Want to prove that the Hilbert transform of a $C^1(\mathbb T)$ function is the principal value of the convolution with $\cot(\pi x)$

Question

So here is my problem,

Let $L^2_0:=\{f\in L^2: \hat{f}(0)=0\}$ and consider the Hilbert transform given by the following map

$$H:L^2_0([0,1])\rightarrow L^2_0([0,1])$$ $$f\mapsto (\mathcal F^{-1}M\mathcal F)(f)$$ where $$\mathcal F:L^2([0,1])\rightarrow \ell^2(\mathbb Z,\mathbb C)$$ $$f\mapsto (\hat{f}(n))_{n\in\mathbb Z}$$ and $$M:\ell^2(\mathbb Z,\mathbb C)\rightarrow \ell^2(\mathbb Z,\mathbb C)$$ $$(x_n)_{n\in\mathbb Z}\mapsto -i\cdot sign(n)x_n$$

Now I would like to show that for $f\in C^1(\mathbb T)$ $$H(f)=p.v\int_{[0,1]}f(y)\cot(\pi(x-y))dy$$ But I have really no idea how to start. Could someone help me by giving me a hint or something like a recipe for a proof? Or does anybody know a link in which I can find the theory concerning my problem?

Thanks in advance!

Answer 1

From Harmonic Analysis by Henry Helson, Chapter 4: The Conjugate Function as a Singular Integral.

Theorem 16 Let $f$ be continuous and satisfy a uniform Lipschitz condition on $\mathbb{T}$. Then $\tilde{f}$ is continous, is the sum of its Fourier series at each point, and has the representation $$ \tilde{f}(e^{ix})=\lim_{\epsilon\downarrow 0}\int_{|t|>\epsilon}f(e^{i(x-t)})\cot\left(\frac{t}{2}\right)d\sigma(t). $$

The region of integration is assumed to be $(-\pi,\pi)$ with $(-\epsilon,\epsilon)$ excluded. Here, $d\sigma$ is normalized Lebesgue measure on $(-\pi,\pi)$ and $\tilde{f}$ is, effectively, the Hilbert transform. $\tilde{f}$ is also the conjugate function. Helson proves this theorem by looking at the conjugate function of the Dirichlet kernel for the Fourier Series.