0

There is no doubt that clear examples consolidate the understanding of concepts being learnt. I am new to Analysis of sequences, I learnt theoretically that suppose in the sequence space, $s = \{ x: \mathbb{N} \to \mathbb{R}\}$, we have that

$$\ell^r, \ell^s = \{x: \mathbb{N} \,\to\, \mathbb{R} \Big| \ \sum\limits_{i=1}^\infty |x|^r < \infty \}, \quad \ell^\infty = \{x: \mathbb{N} \,\to\, \mathbb{R} \big| \ x =(x_i) < \infty \} \\ \mathbf{c} = \{x: \mathbb{N} \,\to\, \mathbb{R} \Big| \ x = (x_i) \ \text{is a convergent sequence} \}, \qquad \mathbf{c}_0 = \{x: \mathbb{N} \,\to\, \mathbb{R} \big| \ x_i \,\to\, 0 \ \text{ as } i \,\to\, \infty \}$$ then this inequality: $$ \ell^r \subsetneq \ell^s \subsetneq \mathbf{c}_0 \subsetneq \mathbf{c} \subsetneq \ell^\infty \subsetneq \mathbf{s} \quad \text{ for } 1 \le r<s<\infty \qquad (1). $$ is strictly satisfied.

After going through https://math.stackexchange.com/ to try and get more knowledge, I read from How do you show monotonicity of the $\ell^p$ norms?, Inequality between $\ell^p$-norms, An example of a sequence which satisfies a number of conditions, and here What sequences could satisfy these requirements? just to catch a glimpse of an example but there wasn't a clear cut example.

Assuming without admitting that the inequality was just an improper subset $"\subseteq"$, then the sequence would have been as easy as $(1, 0, 1, 0, \ldots)$. Now my task is finding that example satisfying (1). Benevolent contributors, I rely on you for assistance.

holala
  • 733

1 Answers1

1

$\ell^r\subsetneq \ell^s$ for $r<s$:

We will use the fact that for $\alpha\in \mathbb{R}$, $$\sum_{n=1}^\infty n^{-\alpha}$$ converges if and only if $\alpha >1$ (which can be shown by the integral test). Pick $1/s<t<1/r$. Then $x(n)=n^{-t}$ is in $\ell^s$ but not $\ell^r$.

$\ell^s\subsetneq \mathbf{c}_0$ for any $s$:

We can construct a sequence that works for all $s$. Consider the sequence $x(n)=1/\ln n$ (for $n>1$, it doesn't matter what happens at $n=1$, since we only need the limiting behaviour). To prove this, it suffices to show that for sufficiently large $n$, $$\ln n<n^{1/s},\tag{*}$$for then we use a comparison with the harmonic series to show that $$\sum_{n=1}^\infty \frac{1}{(\ln n)^s}$$ diverges. To prove (*), we compute the limit $$\lim_{t\to\infty}\frac {\ln t}{t^{1/s}}=\lim_{t\to\infty}\frac{1/t}{(1/s)t^{1/s-1}}=\lim_{t\to\infty}\frac{s}{t^{1/s}}=0,$$ where the first equality follows from l'hopital's rule.

The strictness of the other inclusions are relatively straightforward. For $\mathbf{c}_0\subsetneq \mathbf{c}$, pick your favourite sequence converging to $0$ and then adding some non-zero constant will produce an example. For $\mathbf{c}\subsetneq \ell^\infty$, examples include $x(n)=(-1)^n$.

Tipping Octopus
  • 1,614
  • Can we find just ONE sequence satisfying $ (1)$? e.g., Let $x \in s$ such that $\ x \in \ell^r \subsetneq \ell^s \subsetneq \mathbf{c}_0 \subsetneq \mathbf{c} \subsetneq \ell^\infty \subsetneq \mathbf{s} \quad \text{ for } 1 \le r<s<\infty $? – holala Jan 30 '21 at 15:31
  • @holala $x\in A\subsetneq B $ is not a well-defined notion. As a shorthand, people will sometimes write $x\in A\subseteq B$ to mean something like "we have established $x\in A $ and $A\subseteq B$ and now we can conclude $x\in B$". If you intend $x\in A\subsetneq B $ to mean an example of $x\in B$ but $x\notin A$, then by definition there could not be one single example, for that will require $x$ to simultaneously be in and not in, say, $\ell^s$. – Tipping Octopus Feb 01 '21 at 19:06
  • Is there any trick available for constructing sequence(s) or you have to brainstorm to make it? – holala Feb 03 '21 at 01:38
  • I think it is helpful to be familiar with the "standard" examples (like $\sum n^{-\alpha}$ or geometric series) and have some idea about the relative growth rate of functions (for example, $\ln x$ grows slower than $x^\alpha$ for all $\alpha>0$ and $\exp x$ grows faster than any polynomials) – Tipping Octopus Feb 04 '21 at 06:34