Consider the operator $C$, acting on functions $f$ on the unit circle $S^1 = \left\{ z \in \mathbb C \mid |z| = 1 \right\}$ by the rule $$ (Cf)(z) = \frac{1}{2\pi i} \int\limits_{|\zeta|=1}\frac{f(\zeta)}{\zeta-z(1+0)}d\zeta \equiv \lim\limits_{\epsilon \to + 0}\frac{1}{2\pi i}\int\limits_{|\zeta|=1}\frac{f(\zeta)}{\zeta-z(1+\epsilon)}d\zeta, \quad |z| = 1. $$ The task is to show that $C$ is a linear continuous operator in $L_p(S^1)$ for any $p \in (1,\infty)$.
I have no idea how to deal here with arbitrary $L_p$-functions, but for functions $f_c$, holomorphic in some neigborhood of $S^1$ with the Laurent expansion $$ f_c(z) = \sum\limits_{n=-\infty}^{+\infty} c_n z^n $$ in this neighborhood the operator $C$ just takes the main part of the Laurent expansion with the sign "$-$": $$ (Cf_c)(z)=-\sum\limits_{n=-\infty}^{n=-1} c_n z^n, $$ so $C$ is a linear bounded operator in the space of functions, holomorphic in some neighborhood of $S^1$ (fixed). But why this is also true for the space $L_p$, $p \in (1,\infty)$? I think that the question may be reformulated: if we consider the space of functions, holomorphic in some fixed neighborhood of the unit circle $S^1$ and the operator taking the main part of the Laurent expansion of these functions then this operator may be continuously continued to the space $L_p(S^1)$, $p \in (1,\infty)$.
Consider $p=2$ first. Up to a constant factor, the Hilbert transform acts on $L^2$ by reversing the sings of Fourier coefficients $c_n$ with $n<0$. To see this, recall that the transform sends harmonic functions to conjugate harmonic functions, which implies that $\sum c_{n\in\mathbb Z} e^{in\theta}$ is sent to $-i\sum_{n\in\mathbb Z} (\operatorname{sign} n) c_n e^{in\theta}$.
So, $f\mapsto \frac12(f-iHf)$ is the operator that kills off all terms with $n>0$, and leaves the terms with $n<0$ untouched. This is the operator you described. Since the Hilbert transform is bounded on $L^p$ for $1<p<\infty$, so is this one.