# Proving $\ell^p$ is complete

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.

• That $d$ is not a metric for $p<1$. – user99914 May 11 '15 at 1:14
• The triangle inequality is the Minkowski inequality, isn't it? – user99914 May 11 '15 at 1:22
• Whether or not you need to prove Minkowski inequality depends on you (or the course you are taking......) – user99914 May 11 '15 at 1:25
• Ok then the metric part is done right ? How about the completeness part ? What are your toughts on that ? Where are you having difficulties ? – Alonso Delfín May 11 '15 at 1:43
• OK then, let me try to provide a simple and clear proof of the fact that $\ell^p$ is a complete metric space. – Alonso Delfín May 11 '15 at 1:52

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 $m,n >N$ $$\sum_{j=1}^{k} \left|x^{(m)}_j-x^{(n)}_j\right|^p \leq \sum_{j=1}^{\infty} \left|x^{(m)}_j-x^{(n)}_j\right|^p = \|x^{(m)}-x^{(n)}\|^p_p<\varepsilon^p \tag{1}$$ Letting $n \to \infty$ in (1), we get that for $m>N$ $$\sum_{j=1}^{k}\left|x^{(m)}_j-x_j\right|^p < \varepsilon^p \tag{2}$$ 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 $1\leq p < \infty$.
• 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. – Alonso Delfín May 11 '15 at 2:25