Test for the completeness of the followng normed vector space: $(C^0([0,1],\mathbb{R}),N_1)$ where $N_1(f)=\int_{0}^{1}|f(t)|\mathrm dt $
I know that a metric space M is complete iff every Cauchy sequence in M has a limit in this space. But how can I prove this, for any arbitrary Cauchy sequence?
You can't, because it is not true. If, for each natural $n$, if you define$$\begin{array}{rccc}f_n\colon&[0,1]&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}1&\text{ if }x>\frac12+\frac1{2n}\\2n\left(x-\frac12\right)&\text{ if }x\in\left[\frac12-\frac1{2n},\frac12+\frac1{2n}\right]\\-1&\text{ if }x<\frac12-\frac1{2n},\end{cases}\end{array}$$then $(f_n)_{n\in\mathbb N}$ is a Cauchy sequence which does not converge in $\mathcal{C}^0\bigl([0,1]\bigr)$. It could only converge to a function $f\in\mathcal{C}^0\bigl([0,1]\bigr)$ which would be equl to $-1$ when $x<\frac12$ and equal to $1$ when $x>\frac12$, but no such continuous function exists.