Consider the sequence, $f_n(x)= \begin{cases} (2x)^n & 0 \leq x\leq \frac{1}{2} \\ 1 & \frac{1}{2} \leq x \leq 1\\ \end{cases}$
Then we need to show that $\{ f_n\}$ is Cauchy in the following metric: $d_p = ||f-g||_p= (\int_a^b |f(x) - g(x)|^pdx)^{\frac{1}{p}}$
My attempt:
Without loss of generality we assume that $m > n$. Thus we consider the integral part of $d_p$ and attempt to bound this by $\epsilon$.
$\int_{0}^{\frac{1}{2}} |(2x)^n-(2x)^m|^p dx$ ; at this point I am not sure where to go?
Observe that $f_n \to f = \begin{cases} 0 , & x \in (0,\frac{1}{2}) \\ 1 , & x \in (\frac{1}{2},1) \end{cases}$ in $L^p(0,1)$ because $$\|f_n -f\|^p_p= 2^{np} \int^{1/2}_0 x^{np}\,dx = 2^{np} \cdot \frac{(\frac{1}{2})^{np+1}}{np+1} = \frac{2^{-1}}{np+1} \to 0$$ Since the sequence is convergent, it must also be Cauchy.