Proving a sequence is Cauchy in metric

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?

• I assume $(a,b)$ should be $(0,1)$. Have you tried computing $\|f_n - f_m\|_p$? – Rhys Steele Apr 4 at 10:50
• @RhysSteele I have updated my attempt thus far, I get stuck at this point and I ma not sure where to go? – babynewton Apr 4 at 11:03
• Is it assumed that $p \geq 1$? – Jason Knapp Apr 4 at 13:21
• @JasonKnapp Yes – babynewton Apr 4 at 21:27

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.