# [ilmath]C([0,1],\mathbb{R})[/ilmath] is not complete when considered with [ilmath]L^1[/ilmath] norm

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Demote once entire proof is here, even if not neat, just has to be readable Alec (talk) 12:01, 19 April 2017 (UTC)

## Statement

Define [ilmath](f_n)_{n\in\mathbb{N} }\subseteq C([0,1],\mathbb{R})[/ilmath] as follows:

• [ilmath]f_n:[0,1]\rightarrow\mathbb{R} [/ilmath] by [ilmath]f_n:x\mapsto\left\{\begin{array}{lr}1 & \text{if }x\in[0,\frac{1}{2}] \\ 1-n(x-\frac{1}{2}) & \text{if }x\in[\frac{1}{2},\frac{1}{2}+\frac{1}{n}] \\ 0 & \text{if }x\in[\frac{1}{2}+\frac{1}{n},1]\end{array}\right. [/ilmath]

We will show:

1. [ilmath](f_n)_n[/ilmath] is a Cauchy sequence with the metric: $d(f,g):\eq\Vert f-g\Vert:\eq\int^1_0\vert f(t)-g(t)\vert\mathrm{d}t$
2. If we assume [ilmath]\big(C([0,1],\mathbb{R}),d\big)[/ilmath] is a complete metric space then $\exists f\in C([0,1],\mathbb{R})\left[\mathop{\text{lim} }_{n\rightarrow\infty}\Big(d(f_n,f)\Big)\eq 0\right]$
3. We then show that such an [ilmath]f[/ilmath] cannot be continuous if it is indeed this limit

## Proof outline

We want:

• $\forall (f_n)_{n\in\mathbb{N} }\subseteq C([0,1],\mathbb{R})\exists f\in C([0,1],\mathbb{R})\left[\left(\mathop{\text{lim} }_{n\rightarrow\infty}\left(\Vert f_n-f\Vert_1:\eq\int^1_0\vert f_n(t)-f(t)\vert\ \mathrm{d}t\right)\eq 0\right)\wedge \forall g\in C([0,1],\mathbb{R})\left[\left(\right)\implies g\eq f\right]\right]$
• $\forall (f_n)_{n\in\mathbb{N} }\subseteq C([0,1],\mathbb{R})\Big[\underbrace{\forall\epsilon>0\exists N\in\mathbb{N}\forall n,m\in\mathbb{N}\big[\big((n>m)\wedge(m>N)\big)\implies d(f_n,f_m)<\epsilon\big]}_{(f_n)_n\text{ is a Cauchy sequence} }\iff\underbrace{\exists f\in C([0,1],\mathbb{R})\Big[\overbrace{\mathop{\text{lim} }_{n\rightarrow\infty}\Big( f_n\Big)\eq f}^{\text{in the sense of the metric} }\Big]}_{\text{there exists a point the sequence converges to} }\Big]$