$f_n(x)=\frac{nx}{1+nx}$ in $C[0,1]$ with usual sup-norm metric.

Question

I need to check whether it is Cauchy or not:

$f_n(x)=\frac{nx}{1+nx}$ in $C[0,1]$ with usual sup-norm metric.

As,I know the sequence is Cauchy if for any $\epsilon>0$ there is a no $N$ such that $d(f_m,f_n)<\epsilon$ for all $m,n>N$. So I was calculating $d(f_m,f_n)$ which I got is this: $$d(f_m,f_n)=(m-n)\sup_{x\in[0,1]}\frac{x}{(mx+1)(nx+1)}$$ Now I am not understanding how to deal with this.

Answer 1

Let $m=n^2$, then, as $n$ goes to infinity, we have that $$d(f_{n^2},f_n)=(n^2-n)\sup_{x\in[0,1]}\frac{x}{(n^2x+1)(nx+1)}\\ \stackrel{x=1/n}{\geq} (n^2-n)\frac{(1/n)}{(n^2(1/n)+1)(n(1/n)+1)} =\frac{n-1}{2(n+1)}\to \frac{1}{2}\not=0.$$ Hence $(f_n)_{n\geq 1}$ is not a Cauchy sequence (and the sequence is not (uniformly) convergent in $C[0,1]$).

P.S. Note that the pointwise limit is the function which is $1$ in $(0,1]$ and it is 0 at 0. It is not continuous in $[0,1]$.