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We now have a sequence of functions \[ f_n = \frac{\sin{nx}\sin{x}}{x^2}, \] and we need to show that $\lVert f_n \rVert_1 \to \infty$ as $n \to \infty$, which means \[ \lim_{n \to \infty}\int_{-\infty}^{\infty}|f_n(x)|dx = 0. \]

How could we do that? It's easy to see that $\int_{-\infty}^{\infty}f_ndx=\pi$, but there is an absolute value function. I managed to compute \[ J_n = \int_{0}^{2\pi}|\sin{nx}\sin{x}|dx=\frac{2n}{n^2-1}\frac{\sin\frac{(n+1)\pi}{2n}}{\sin\frac{\pi}{2n}} \] and wanted to multiply it with $\sum_{j=1}^{\infty}\frac{1}{(2\pi j)^2}$, but the sequence $\{J_n\}$ converges (with respect to $n$), so if my calculation is not wrong, this estimation is too floopy.

For those who are interested about the source of this problem, you can find it on W. Rudin's Real and Complex Analysis, Chapter 9, exercise 2. It is the last step of proving the range of the Fourier transform is a proper subset of $C_0$. Define $g_n = \chi_{[-n,n]}$, we can show that $g_n \ast g_1$ is the Fourier transform of $f_n$. It remains to prove that $\lVert f_n \rVert_1 \to \infty$ as $n \to \infty$. And this is where I stuck.

Update: think I find the solution in another question: A proof of the fact that the Fourier transform is not surjective. So I'm OK now. But I think random people in the future may search this function so I think this title may help people. Still if you have good method it's a good idea to put it down.

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$$\int_{-\infty}^{\infty} |\frac {\sin (nx) \sin x}{ x^{2}}|dx$$ $$\geq \int_0^{1} |\frac {\sin (nx) \sin x}{ x^{2}}|dx$$ $$ \geq \frac 2 {\pi} \int_0^{1} |\frac {\sin (nx) }{ x}|dx $$ since $\sin x\geq \frac 2 {\pi} x$ for $0 <x<1$. Now put $y=nx $ and use the fact that $\int_0^{\infty} |\frac {\sin y} y| dy =\infty$.

Kavi Rama Murthy
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