1
0
Fork 0
mathematics-physics-wiki/docs/en/mathematics/functional-analysis/inner-product-spaces/inner-product-spaces.md

4.3 KiB

Inner product spaces

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

  1. \forall x \in X: \langle x, x \rangle \geq 0,
  2. \langle x, x \rangle = 0 \iff x = 0,
  3. \forall x, y \in X: \langle x, y \rangle = \overline{\langle y, x \rangle},
  4. \forall x, y \in X, \alpha \in F: \langle \alpha x, y \rangle = \alpha \langle x, y \rangle,
  5. \forall x, y, z \in X: \langle x + y, z \rangle = \langle x, z \rangle + \langle y, z \rangle.

Similar to the case in normed spaces we have the following proposition.

Proposition 1: an inner product \langle \cdot, \cdot \rangle on a vector space X defines a norm \|\cdot\| on X given by

|x| = \sqrt{\langle x, x \rangle},

for all x \in X and is called the norm induced by the inner product.

??? note "Proof:"

Will be added later.

Which makes an inner product space also a normed space as well as a metric space, referring to proposition 1 in normed spaces.

Definition 2: a Hilbert space H is a complete inner product space with its metric induced by the inner product.

Definition 2 makes a Hilbert space also a Banach space, using proposition 1.

Properties of inner product spaces

Proposition 2: let (X, \langle \cdot, \cdot \rangle) be an inner product space, then

| x + y |^2 + | x - y |^2 = 2\big(|x|^2 + |y|^2\big),

for all x, y \in X.

??? note "Proof:"

Will be added later.

Proposition 2 is also called the parallelogram identity.

Lemma 1: let (X, \langle \cdot, \cdot \rangle) be an inner product space, then

  1. \forall x, y \in X: |\langle x, y \rangle| \leq \|x\| \cdot \|y\|,
  2. \forall x, y \in X: \|x + y\| \leq \|x\| + \|y\|.

??? note "Proof:"

Will be added later.

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.

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

\lim_{n \to \infty} \langle x_n, y_n \rangle = \langle x, y \rangle.

??? note "Proof:"

Will be added later.

Completion

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

\langle Tx, Ty \rangle_{\tilde X} = \langle x, y \rangle_X,

for all x, y \in X.

As a first application of lemma 2, let us prove the following.

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

  1. W is an inner product space isomorphic with X.
  2. W is dense in X.

??? note "Proof:"

Will be added later.

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.

Proposition 3: let Y be a subspace of a Hilbert space X, then

  1. Y is complete \iff Y is closed in X,
  2. if Y is finite-dimensional, then Y is complete,
  3. Y is separable if X is separable.

??? note "Proof:"

Will be added later.

Orthogonality

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

\langle x, y \rangle = 0,

and we write x \perp y.

Furthermore, we can also say that x and y are orthogonal.

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

\langle x, y \rangle = 0,

and we write A \perp B.

Similarly, we may state that A and B are orthogonal.