Subjects: Metric and Normed Spaces
Links: Normed Vector Spaces, Vector Spaces
We can look at the space of sequences such that for $p\in [1,\mathrm{\infty })$, in general we can denote $x_{\bullet }$ to be a sequnce of real numbers (notation Wikipedia)

This actually forms a infinite dimensional vector space.

converges, and it is a vector space, and it is denoted as ${\ell }_p$ or $\text{ell { <a class="tag" onclick="toggleTagSearch(this)" data-content="\#p">\#p</a>}}$ denpending of who is writing, and we can define

defines a norm in ${\ell }^p$.

In the case where $p=\mathrm{\infty }$, the space ${\ell }^{\mathrm{\infty }}$ is the set of all bounded sequences of real numbers, is a vector space and

is a norm in ${\ell }^{\mathrm{\infty }}$.

Since they are norms, we have that

Hölder’s Inequalities for Series

We have that for $p,q$ harmonic conjuagtes. Then if $x_{\bullet }\in {\ell }^p$ and $y_{\bullet }\in {\ell }^q$. Then

Minkowski’s Inequalities for Series

We can compare how does the ${\ell }^p$ spaces when varying $p$. We get that

• Let $1\le s<r\le \mathrm{\infty }$, then $\text{ell^s subset ell { <a class="tag" onclick="toggleTagSearch(this)" data-content="\#r">\#r</a>}}$ , and for any $x_{\bullet }\in {\ell }^s$, we get that $\parallel x_{\bullet }{\parallel }_r\le \parallel x_{\bullet }{\parallel }_s$

• For any $1\le p<\mathrm{\infty }$ and $x_{\bullet }\in {\ell }^p$, we have that

  $$\parallel x_{\bullet }{\parallel }_{\mathrm{\infty }}=\underset{r\to \mathrm{\infty }}{lim}\parallel x_{\bullet }{\parallel }_r$$

${\ell }^p(\mathbb{Z})$

This are the two sided sequences

${\ell }^2(\mathbb{Z})$

The vector space ${\ell }^2(\mathbb{Z})$ over $\mathbb{C}$ is the set of all (two-sided) infinite sequences of complex numbers $$(\dots, a_{-n}, \dots, a_{-1}, a_0, a_1, \dots, a_n, \dots)$$such that $$\sum_{n \in \Bbb Z}|a_n|^2 < \infty$$That is, the series converges. Addition is defined componentwise, and so is scalar multiplication. The inner product between two vectors $A=(\dots ,a_{-1},a_0,a_1,\dots )$ and $B=(\dots ,b_{-1},b_0,b_1,\dots )$ is defined by the absolutely convergent series $$\langle A, B\rangle := \sum_{n \in \Bbb Z}a_n \overline{b_n}$$Thus the norm of $A$ is given by $$|A| = \left(\sum_{n \in \Bbb Z} |a_n|^2\right)^{1/2}$$