TMA4230: Functional Analysis
# Introduction and preliminaries
We begin by reviewing some elementary definitions and theorems.
> **Definition 1**. A *normed space* is a real or complex vector space $X$, together with a real-valued function $x \mapsto \|x\|$, defined for all $x \in X$, such that
> 1. $\|x\| \geq 0$ for all $x \in X$, and $\|x\| = 0$ iff. $x = 0$.
> 2. $\|\alpha x\| = \vert \alpha \vert\|x\|$, $\alpha$ scalar
> 3. $\|x + y| \leq \|x\| + \|y\|$
If we wish to emphasize that we're refering to the norm *of $X$*, we write $\|x\|$_{X}.
> **Definition 2**. A *metric space* is a set $X$ with a map $d: X \times X \rightarrow \mathbb{R}$, defined for all $x, y \in X$ such that
>
> 1. $d(x, y) \geq 0$ for all $x, y \in X$, with equality only when $x = y$.
> 2. $d(x, y) = d(y, x)$ for all $x, y \in X$
> 3. $d(x, z) \leq d(x, y) + d(y, z)$ for all $x,y,z \in X$
We sometimes write a metric space $X$ with map $d$ as $(X, d)$.
As might be clear, every normed space is *metrizable*, i.e. we can introduce a metric on the normed space, defined by
$$
d(x, y) = \|x-y\|.
$$
We usually refer to $\|x-y\|$ as the *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 $X$ by $(x_n)_{n=1}^{\infty}$.
>> **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 a **cauchy sequence** if 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 $X$. Take for instance the sequence of rational numbers that is the increasing decimal expansion of $\sqrt{2}: 1, 1.4, 1.41, 1.414, 1.4142, \ldots$. The sequence is obviously a cauchy sequence, but converges to $\sqrt{2}$, which is not a part of the rational numbers.
This motivates the definition of **complete metric spaces**.
>> **Definition 5**. A **complete metric space** is 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**. A **Banach space** is 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 $T: X \rightarrow Y$, where $X, Y$ typically are normed (Banach) spaces. The linearity requirement says that $T(x+y) = Tx + Ty,\>T(\alpha x) = \alpha Tx$.
>> **Definition 7**. A linear map $T: X \rightarrow Y$ is said to be **bounded** if 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 $X$ unproportionally far away from each other in $Y$.
>> **Proposition 8**. Let $T$ be a linear map between normed spaces, $T: X \rightarrow Y$. Then the following are equivalent:
>>
>> 1. $T$ is continuous
>> 2. $T$ is continuous at zero,
>> 3. $T$ 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 $\mathcal{L}(X,Y)$ is typically called an operator. And the value $\|T\|$ for any $T \in \mathcal{L}$ is called the **operator norm** on $\mathcal{L}(X,Y)$. It can be verified that the operator norm indeed is a norm under reasonable assumptions of the vector operations.
> **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 an **isometry**. 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 an **isomorphism**. A map that satisfies both this and Definition 11 is called an **isometric isomorphism**.
Normed spaces $X$ and $Y$ as in the previous definitions are said to be **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 $\mathbb{C}$). The **dual space** of $X$ 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 $(x_n)_{n=1}^{\infty}$ converges, we know that for $\epsilon > 0$ there exists sufficiently large $N$ such that for any $n > N$, $d(x, x_n) < \epsilon$.
Now let $M = 2N$, and let $n,m > N$. Then,
$$
d(x_n, x_m) < d(x_n, x) + + d(x, x_m) < 2N = M
$$
> **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 $T$ be an isomorphism between $X$ and $Y$. Recall from the definiton that $T$ is a continuous linear bijection with continuous inverse.
Fix an element $x$ of $X$. Because $T$ is continuous, it is also bounded. Then $\|Tx\| \leq K_1\|x\|$ for some $K_1$. Let $c_2 = K_1$. Let then $Tx$ be an element of $Y$. As $T^{-1}$ is continuous, it is also bounded. Then $T^{-1}(Tx) \leq K_2\|x\|$ for some $K_2$. As $T$ is a bijection, $T^{-1}(Tx) = x$. Let $c_1 = K_2$, and the inequality follows.
We now prove the other direction. As $T$ is a surjection, we know that the image of $X$ is $Y$. Then let $A \subset X$ be the preimage of any $y \in Y$. We will show that $A$ contains only one element $x$, proving that $T$ is a bijection. Let $x, z$ be two elements in $A$ such that $Tx = Tz = y \in Y$.
Then
$$
c1\|x\| \leq \|y| \leq c_2\|x\| \\
c1\|z\| \leq \|y| \leq c_2\|z\|
$$
But this means that $c1\|x\| \leq c_2\|z\|$ and $c1\|z\| \leq c_2\|x\|$. Thus $x = z$, and $T$ is a bijection. It is also continuous with a continuous inverse as it is bounded due to the inequality, by following steps similar to the first part of the solution. This concludes the proof.
# Sequences and sequence spaces
A sequence is an infinite tuple of numbers, index by a $i \in \mathbb{N}$. We typically write a sequence as $(x_0, x_1, x_2, \ldots, x_n,\ldots)$. Another equivalent way of defining a sequence is as a function with domain $\mathbb{N}$.
We define the following norms for sequences:
$$
\begin{align}
\|x\|_1 &= \sum_{i=1}^{\infty} x_i \\
\|x\|_2 &= \left(\sum_{i=1}^{\infty} x_i^2 \right)^{1/2} \\
\|x\|_p &= \left(\sum_{i=1}^{\infty} x_i^p \right)^{1/p} \\
\|x\|_{\infty} &= sup_{i \in \mathbb{N}} \>x_i
\end{align}
$$
> **Definition 15.** The **set 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 the **set 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](http://www.maths.manchester.ac.uk/~nikita/31002/lp-dual.pdf).
> **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 $\ell_{\infty}^*$ is $l_1$. However, this is not the case. However, we can make the following inference:
> **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 $q$ are **conjugate exponents** if 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*. $1 \implies 2$ is obvious. Lets prove $2 \implies 3$:
Let $T$ be continuous at zero. Assume $T$ is unbounded. Then for any $n \in \mathbb{N}$ there exists $\|x_n\| \leq 1$ such that $\|Tx_n\| \geq n$. We then have
$$
\|\frac{x_n}{n}\| \leq \frac{1}{n}
$$
Now let $n$ go to infinity, then $\lim_{n\to\infty} \|\frac{x_n}{n}\| = 0$, and consequently $\lim_{n\to\infty} \frac{x_n}{n} = 0$. By continuity at zero, we must have that
$$
\lim_{n\to\infty} \|T\frac{x_n}{n}\| = \|T(0)\| = 0
$$.
But $\|Tx_n\| \geq n \implies \|T(\frac{x_n}{n}\| \geq 1, n \in \mathbb{N}$, which is a contradiction. Therefore, $T$ must be bounded.
$3 \implies 1$:
As $T$ is bounded, we have $\|T(x)\| \leq \|T\|\|x\|$. This implies that $\|T(\frac{x}{\|x\|})\| < \|T\|$, as $\|\frac{x}{\|x\|} = 1$. Using this, we can use linearity to get:
$$
\|Tx\| = \|T(\frac{x}{\|x\|})\cdot\|x\|\| \leq \|T\|\|x\|,\quad x \in X
$$
Thus, substituting $x-y$ for $x$ (legal as of linearity), we get
$$
\|Tx-Ty\| \leq \|T\|\|x-y\|,
$$
This proves the proposition.
In the above proof, we prove the slightly stronger statement, namely that $T$ is **Lipschitz continuous**. For a proof that Lipschitz continuity implies continuity, see [this link](http://www.tormodhaugland.com/some-proofs-of-interest/).