1-Sequence Space is Proper Subset of 2-Sequence Space
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Theorem
Let $\ell^1$ and $\ell^2$ be the $1$-sequence space and $2$-sequence space respectively.
Then $\ell^1$ is a proper subset of $\ell^2$.
Proof
$\ell^1$ is a subset of $\ell^2$
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $\ell^1$.
By definition of $1$-sequence space:
- $\ds \sum_{n \mathop = 0}^\infty \size {x_n} < \infty$
By Terms in Convergent Series Converge to Zero:
- $\ds \lim_{n \mathop \to \infty} \size {x_n} = 0$
By definition of convergent sequence:
- $\exists N \in \N : \forall n > N : \size {x_n} < 1$
Furthermore:
- $\forall \size {x_n} < 1 : \size {x_n}^2 < \size {x_n}$
By Comparison Test:
- $\ds \sum_{n \mathop = 0}^\infty \size {x_n}^2 < \infty$
Hence:
- $\sequence {x_n}_{n \mathop \in \N} \in \ell^2$
That is:
- $\ell^1 \subseteq \ell^2$
$\Box$
$\ell^2$ is not a subset of $\ell^1$
Let $x := \tuple {1, \dfrac 1 2, \ldots, \dfrac 1 n, \ldots}$ with $n \in \N_{>0}$.
From Basel Problem:
- $\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^2} < \infty$
From Harmonic Series is Divergent:
- $\ds \sum_{n \mathop = 1}^\infty \frac 1 n$
Hence:
- $x \in \ell^2$
but:
- $x \notin \ell^1$
That is:
- $\ell^2 \nsubseteq \ell^1$
$\Box$
By definition, $\ell^1$ and $\ell^2$ are not equal.
Therefore:
- $\ell^1 \subsetneqq \ell^2$
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces