Completeness of continuous functions space under $L^{1}$ norm Prove that the space of continuous functions $C([0,1])$ is not complete under the norm
$\|\cdot\|_{L^{1}([0,1])}$.
I have tried the following sequence of functions
$f_{n}(x)=\left\{\begin{array}{cl}1 & \text { if } & 0 \leq x \leqslant \frac{1}{2} \\ 1-n\left(x-\frac{1}{2}\right) & \text { if } \frac{1}{2} \leqslant x \leqslant \frac{1}{2}+\frac{1}{n} \\ 0, & \text { if } \frac{1}{2}+\frac{1}{n} \leqslant x \leqslant 1\end{array}\right.$
I got
$\int_{\frac{1}{2}}^{\frac{1}{2}+\frac{1}{n}}\left|f_{n}(x)-f_{m}(x)\right| d x .$
$=\int_{\frac{1}{2}}^{\frac{1}{2}+\frac{1}{n}}\left|\left(1-n\left(x-\frac{1}{2}\right)\right)-\left(1-m\left(x-\frac{1}{2}\right)\right)\right| d x=\left|\frac{m-n}{2 n^{2}}\right|$
Is the last term bounded by $\frac{1}{n}$? why? and is it ok that my integration limits deals with $n$ and not also $m$?
As last step I thought of taking limit of n to $\infty$ and since 2 limits are obtained - $0,1$ this space is not complete.
 A: Say $m>n$. Then
$$\int_0^1 |f_n-f_m|=\int_{\frac{1}{2}}^{\frac{1}{2}+\frac{1}{m}}|f_n-f_m|+\int_{\frac{1}{2}+\frac{1}{m}}^{\frac{1}{2}+\frac{1}{n}} |f_n-f_m|$$
You've already dealt with the first integral on the RHS. To deal with the second one, note that
$$ \int_{\frac{1}{2}+\frac{1}{m}}^{\frac{1}{2}+\frac{1}{n}} |f_n-f_m|=\int_{\frac{1}{2}+\frac{1}{m}}^{\frac{1}{2}+\frac{1}{n}} |f_n|\le \frac{1}{n}-\frac{1}{m}$$
because $|f_m(x)|=0$ whenever $\frac{1}{2}+\frac{1}{m}\le x\le \frac{1}{2}+\frac{1}{n}$ and $|f_n|\le 1$ everywhere.
This shows that $\{ f_n\} _{n=1}^\infty$ is Cauchy in $(C([0,1]),\|\cdot\|_1)$.
Suppose, towards a contradiction, that $f_n\to f$ in the $L^1$ norm for some $f\in C([0,1])$. Let $g$ be the characteristic function of $[0,\frac{1}{2}]$. You can check that $f_n\to g$ in the space $(L^1,\|\cdot\|_1)$, so that $f=g$ a.e. But this is impossible (do you see why?).
A: Assuming $m<n$, we first show $(f_n)$ is cauchy's sequence.
$\|f_n-f_m\|= \int_0^1 |f_n-f_m|dx$
RHS is equal to
$\int_{\frac{1}{2}}^{\frac{1}{2}+\frac{1}{n}}|m(x-\frac{1}{2})-n(x-\frac{1}{2})|dx+$
$\int_{\frac{1}{2}+\frac{1}{n}}^{\frac{1}{2}+\frac{1}{m}}|1-m(x-\frac{1}{2})|$
Try to simplify, if needed you can use $\frac{n-m}{2n^2} \leq \frac{n+m}{2n^2} <\frac{n+n}{2n^2}=\frac{1}{n}$.
Once you prove it is cauchy's, you can easily show it does not coverge in given space.
A: With your idea, you get a sequence $f_n$ on $[0,1]$ such that on $[0,\frac{1}{2}]$ it converges (in the $L^1$ norm) to a function $g$ ( constant $1$), on $[\frac{1}{2},1]$ it converges to a function $h$ ( constant $0$). Note that the $L^1$ norm on $[0,1]$ is the sum of the $L^1$ norms on $[0,\frac{1}{2}]$ and $[\frac{1}{2},1]$. So the sequence $f_n$ will be Cauchy. However, $f_n$ does not converge to a continuous function $f$. Otherwise, $f$ would equal $g$ on $[0,\frac{1}{2}]$, and $h$ on $[\frac{1}{2},1]$, not possible.
A: If $m,n \ge N$ then $f_n(x) = f_m(x)$ if $x \notin [\frac12,\frac12+{1 \over N}]$ and so $\|f_m-f_n\|_1 \le {1 \over N}$. It follows that $f_n$ is Cauchy. If we let $f=1_{[1,\frac12]}$ we see that $f_n(x) \to f(x)$ pointwise. In a similar manner, it is straightforward to see that $\|f_n-f\|_1 \le {1 \over n}$.
To finish we need to show that the equivalence class $[f]$ contains no continuous function. Suppose $g \sim f$, then there are $x_n \uparrow \frac12, y_n \downarrow \frac12$ such that $g(x_n)=f(x_n) = 1$ and $g(y_n) = f(y_n) = 0$. In particular, $g$ is not continuous.
