$f_n(x)=1-nx $ if $x\in [0,\frac{1}{n}],=0$ if $x\in [\frac{1}{n},1]$ in $C[0,1]$ equipped with $d(f,g):=\int_0^1|f(t)-g(t)|dt$

Question

I am dealing with the sequence,

$$f_n(x)=\begin{cases} 1-nx &\text{ if } x\in [0,\frac{1}{n}]\\ 0 &\text{ if } x\in [\frac{1}{n},1] \end{cases}, $$ in $C[0,1]$ equipped with $d(f,g):=\int_0^1|f(t)-g(t)|dt$

I need to verify it is Cauchy sequence and also convergent or not?

I want to show it is uniformly convergent. When I am trying to find its pointwise limit I can not understand its nature close to $0$.As, $n\to \infty$, $x\to 0$ then where $nx$ is going? If I try to show it cauchy from direct definition $d(f_m,f_n)=\int_0^1(n-m)xdx=(n-m)/2$ how it will go to $0$ as $n,m \to \infty$

Answer 1

If $m > n$, then as $m,n \to \infty$,

$$d(f_n,f_m) = \int_0^{1/m}|m-n|x \,dx + \int_{1/m}^{1/n}|1 - nx| \, dx \\ \leqslant \frac{m}{2m^2} + \left(\frac{1}{n} - \frac{1}{m}\right) \to 0$$

Answer 2

Notice that $\int_0^1|f_n-0|=\frac{1}{2n} \rightarrow 0$. Thus $f_n \rightarrow 0$. Convergent sequences are always Cauchy. Can you prove this?