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.

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

Appendix