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

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



Lemma 1:

  • Let [ilmath](\mathbf{x}_n)_{n\in\mathbb{N} }\subseteq\ell^p(\mathbb{C})[/ilmath] be given
    • Suppose [ilmath](\mathbf{x}_n)_{n\in\mathbb{N} } [/ilmath] is a Cauchy sequence
      • Write them out as follows:
        [ilmath]\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}[/ilmath]
      • Now consider the [ilmath]k^\text{th} [/ilmath] column of this table, this gives us the sequence: [ilmath](x_m^k)_{m\in\mathbb{N} } [/ilmath]
        • We claim that [ilmath](x_m^k)_{m\in\mathbb{N} } [/ilmath] is a Cauchy sequence itself

Lemma 2:

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

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

Proof body

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