notes/docs/mathematics/functional-analysis/inner-product-spaces/direct-sums.md

Direct sums

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

\delta = \inf_{\tilde y \in M} d(x,\tilde y).

In a normed space (X, \|\cdot\|) this becomes

\delta = \inf_{\tilde y \in M} |x - \tilde y|.

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

l = {z \in X ;|; \exists \alpha \in [0,1]: z = \alpha x + (1 - \alpha) y}.

Using definition 2, we may define the following.

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.

This definition is true for projections of convex lenses which have been discussed in optics.

We can now provide the main theorem in this section.

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

\delta = \inf_{\tilde y \in M} |x - \tilde y| = |x - y|,

if M is a complete subspace Y of X, then x - y is orthogonal to X.

??? note "Proof:"

Will be added later.

Now that the foundation is set, we may introduce direct sums.

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

x = y + z,

for y \in Y and z \in Z.

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.

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.

Proposition 1: let Y \subset X be any closed subspace of a Hilbert space X, then

X = Y \oplus Y^\perp,

with Y^\perp = \{x\in X \;|\; x \perp Y\} the orthogonal complement of Y.

??? note "Proof:"

Will be added later.

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.

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

1. P is a bounded linear operator,
2. \|P\| = 1,
3. \mathscr{N}(P) = \{x \in X \;|\; Px = 0\}.

??? note "Proof:"

Will be added later.

Lemma 2: if Y is a closed subspace of a Hilbert space X, then Y = Y^{\perp \perp}.

??? note "Proof:"

Will be added later.

Then it follows that X = Y^\perp \oplus Y^{\perp \perp}.

??? note "Proof:"

Will be added later.

Lemma 3: for every non-empty subset M \subset X of a Hilbert space X we have

\mathrm{span}(M) \text{ is dense in } X \iff M^\perp = {0}.

??? note "Proof:"

Will be added later.