The [ilmath]\ell^p(\mathbb{C})[/ilmath] spaces are complete
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Contents
Statement
Let [ilmath]p\in[1,+\infty]\subseteq[/ilmath][ilmath]\overline{\mathbb{R} } [/ilmath] be given and consider [ilmath]\ell^p(\mathbb{C})[/ilmath] then we claim[1]:
- For [ilmath]p\in\mathbb{R}_{\ge 1} [/ilmath] that [math]\ell^p(\mathbb{C}):\eq\left\{(x_n)_{n\in\mathbb{N} }\subseteq\mathbb{C}\ \left\vert\ \sum^\infty_{n\eq 1}\vert x_n\vert^p<+\infty\right\}\right. [/math] is a complete metric space with respect to the metric induced by the norm: [math]\Vert(x_n)_{n\in\mathbb{N} }\Vert_p:\eq\left(\sum^\infty_{n\eq 1}\vert x_n\vert^p\right)^\frac{1}{p} [/math]
- For [ilmath]p\eq+\infty[/ilmath][Note 1] that [math]\ell^{\infty}(\mathbb{C}):\eq\left\{(x_n)_{n\in\mathbb{N} }\subseteq\mathbb{C}\ \left\vert\ \mathop{\text{Sup} }_{n\in\mathbb{N} }(\vert x_n\vert)<+\infty \right\}\right. [/math] is a complete metric space with respect to the metric induced by the norm: [math]\Vert(x_n)_{n\in\mathbb{N} }\Vert_\infty:\eq\mathop{\text{Sup} }_{n\in\mathbb{N} }(\vert x_n\vert)[/math]
Proof of claims
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Notes
- ↑ Herein just [ilmath]\infty[/ilmath] as only one is in the relevant range of [ilmath]p[/ilmath]
References