2
$\begingroup$

I am trying to learn about the $l^p$ space for $1\leq p < \infty$. I have read some articles and such but I still don't have any kind of understanding about what sequences are in the $l^p$ space. Or what sequences aren't. Does anyone have a good way to explain this intuitively to me?

Thanks.

$\endgroup$
2
$\begingroup$

The sequences $(x_1,x_2,x_3,\ldots)$ in $l^p$ are those such that the series$$\sum_{n=1}^\infty|x_n|^p$$converges. All other sequences don't belong to $l^p$.

$\endgroup$
  • $\begingroup$ Does that mean that any sequences that don't absolutely converge would not be in $l^p$? $\endgroup$ – MathIsHard Aug 24 '17 at 18:39
  • 1
    $\begingroup$ @Math4Life No, it does not. For instance, the series $\sum_{n=1}^\infty\frac1n$ doesn't converge absolutely, but the sequence $\left(\frac1n\right)_{n\in\mathbb N}$ belongs to $l^2$. $\endgroup$ – José Carlos Santos Aug 24 '17 at 18:48
  • $\begingroup$ how is $\frac{1}{n}$ in $l^p$ if it doesn't converge? $\endgroup$ – MathIsHard Aug 24 '17 at 18:53
  • 1
    $\begingroup$ @Math4Life What?! The series $\sum_{n=1}^\infty\left(\frac1n\right)^2=\sum_{n=1}^\infty\frac1{n^2}$ converges! $\endgroup$ – José Carlos Santos Aug 24 '17 at 18:55
  • $\begingroup$ Ohhh I see. Thank you :) $\endgroup$ – MathIsHard Aug 24 '17 at 18:56
2
$\begingroup$

$\ell^p$ is a sequences space. A sequence in $\ell^p$ is a sequence of p-summable sequences.

More in detail

$$\ell^p:=\{(x_n)_{n \in \mathbb{N}}: \sum\limits_{i=1}^\infty |x_n|^p<+\infty\}$$

This space is a normed space with the norm defined by

$$ \|x\|_p:=\left(\sum\limits_{i=1}^\infty |x_n|^p<+\infty \right)^{\frac{1}{p}}$$

Observations:

  • Every $(\ell^p,\|f\|_p)$ is a Banach space (complete normed space).
  • $\ell^1$ is the space of sequences which have absolutely convergent series.
  • Only $\ell^2$ is an Hilbert space (the norm $\|x\|_2$ is induced by an inner product).
  • $\ell^\infty$ is the space of bounded sequences
$\endgroup$
1
$\begingroup$

$$\ell^p=\left\{(x_n)_{n\in\mathbb N}\mid \sum_{n\geq 1}|x_n|^p<\infty \right\}.$$

Therefore if $p>1$ then $\left(\sqrt[p]\frac{1}{n}\right)_n\notin\ell^p$ and $\left(\frac{1}{n}\right)_n\in \ell^p$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.