One Particular Banach Space
Question
Is \((\ell^p, \|\cdot\|_{\ell^q})\) Banach for \( 1 \lt p \lt q \) ?
Note
This proof is wrong. There is a simple counterexample. If anyone knows where I messed up, please email me at (jromanoday) at gmail.com.
Terminology
\(\ell^p\) is the space of sequences \((x_n)_{n \in \mathbb{N}}, x_n \in \mathbb{F}\), where \(\mathbb{F}\) is some field, and
$$ \|(x_n)\|_{\ell^p} := \left( \sum_{n=1}^\infty |x_n|^p \right)^{1 \over p} \lt \infty $$\( (\ell_p, \| \cdot \|_{\ell^q}) \) is the space of \( \ell^p \) sequences, equipped with the \( \ell^q \) norm.
We define element-wise addition and s-multiplication as typical for infinite vectors.
A Banach space is a normed complete linear space. Complete means that every Cauchy sequence in the space converges to an element in the space.
Proof
Following the proof that \(\ell^p\) is Banach here
We start with the definition of a Cauchy sequence \( (x_n)_k \) in \((\ell^p, \|\cdot\|_{\ell^q})\):
$$ \forall \varepsilon \gt 0: \ \exists N \gt 0: \ \forall k,s \gt N: \ \| (x_n)_k - (x_n)_s \|_{\ell^q} \lt \varepsilon $$Inserting the definition of \( \|\cdot\|_{\ell^q} \), we have (for all \(k,s\) greater than some \(N\)):
$$ \left( \sum_{n=1}^\infty |(x_n)_k - (x_n)_s|^q \right)^{1 \over q} \lt \varepsilon $$Raising both sides to the \(q\) power and examining each \(n\)th term, it is immediately clear that
$$ \forall n: \ |(x_n)_k - (x_n)_s|^q \le \sum_{n=1}^\infty |(x_n)_k - (x_n)_s|^q $$(on the left, we take the norm of two particular terms in the two sequences \((x_n)_k\) and \((x_n)_s\); on the right we sum over all the terms in the two sequences).
Now we can raise each term in the inequality to the \(1/q\) power. The inequality stays the same because \(1/q \gt 0\). The term on the right becomes the \(\ell^q\)-norm of \((x_n)_k - (x_n)_s\).
$$ \forall n: \ |(x_n)_k - (x_n)_s| \le \left(\sum_{n=1}^\infty |(x_n)_k - (x_n)_s|^q\right)^{1 \over q} \lt \varepsilon $$Now we note that, for each choice of \(n \in \mathbb N\), we have a Cauchy sequence \(((x_n)_k)_{n \in \mathbb N}\). Since our field \(\mathbb F\) is complete, this sequence converges as \(k \to \infty\) to some number:
$$ \forall n: \ \lim_{k \to \infty} (x_n)_k = x^*_n \in \mathbb F $$We will argue that this sequence \(x^*_n\) is:
- An element of \(\ell^p\)
- The sequence to which \((x_n)_k\) converges.
To prove (1), we note that the sequence \(((x_n)_k)_{k \in N}\) is a Cauchy sequence in a metric space and is therefore bounded (per Wikipedia.). This means that, the norm will not exceed some \(M\) for any element:
$$ \exists M: \ \forall k: \ \| (x_n)_k \|_{\ell^q} \lt M $$Raising both sides to the \(q\) power:
$$ \forall k: \ \sum_{n=1}^\infty | (x_n)_k |^q \lt M^q $$Since \(p \lt q\),
$$ \forall k: \ \sum_{n=1}^\infty | (x_n)_k |^p \le \sum_{n=1}^\infty |(x_n)_k|^q \lt M^q $$Raising both sides to the \(1/p\) power:
$$ \forall k: \ \| (x_n)_k \|_{\ell^p} \le \left(\sum_{n=1}^\infty |(x_n)_k|^q \right)^{1\over p} \lt M^{q \over p} $$In other words, although \((x_n)_k\) might not be Cauchy in \(\ell^p\), the fact that it’s bounded in \((\ell^p, \|\cdot\|_{\ell^q})\) implies that the seqence is also bounded in \(\ell^p\).
Since \(\|(x_n)_k\|_{\ell^p} \lt M^{q \over p}\), for all \(k\), we can replace our infinite sum with a finite sum and retain the inequality:
$$ \forall k, j \in \mathbb N: \ \sum_{n = 1}^j |(x_n)_k|^p \lt M^q $$Taking the limit as \(k \to \infty\):
$$ \forall j : \ \sum_{n=1}^j |x^*_n|^p \lt M^q \implies \sum_{n=1}^\infty |x^*_n|^p \lt M^q \implies \|x^*\|_{\ell^p} \lt M^{q \over p} \lt \infty $$meaning that our sequence \(x^*_n\) is an element of \(\ell^p\), proving (1).
To prove (2), we’ll use the Cauchy condition in \(\|\cdot\|_{\ell^q}\):
$$ \forall s > k > N: \ \sum_{n=1}^\infty |(x_n)_k - (x_n)_s|^q < \varepsilon^q $$For safety, let’s replace that infinite sum with an arbitrary finite sum:
$$ \forall j \in \mathbb{N}: \ \sum_{n=1}^j |(x_n)_k - (x_n)_s|^q \lt \varepsilon^q $$Taking the limit as \(s \to \infty\) and then \(j \to \infty\):
$$ \forall k > N: \sum_{n=1}^j |(x_n)_k - x^*_n|^q \lt \varepsilon^q \implies \sum_{n=1}^\infty |(x_n)_k - x^*_n| \lt \varepsilon^q $$meaning that
$$ \forall k > N: \|(x_n)_k - x^*_n\|_{\ell^q} \lt \varepsilon $$and the sequence \((x_n)_k\) converges to \(x^*_n\) in the \(\ell^q\) norm, as desired.