$\sin(nx)$ does not contain Cauchy subsequence in $L^p([0,2\pi]) $ for $1\leq p < \infty$

My attempt:

Set $f_n(x) = \sin(nx)$.

Argue by contradiction, suppose there exists a Cauchy subsequence $f_{n_k}$ in $L^p$ for some fixed $p$, then $f_{n_k}$ converges strongly to some $f \in L^p$, this $f$ must be the zero function since we know that $f_{n_k}$ converges weakly to zero in $L^p$.

But $f\equiv 0$ is impossible since $||f_{n_k}||_p$ is bounded below by a positive number. To see this, we look at the preimage $$\{{|f_{n_k}|}^p \geq \big( \frac{\sqrt 2}{2}\big)^p \} = \{{|f_{n_k}|}\geq \frac{\sqrt 2}{2} \} $$ the measure of this set is bounded below by $\pi$. By Chebyshev's Inequality ${||f_{n_k}||_p}^p$ is bounded below by $\pi\big( \frac{\sqrt 2}{2}\big)^p $.

Is this okay? thank you very much!

  • 1
    $\begingroup$ Your argument is true, but need some more detail for "$f_{n_k}$ should converge weakly to $0$". $\endgroup$ – Golbez Aug 27 '14 at 6:27

Your argument is fine. You can also do a direct variant of it like this. If $\{f_{n_k}\}_{k=1}^{\infty}$ were a Cauchy sequence, then $\lim_{k \rightarrow 0} ||f_{n_{k+1}} - f_{n_k}||_{L^p[0,2\pi]}$ would be zero. But by Holder's inequality $$\int_0^{2\pi}|f_{n_{k+1}} - f_{n_k}|^2 \leq||f_{n_{k+1}} - f_{n_k}||_{L^p[0,2\pi]}\,\,\,\,||f_{n_{k+1}} - f_{n_k}||_{L^{p'}[0,2\pi]}$$ But $||f_{n_{k+1}} - f_{n_k}||_{L^{p'}[0,2\pi]}$ is bounded by a fixed constant $C$ just by taking absolute values of the integrand and integrating. So you have $$\int_0^{2\pi}|f_{n_{k+1}} - f_{n_k}|^2 \leq C||f_{n_{k+1}} - f_{n_k}||_{L^p[0,2\pi]}$$ Taking limits as $k \rightarrow \infty$ gives zero on the right hand side, but you can directly compute the left-hand integral to be $2\pi$ for all $k$, giving a contradiction.


The given arguments are good. I agree with Golbez, the fact that if $(f_{n_k})_k$ converge weakly to something, then it converges to $0$ has to be detailed, for example noticing that $\int_{[0,2\pi]} f_{n_k}g\mathrm d\lambda\to 0$ for each continuous function $g$ and concluding by a density argument.


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