Does $\{\sin (nx)\}_1^\infty$ converge in the $L^1$ norm on $[0,2\pi]$?

Question

This is a homework question from a problem set in an undergraduate-level real analysis course (coming from merely an intro to analysis course) about $L^p$ spaces.

Show that $\{\sin (nx)\}_{n=1}^\infty$ converges in the $L^1$ norm on $[0,2\pi]$

I showed that, for $f_n(x)=\sin(nx)$, the sequence of norms $\lVert f_n \rVert$ converges, but apparently I was supposed to show that $\lVert f-f_n\rVert\to0$, which I'm not really sure how to do. I'm probably missing something relatively simple, but I would appreciate the help.

Answer 1

No, the sequence doesn't converge in $L^1$. You can show that $\sin(nx)$ converge weakly to $f=0$ on $L^1(0,1)$. If $\sin(nx)$ converge strongly in $L^1(0,1)$, it must converge to $f=0$ since strong limit is weak limit. Then show $\|\sin(nx)\|_{L^1}$ does not converge to $0$.

Answer 2

What you want to show is false. By the Riemann Lebesgue Lemma, $$ \int_0^{2\pi} \sin(nx) g( x)\,dx \to 0 $$ as $n\to \infty$ for every $g \in C([0,2\pi])$. Thus, if $\sin(nx) \to f$ in $L^1$, then $\int_0^{2\pi} g( x)f(x)\,dx=0$ for all $g\in C([0,2\pi])$, which implies $f=0$ almost everywhere.

But it is not hard to show (you apparently did that already) that $\|\sin(nx)\|_{L^1}\not\to 0$.