= Sine sequence is not Cauchy in the integral norm
{title2=$f_n(x)=\sin(nx)$}
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 $\int_0^1d_n^2\,dx=1-\sin(2n)/(4n)-\sin(4n)/(8n)-\sin n/n+\sin(3n)/(3n)$ tends to $1$. Since $|d_n|\leq2$, one has $\int_0^1|d_n|\,dx\geq\frac12\int_0^1d_n^2\,dx$, 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>.
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