 Sine sequence is not Cauchy in the integral norm

On $[0,1]$, the continuous functions $f_n(x)=\sin (nx)$ do not form a Cauchy sequence in the Lp space with $p=1$. Put $d_n=f_n-f_{2n}$. The identity ${\int }_0^1{d}_n^{2}dx=1-\sin (2n)/(4n)-\sin (4n)/(8n)-\sin n/n+\sin (3n)/(3n)$ tends to $1$. Since $|d_n|\le 2$, one has ${\int }_0^1|d_n|dx\ge \frac{1}{2}{\int }_0^1{d}_n^{2}dx$, which stays bounded away from zero. Thus arbitrarily late pairs are separated in the integral norm; in particular the sequence cannot converge in that norm.