4.3 KiB
Inner product spaces
Definition 1: a vector space
Xover a fieldFis an inner product space if an inner product\langle \cdot, \cdot \rangle: X \times X \to Fis defined onXsatisfying
\forall x \in X: \langle x, x \rangle \geq 0,\langle x, x \rangle = 0 \iff x = 0,\forall x, y \in X: \langle x, y \rangle = \overline{\langle y, x \rangle},\forall x, y \in X, \alpha \in F: \langle \alpha x, y \rangle = \alpha \langle x, y \rangle,\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 \rangleon a vector spaceXdefines a norm\|\cdot\|onXgiven by
|x| = \sqrt{\langle x, x \rangle},for all
x \in Xand 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
His 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
\forall x, y \in X: |\langle x, y \rangle| \leq \|x\| \cdot \|y\|,\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 inX, if we havex_n \to xandy_n \to yasn \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
Tof an inner product space(X, \langle \cdot, \cdot \rangle)_Xonto an inner product space(\tilde X, \langle \cdot, \cdot \rangle)_{\tilde X}over the same fieldFis a bijective linear operatorT: X \to \tilde Xwhich 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)_Xthere exists a Hilbert space(\tilde X, \langle \cdot, \cdot \rangle)_{\tilde X}that contains a subspaceWthat satisfies the following conditions
Wis an inner product space isomorphic withX.Wis dense inX.
??? 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
Ybe a subspace of a Hilbert spaceX, then
Yis complete\iffYis closed inX,- if
Yis finite-dimensional, thenYis complete,Yis separable ifXis separable.
??? note "Proof:"
Will be added later.
Orthogonality
Definition 4: let
(X, \langle \cdot, \cdot \rangle)be an inner product space, a vectorx \in Xis orthogonal to a vectory \in Xif
\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 letA, B \subset Xbe subspaces ofX. ThenAis orthogonal toBif for everyx \in Aandy \in Bwe have
\langle x, y \rangle = 0,and we write
A \perp B.
Similarly, we may state that A and B are orthogonal.