$f: S^1 \to \mathbb{C}$ is a function with $f \in L^1(S^1)$, i.e. $$ \int^{2\pi}_0|f(e^{ix})|\,dx <\infty.$$ If $$ \int_{-\pi}^{\pi} \left|\frac{f(e^{ix})}{x}\right|dx <\infty,$$ then why is $g:S^1 \to \mathbb{C}$ defined by $$ g(e^{ix}) = \frac{f(e^{2ix})}{\sin(x)}$$ in $L^1(S^1)$?
Thanks.
Clearly $g$ is measurable.
Use the following inequalities:
The first one is seen geometrically by considering the line through the points $(0,0)$ and $(\pi/2,\sin(\pi/2))$.
The second one by considering the line through the points $(\pi/2,\sin(\pi/2))$ and $(\pi,\sin\pi)$.
We have \begin{align} \int_{-\pi}^{\pi} \left|\frac{f(e^{2ix})}{\sin x}\right|\,dx &=\int_{-\pi}^{-\pi/2} \left|\frac{f(e^{2ix})}{\sin x}\right|\,dx + \int_{-\pi/2}^{\pi/2} \left|\frac{f(e^{2ix})}{\sin x}\right|\,dx + \int_{\pi/2}^{\pi} \left|\frac{f(e^{2ix})}{\sin x}\right|\,dx \\ &=: I_1+I_2+I_3 \end{align} Let's show that $I_1,I_2,I_3<\infty$.
First, $I_1$ and $I_3$ are treated similarly so let's show $I_1<\infty$ : \begin{align} I_1 &\leq \int_{-\pi}^{-\pi/2} \frac{|f(e^{2ix})|}{-|2x|/\pi+2}\,dx\\ &=\frac{\pi}{2}\int_{-2\pi}^{-\pi}\frac{|f(e^{ix})|}{x+2\pi}\,dx\\ &=\frac{\pi}{2}\int_0^{\pi}\left|\frac{f(e^{ix})}{x}\right|\,dx\\ &\leq\frac{\pi}{2}\int_{-\pi}^{\pi}\left|\frac{f(e^{ix})}{x}\right|\,dx\\ &<\infty \end{align}
Finally, \begin{align} I_2 &\leq \pi \int_{-\pi/2}^{\pi/2}\left|\frac{f(e^{2ix})}{2x}\right|\,dx\\ &=\frac{\pi}{2}\int_{-\pi}^{\pi}\left|\frac{f(e^{ix})}{x}\right|\,dx\\ &<\infty \end{align}