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How does one show that $(C[0,1], ||.||_{2})$ where $||.||_2=(\int_{0}^{1}|f(t)|^2dt)^{1/2}$ and $C[0,1]$ is the space of all continous function which are mapped from $[0,1]\rightarrow \mathbb{R}$, is not complete?

First to show that $(C[0,1], ||.||_{2})$ is a normed vector space:

$||.||_2$ fulfills $||f(t)||=0 \Rightarrow f(t)=0$

$||.||_2$ fulfills the scalar multiplication property: $$||\lambda f(t)||_{2} = (\int_{0}^{1}|\lambda f(t)|^2dt)^{1/2}=|\lambda|(\int_{0}^{1}| f(t)|^2dt)^{1/2}=|\lambda|||f(t)||_2$$

Thirdly, it also fulfills the triangle inequality:$$||x(t)+y(t)||_2^2 = \int_{0}^{1}|x(t)+y(t)|^2dt \le \int_{0}^{1}|x(t)|^2dt + 2\int_{0}^{1}|x(t)y(t)|dt +\int_{0}^{1}|y(t)|^2dt \le ||x(t)||_2^2 + 2||x(t)||_2||y(t)||_2+||y(t)||_2^2 = (||x(t)||_2+||(y(t)||_2)^2$$

this also shows the closure under addition in $C[0,1]$

so , $(C[0,1],||.||_2)$ is a normed vector space , but how can we show that it is not complete?

Rudy the Reindeer
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bakabakabaka
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1 Answers1

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You can prove that a metric or normed space is not complete by finding a Cauchy sequence whose limit is not in the space.

For example consider the sequence $f_n$ where $f_n = 0$ on $[0, \frac12]$, $1$ on $[\frac12 + \frac1n, 1]$ and linear on $[ \frac12, \frac12 + \frac1n]$. Then $f_n$ is Cauchy with respect to $L^2$ (prove it) but its limit $f$ which is $0$ on $[0, \frac12]$ and $1$ on $[\frac12,1]$ is not in $C[0,1]$.

Rudy the Reindeer
  • 46,359