Lecture 05 - Linear Spans

#math

We define the following two spaces:

L^{2} (Ω) = {f : Ω \to R (or C) s . t . \int_{Ω} | f |^{2} < \infty}

L^{1} (Ω) = {f : Ω \to R (or C) s . t . \int_{Ω} | f |^{1} < \infty}

In general:

L^{p} (Ω) = {f : Ω \to R (or C) s . t . \int_{Ω} | f |^{p} < \infty}

One could verify that:

| | f | |_{1} \leq c_{2} | | f | |_{2} \leq . . . \leq c_{p} | | f | |_{p}

Linear Spans

Definition

Let $A$ be a subset of a normed space $V$ . The linear span of $A$ , denoted by $[A] = l i n (A)$ is the intersection of all linear subspaces of $V$ containing $A$ .
The closed linear subspaces of $V$ containing $A$ is denoted by $\overset{―}{l i n A}$ is the intersection of all closed linear subspaces of $V$ containing $A$ .

Problem

Prove that the intersection of all closed linear subspaces is simply the closure of the intersection of linear subspaces.

Proof

Since $\overset{―}{l i n A}$ is closed this implies (by definition of the intersection):

⋂_{A \subset V} V \subset \overset{―}{l i n A}

where $V$ is a closed linear subspace. Now let $x \in \overset{―}{l i n A}$ and assume that $x \notin ⋂_{A \subset V} V$ then $\exists V$ closed such that $x \in V^{c}$ open that is there exists an open set such that $x \notin V^{c} \cap l i n A$ . $\Rightarrow \Leftarrow$ .

Lemma

(Inner product limit) Assume that $H$ is an inner product space and that

\begin{array}{r} {x_{n}} \to x \\ {y_{n}} \to y \end{array}

Then $⟨ x_{n}, y_{n} ⟩ \overset{n \to \infty}{\to} ⟨ lim_{n \to \infty} x_{n}, lim_{n \to \infty} y_{n} ⟩ = ⟨ x, y ⟩$

Proof

$\begin{aligned} | ⟨ x, y ⟩ - ⟨ x_{n}, y_{n} ⟩ | & = | ⟨ x - x_{n}, y_{n} ⟩ + ⟨ x, y - y_{n} ⟩ | \\ \leq | ⟨ x - x_{n}, y_{n} ⟩ | + | ⟨ x, y - y_{n} ⟩ | \\ \leq | | x - x_{n} | | | | y_{n} | | + | | x | | | | y - y_{n} | | (Cauchy Schwarz) \end{aligned}$

Orthogonality

Definition

We say $x$ is orthogonal to $y$ , denoted by $x ⊥ y$ if and only if $⟨ x, y ⟩ = 0$ .
An orthogonal sequence (or $⊥$ system) ${e_{n}}$ if $⟨ e_{n}, e_{m} ⟩ = 0$ whenever $n \neq m$ and $⟨ e_{n}, e_{n} ⟩ = | | e_{n} | |^{2}$
For orthonormal sequences we have $| | e_{n} | |^{2} = 1$ .

Problem

Show that if all vectors of an orthogonal system are non zero then they are linearly independent.

Proof

Assume we have vectors $v^{i}$ for $i \geq 0$ .Assume otherwise that they are linearly dependent. Then there exists a linear combination such that:

a_{i} v^{i} = v^{0}, i \geq 1

However $⟨ v^{0}, v^{0} ⟩ = ⟨ a_{i} v^{i}, v^{0} ⟩ = a_{i} ⟨ v^{i}, v^{0} ⟩ = 0$
Thus $v^{0} = 0 \Rightarrow \Leftarrow$

Bessel's Inequality

Let $M$ be a subspace of a Hilbert space $H$ and a point $x \in H$ be fixed. Then $z \in M$ is the nearest point to $x$ iff $x - z ⊥ v, \forall v \in M$ .