123 lines
4.3 KiB
Markdown
123 lines
4.3 KiB
Markdown
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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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> $$
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> \|x\| = \sqrt{\langle x, x \rangle},
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> $$
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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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> $$
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> \| x + y \|^2 + \| x - y \|^2 = 2\big(\|x\|^2 + \|y\|^2\big),
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> $$
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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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