - \alpha) y\}.
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> $$
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Using definition 2, we may define the following.
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> *Definition 3*: a subset $M \subset X$ of a vector space $X$ is **convex** if for all $x, y \in M$ the line segment between $x$ and $y$ is contained in $M$.
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This definition is true for projections of convex lenses which have been discussed in [optics]().
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We can now provide the main theorem in this section.
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> *Theorem 1*: let $X$ be an inner product space and let $M \subset X$ be a complete convex subset of $X$. Then for every $x \in X$ there exists a unique $y \in M$ such that
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>
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> $$
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> \delta = \inf_{\tilde y \in M} \|x - \tilde y\| = \|x - y\|,
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> $$
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>
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> if $M$ is a complete subspace $Y$ of $X$, then $x - y$ is orthogonal to $X$.
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??? note "*Proof*:"
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Will be added later.
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Now that the foundation is set, we may introduce direct sums.
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> *Definition 4*: a vector space $X$ is a **direct sum** $X = Y \oplus Z$ of two subspaces $Y \subset X$ and $Z \subset X$ of $X$ if each $x \in X$ has a unique representation
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>
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> $$
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> x = y + z,
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> $$
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>
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> for $y \in Y$ and $z \in Z$.
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Then $Z$ is called an *algebraic complement* of $Y$ in $X$ and vice versa, and $Y$, $Z$ is called a *complementary pair* of subspaces in $X$.
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In the case $Z = \{z \in X \;|\; z \perp Y\}$ we have that $Z$ is the *orthogonal complement* or *annihilator* of $Y$. Also denoted as $Y^\perp$.
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> *Proposition 1*: let $Y \subset X$ be any closed subspace of a Hilbert space $X$, then
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>
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> $$
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> X = Y \oplus Y^\perp,
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> $$
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>
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> with $Y^\perp = \{x\in X \;|\; x \perp Y\}$ the orthogonal complement of $Y$.
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??? note "*Proof*:"
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Will be added later.
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We have that $y \in Y$ for $x = y + z$ is called the *orthogonal projection* of $x$ on $Y$. Which defines an operator $P: X \to Y: x \mapsto Px \overset{\mathrm{def}}= y$.
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> *Lemma 1*: let $Y \subset X$ be a subset of a Hilbert space $X$ and let $P: X \to Y$ be the orthogonal projection operator, then we have
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>
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> 1. $P$ is a bounded linear operator,
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> 2. $\|P\| = 1$,
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> 3. $\mathscr{N}(P) = \{x \in X \;|\; Px = 0\}$.
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??? note "*Proof*:"
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Will be added later.
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> *Lemma 2*: if $Y$ is a closed subspace of a Hilbert space $X$, then $Y = Y^{\perp \perp}$.
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??? note "*Proof*:"
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Will be added later.
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Then it follows that $X = Y^\perp \oplus Y^{\perp \perp}$.
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??? note "*Proof*:"
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Will be added later.
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> *Lemma 3*: for every non-empty subset $M \subset X$ of a Hilbert space $X$ we have
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>
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> $$
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> \mathrm{span}(M) \text{ is dense in } X \iff M^\perp = \{0\}.
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> $$
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??? note "*Proof*:"
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Will be added later.
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# Inner product spaces
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> *Definition 1*: a vector space $X$ over a field $F$ is an **inner product space** if an **inner product** $\langle \cdot, \cdot \rangle: X \times X \to F$ is defined on $X$ satisfying
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>
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> 1. $\forall x \in X: \langle x, x \rangle \geq 0$,
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> 2. $\langle x, x \rangle = 0 \iff x = 0$,
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> 3. $\forall x, y \in X: \langle x, y \rangle = \overline{\langle y, x \rangle}$,
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> 4. $\forall x, y \in X, \alpha \in F: \langle \alpha x, y \rangle = \alpha \langle x, y \rangle$,
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> 5. $\forall x, y, z \in X: \langle x + y, z \rangle = \langle x, z \rangle + \langle y, z \rangle$.
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Similar to the case in normed spaces we have the following proposition.
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> *Proposition 1*: an inner product $\langle \cdot, \cdot \rangle$ on a vector space $X$ defines a norm $\|\cdot\|$ on $X$ given by
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>
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> $$
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> \|x\| = \sqrt{\langle x, x \rangle},
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> $$
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>
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> for all $x \in X$ and is called the **norm induced by the inner product**.
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??? note "*Proof*:"
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Will be added later.
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Which makes an inner product space also a normed space as well as a metric space, referring to proposition 1 in normed spaces.
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> *Definition 2*: a **Hilbert space** $H$ is a complete inner product space with its metric induced by the inner product.
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Definition 2 makes a Hilbert space also a Banach space, using proposition 1.
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## Properties of inner product spaces
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> *Proposition 2*: let $(X, \langle \cdot, \cdot \rangle)$ be an inner product space, then
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>
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> $$
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> \| x + y \|^2 + \| x - y \|^2 = 2\big(\|x\|^2 + \|y\|^2\big),
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> $$
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>
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> for all $x, y \in X$.
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??? note "*Proof*:"
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Will be added later.
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Proposition 2 is also called the parallelogram identity.
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> *Lemma 1*: let $(X, \langle \cdot, \cdot \rangle)$ be an inner product space, then
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>
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> 1. $\forall x, y \in X: |\langle x, y \rangle| \leq \|x\| \cdot \|y\|$,
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> 2. $\forall x, y \in X: \|x + y\| \leq \|x\| + \|y\|$.
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??? note "*Proof*:"
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Will be added later.
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Statement 1 in lemma 1 is known as the Schwarz inequality and statement 2 is known as the triangle inequality and will be used throughout the section of inner product spaces.
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> *Lemma 2*: let $(X, \langle \cdot, \cdot \rangle)$ be an inner product space and let $(x_n)_{n \in \mathbb{N}}$ and $(y_n)_{n \in \mathbb{N}}$ be sequences in $X$, if we have $x_n \to x$ and $y_n \to y$ as $n \to \infty$, then
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>
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> $$
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> \lim_{n \to \infty} \langle x_n, y_n \rangle = \langle x, y \rangle.
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> $$
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??? note "*Proof*:"
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Will be added later.
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## Completion
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> *Definition 3*: an **isomorphism** $T$ of an inner product space $(X, \langle \cdot, \cdot \rangle)_X$ onto an inner product space $(\tilde X, \langle \cdot, \cdot \rangle)_{\tilde X}$ over the same field $F$ is a bijective linear operator $T: X \to \tilde X$ which preserves the inner product
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>
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> $$
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> \langle Tx, Ty \rangle_{\tilde X} = \langle x, y \rangle_X,
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> $$
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>
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> for all $x, y \in X$.
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As a first application of lemma 2, let us prove the following.
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> *Theorem 1*: for every inner product space $(X, \langle \cdot, \cdot \rangle)_X$ there exists a Hilbert space $(\tilde X, \langle \cdot, \cdot \rangle)_{\tilde X}$ that contains a subspace $W$ that satisfies the following conditions
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>
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> 1. $W$ is an inner product space isomorphic with $X$.
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> 2. $W$ is dense in $X$.
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??? note "*Proof*:"
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Will be added later.
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Somewhat trivially, we have that a subspace $M$ of an inner product space $X$ is defined to be a vector subspace of $X$ taken with the inner product on $X$ restricted to $M \times M$.
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> *Proposition 3*: let $Y$ be a subspace of a Hilbert space $X$, then
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>
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> 1. $Y$ is complete $\iff$ $Y$ is closed in $X$,
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> 2. if $Y$ is finite-dimensional, then $Y$ is complete,
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> 3. $Y$ is separable if $X$ is separable.
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??? note "*Proof*:"
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Will be added later.
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## Orthogonality
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> *Definition 4*: let $(X, \langle \cdot, \cdot \rangle)$ be an inner product space, a vector $x \in X$ is **orthogonal** to a vector $y \in X$ if
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>
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> $$
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> \langle x, y \rangle = 0,
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> $$
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>
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> and we write $x \perp y$.
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Furthermore, we can also say that $x$ and $y$ *are orthogonal*.
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> *Definition 5*: let $(X, \langle \cdot, \cdot \rangle)$ be an inner product space and let $A, B \subset X$ be subspaces of $X$. Then $A$ is **orthogonal** to $B$ if for every $x \in A$ and $y \in B$ we have
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>
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> $$
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> \langle x, y \rangle = 0,
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> $$
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>
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> and we write $A \perp B$.
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Similarly, we may state that $A$ and $B$ *are orthogonal*.
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# Operator classes
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## Hilbert-adjoint operator
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> *Definition 1*: let $(X, \langle \cdot, \cdot \rangle_X)$ and $(Y, \langle \cdot, \cdot \rangle_Y)$ be Hilbert spaces over the field $F$ and let $T: X \to Y$ be a bounded linear operator. The **Hilbert-adjoint operator** $T^*$ of $T$ is the operator $T^*: Y \to X$ such that for all $x \in X$ amd $y \in Y$
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>
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> $$
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> \langle Tx, y \rangle_Y = \langle x, T^* y \rangle.
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> $$
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We should first prove that for a given $T$ such a $T^*$ exists.
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> *Proposition 1*: the Hilbert-adjoint operator $T^*$ of $T$ exists is unique and is a bounded linear operator with norm
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>
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> $$
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> \|T^*\| = \|T\|.
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> $$
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??? note "*Proof*:"
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Will be added later.
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The Hilbert-adjoint operator has the following properties.
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> *Proposition 2*: let $T,S: X \to Y$ be bounded linear operators, then
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>
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> 1. $\forall x \in X, y \in Y: \langle T^* y, x \rangle_X = \langle y, Tx \rangle_Y$,
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> 2. $(S + T)^* = S^* + T^*$,
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> 3. $\forall \alpha \in F: (\alpha T)^* = \overline \alpha T^*$,
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> 4. $(T^*)^* = T$,
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> 5. $\|T^* T\| = \|T T^*\| = \|T\|^2$,
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> 6. $T^*T = 0 \iff T = 0$,
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> 7. $(ST)^* = T^* S^*, \text{ when } X = Y$.
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??? note "*Proof*:"
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Will be added later.
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## Self-adjoint operator
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> *Definition 2*: a bounded linear operator $T: X \to X$ on a Hilbert space $X$ is **self-adjoint** if
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>
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> $$
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> T^* = T.
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> $$
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If a basis for $\mathbb{C}^n$ $(n \in \mathbb{N})$ is given and a linear operator on $\mathbb{C}^n$ is represented by a matrix, then its Hilbert-adjoint operator is represented by the complex conjugate transpose of that matrix (the Hermitian).
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Proposition 3, 4 and 5 pose some interesting results of self-adjoint operators.
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> *Proposition 3*: let $T: X \to X$ be a bounded linear operator on a Hilbert space $(X, \langle \cdot, \cdot \rangle_X)$ over the field $\mathbb{C}$, then
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>
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> $$
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> T \text{ is self-adjoint} \iff \forall x \in X: \langle Tx, x \rangle \in \mathbb{R}.
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> $$
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??? note "*Proof*:"
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Will be added later.
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> *Proposition 4*: the product of two bounded self-adjoint linear operators $T$ and $S$ on a Hilbert space is self-adjoint if and only if
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>
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> $$
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> ST = TS.
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> $$
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??? note "*Proof*:"
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Will be added later.
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Commuting operators therefore imply self-adjointness.
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> *Proposition 5*: let $(T_n)_{n \in \mathbb{N}}$ be a sequence of bounded self-adjoint operators $T_n: X \to X$ on a Hilbert space $X$. If $T_n \to T$ as $n \to \infty$, then $T$ is a bounded self-adjoint linear operator on $X$.
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??? note "*Proof*:"
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Will be added later.
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## Unitary operator
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> *Definition 3*: a bounded linear operator $T: X \to X$ on a Hilbert space $X$ is **unitary** if $T$ is bijective and $T^* = T^{-1}$.
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A bounded unitary linear operator has the following properties.
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> *Proposition 6*: let $U, V: X \to X$ be bounded unitary linear operators on a Hilbert space $X$, then
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>
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> 1. $U$ is isometric,
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> 2. $\|U\| = 1 \text{ if } X \neq \{0\}$,
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> 3. $UV$ is unitary,
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> 4. $U$ is normal, that is $U U^* = U^* U$,
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> 5. $T \in \mathscr{B}(X,X)$ is unitary $\iff$ $T$ is isometric and surjective.
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??? note "*Proof*:"
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Will be added later.
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# Orthonormal sets
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> *Definition 1*: an **orthogonal set** $M$ in an inner product space $X$ is a subset $M \subset X$ whose elements are pairwise orthogonal.
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Pairwise orthogonality implies that $x, y \in M: x \neq y \implies \langle x, y \rangle = 0$.
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> *Definition 2*: an **orthonormal set** $M$ in an inner product space $X$ is an orthogonal set in $X$ whose elements have norm 1.
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That is for all $x, y \in M$:
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$$
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\langle x, y \rangle = \begin{cases}0 &\text{if } x \neq y, \\ 1 &\text{if } x = y.\end{cases}
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$$
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> *Lemma 1*: an orthonormal set is linearly independent.
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??? note "*Proof*:"
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Will be added later.
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In the case that an orthogonal or orthonormal set is countable it can be arranged in a sequence and call it can be called an *orthogonal* or *orthonormal sequence*.
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> *Theorem 1*: let $(e_n)_{n \in \mathbb{N}}$ be an orthonormal sequence in an inner product space $(X, \langle \cdot, \cdot \rangle)$, then
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>
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> $$
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> \sum_{n=1}^\infty |\langle x, e_n \rangle|^2 \leq \|x\|^2,
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> $$
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>
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> for all $x \in X$.
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??? note "*Proof*:"
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Will be added later.
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Theorem 1 is known as the Bessel inequality, and we have that $|\langle x, e_n \rangle|$ are called the Fourier coefficients of $x$ with respect to the orthonormal sequence $(e_n)_{n \in \mathbb{N}}$.
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## Orthonormalisation process
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Let $(x_n)_{n \in \mathbb{N}}$ be a linearly independent sequence in an inner product space $(X, \langle \cdot, \cdot \rangle)$, then we can use the **Gram-Schmidt process** to determine the corresponding orthonormal sequence $(e_n)_{n \in \mathbb{N}}$.
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Let $e_1 = \frac{1}{\|x_1\|} x_1$ be the first step and let $e_n = \frac{1}{\|v_n\|} v_n$ be the $n$th step with
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$$
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v_n = x_n - \sum_{k=1}^{n-1} \langle x_n, e_k \rangle e_k.
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$$
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## Properties
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> *Proposition 1*: let $(e_n)_{n \in \mathbb{N}}$ be an orthonormal sequence in a Hilbert space $(X, \langle \cdot, \cdot \rangle)$ and let $(\alpha_n)_{n \in \mathbb{N}}$ be a sequence in the field of $X$, then
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>
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> 1. the series $\sum_{n=1}^\infty \alpha_n e_n$ is convergent in $X$ $\iff$ $\sum_{n=1}^\infty | \alpha_n|^2$ is convergent in $X$.
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> 2. if the series $\sum_{n=1}^\infty \alpha_n e_n$ is convergent in $X$ and $s = \sum_{n=1}^\infty \alpha_n e_n$ then $a_n = \langle s, e_n \rangle$.
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> 3. the series $\sum_{n=1}^\infty \alpha_n e_n = \sum_{n=1}^\infty \langle s, e_n \rangle e_n$ is convergent in $X$ for all $x \in X$.
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??? note "*Proof*:"
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Will be added later.
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Furthermore, we also have that.
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> *Proposition 2*: let $M$ be an orthonormal set in an inner product space $(X, \langle \cdot, \cdot \rangle)$, then any $x \in X$ can have at most countably many nonzero Fourier coefficients $\langle x, e_k \rangle$ for $e_k \in M$ over the uncountable index set $k \in I$ of $M$.
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??? note "*Proof*:"
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Will be added later.
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# Representations of functionals
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> *Lemma 1*: let $(X, \langle \cdot, \cdot \rangle)$ be an inner product space, if
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>
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> $$
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> \forall z \in X: \langle x, z \rangle = \langle y, z \rangle \implies x = y,
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> $$
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>
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> and if
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>
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> $$
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> \forall z \in X: \langle x, z \rangle = 0 \implies x = 0.
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> $$
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??? note "*Proof*:"
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Will be added later.
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Lemma 1 will be used in the following theorem.
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> *Theorem 1*: for every bounded linear functional $f$ on a Hilbert space $(X, \langle \cdot, \cdot \rangle)$, there exists a $z \in X$ such that
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>
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> $$
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> f(x) = \langle x, z \rangle,
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> $$
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>
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> for all $x \in x$, with $z$ uniquely dependent on $f$ and $\|z\| = \|f\|$.
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??? note "*Proof*:"
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Will be added later.
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## Sequilinear form
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> *Definition 1*: let $X$ and $Y$ be vector spaces over the field $F$. A **sesquilinear** form $h$ on $X \times Y$ is an operator $h: X \times Y \to F$ satisfying the following conditions
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>
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> 1. $\forall x_{1,2} \in X, y \in Y: h(x_1 + x_2, y) = h(x_1, y) + h(x_2, y)$.
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> 2. $\forall x \in X, y_{1,2} \in Y: h(x, y_1 + y_2) = h(x_1, y_1) + h(x_2, y_2)$.
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> 3. $\forall x \in X, y \in Y, \alpha \in F: h(\alpha x, y) = \alpha h(x,y)$.
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> 4. $\forall x \in X, y \in Y, \beta \in F: h(x, \beta y) = \overline \beta h(x,y)$.
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|
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Hence, $h$ is linear in the first argument and conjugate linear in the second argument. Bilinearity of $h$ is only true for a real field $F$.
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> *Definition 2*: let $X$ and $Y$ be normed spaces over the field $F$ and let $h: X \times Y \to F$ be a sesquilinear form, then $h$ is a **bounded sesquilinear form** if
|
||||
>
|
||||
> $$
|
||||
> \exists c \in F: |h(x,y)| \leq c \|x\| \|y\|,
|
||||
> $$
|
||||
>
|
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> for all $(x,y) \in X \times Y$ and the norm of $h$ is given by
|
||||
>
|
||||
> $$
|
||||
> \|h\| = \sup_{\substack{x \in X \backslash \{0\} \\ y \in Y \backslash \{0\}}} \frac{|h(x,y)|}{\|x\| \|y\|} = \sup_{\|x\|=\|y\|=1} |h(x,y)|.
|
||||
> $$
|
||||
|
||||
For example, the inner product is sesquilinear and bounded.
|
||||
|
||||
> *Theorem 2*: let $(X, \langle \cdot, \cdot \rangle_X)$ and $(Y, \langle \cdot, \cdot \rangle_Y)$ be Hilbert spaces over the field $F$ and let $h: X \times Y \to F$ be a bounded sesquilinear form. Then there exists a bounded linear operators $T: X \to Y$ and $S: Y \to X$, such that
|
||||
>
|
||||
> $$
|
||||
> h(x,y) = \langle Tx, y \rangle_Y = \langle x, Sy \rangle_X,
|
||||
> $$
|
||||
>
|
||||
> for all $(x,y) \in X \times Y$, with $T$ and $S$ uniquely determined by $h$ with norms $\|T\| = \|S\| = \|h\|$.
|
||||
|
||||
??? note "*Proof*:"
|
||||
|
||||
Will be added later.
|
||||
|
|
@ -0,0 +1,58 @@
|
|||
# Total sets
|
||||
|
||||
> *Definition 1*: a **total set** in a normed space $(X, \langle \cdot, \cdot \rangle)$ is a subset $M \subset X$ whose span is dense in $X$.
|
||||
|
||||
Accordingly, an orthonormal set in $X$ which is total in $X$ is called a total orthonormal set in $X$.
|
||||
|
||||
> *Proposition 1*: let $M \subset X$ be a subset of an inner product space $(X, \langle \cdot, \cdot \rangle)$, then
|
||||
>
|
||||
> 1. if $M$ is total in $X$, then $M^\perp = \{0\}$.
|
||||
> 2. if $X$ is complete and $M^\perp = \{0\}$ then $M$ is total in $X$.
|
||||
|
||||
??? note "*Proof*:"
|
||||
|
||||
Will be added later.
|
||||
|
||||
## Total orthornormal sets
|
||||
|
||||
> *Theorem 1*: an orthonormal sequence $(e_n)_{n \in \mathbb{N}}$ in a Hilbert space $(X, \langle \cdot, \cdot \rangle)$ is total in $X$ if and only if
|
||||
>
|
||||
> $$
|
||||
> \sum_{n=1}^\infty |\langle x, e_n \rangle|^2 = \|x\|^2,
|
||||
> $$
|
||||
>
|
||||
> for all $x \in X$.
|
||||
|
||||
??? note "*Proof*:"
|
||||
|
||||
Will be added later.
|
||||
|
||||
> *Lemma 1*: in every non-empty Hilbert space there exists a total orthonormal set.
|
||||
|
||||
??? note "*Proof*:"
|
||||
|
||||
Will be added later.
|
||||
|
||||
> *Theorem 2*: all total orthonormal sets in a Hilbert space have the same cardinality.
|
||||
|
||||
??? note "*Proof*:"
|
||||
|
||||
Will be added later.
|
||||
|
||||
This cardinality is called the Hilbert dimension or the orthogonal dimension of the Hilbert space.
|
||||
|
||||
> *Theorem 3*: let $X$ be a Hilbert space, then
|
||||
>
|
||||
> 1. if $X$ is separable, every orthonormal set in $X$ is countable.
|
||||
> 2. if $X$ contains a countable total orthonormal set, then $X$ is separable.
|
||||
|
||||
??? note "*Proof*:"
|
||||
|
||||
Will be added later.
|
||||
|
||||
> *Theorem 4*: two Hilbert spaces $X$ and $\tilde X$ over the same field are isomorphic if and only if they have the same Hilbert dimension.
|
||||
|
||||
??? note "*Proof*:"
|
||||
|
||||
Will be added later.
|
||||
|
||||
Loading…
Add table
Add a link
Reference in a new issue