Proving that $\sin(x)\sin(nx)/x^2$ has $L^1$ norm tending to infinity

Question

This is taken as a side question from Rudin's book on Real and Complex Analysis. I need to prove that

$$f_n(x)=\frac{\sin{(x)}\sin{(nx)}}{x^2}$$

has an $L^1$ norm that tends to infinity as $n\to\infty$. Unfortunately since $L^1$ norm means taking the absolute value of $\lvert\sin(x)\sin(nx)\rvert$ it got messy. My best evaluations so far used $$ \lvert\sin(x)\sin(nx)\rvert \geq \sin^2(x)\sin^2(nx) $$ And then I used a substitution $nx=u$ to get: $$ \int\limits_{-\infty}^{\infty} \frac{\lvert\sin(x)\sin(nx)\rvert}{x^2} \, \mathrm{d}x \geq n\int\limits_{-\infty}^{\infty} \frac{\sin^2(u)\sin^2\left(\frac{u}{n}\right)}{u^2} \, \mathrm{d}x $$ Even then I had some hope since writing $\sin^2\left(\frac{u}{n}\right) = -\frac{1}{4}e^{\frac{2iu}{n}}-\frac{1}{4}e^{-\frac{2iu}{n}}+\frac{1}{2}$ turns this integral into the Fourier transform of $\frac{\sin^2(u)}{u^2}$, but unfortunately I get that the bound is independent of $n$ so I can't show it tends to infinity.

I've given everything I can into this, I guess I'm missing something because the excercise seemed easy by the way it was put as a side note.

Answer 1

$\int |f_n(x)|dx=n\int \frac{|sin y| |\sin (\frac y n)| } {y^{2}}dy$ by the substitution $y=nx$. Hence $\int |f_n(x)|dx \geq n\int_{n-1}^{n+1} \frac{|\sin y| |sin (\frac y n)| } {y^{2}}dy$. Since $|sin (\frac y n)|$ is bounded below on $(n-1,n+1)$ it is enough to show that $n\int_{n-1}^{n+1} \frac{|\sin y| } {y^{2}}dy \to \infty$. But $n\int_{n-1}^{n+1} \frac{|\sin y| } {y^{2}} dy \geq\frac n {n+1}\int_{n-1}^{n+1} \frac{|\sin y| } {|y|}dy $ and it is well known that $\int_{\mathbb R} |\frac {\sin y} y| dy=\infty$.

Answer 2

Integrability of $f_n(x)$ is clear since $|f_n(x)|\leq n\mathbb{1}_{[-1,1]}(x)+\frac{1}{x^2}\mathbb{1}_{[-1,1]^c}(x)$. To estimate the limiting behavior of $f_n$ we may proceed as follows:

\begin{aligned} \int^\infty_{-\infty} |f_n(x)|dx &\stackrel{y=nx}{=} 2n\int^\infty_0\frac{|\sin y| |\sin (\frac y n)| } {y^{2}}dy = 2\int^\infty_0\Big|\frac{\sin y}{y}\Big|\Big|\frac{\sin(y/n)}{y/n}\Big|\,dy \\ &\geq 2\int^n_0\Big|\frac{\sin y}{y}\Big|\Big|\frac{\sin(y/n)}{y/n}\Big|\,dy\geq 2\sin(1)\int^n_0\Big|\frac{\sin x}{x}\Big|\,dx\xrightarrow{n\rightarrow\infty}\infty \end{aligned} Here we have used that $\sin(x)/x$ is decreasing (and positive) on $[0,\pi]$ and the well known fact that $\frac{\sin y}{y}\notin\mathcal{L}_1[0,\infty)$.