# TMA4230: Functional Analysis

# Introduction and preliminaries

We begin by reviewing some elementary definitions and theorems.

Definition 1. Anormed spaceis a real or complex vector space$X$ , together with a real-valued function$x \mapsto \|x\|$ , defined for all$x \in X$ , such that

$\|x\| \geq 0$ for all$x \in X$ , and$\|x\| = 0$ iff.$x = 0$ .$\|\alpha x\| = \vert \alpha \vert\|x\|$ ,$\alpha$ scalar$\|x + y| \leq \|x\| + \|y\|$

If we wish to emphasize that we're refering to the norm *of $X$*, we write

Definition 2. Ametric spaceis a set$X$ with a map$d: X \times X \rightarrow \mathbb{R}$ , defined for all$x, y \in X$ such that

$d(x, y) \geq 0$ for all$x, y \in X$ , with equality only when$x = y$ .$d(x, y) = d(y, x)$ for all$x, y \in X$ $d(x, z) \leq d(x, y) + d(y, z)$ for all$x,y,z \in X$

We sometimes write a metric space

As might be clear, every normed space is *metrizable*, i.e. we can introduce a metric on the normed space, defined by

We usually refer to *displacement or distance between the elements $x$ and $y$*.

Large portions of functional analysis studies the behaviour of sequences, and consequently it is handy to have a concise notation for sequences. We denote a (possibly arbitrary sequence of a set

Definiton 3. Let$(x_n)_{n=1}^{\infty}$ be a sequence of a metric space$(X,d)$ . The sequence is said to converge to a point$x$ in$X$ , if for any$\epsilon > 0$ , we can find an$N \in \mathbb{N}$ such that whenever$n \geq N$ ,$d(x, x_n) < \epsilon$ .

Definition 4. Let$(x_n)_{n=1}^{\infty}$ be a sequence of a metric space$(X,d)$ . The sequence is acauchy sequenceif for any$\epsilon > 0$ , there is an$N \in \mathbb{N}$ , such that whenever$n,m > N$ ,$d(x_n, x_m) < \epsilon$ .

Note the similarity between the definitions of convergence and cauchy sequences. The inclusion only works one way though; all converging sequences are cauchy sequences. However, some cauchy sequences might converge to a point which is not part of the original metric space

This motivates the definition of **complete metric spaces**.

Definition 5. Acomplete metric spaceis a metric space$(X,d)$ in which every cauchy sequence converge.

A complete metric space is a metric space that does not have "*holes*" in it.

Definition 6. ABanach spaceis a normed space$X$ which is also a complete metric space with the metric$d(x,y) = \|x-y\|$ .

Of particular interest are **linear maps**, i.e. maps

Definition 7. A linear map$T: X \rightarrow Y$ is said to beboundedif there exists a scalar$M \geq 0$ such that$$\|Tx\|_{Y} \leq M \|x\|_{X},$$ for all$x$ . Equivalently, we can write$\|Tx\| \leq M$ , whenever$\|x\| \leq 1$ . The smallest such$M$ , i.e.$\sup \{\|Tx\| \mid \|x\| \leq 1\}$ is denoted$\|T\|$ .

Intuitively and unformally, we can read the definiton of boundedness as the requirement of not sending two points in

Proposition 8. Let$T$ be a linear map between normed spaces,$T: X \rightarrow Y$ . Then the following are equivalent:

$T$ is continuous$T$ is continuous at zero,$T$ is bounded

The proof is pretty straight-forward and is given in the appendix.

Definition 9. The set$\mathcal{L}(X,Y)$ is the set of all bounded linear maps from$X$ to$Y$ .

This set is itself a vector space. A member of **operator norm** on

Proposition 10. If$X$ is a normed space, and$Y$ is a Banach space, then$\mathcal{L}(X,Y)$ is a Banach space.

Definition 11. If a map$T$ preserves distances for each$x$ , i.e$\|Tx - Ty\| = \|x -y\|, \forall x,y \in X$ , then$T$ is called anisometry. If$T$ is also linear, then$\|Tx\| = \|x\|, \forall x \in X$ .

Definition 12. If a map$T$ is bijective, linear, and both it and its inverse is linear, we say that$T$ is anisomorphism. A map that satisfies both this and Definition 11 is called anisometric isomorphism.

Normed spaces **isometric** and **isomorphic**, respectively.

Definition 13. Two norms$\|\cdot\|_a, \|\cdot\|_b$ are equivalent if there exists constants$c$ and$d$ such that$$ c\|\cdot\|_a \leq \|\cdot\|_b \leq d\|\cdot\|_a,\quad \forall x \in X$$

Definition 14. Let$X$ be a normed space over$\mathbb{K}$ (either$\mathbb{R}$ or$\mathbb{C}$ ). Thedual spaceof$X$ is the space of all bounded linear functionals from$X$ to$\mathbb{K}$ .

Example 1.Let$X$ be a metric space, and let$(x_n)_{n=1}^{\infty}$ be a convergent sequence. Show that$(x_n)_{n=1}^{\infty}$ is a cauchy sequence.

*Solution.*
As

Example 2.Let X and Y be normed vector spaces and suppose$T : X \rightarrow Y$ is a linear surjection. Show that T is an isomorphism if and only if there are constants$c_1 > 0$ and$c_2 > 0$ such that:

$$ c1\|x\| \leq \|Tx\| \leq c_2\|x\| $$

*Solution.*
We first show that an isomorhism implies the inequality. So let

Fix an element

We now prove the other direction. As

Then

But this means that

# Sequences and sequence spaces

A sequence is an infinite tuple of numbers, index by a

We define the following norms for sequences:

Definition 15.Theset of$\ell_p$ -summable sequences,$p \in [1, \infty)$ , are all sequences$\xi$ for which$\|\xi\|_p$ is a finite value, i.e. less than$\infty$ . Similarly defined is the set$\ell_{\infty}$ , also called theset of bounded sequences.

The next lemma is important, though its proof is a bit too long to be articulated here. We refer the interested reader to this article.

Lemma 16.For$p \in (1, \infty)$ , the dual space$\ell_p^*$ can be identified with$\ell_q$ , where$\frac{1}{p} + \frac{1}{q} = 1$ .

Lemma 17.The space$\ell_1^*$ can be identified with$\ell_{\infty}$ .

Extrapolating the two previous lemmas, one might be inclined to think that

Lemma 18.Whenever$p, q \in \mathbb{N} \div \{1\}$ ,$\ell_p^* = \ell_q$ and$\ell_q^* = \ell_p$ .

Definition 19.We say that$p$ and$q$ areconjugate exponentsif they obey the equality$\frac{1}{p} + \frac{1}{q} = 1$ .

Theorem 20.The dual of the space of all convergent sequences$c = \left\{(\xi_n)_n : \lim_{n \to \infty} \xi_n exists\right\}$ can be identified with$\ell_1$ .

Proof of Lemma 17 and Theorem 20 are equivalent to the proof of lemma 16.

# Function spaces

## Preliminaries

Unless you've had the course real analysis, some preliminary defintions are in order.

# Appendix

## Theorems

## Extra proofs

*Proof of Proposition 8*.

Now let

But

Thus, substituting

This proves the proposition.

In the above proof, we prove the slightly stronger statement, namely that **Lipschitz continuous**. For a proof that Lipschitz continuity implies continuity, see this link.