Functional Analysis Lecture 1

difference: size = dimension. We have finite vector space and infinite vector space.

finite vector space

$V$ is finite if every linearly independent set is finite.

ie, if for every set $E \subseteq V$ such that $\forall v_{1}, \dots, v_{N} \in E$

\sum_{i = 1}^{N} a_{i} v_{i} = 0 ⟹ a_{1} = \dots = a_{N} = 0

and $E$ has finite cardinality, then $V$ is finite-dimensional. Otherwise, $V$ is infinite-dimensional.

Example

$C ([0, 1])$ is infinite dimensional because

E = {f_{n} (x) = x^{n} | n \in N \cup {0}}

is a linearly independent set with non-finite cardinality.

notion of "length" in a vector space

norm

A norm on a [[vector space]] $V$ is a function $| | \cdot | | : V \to [0, \infty)$ such that

$| | v | | = 0 ⟺ v = 0$ definiteness
$| | λ v | | = | λ | \cdot | | v | | \forall λ \in K, v \in V$ homogeneity
$\forall v_{1}, v_{2} \in V | | v_{1} + v_{2} | | \leq | | v_{1} | | + | | v_{2} | |$ [[triangle inequality]]

A [[vector space]] with a norm $| | \cdot | |$ is called a normed linear space.

semi-norm

A semi-norm is a function $| | \cdot | | V \to [0, \infty)$ satisfying homogeneity and the [[triangle inequality]], but not necessarily definiteness.

Metric

If $X$ is a set then $d : X \times X \to [0, \infty)$ is a metric if

(identifiability) $d (x, y) = 0 ⟺ x = y$
(symmetry) $\forall x, y \in X, d (x, y) = d (y, x)$
([[triangle inequality]]) $\forall x, y, z \in X, d (x, y) \leq d (x, z) + d (z, y)$

Proof

Definiteness for the [[norm]] implies $d (x, y) = 0 ⟺ x = y$
By homogeneity we get symmetry:

d (v, w) = | | v - w | | = | | (- 1) (w - v) | | = | - 1 | \cdot | | w - v | | = | | w - v | | = d (w, v)

triangle inequality follows immediately since $x - y = (x - z) + (z - y)$

Example

Multi-column

Blank

$R^{n}$ or $C^{n}$ with the euclidean norm

\begin{aligned} | | x | |_{2} & = {(\sum_{i = 1}^{n} | x_{i} |^{2})}^{1 / 2} \\ | | x | |_{\infty} & = max_{1 \leq i \leq n} | x_{i} | \\ | | x | |_{p} & = {(\sum_{i = 1}^{n} | x_{i} |^{p})}^{1 / p} 1 \leq p < \infty \end{aligned}

Intuitively, what does it mean to take a L^p norm for p > 2? : r/math

C_{\infty} (X)

If $X$ is a [[vector space]], the vector space

C_{\infty} (X) = {f : X \to C | f is continuous and bounded}

Example

$C_{\infty} ([0, 1]) = C ([0, 1])$ since we know the set $X$ is bounded

Proposition

Then $C_{\infty} (X)$ is a [[vector space]] and we can define the [[norm]]

| | u | |_{\infty} = sup_{x \in X} | u (x) | $ $ o n $ C_{\infty} (X) $

Proof

NTS that this is indeed a norm by verifying each of the properties. identifiability and homogeneity are satisfied from the definition of the norm. It suffices then to show that the [[triangle inequality]] holds.

If $(u, v) \in C_{\infty} (X)$ then for all $x \in X$

\begin{aligned} | u (x) + v (x) | & \leq | u (x) | + | v (x) | by △ inequality \\ \leq | | u | |_{\infty} + | | v | |_{\infty} \\ ⟹ | | u + v | |_{\infty} & = sup_{x \in X} | u (x) + v (x) | \\ \leq | | u | |_{\infty} + | | v | |_{\infty} \end{aligned}

Note

Convergence in this norm means

\begin{aligned} u_{n} \to u in C_{\infty} (X) & ⟺ | | u_{n} - u | |_{\infty} \to 0 as n \to \infty \\ ⟺ \forall ϵ > 0, \exists N \in N s.t. \forall n \geq N, \forall x \in X, | u_{n} (x) - u (x) | < ϵ \\ ⟺ u_{n} \to u uniformly on X \end{aligned}

Thus convergence in this metric is uniform convergence when the functions $u$ are bounded and continuous.

ℓ^{p}

space

$ℓ^{p} = {a = {a_{j}}_{j = 1}^{\infty} | | | a | |_{p} < \infty} $ $ i s t h e s p a c e o f (i n f i n i t e) s e q u e n c e s . W e d e f i n e t h e $ ℓ^{p} $ n o r m a s $ $ | | a | |_{p} := {\begin{cases} (\sum_{i = 1}^{\infty} | a_{i} |^{p})^{1 / p} 1 \leq p < \infty \\ sup_{1 \leq j < \infty} | a_{j} | p = \infty \end{cases}$

Example

${\frac{1}{j}}_{j = 1}^{\infty} \in ℓ^{p} \forall p > 1$ but not for $p = 1$

Banach Space

A normed space is a Banach space if it is complete with respect to the metric induced by the [[norm]].

ie, Cauchy sequences in the space converge in the space.

Example

$Q$ is not complete, but $R$ is. $R^{n}, C^{n}$ are complete wrt any of the $| | \cdot | |_{p}$ norms.

Theorem

If $X$ is a complete [[metric space]], then the [[continuous bounded function space]] $C_{\infty} (X)$ is a [[Banach space]].

Proof

NTS every Cauchy sequence ${u_{n}} \in C_{\infty} (X)$ has a limit in $C_{\infty} (X)$ .

Let ${u_{n}}$ be a Cauchy sequence in $C_{\infty} (X)$ . Then for all $ϵ > 0$ there exists $N \in N$ such that for all $n, m \geq N$ , $| | u_{n} - u_{m} | |_{\infty} < ϵ$ .

In particular, $\exists N_{0} \in N$ such that for all $m, n \geq N_{0}, | | u_{n} - u_{m} | |_{\infty} < 1$ . Then for all $n \geq N_{0}$ , we have

\begin{aligned} | | u_{n} | |_{\infty} & \leq | | u_{n} - u_{N_{0}} | |_{\infty} + | | u_{N_{0}} | |_{\infty} \\ < 1 + | | u_{N_{0}} | | \end{aligned}

And therefore for all $n \in N$ , we have

\begin{aligned} | | u_{n} | |_{\infty} & \leq | | u_{1} | |_{\infty} + | | u_{2} | |_{\infty} + \dots + | | u_{N_{0}} | |_{\infty} + 1 \\ = B \end{aligned}

Since for all $x \in X$ we have $| u_{n} (x) - u_{m} (x) | \leq | | u_{n} - u_{m} | |_{\infty}$ , we have that
for each $x \in X$ , the sequence ${u_{n} (x)}$ is a Cauchy sequence. And since $C$ is complete, its limit is in $C$ .

(define a candidate limiting function)
Define $u : X \to C, u (x) = lim_{n \to \infty} u_{n} (x)$
Then for all $x \in X$ ,

| u_{n} (x) | = lim_{n \to \infty} | u_{n} (x) | \leq B ⟹ sup_{x \in X} | u (x) | \leq B

thus $u$ is a bounded function.

Finally, we need to show continuity and convergence.

| | u - u_{n} | |_{\infty} \to 0

Let $ϵ > 0$ . Since ${u_{n}}$ is Cauchy in $C_{\infty} (X)$ , there exists some $N_{1} \in N$ such that $\forall n, m \geq N$ we have $| | u_{n} - u_{m} | |_{\infty} < \frac{ϵ}{2}$ .

So for any fixed $x \in X$ , we have

| u_{n} (x) - u_{m} (x) | \leq | | u_{n} - u_{m} | |_{\infty} < \frac{ϵ}{2}

Thus as $m \to \infty$ , we have for all $n \geq N_{1}$ and each $x \in X$ ,

| u_{n} (x) - u (x) | \leq \frac{ϵ}{2} < ϵ

Thus we have $| | u_{n} - u | |_{\infty} \to 0$ and this implies that $u_{n} \to u$ uniformly on $X$ . And because each $u_{n}$ is continuous, this implies that the limit $u$ is also continuous.

Thus, $u \in C_{\infty} (X)$ and $u_{n} \to u \in C_{\infty} (X)$ , thus $C_{\infty} (X)$ is complete and therefore a [[Banach space]].

Note

the approach for this proof is basically the same as any proof to show that something is a Banach space.

choose a candidate for the limit
show that the limit is in the space

Exercise

show $c_{0} = {a \in ℓ^{\infty} | lim_{j \to \infty} a_{j} = 0}$ is also a [[Banach space]] with norm $| | a | |_{\infty} = sup_{1 \leq j \leq \infty} | a_{j} |$

review

What three properties must a norm satisfy?
-?-
Definiteness/nonnegativity, homogeneity, [[triangle inequality]]

What is a Banach space?
-?-
A complete normed/metric space. ie, all convergent sequences in the space converge in the space.

1. Normed and Banach Spaces

notion of "length" in a vector space

review