Suppose the sequence spaces

$$d \colon=\left\lbrace \left\lbrace x_n\right\rbrace_{n \in \mathbb{N}} \in \mathbb{K}^{\mathbb{N}} \colon x_n=0 \ \text{for almost all} \ n \right\rbrace$$


$$\ell^p \colon= \left\lbrace x=\left\lbrace x(n)\right\rbrace_{n \in \mathbb{N}} \in \mathbb{K}^{\mathbb{N}}\colon \sum_{n=1}^{\infty} |x(n)|^p < \infty \right\rbrace,$$

with $$\|x\|_p \colon = \left( \sum_{n=1}^{\infty} |x(n)|^p \right)^{\frac{1}{p}} \text{for} \ x=\left\lbrace x(n)\right\rbrace_{n \in \mathbb{N}} \in \ell^p.$$

Then it holds that $\overline{d}=\ell^p$.

I have constructed and example which confuses me:

Let $x_n \colon = \{1,2,3,\ldots,n,0,0, \ldots \} \in d$. Then it follows that $$ \lim_{n \to \infty} x_n =\{n\}_{n \in \mathbb{N}}$$

and so $\{n\}_{n \in \mathbb{N}} \in \overline{d}$ but $\{n\}_{n \in \mathbb{N}} \notin \ell^p$.

Where is my mistake?

  • 1
    $\begingroup$ $x_n$ does not converge to $\{n\}$ with respect to the $\ell^p$ norm. Just compute $$\|x_n -\{n\}\|_p$$ $\endgroup$ – Prahlad Vaidyanathan Apr 21 '14 at 15:52
  • $\begingroup$ How are you defining $\bar d$? $\endgroup$ – Umberto P. Apr 21 '14 at 15:52
  • $\begingroup$ @Prahlad Vaidyanathan: Whats the limit instead? $\endgroup$ – user138765 Apr 21 '14 at 15:58
  • $\begingroup$ @Umberto P.: I thought $\overline{d}$ is the closure of $d$ and if I take a sequence in $d$ it's limit is in $\overline{d}$. $\endgroup$ – user138765 Apr 21 '14 at 15:58

Perhaps what is confusing here is that there are two types of convergence for sequences :

  1. Pointwise convergence : ie. $x_n \to x$ iff $x_n(j) \to x(j)$ for all $j$.

  2. Uniform convergence (with respect to the $p$-norm): ie. $$ \|x_n - x\|_p \to 0 $$ This is strictly stronger than pointwise convergence (Note that something similar happens with uniform convergence of functions - sequences are just functions after all!).

In your case, the sequence $x_n$ does not converge uniformly to anything - it is simply not Cauchy. Just check what $$ \|x_n - x_{n-1}\|_p $$ is. However, it does converge to $x = \{n\}$ pointwise. Therein lies the rub.


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