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I need to show that $C[0,1]$ is not complete under the norm $||f|| = \int^{1}_{0}|f|$(usual Riemann Integral). For this I had the idea that I need to somehow make the limit discontinuous at atleast a point to get a contradiction. I considered the following candidate:

\begin{align*} f_n(x) = -nx+1, x\leq 1/n\\ =0, x>1/n \end{align*} Then we have($n>m$): \begin{align*} ||f_n-f_m|| = \int^1_{0} |f_n-f_m| = \int^{1/m}_{0}|f_n-f_m| \leq \frac{1}{m} \end{align*} Hence given $\epsilon>0$, we can find an $N$ so that $||f_n-f_m||<\epsilon, \forall n>m>N$. But if we take the limit: \begin{align*} \lim_{n\rightarrow \infty} f_n(x) &= 0, x>0\\ &= 1, x=0 \end{align*} which is clearly discontinuous.

Q1: Is the following example sufficient to conclude that the space is incomplete?

Q2: How bad can we make the example? Is it possible to get a limit function so discontinuous that the limit is not integrable in any neighbourhood?

Rohan Didmishe
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    This is not the right approach. In fact, this sequence converges in the $L^1$ norm to the zero function! – diracdeltafunk Dec 27 '22 at 07:14
  • @diracdeltafunk Could you please elaborate, I thought that the limit function should be inside the vector space ...? – Rohan Didmishe Dec 27 '22 at 07:16
  • "Convergence in $L_1$" is a different criterion from "Pointwise convergence". So you should not expect the results to be the same. You have found the pointwise limit, and claim it is the only $L_1$ limit. – Arthur Dec 27 '22 at 07:17
  • @Rohan First, a question that might help you think this through: what does "complete under the $L^1$ norm" mean? – diracdeltafunk Dec 27 '22 at 07:18
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    Oh ok, I understand, if $f_n$ converges with respect to $L_1$ to $f$ then $||f_n-f||\rightarrow 0$ as n\rightarrow \infty$, may not converge pointwise – Rohan Didmishe Dec 27 '22 at 07:20
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    That make sense, so in fact this sequence is convergent in $L_1$ to $0$...thanks a lot – Rohan Didmishe Dec 27 '22 at 07:21
  • Right! Still, you can try to make something like this work. The main problem with your sequence is that the pointwise limit is "almost continuous" -- it only needs to be modified at a single point to become continuous, and this doesn't change its integral. Instead, try to find a sequence whose pointwise limit is a relatively simple function, but discontinuous in a more serious way! – diracdeltafunk Dec 27 '22 at 07:22
  • The completion is $L^{1}[0,1]$ and this tells us how bad the limit can be. – Kavi Rama Murthy Dec 27 '22 at 07:29
  • See also https://math.stackexchange.com/questions/4564347/ and its linked posts – Anne Bauval Dec 27 '22 at 07:30
  • If you modify your example by shifting the discontinuity in the limit to the middle of the interval: $$ f(x)=\cases{1&for $\ 0\le x\le \frac{1}{2}$\ 1+n\left(\frac{1}{2}-x\right)&for $\ \frac{1}{2}<x<\frac{n+2}{2n}$\ 0&for $\ \frac{n+2}{2n}\le x\le1$} $$ that should give you a sequence which does what you're trying to achieve. – lonza leggiera Dec 27 '22 at 07:35
  • Thank you very much for your responses! – Rohan Didmishe Dec 27 '22 at 07:39
  • @Ionza Is the logic the following: Suppose there exists a limit function in $C[0,1]$. Then if it is continuous at every $x \geq 1/2$, and is nonzero, there must be a small neighbourhood where it is nonzero; then we can estimate $||f_n-f||$ and claim it cannot be arbitrarily small. Hence it must be $0$ for $x\geq 1/2$. On the other hand we can show that it must be $1$ in the left half by the same logic, and thus we cannot get a continuous limit with respect to $L^1$? – Rohan Didmishe Dec 27 '22 at 07:45
  • @Rohan Didmishe The argument is simpler than that. First I note that there's a typo in my previous comment—$\ f(x)\ $ should have been $\ f_n(x)\ $. With $\ f_n(x)\ $ thus defined and $$ f(x)=\cases{1&for $\ 0\le x\le\frac{1}{2}$\ 0&for $\ \frac{1}{2}<x\le1$} $$ you should be able to show that $\ \lim_\limits{n\rightarrow\infty}f_n=f\ $ with respect to the $\ L^1\ $ norm. – lonza leggiera Dec 27 '22 at 08:26
  • Since $\ f_n\in C[0,1]\ $ for all $\ n\ $ but $\ f\not\in C[0,1]\ $ this means that $\ f\in$$,\text{cl}_{L^1}(C[0,1])\setminus C[0,1]\ $, so $\ C[0,1]\ $ is not closed and hence not complete with respect to the $\ L^1\ $ norm. – lonza leggiera Dec 27 '22 at 08:26

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