Difference between revisions of "C( )0,1( ,R) is not complete when considered with the L^1 norm"
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# If we assume {{M|\big(C([0,1],\mathbb{R}),d\big)}} is a [[complete metric space]] then {{MM|\exists f\in C([0,1],\mathbb{R})\left[\mathop{\text{lim} }_{n\rightarrow\infty}\Big(d(f_n,f)\Big)\eq 0\right] }} | # If we assume {{M|\big(C([0,1],\mathbb{R}),d\big)}} is a [[complete metric space]] then {{MM|\exists f\in C([0,1],\mathbb{R})\left[\mathop{\text{lim} }_{n\rightarrow\infty}\Big(d(f_n,f)\Big)\eq 0\right] }} | ||
# We then show that such an {{M|f}} cannot be continuous if it is indeed this limit | # We then show that such an {{M|f}} cannot be continuous if it is indeed this limit | ||
+ | ==Proof outline== | ||
+ | We want: | ||
+ | * {{MM|\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]}} | ||
+ | * {{MM|\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]}} | ||
==References== | ==References== | ||
<references/> | <references/> | ||
{{Theorem Of|Functional Analysis|Analysis}} | {{Theorem Of|Functional Analysis|Analysis}} |
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Contents
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:
- [ilmath](f_n)_n[/ilmath] is a Cauchy sequence with the metric: [math]d(f,g):\eq\Vert f-g\Vert:\eq\int^1_0\vert f(t)-g(t)\vert\mathrm{d}t[/math]
- If we assume [ilmath]\big(C([0,1],\mathbb{R}),d\big)[/ilmath] is a complete metric space then [math]\exists f\in C([0,1],\mathbb{R})\left[\mathop{\text{lim} }_{n\rightarrow\infty}\Big(d(f_n,f)\Big)\eq 0\right] [/math]
- We then show that such an [ilmath]f[/ilmath] cannot be continuous if it is indeed this limit
Proof outline
We want:
- [math]\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][/math]
- [math]\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][/math]
References