Notes:Ell^p(C) is complete for p between one and positive infinity inclusive

Statement

$\forall p\in[1,+\infty]\subseteq\overline{\mathbb{R} }$ the space $\ell^p(\mathbb{C})$ is complete.

Proof

Leammas

Lemma 1:

• Let $(\mathbf{x}_n)_{n\in\mathbb{N} }\subseteq\ell^p(\mathbb{C})$ be given
  • Suppose $(\mathbf{x}_n)_{n\in\mathbb{N} }$ is a Cauchy sequence
    • Write them out as follows:

      $\begin{array}{ccccc} \mathbf{x}_1 & \eq & (x_1^m)_{m\in\mathbb{N} } & \eq & x_1^1,\ x_1^2,\ \ldots ,\ x_1^k,\ \ldots \\ \mathbf{x}_2 & \eq & (x_2^m)_{m\in\mathbb{N} } & \eq & x_2^1,\ x_2^2,\ \ldots,\ x_2^k,\ \ldots \\ \vdots \end{array}$

    • Now consider the $k^\text{th}$ column of this table, this gives us the sequence: $(x_m^k)_{m\in\mathbb{N} }$
      • We claim that $(x_m^k)_{m\in\mathbb{N} }$ is a Cauchy sequence itself

Lemma 2:

Let $(X,d)$ be a metric space, then $A\in\mathcal{P}(X)$ is bounded in $X$ if and only if:

• $\forall x\in X\exists C\in\mathbb{R}_{>0}\forall a\in A[d(a,x)\le C]$

Proof body

• Let $p\in[1,+\infty]\subseteq\overline{\mathbb{R} }$ be given.
  • Let $(\mathbf{x}_n)_{n\in\mathbb{N} }\subseteq\ell^p(\mathbb{C})$ be a Cauchy sequence in $\ell^p(\mathbb{C})$
    • By lemma 1, we obtain, for each $k\in\mathbb{N}$ the sequence: $(x_m^k)_{m\in\mathbb{N} }$ and we know this is Cauchy
    • As $\mathbb{C}$ is a complete metric space -
      TODO: link to this
      - we see that $(x_m^k)_m$ converges
      • For $k\in\mathbb{N}$ define: $$t_k:\eq\lim_{m\rightarrow\infty}(x^k_m)$$ which we have just established exists
        • Define $\mathbf{t}:\eq(t_1,\ldots,t_k,\ldots)$ - a sequence
          • We claim that:
            1. $\mathbf{t}\in\ell^p(\mathbb{C})$
            2. $$\lim_{n\rightarrow\infty}(\mathbf{x}_n)\eq\mathbf{t}$$
          • Proof:
            1. That $\mathbf{t}\in\ell^p(\mathbb{C})$
               • Let $\epsilon>0$ be given
                 • Since $(\mathbf{x}_n)_n$ is Cauchy there exists $N'\in\mathbb{N}$ such that $\forall n,m\in\mathbb{N}[n>m\ge N\implies d(\mathbf{x}_m,\mathbf{x}_n)<\epsilon]$ Caveat:Or equivalent definition of Cauchy sequence - I initially used $(m\ge N\wedge n\ge N)\implies d(\ldots$
                 • Warning:"Also $(\mathbf{x}_n)_n$ is bounded - that's all I've written.
                   • Thus, by lemma 2, $\exists R\in\mathbb{R}_{>0}\forall n\in\mathbb{N}[d(x_n,(0,\ldots,0,\ldots))<R]$
                     • We now show if $p\eq +\infty$ that $\mathbf{t}\in\ell^\infty(\mathbb{C})$
                       • In this case we see that $$d(\mathbf{x}_n,\mathbf{0})<R\iff\left[\mathop{\text{Sup} }_{i\in\mathbb{N} }\left(\vert x^i_n\vert\right)<R\right]$$
                         • Hence by definition of $\mathbf{t}$'s elements:
                           • $$\forall i\in\mathbb{N}\left[\vert t_i\vert\eq \lim_{n\rightarrow\infty}\left(\vert x_n^i\vert\right)\le R\right]$$
                             • "Thus $\mathbf{t}\in\ell^\infty(\mathbb{C})$" - all I've written.
            2. We now wish to show limit bit.
               • Notice that $d(\mathbf{x}_n,\mathbf{x}_m)<\epsilon\iff \big\vert x_n^i-x_m^i\vert<\epsilon$
                 • Fix $m\ge N$, then let $n\rightarrow\infty$ to obtain: $$\forall i\in\mathbb{N}[\vert t_i-x_m^i\vert\le\epsilon]$$
                   • Since $m\ge N$ was arbitrary we see: $$\forall m\in\mathbb{N}\left[m\ge N\implies \mathop{\text{Sup} }_{i\in\mathbb{N} }(\vert t_i-x_m^i\vert)\le\epsilon\right]$$
                     • Since $\epsilon>0$ was arbitrary it follows that $(\mathbf{x}_n)$ converges in $\ell^\infty(\mathbb{C})$ to $\mathbf{t} \in\ell^\infty(\mathbb{C})$