How do I prove completeness / incompleteness on a set of functions? I'm asked to show that 
$$((C[0,1], \mathbb{R}), ||f||_1), \quad ||f||_1 = \int_0^1{|f(x)|dx}$$ 
is not a Banach space. 
$||f||_1$ is a norm on this space so therefore I need to show that the space is not complete. 
How do I prove completeness / incompleteness on a space of functions? How do I even construct a cauchy sequence of functions? What does it mean for a sequence of functions to converge? Is completeness defined in another way?
 A: You can, for instance, define$$\begin{array}{rccc}f_n\colon&[0,1]&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}0&\text{ if }x<\frac12-\frac1{2n}\\2n\left(x-\frac12\right)&\text{ if }x\in\left[\frac12-\frac1{2n},\frac12+\frac1{2n}\right]\\1&\text{ if }x>\frac12+\frac1{2n}.\end{cases}\end{array}$$It's graph is like this:

Now:


*

*Prove that this is a Cauchy sequence.

*Prove that it doesn't converge in your space.

A: Given a norm $||\cdot||$, you can define a metric $d$ on the space as $d(x,y) := ||x-y||$. 
Here, the norm is given by $||f||_{1} = \int_0^1 |f(x)| dx$. Thus, the metric is $d(f,g) = \int_0^1 |f(x)-g(x)| dx$. 
A sequence of functions $\{f_n\}$ is said to be cauchy, if for every $\epsilon>0$, there exists $N \in \mathbb{N}$ such that $d(f_n,f_m) = \int_0^1 |f_n(x) - f_m(x)| < \epsilon$ for all $n,m \geq N$.
A sequence of functions $\{f_n\}$ is said to converge to $f$, if for every
$\epsilon>0$, there exists $N \in \mathbb{N}$ such that $d(f_n,f) = \int_0^1 |f_n(x) - f(x)| < \epsilon$ for all $n \geq N$.
$(C[0,1], \mathbb{R}), || \cdot||_1)$ is not Banach by the example in the answer above. 
