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I am trying to prove that the Euclidean Norm/inner product on $C([0,1])$ does not give rise to a complete metric space. To do this I am trying to find a Cauchy Sequence which does not converge in $C([0,1])$. As a template (on $C([a,b])$) I am led to believe the following is Cauchy:

$$f_n (x) = \begin{cases} 1 &\mbox{if } 0 \leq x \leq \frac{1}{2} \\ 1 - 2n(x-\frac{1}{2}) & \mbox{if } \frac{1}{2}\leq x \leq \frac{1}{2n} + \frac{1}{2}\\ 0 & \mbox{if } \frac{1}{2n} + \frac{1}{2} \leq x \leq 1 \end{cases} $$

But I am struggling to show this is Cauchy, I have tried integrating from 0 to 1 but this is giving a very nasty integral and I was wondering if anyone has a better method?

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  • $\begingroup$ what is the Euclidean Norm/inner product on $C([0,1])$? You mean the $L^1$-norm? $\endgroup$ – Emanuele Paolini Mar 8 '14 at 9:18
  • $\begingroup$ Hint: Let say $m>n$, then $f_m$ and $f_n$ differs on a small set (What is that set?) $\endgroup$ – user99914 Mar 8 '14 at 9:22
  • $\begingroup$ Sorry (mod 2) came from the latex template I was using. And I mean $L^2$, is this not the Euclidean norm? I will edit the above if so thank you $\endgroup$ – user132548 Mar 8 '14 at 9:23
  • $\begingroup$ You can make things easy and approximate $\Vert f_n- f_m\Vert_2^2\le 4\cdot |1/(2n)-1/(2m)|$. (Note $f_n-f_m$ is $0$ of the interval with endpoints $1/(2n)+1/2$ and $1/(2m)+1/2$.) $\endgroup$ – David Mitra Mar 8 '14 at 10:17
  • $\begingroup$ Great thanks David got it! $\endgroup$ – user132548 Mar 8 '14 at 18:24
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You can make things easy and approximate $$\Vert f_n- f_m\Vert_2^2\le 4\cdot |1/(2n)-1/(2m)|$$ Note $f_n-f_m$ is $0$ of the interval with endpoints $1/(2n)+1/2$ and $1/(2m)+1/2$.

-- David Mitra

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