Sequence Spaces (ℓp spaces)
Definition 1
For $1 \le p < \infty$, distance space $( \ell^{p} , d^{p} )$ is defined as follows:
(i) Set of converging sequences:
$$ \ell^{p} := \left\{ \left\{ x_{n} \right\}_{n \in \mathbb{N}} \subset \mathbb{C} \left| \left( \sum_{i=1}^{\infty} | x_{i} |^{p} \right)^{{1} \over {p}} < \infty \right. \right\} $$
(ii) Distance function:
$$ d^{p} ( x_{n} , y_{n} ) := \left( \sum_{i = 1}^{\infty} | x_{i} - y_{i} |^{p} \right)^{ {{1} \over {p}} },\quad \left\{ x_{n} \right\} , \left\{ y_{n} \right\} \in \ell^{p} $$
For $p = \infty$, distance space $( \ell^{\infty} , d^{\infty} )$ is defined as follows:
(i)’ Set of bounded sequences:
$$ \ell^{\infty} := \left\{ \left\{ x_{n} \right\}_{n \in \mathbb{N}} \ \left| \ \sup_{i \in \mathbb{N}} | x_{i} | < \infty \right. \right\} $$
(ii)’ Distance function:
$$d^{\infty} ( x_{n} , y_{n} ) := \sup_{i \in \mathbb{N}} | x_{i} - y_{i} |,\quad \left\{ x_{n} \right\} , \left\{ y_{n} \right\} \in \ell^{\infty} $$
Explanation
$\ell^{p}$ is called a sequence space, and $\ell^{p}$ is pronounced as [ell_p]. The TeX code for $\ell$ is \ell.
The difference between space $\ell^{p}$ and $L^{p}$ space is merely whether it’s about sequences or functions, series or integrals. This includes Young’s inequality, Cauchy-Schwarz inequality, Hölder’s inequality, Minkowski’s inequality, as well as completeness. Since the facts themselves are similar, the methods of proof are also largely the same, so if one is thoroughly studied, there is no real need to study the other.
On the other hand, $\ell^{\infty}$ is, in reality, the same as when $p \to \infty$, proven to be unnecessary to define separately. Most properties of $\ell^{p}$ and $\ell^{\infty}$ are almost the same, so there’s no need to think of them separately.
However, a notable exception is divisibility.
Theorem 1
Let’s say $1 \le p_{0} < \infty$.
- (a): $\ell^{p_{0}}$ is a divisible space.
- (b): $\ell^{\infty}$ is an indivisible space.
This difference occurs because $\ell^{p_{0}}$ has convergence as a condition, whereas $\ell^{\infty}$ only has boundedness as a condition.
Proof
(a)
Strategy: A converging sequence always allows to select $i_{0}$ so that $\displaystyle \left| \sum_{i = i_{0}}^{\infty} a_{i} \right|$ becomes sufficiently small. Based on this $i_{0}$, it divides into finite and infinite parts, and then explicitly finds a subset that makes $l^{p_{0}}$ a divisible space by utilizing the fact that any finite sequence must converge to a point.
Claim: A countable set $M \subset \ell^{p_{0}}$ that satisfies $\overline{M} = \ell^{p_{0}}$ exists.
Consider a set of complex sequences that repeat only $0$ from a certain $j_{0}$
$$ M : = \left\{ \left\{ m_{j} \right\} \in \ell^{p_{0}} \ | \ m_{j} \in \mathbb{Q} + i \mathbb{Q} , m_{j} = 0 , \ j>j_{0} , \ j_{0} \in \mathbb{N} \right\} $$
Since $M$ is a countable set and $\overline{M} \subset \ell^{p_{0}}$ obviously holds, it is sufficient to show $\ell^{p_{0}} \subset \overline{M}$. According to the definition of $l^{p_{0}}$, every sequence $x : = ( x_{1} , x_{2} , \cdots ) \in \ell^{p_{0}}$ must have
$$ \left( \sum_{j > N} | x_{j} |^{p_{0}} \right)^{ {{1} \over {p_{0}}} } < {{ \varepsilon } \over {2}} $$
that satisfies for any $\varepsilon > 0$. Then, for each $x$
$$ \left( |x_{1} - m_{1}|^{p_{0}} + \cdots + |x_{N} - m_{N}|^{p_{0}} \right)^{ {{1} \over {p_{0}}} } < {{\varepsilon} \over {2}} $$
that satisfies also exists
$$ d^{p_{0}} ( x, m) = \left( \sum_{j \le N} |x_{j} - m_{j}|^{p_{0}} + \sum_{j > N} |x_{j}|^{p_{0}} \right)^{{1} \over {p_{0}}} < {{ \varepsilon} \over {2}} + {{ \varepsilon} \over {2}} = \varepsilon $$
for all $\varepsilon >0$, $B^{d^{p_{0}}} (x ; \varepsilon ) \cap M \ne \emptyset$, so $x \in \overline{M}$
■
(b)
Strategy: Define manageable bounded function $e_{I} \in \ell^{\infty}$ and functions $\psi$ on them, and use their injectiveness to calculate cardinality.
Claim: No countable set $M \subset \ell^{p_{0}}$ satisfying $\overline{M} = \ell^{\infty}$ exists.
It is sufficient to show that all $M \subset \ell^{\infty}$ satisfying $\overline{ M} = \ell^{\infty}$ are uncountable.
Part 1.
Let’s define function $e_{I} : I \to \left\{ 0, 1 \right\}$ with domain $I \subset \mathbb{N}$ as
$$ e_{I} (j) := \begin{cases} 1 & , j \in I \\ 0 & , j \in ( \mathbb{N} \setminus I ) \end{cases} $$
For example, if $I = 2 \mathbb{N} = \left\{ 2, 4, 6 , \cdots \right\} $, then the function value appears as $e_{2 \mathbb{N}} (1) = 0$, $e_{2 \mathbb{N}} (2) = 1$, $e_{2 \mathbb{N}} (3) = 0$, $e_{2 \mathbb{N}} (4) = 1 $.
Considering the set of these functions $A: = \left\{ e_{I} \ | \ I \subset \mathbb{N} \right\}$, since the function value cannot exceed $[0,1]$, $A \subset \ell^{\infty}$ holds.
Part 2.
Define function $\phi : \mathscr{P} ( \mathbb{N} ) \to A$ as $\phi (I) : =e_{I}$. Then, if $I , I’ \subset \mathbb{N}$ is $I \ne i '$, $\phi (I) = e_{I} \ne e_{I’} = \phi (I’)$ holds, so $\phi$ is injective, and therefore
$$ |A| \ge | \mathscr{P} ( \mathbb{ N} ) | = 2^{\aleph_{0}} = \aleph_{1} $$
For any $x \in \ell^{\infty} = \overline{M}$ and $\varepsilon >0$, $B_{d^{\infty}} (x ; \varepsilon ) \cap M \ne \emptyset$ holds
$$ B_{d^{\infty}} \left( e_{I} ; {{1} \over {3}} \right) \cap M \ne \emptyset $$
Part 3.
Since $\displaystyle B_{d^{\infty}} \left( e_{I} ; {{1} \over {3}} \right) \cap M \ne \emptyset$ holds
$$ \psi ( e_{I} ) \in \left( B_{d^{\infty}} \left[ e_{I} ; {{1} \over {3}} \right] \cap M \right) $$
we can define function $\psi : A \to M$ that satisfies. Assuming $\psi$ is not injective, for $\psi ( e_{I}) = \psi ( e_{I’ })$, we have
$$ \psi ( e_{I}) = \psi ( e_{I’ }) \in \left[ B_{d^{\infty}} \left( e_{I} ; {{1} \over {3}} \right) \cap B_{d^{\infty}} \left( e_{I’} ; {{1} \over {3}} \right) \right] $$
According to the triangle inequality
$$ 1 = d^{\infty} ( e_{I} , e_{I’} ) \le d^{\infty} ( e_{I} , \psi (e_{I}) ) + d^{\infty} ( \psi (e_{I}) , e_{I’} ) \le {{1} \over {3}} + {{1} \over {3}} = {{2} \over {3}} $$
It concludes $\displaystyle 1 \le {{2} \over {3}}$, which is a contradiction, hence $\psi$ is injective. Also, since $\psi : A \to M$ is injective, $|M| \ge |A| = \aleph_{1}$, and $M$ cannot be countable.
■