Functional Analysis Lecture 1

[[lecture-data]]

Lecture 2 →

Lecture Notes: Rodriguez, page 1

Introduction LA/Calc: finite vars FA: vector spaces w infinite vars

1. Normed and Banach Spaces

normed linear space

Vector Space

Let $V$ be a vector space over $\mathbb{R}$ or $\mathbb{C}$ (call it $\mathbb{K}$). Then $V$ comes with two operations

• addition $+:V\times V\to V,(v_1,v_2)\mapsto v_1+v_2$
• scalar multiplication $\cdot :\mathbb{K}\times V\to V,(\alpha ,v)\mapsto \alpha \cdot v$

Example

• $\mathbb{R},{\mathbb{R}}^2,{\mathbb{R}}^n,\mathbb{C},{\mathbb{C}}^n$
• $C([0,1])=\{f:[0,1]\to \mathbb{C}|f\text{ is continuous}\}$

see vector space

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 $\forall v_1,\dots ,v_N\in E$ ${\sum }_{i=1}^N{a}_i{v}_i=0⟹a_1=\cdots =a_N=0$ and $E$ has finite cardinality, then $V$ is finite-dimensional. Otherwise, $V$ is infinite-dimensional.

Example

$C([0,1])$ is infinite dimensional because $E=\{f_n(x)=x^n|n\in \mathbb{N}\cup \{0\}\}$ is a linearly independent set with non-finite cardinality.

see finite vector space

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

1. $||v||=0⟺v=0$ definiteness
2. $||\lambda v||=|\lambda |\cdot ||v||\forall \lambda \in \mathbb{K},v\in V$ homogeneity
3. $\forall v_1,v_2\in V||v_1+v_2||\le ||v_1||+||v_2||$ triangle inequality

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

see norm

semi-norm

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

see semi-norm

Recall:

Metric

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

1. (identifiability) $d(x,y)=0⟺x=y$
2. (symmetry) $\forall x,y\in X,d(x,y)=d(y,x)$
3. (triangle inequality) $\forall x,y,z\in X,d(x,y)\le d(x,z)+d(z,y)$

see metric

Theorem

If $||\cdot ||$ is a norm on a vector space $V$, then $d(v,w):=||v-w||$ defines a metric on $V$ (and this metric is induced by the norm $||\cdot ||$)

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 $x-y=(x-z)+(z-y)$

see norm spaces induce metric spaces

Example

Multi column

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

\lvert \lvert x \rvert \rvert {2} &= \left( \sum{i=1}^n \lvert x_{i} \rvert^2 \right)^{1/2} \ \lvert \lvert x \rvert \rvert_{\infty} &= \max_{1 \leq i \leq n} \lvert x_{i} \rvert \ \lvert \lvert x \rvert \rvert_{p} &= \left( \sum_{i=1}^n \lvert x_{i} \rvert^p \right)^{1/p}\quad 1 \leq p < \infty \end{align}$$

$C_{\infty }(X)$

If $X$ is a vector space, the vector space $C_{\infty }(X)=\{f:X\to \mathbb{C}|f\text{ is continuous and bounded}\}$

Example

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

see continuous bounded function space

Proposition

Then $C_{\infty }(X)$ is a vector space and we can define the norm $||u|{|}_{\infty }={\sup }_{x\in X}|u(x)|$on $C_{\infty }(X)$

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 $x\in X$

\lvert u(x)+v(x) \rvert &\leq \lvert u(x) \rvert + \lvert v(x) \rvert \quad \text{by }\triangle \text{ inequality} \\ &\leq \lvert \lvert u \rvert \rvert_{\infty} + \lvert \lvert v \rvert \rvert_{\infty} \\ \implies \lvert \lvert u + v \rvert \rvert_{\infty } &=\sup_{x \in X} \lvert u(x) + v(x) \rvert \\ &\leq \lvert \lvert u \rvert \rvert_{\infty} + \lvert \lvert v \rvert \rvert_{\infty} \end{align}$$

NOTE

Convergence in this norm means

u_{n} \to u \text{ in }C_{\infty}(X) &\iff \lvert \lvert u_{n} -u \rvert \rvert_{\infty} \to 0 \text{ as } n \to \infty \\ &\iff \forall \epsilon > 0, \, \exists N \in \mathbb{N} \text{ s.t. } \forall n \geq N, \forall x \in X, \lvert u_{n}(x) - u(x) \rvert < \epsilon \\ &\iff u_{n} \to u \text{ uniformly on } X \end{align}$$ Thus convergence in this [[distance|metric]] is uniform convergence when the functions $u$ are bounded and continuous.

see infinity norm for continuous bounded function space

m o r e e x a m p l e s of normed vector spaces

${\ell }^p$ space

${\ell }^p=\{a=\{a_j{\}}_{j=1}^{\infty }|||a|{|}_p<\infty \}$ is the space of (infinite) sequences. We define the ${\ell }^p$ norm as

(\sum_{i=1}^\infty \lvert a_{i} \rvert^p)^{1/p}\quad\quad 1 \leq p <\infty \\ \sup_{1 \leq j < \infty} \lvert a_{j} \rvert\qquad\,\,\; p = \infty \end{cases} $$

Example

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

see l-p vector space

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

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

see Banach space

Theorem

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

see complete metric spaces have banach continuous bounded function spaces

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 $N\in \mathbb{N}$ such that for all $n,m\ge N$, $||u_n-u_m|{|}_{\infty }<ϵ$.

In particular, $\exists N_0\in \mathbb{N}$ such that for all $m,n\ge N_0,||u_n-u_m|{|}_{\infty }<1$. Then for all $n\ge N_0$, we have

\lvert \lvert u_{n} \rvert \rvert _{\infty}&\leq \lvert \lvert u_{n} - u_{N_{0}} \rvert \rvert _{\infty}+ \lvert \lvert u_{N_{0}} \rvert \rvert _{\infty} \\ &< 1 + \lvert \lvert u_{N_{0}} \rvert \rvert \end{align}$$ And therefore for all $n \in \mathbb{N}$, we have $$\begin{align} \lvert \lvert u_{n} \rvert \rvert _{\infty}&\leq \lvert \lvert u_{1} \rvert \rvert_{\infty} + \lvert \lvert u_{2} \rvert \rvert _{\infty}+ \dots + \lvert \lvert u_{N_{0}} \rvert \rvert_{\infty} + 1 \\ &= B \end{align}$$ Since for all $x \in X$ we have $\lvert u_{n}(x)-u_{m}(x) \rvert \leq \lvert \lvert u_{n} - u_{m} \rvert \rvert_{\infty}$, we have that for each $x \in X$, the sequence$\{ u_{n}(x) \}$ is a Cauchy sequence. And since $\mathbb{C}$ is complete, its limit is in $\mathbb{C}$. (*define a candidate limiting function*) Define $u: X \to \mathbb{C}, \quad u(x) = \lim_{ n \to \infty } u_{n}(x)$ Then for all $x \in X$, $$\lvert u_{n}(x) \rvert = \lim_{ n \to \infty } \lvert u_{n}(x) \rvert \leq B \implies \sup_{x \in X}\lvert u(x) \rvert \leq B$$ thus $u$ is a bounded function. Finally, we need to show continuity and convergence. $$\lvert \lvert u -u_{n} \rvert \rvert_{\infty} \to 0$$ Let $\epsilon >0$. Since $\{ u_{n} \}$ is Cauchy in $C_{\infty}(X)$, there exists some $N_{1} \in \mathbb{N}$ such that $\forall n,m \geq N$ we have $\lvert \lvert u_{n} - u_{m} \rvert \rvert_{\infty} < \frac{\epsilon}{2}$. So for any fixed $x \in X$, we have $$\lvert u_{n}(x) -u_{m}(x) \rvert \leq \lvert \lvert u_{n} - u_{m} \rvert \rvert_{\infty} < \frac{\epsilon}{2}$$ Thus as $m \to \infty$, we have for all $n \geq N_{1}$ and each $x \in X$, $$\lvert u_{n}(x) - u(x) \rvert \leq \frac{\epsilon}{2} < \epsilon$$ Thus we have $\lvert \lvert u_{n} - u \rvert \rvert_{\infty} \to 0$ 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

example ${\ell }^p$ is a Banach space for all $1\le p\le \infty$.

Exercise

show $c_0=\{a\in {\ell }^{\infty }|{\lim }_{j\to \infty }{a}_j=0\}$ is also a Banach space with norm $||a|{|}_{\infty }={\sup }_{1\le j\le \infty }|a_j|$

idea

• each “point” in the space is sequence

review

functional

What two operations are required for a vector space? -?-

1. addition
2. scalar multiplication

What makes a vector space finite? -?- Every linearly independent set is finite

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.

What is the general approach to show that a space is Banach? -?-

1. choose a candidate for the limit of a sequence
2. show that the limit is in the space

Lecture 2 →

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