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Let be $1\leq p\in\mathbb{R}$, denote: $$\ell^p(\mathbb {R})=\left\{(x_n)\subset \mathbb{R}: (x_n) \mbox{ is a sequence with } \displaystyle\sum_{n=1}^{\infty}|x_n|^p<\infty \right\}$$ Prove that:

  • The function: $d_p:\ell^p(\mathbb{R})\times\ell^p(\mathbb{R})\to \mathbb{R}$ is a metric for $\ell^p(\mathbb{R})$ where $d_p(x_n,y_n)= \left| \displaystyle\sum_{n=1}^\infty |x_n-y_n|^p \right|^\frac{1}{p}$ (Only triangular inequality, I work in $\mathbb{R}$, should I assume Minkowski inequality and its done?)
  • $\ell^p(\mathbb{R})$ is a complete metric space.

it is right? I mean $p\in\mathbb{R}$? I've never work with $\ell^p$ spaces, this is a question from introduction to topology.

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    $\begingroup$ That $d$ is not a metric for $p<1$. $\endgroup$
    – user99914
    May 11, 2015 at 1:14
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    $\begingroup$ The triangle inequality is the Minkowski inequality, isn't it? $\endgroup$
    – user99914
    May 11, 2015 at 1:22
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    $\begingroup$ Whether or not you need to prove Minkowski inequality depends on you (or the course you are taking......) $\endgroup$
    – user99914
    May 11, 2015 at 1:25
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    $\begingroup$ Ok then the metric part is done right ? How about the completeness part ? What are your toughts on that ? Where are you having difficulties ? $\endgroup$ May 11, 2015 at 1:43
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    $\begingroup$ OK then, let me try to provide a simple and clear proof of the fact that $\ell^p$ is a complete metric space. $\endgroup$ May 11, 2015 at 1:52

1 Answer 1

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Let $\left( x^{(n)}\right)_{n=1}^{\infty} \subset \ell^p$ be a Cauchy sequence. Since I see you have troubles with your notations of sequence of sequences, this is the notation that I will use for each element $x^{(n)}$ in the sequence: $$ x^{(n)} = \left( x_j^{(n)}\right)_{j=1}^{\infty} = \left( x_1^{(n)},x_2^{(n)}, \cdots \right)\in \ell^p $$

For $x= \left( x_j\right)_{j=1}^{\infty} , y= \left( y_j\right)_{j=1}^{\infty} \in \ell^p$, lets define the $p$-norm $\| \cdot \|_p$ as the one who induces $d_p$, that is $\|x-y\|_p=d_p(x,y)$. Precisely $$ \|x-y\|_p= \left(\sum_{j=1}^{\infty} \left|x_j-y_j\right|^p\right)^{1/p} $$

Now lets continue, take $\varepsilon>0$, then there exist a $N=N(\varepsilon) \in \mathbb{N}$, such that if $m,n >N$ then $$ \|x^{(m)}-x^{(n)}\|_p<\varepsilon. $$ Thus, for any $j \in \mathbb{N}$, it follows that $$ \left|x^{(m)}_j-x^{(n)}_j\right|^p \leq \sum_{j=1}^{\infty} \left|x^{(m)}_j-x^{(n)}_j\right|^p = \|x^{(m)}-x^{(n)}\|^p_p<\varepsilon^p $$ that is, for any $j \in \mathbb{N}$ the sequence $\left( x^{(n)}_j\right)_{n=1}^{\infty} \subset \mathbb{R}$ is a Cauchy one. Since $\mathbb{R}$ is complete, for each $j$ there exist a $x_j \in \mathbb{R}$ such that $$ \lim_{n \to \infty} x^{(n)}_j = x_j $$ Lets fix $k \in \mathbb{N}$, then in a similar way for $m,n >N$ \begin{equation} \sum_{j=1}^{k} \left|x^{(m)}_j-x^{(n)}_j\right|^p \leq \sum_{j=1}^{\infty} \left|x^{(m)}_j-x^{(n)}_j\right|^p = \|x^{(m)}-x^{(n)}\|^p_p<\varepsilon^p \tag{1} \end{equation} Letting $n \to \infty$ in (1), we get that for $m>N$ \begin{equation} \sum_{j=1}^{k}\left|x^{(m)}_j-x_j\right|^p < \varepsilon^p \tag{2} \end{equation} Then by the usual triangle inecuality ( Minkowski's inequality for $\|\cdot\|_p$ in $\mathbb{R}^k$) we get that if $m>N$ $$ \left( \sum_{j=1}^{k}|x_j|^p \right)^{1/p} \leq \left( \sum_{j=1}^{k}\left|x^{(m)}_j-x_j\right|^p \right)^{1/p} + \left( \sum_{j=1}^{k} \left|x^{(m)}_j \right| \right)^{1/p} < \varepsilon + \left( \sum_{j=1}^{k} \left|x^{(m)}_j \right| \right)^{1/p} $$ by letting $k \to \infty$, we get $\|x\|_p\leq \varepsilon + \|x^{(m)}\|_p$, which is the same as getting that $x=\left( x_j\right)_{j=1}^{\infty} \in \ell^p$. Again, letting $k \to \infty$ in (2), we obtain that if $m>N$ $$ \|x^{(m)}-x\|_p^p= \sum_{j=1}^{\infty}\left|x^{(m)}_j-x_j\right|^p < \varepsilon^p $$ thus $$ \lim_{m \to \infty} \|x^{(m)}-x\|_p= 0 $$ so indeed, $\left( x^{(m)}\right)_{m=1}^{\infty} \subset \ell^p$, is a convergent sequence who converges to $x \in \ell^p$. We conclude then that $\ell^p$ is a complete metric space for $1\leq p < \infty$.

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    $\begingroup$ Ahora entiendo mejor la manera de trabajar en pruebas sobre completitud de espacios métricos, gracias! *Now I can start proving myself the rest of exercices from this chapter, thank you! $\endgroup$
    – L F
    May 11, 2015 at 2:18
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    $\begingroup$ Hola @LuisFelipeVillavicencioLopez veo que hablas español. Acabo de editar mi respuesta para que quede muy clara la notación pues en tu pregunta vi que tienes ciertos problemas para identificar las sucesiones de elementos en $\ell^p$, espero lo entiendas y cualquier cosa que no entiendas me avisas y espero poderte ayudar. English translate: I have just edited my answer so you can understand the notation without problems, since I noted that you have some troubles denoting sequence of elements in $\ell^p$, I hope you understand it and let me know if I can help with anything else. $\endgroup$ May 11, 2015 at 2:25
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    $\begingroup$ Sí, gracias, ahora me queda más claro. Estaba perdido en estas cosas pero ya veo la luz translate: Yes, thank you, now is clear for me. I was lost in this theme but now I can see with clarity. $\endgroup$
    – L F
    May 11, 2015 at 2:28
  • $\begingroup$ @AlonsoDelfín Can you explain the part where you let $n$ tend to $\infty$? I am not able to prove it. $\endgroup$
    – Babai
    Sep 16, 2022 at 2:15
  • $\begingroup$ @Babai take the limit as $n \to \infty$ on both ends of equation (1). The LHS end does depend on $n$ and the limit enters through the finite sum and the continuous function $|-|^p$. Further, $x_j^{(n)}$ converges to $x_j$ and therefore the LHS on equation (1) converges to the LHS in equation (2) as $n \to \infty$. Finally, the RHS end is independent of $n$ and therefore is equal to $\varepsilon$ as $n \to \infty$, yielding equation (2). $\endgroup$ May 26, 2023 at 15:12

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