Is the space $C[0,1]$ equipped with the 1-norm complete? Proof or disproof that : $(C[0,1],\|\cdot\|_1)$ is complete.
I think I can prove that $C[0,1]$ is complete when it is equipped with sup-norm. And are norms really equivalent here? If it is wrong, is there any counter example? I sorta can't tell right now.
 A: If $\|f\|_1:=\int |f(x)|\,dx$, then $C[0,1]$ is not complete.
Hint: To find a counterexample, you need a sequence $\{f_n\}$ of continuous functions that "get closer and closer to each other" (i.e., are Cauchy) in norm, yet their limit is not continuous. To imagine a sequence of functions that converge to a discontinuous limit, think of functions that are constant 0 in one region, constant 1 in another region, and connected by a line segment in between. If you define your $\{f_n\}$ carefully, the norm difference between $f_n$ and $f_m$ (which will be the area between the two function's curves) can be calculated easily using simple geometry.
Note: You'll also need to show that the $\|\cdot\|$ distance between $f_n$ and this discontinuous limit $f$ tends to zero, which should be straightforward. And this limit $f$ should not have its only discontinuity at zero, or at one either. (Why not?)
A: Here's a counterexample: Let $$f_n(x)=\begin{cases} 0 &\text{ if } x\in [0,\frac{1}{2}]\\
n(x-1/2) &\text{ if }x\in[\frac{1}{2}, \frac{1}{2}+\frac{1}{n}]\\
1 &\text{ if }x\in [\frac{1}{2}+\frac{1}{n},1]\end{cases}$$$f_n\in C[0,1]$ since it is piecewise continuous. Try drawing a graph of this function. As $n$ gets large, this is essentially becoming a step function at $x=\frac{1}{2}$. One can show this Cauchy in this norm (just choose $n,m$ and evaluate $||f_n-f_m||_1$. There's only a very small region you have to worry about, and in this case the value of the integral can be made small. You can show this algebraically, but this is also clear by looking at the graphs, since the area of intersection of $f_n,f_m$ will get very small).
You can show that $f_n$ actually converges to $\chi_{[\frac{1}{2},1]}$, which is not continuous, so $\lim_n f_n\not\in C[0,1]$.
Note: $L^1[0,1]$ is the completion of $C[0,1]$ with this norm. In this example, $\chi_{[\frac{1}{2},1]}\in L^1[0,1]$ obviously.
