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Show that $(C[0,1],\|\cdot\|_2)$ is not complete. For that consider the sequence $(g_n)_{n\in \mathbb{N}}$ defined below and show that this is not a convergent Cauchy-sequence.

$g_n:[0,1]\to \mathbb{R},~g_n(t):=\begin{cases}f_n(t)~~~~~~~ t\in [0,\frac{1}{2}]\\1~~~~~~~~~~~~otherwise \end{cases}$

$f_n:[0,1]\to \mathbb{R},~f_n(t):=\begin{cases}0, &for& 0\leq t<\frac{1}{2}-\frac{1}{n}\\ n[t-(\frac{1}{2}-\frac{1}{n})], &for& \frac{1}{2}-\frac{1}{n}\leq t<\frac{1}{2}\\-n[t-(\frac{1}{2}+\frac{1}{n})], &for& \frac{1}{2}\leq t<\frac{1}{2}+\frac{1}{n} \\ 0, &for& \frac{1}{2}+\frac{1}{n}\leq t\leq 1\end{cases}$

https://i.stack.imgur.com/ew2Pg.jpg

Several hours ago I found this question while browsing through SE and it piqued my interest since $f_n$ looked kind of cool.

Anyway, someone got some hints? Or maybe the original poster found a solution and just deleted his post. If so I hope he sees this and shares his solution because I hate to go to bed with stuff on my mind.

Nicole Parson
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1 Answers1

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Assume $n <m $. Then $$ \|g_m-g_n\|_2^2=\int_{1/2-1/m}^{1/2}|g_m (t)-g_n (t)|^2\,dt\leq\frac4m, $$ so the sequence is $\|\cdot\|_2$-Cauchy. Now show that the limit is the characteristic function of $[1/2,1] $, which is not in $C [0,1] $: $$ \|g_n-1_{[1/2,1]}\|_2^2=\int_{1/2-1/n}^{1/2}n^2\,\left (t-\left (\frac12-\frac1n\right)\right)^2\,dt=\frac {n^2}3\,\frac1 {n^3}=\frac1 {3n}. $$

Martin Argerami
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