On the conjugate Poisson integral and Hilbert transform of Fourier series Let $Q_r(x):=\dfrac{2r\sin x}{1-2r\cos x +r^2} , \forall r\in [0,1) , \forall x \in [-\pi,\pi]$ . Let $f\in L^2[-\pi,\pi]$ , and 
$(f*Q_r)(x):=\dfrac 1{2\pi}\int_{-\pi}^\pi Q_r(x-t)f(t)dt,\forall x \in[-\pi,\pi]$ . Then $(f*Q_r)(x)=\dfrac 1{2\pi}\int_{-\pi}^\pi\dfrac{2r\sin (x-t)}{1-2r\cos (x-t) +r^2}f(t)dt=\sum_{n=-\infty}^\infty -isgn(n)r^{|n|}\hat f (n) e^{inx},\forall x \in [-\pi,\pi]$  
My question is , how to show that : 
$||f*Q_r-H(f)||_{L_2} \to 0$ as $r \to 1$ , where $H(f)(x):=\sum_{n=-\infty}^\infty -isgn(n)\hat f (n) e^{inx} , \forall x \in [-\pi , \pi]$ ?
 A: Since convolution amounts to multiplying Fourier coefficients, it's natural to look at those. 
The Poisson kernel $P_r(x)$ has the Fourier series
$$P_r(x)
= \frac{1-r^2}{1-2r\cos x+r^2} = \sum_{n\in\mathbb{Z}} r^{|n|}e^{inx}
\tag1$$
Indeed, the second formula here can be obtained by summing the series on the right by splitting it into $n\ge 0$ and $n\le 0$, and using the geometric series formula. Similarly, the conjugate Poisson kernel has the Fourier series
$$Q_r(x) = \frac{2r\sin x}{1-2r\cos x +r^2} = -i \sum_{n\in\mathbb{Z}\setminus\{0\}} \operatorname{sgn}(n) r^{|n|}e^{inx}
\tag2$$
which can be verified in the same way, by splitting the sum as 
$$
\sum_{n=1}^\infty  r^{n}e^{inx} - \sum_{n=1}^\infty  r^{n}e^{-inx}
=\frac{re^{ix}}{1-re^{ix}} - \frac{re^{-ix}}{1-re^{-ix}}
= \frac{r (e^{ix}-e^{-ix})}{|1-re^{ix}|^2} = iQ_r(x)
$$
Actually, both of these come from the Cauchy kernel 
$$
\frac{1}{1-z} = \sum_{n=0}^\infty z^n
$$
by separating the real and imaginary parts, but that's beside the point. 
According to (2), you should show that for every $\ell^2$ sequence $(c_n)$, the product $(-i\operatorname{sgn}(n) r^{|n|}c_n)$ converges to $(-i\operatorname{sgn}(n) c_n)$ in $\ell^2$ norm as $r\to 1^-$. The argument is typical: for every $\epsilon$, there is $N$ such that $\sum_{|n|>N} |c_n|^2<\epsilon$ (so the tail doesn't matter), while taking the limit $r\to 1$ in the finite sum over $|n|\le N$ is straightforward.  
