Let
$$f_n(x)=\begin{cases} \left(x+\frac{1}{2}\right)^n,&\text{if }0\le x<\frac{1}{2}\\\\ 1,&\text{if } \frac{1}{2}\le x\le 1\;. \end{cases}$$
Then clearly $(f_n)$ is sequence in $C[0,1]$. I need to show this is cauchy sequence in $C[0,1]$ with $L^1$ norm.
My attempt By definition of cauchy sequence, i think we need establish
$||f_n-f_m||<\epsilon$ for sufficiently large $m, n$
i.e. $\int^{1}_0 |f_n(x)-f_m(x)|dx<\epsilon$ for sufficiently large $m, n$
But,
$\int^{1}_0 |f_n(x)-f_m(x)|dx\\ =\int^{1/2}_0 |(x+1/2)^n-(x+1/2)^m|dx$
(Since on $[1/2,1]$ both $f_n$, $f_m$ are equal for all $m, n$ )
In next step i had used traingle inequality and spit the integral into two parts and obtained,
$\int^{1}_0 |f_n(x)-f_m(x)|dx\\ ≤\int^{1/2}_0 |(x+1/2)^n| dx+\int^{1/2}_0 |(x+1/2)^m| dx$\ $=\frac{1}{n+1}[1-(1/2)^{n+1}]+\frac{1}{m+1}[1-(1/2)^{m+1}]$
But i am unable to obtain conclusion
Further, i like to know is there is any direct approach to conclude it is Cauchy sequence?
Let $\varepsilon > 0$ be fixed. Note that \begin{eqnarray*} \int_{0}^{1/2} \left| (x + 1/2)^n - (x + 1/2)^m \right| \, dx & \leq & \int_{0}^{1/2} \left| (x + 1/2)^n \right| \, dx + \int_{0}^{1/2} \left| (x + 1/2)^n \right| \, dx \\ & = & \left[\frac{1}{n+1} - \frac{(1/2)^{n+1}}{n + 1} \right] + \left[ \frac{1}{n+1} - \frac{(1/2)^{m+1}}{m + 1} \right]\\ & < & \frac{1}{n} + \frac{1}{m}. \end{eqnarray*} From here, can you determine what constraints you must put on $n$ (resp. $m$) so tha $\dfrac{1}{n} < \dfrac{\varepsilon}{2}$?