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Problem: Show that $(C([a,b]),|{cdot}|_2)$ is not a complete normed vector space.

I have tried to proceed along the lines given in the accepted answer, and subsequent comment, to this question: Example of a non complete normed vector space. Although, I am running into difficulties and wondering if the example really generalises so well when we have a general $[a,b]$. Is the Cauchy sequence I have chosen below a valid one, or have I overlooked a much simpler example to use in this case?

Attempt: In order to show that $(C([a,b]),|{cdot}|_2)$ is not a complete normed vector space we need to show that there is some Cauchy sequence that does not converge to it's limit in $C([a,b])$. In particular, it will suffice to show that our chosen Cauchy sequence converges under $|cdot|_2$ to a discontinuous function.

Consider the following sequence of functions, $(f_n)_{n=1}^inftysubset C([a,b])$, given by,

$$f_n(x)=begin{cases}
0,,text{when},,xin[a,frac{b-a}2-frac{2^{-n}}{b-a}] \[2ex]
1,,text{when},,xin[frac{b-a}2+frac{2^{-n}}{b-a},b] \[2ex]
frac{(2x-b+a)(b-a)-2^{-n+1}}{2^{-n+2}},,text{when},,xin[frac{b-a}2-frac{2^{-n}}{b-a},frac{b-a}2+frac{2^{-n}}{b-a}]
end{cases}
$$

The definition of $f_n(x)$ when $xin[frac{b-a}2-frac{2^{-n}}{b-a},frac{b-a}2+frac{2^{-n}}{b-a}]$ is the linear interpolant over that particular subinterval of $[a,b]$.

I claim that this is a Cauchy sequence. We consider:

$$|f_n-f_m|_2^2=int_a^b|f_n(x)-f_m(x)|^2dx$$

$$=int_a^{frac{b-a}2-frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2dx+int_{frac{b-a}2-frac{2^{-n}}{b-a}}^{frac{b-a}2+frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2dx+int_{frac{b-a}2+frac{2^{-n}}{b-a}}^b|f_n(x)-f_m(x)|^2dx$$

$$=0+int_{frac{b-a}2-frac{2^{-n}}{b-a}}^{frac{b-a}2+frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2dx+0$$

The $0$'s for the first and last integrals because we have $f_n(x)-f_m(x)=0-0$ and $f_n(x)-f_m(x)=1-1$ on each respectively. As to the middle integral, here is what I've gotten so far:

$$int_{frac{b-a}2-frac{2^{-n}}{b-a}}^{frac{b-a}2+frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2dx=int_{frac{b-a}2-frac{2^{-n}}{b-a}}^{frac{b-a}2+frac{2^{-n}}{b-a}}|(2x-b-a)(b-a)(2^{n-2}-2^{m-2})|^2dx$$

Based in the example I am following, supposing that $mge n$, I think that I should be able to argue that the whole thing is bound above by $frac{2^{-n}}{b-a}$ which would mean that $|f_n-f_m|_2^2lefrac{2^{-n}}{b-a}to0$ as $ntoinfty$.

I don't quite see how I'm meant to bound it above by $frac{2^{-n}}{b-a}$, as I keep tripping up over applying that $mge n$, leading me to question whether the approach I have employed here is a fruitful one after all.

Use the mean value theorem:

$$int_a^b f(x),dx = (b-a)f(c), text{ for some } cin langle a , brangle$$

In particular, since your $f_n(x) in [0,1]$, we have:

begin{align}
int_{frac{b-a}2-frac{2^{-n}}{b-a}}^{frac{b-a}2+frac{2^{-n}}{b-a}}|f_n(x)-f_m(x)|^2,dx &= left(frac{b-a}2+frac{2^{-n}}{b-a} - frac{b-a}2+frac{2^{-n}}{b-a} right)underbrace{|f_n(c) - f_m(c)|^2}_{le 2^2 = 4}\
&le 8cdot frac{2^{-n}}{b-a} xrightarrow{ntoinfty} 0
end{align}

Therefore, your sequence is Cauchy.