# Convergence of $\frac{\sin(nx)}{1+\vert x\vert^2}$

I am trying to see why $\frac{\sin(nx)}{1+\vert x\vert^2}$ has no convergent subsequence in $L^1(\mathbb{R})$. This is an "optional" homework problem from my disorganized analysis professor which is worth no points, and which I wouldn't do anyway, because I'm terrible at analysis and have no clue how to deal with this.

To me it seems clear though that I want to be looking at the integral $\int\limits_\mathbb{R} \frac{\vert\sin(n_kx)-\sin(n_jx)\vert}{1+\vert(x)\vert^2}$ and finding some kind of constant, maybe in terms of when $n_k\neq n_j$ based on vague memories of undergrad PDE. Any hints?

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So, if everything is so bad - why do you care about this problem? –  AD. Dec 4 '10 at 5:39
Haha. I knew my comments might spark some kind of reaction. I care because this seems like a standard type analysis problem, though I'm not sure, and those sorts of problems are on the qualifiers...right? –  Jon Beardsley Dec 4 '10 at 17:14
This is homework-like so I'll just give some hints. Show that if ${\sin(n_kx) \over 1 + x^2}$ converges in $L^1$ to some $f(x)$, then $\int_R{\sin(n_kx) \over 1 + x^2} g(x)$ converges to $\int_R f(x)g(x)$ for any bounded $g(x)$. Show this contradicts the Riemann-Lebesgue lemma in some way. Note you have to consider the case where $f(x) = 0$ too.