Let $C([0,1])$ be the set of all continuous real valued functions on $I=[0,1]$, with the norm $$||u|| _{L_1 ([0,1])} = \int ^1 _0 |u(x)|dx$$ for any $u \in C([0,1])$.
Consider the sequence of functions $\{u_n\}$ defined by $$u_1(x) = 1$$ for all $x \in [0,1]$
$$u_n(x)=1$$ for $0 \leq x \leq \frac{1}{2}$
$$u_n(x)= 1-(x-\frac{1}{2})n$$ for $\frac{1}{2} < x <\frac{1}{2} +\frac{1}{n}$
$$u_n(x)=0$$for $\frac{1}{2} + \frac{1}{n} \leq x \leq 1$
The Question
Show that if $C([0,1])$ is equipped with the norm (as above) then $\{u_n\}$ is a cauchy sequence in $C([0,1])$
This is the proof
Let $2 \leq n \leq m$
Why is $2 \leq n \leq m$?
$||u_n - u_m||_{L_1 ([0,1])}= \int ^1 _ 0 |u_m (x) - u_n(x)| dx$
$\leq\int^{\frac{1}{2}+\frac{1}{n}}_{\frac{1}{2}} |u_m(x) - u_n(x)|dx \leq\frac{1}{2n}$ $\to 0$
As $n \to \infty$
Why do the integrals become $\leq$ when the integral limits change?
Also where does $\frac{1}{2n}$ come from?
Why do we not consider the integral $\int^\frac{1}{2} _0$?
That is because on $[0,\frac12]$ the functions are both equal to $1$, hence the integral of the difference on that interval is $0$. Similarly, if $n\le m$, the functions are equal to $0$ on $[\frac12+\frac1n,1]$.
The inequality of integrals, when the bounds change, is actually an equality.
Finally $\frac1{2n}$ comes from the fact that $u_m(x)\le u_n(x)$, hence $$\int_\frac12^{\frac12+\frac1n}\lvert u_n(x)-u_m(x)\rvert\,\mathrm d\mkern1mu x\le=\int_\frac12^{\frac12+\frac1n}\lvert u_n(x)\rvert\,\mathrm d\mkern1mu x =\int_\frac12^{\frac12+\frac1n} u_n(x)\,\mathrm d\mkern1mu x=\frac1{2n}$$ (it is the area of a right-angled triangle with legs of length $1$ and $\frac1n$).