Show divergence in $L_1$

Question

Denote $g_n$ by

$$ g_n(x) := \sin(2 \pi x) \sin(2 \pi n x) \frac{1}{x^2} \ . $$

How can we show that $\|g_n\|_{L_1(\mathbb{R})} \rightarrow \infty$, as n tends to $\infty$

Answer 1

Show that if $f$ is continuous on $[a,b]$, then as $n \to \infty$ $$ \int_a^b f(x) |\sin(2\pi n x| \, dx \to \frac2 \pi \int_a^b f(x) \, dx $$ I don't know if the result will be immediately applicable, but the ideas used in the proof will be.

$\frac2\pi$ comes from $\int_0^1 \sin(\pi x) \, dx$. Chop $[a,b]$ into lots of intervals $[k/2n,(k+1)/2n]$ and approximate (or in your case bound from below) the integral over each little interval.