Let $f \in L^1(\Bbb T)$ be Lipschitz-continuous on an interval $I \subset \Bbb T$. Then, $\widetilde f$ is Hölder-continuous with constant $\alpha$ for all $0 < \alpha < 1$ on $I$.
I'm trying to show the above result. $\widetilde f$ is the conjugate function of $f \in L^1(\Bbb T)$, defined as follows. First, define $\widetilde f_r(t) = f \ast Q_r(t)$ where $Q_r(t) = \widetilde{P_r(t)} = \frac{2r\sin t}{1-2r\cos t + r^2}$. We set $\widetilde f = \lim_{r\to 1^-} \widetilde f_r(t)$. We know that this limit exists a.e. and that $\widetilde f \in L^{1,w}(\Bbb T)$.
The hypothesis is that for some constant $C > 0$, $$|f(x) - f(y)| \le C|x-y|$$ for all $x,y \in I$. We want to show, for all $0 < \alpha < 1$, that $|\widetilde f(x) - \widetilde f(y)| \lesssim_\alpha |x-y|^\alpha$ for all $x,y\in I$.
$$\begin{align*} |\widetilde f(x) - \widetilde f(y)| &= \lim_{r\to 1^-} |(f\ast Q_r)(x) - (f\ast Q_r)(y)|\\ &= \lim_{r\to 1^-} \left|\int_{\Bbb T} f(x-t) Q_r(t)\, dt - \int_{\Bbb T} f(y-t) Q_r(t)\, dt \right|\\ &= \lim_{r\to 1^-} \left|\int_{\Bbb T} (f(x-t) - f(y-t)) Q_r(t)\, dt \right|\\ &\le \lim_{r\to 1^-} \int_{\Bbb T} |f(x-t) - f(y-t)| |Q_r(t)|\, dt\\ &\le C |x-y| \lim_{r\to 1^-} \int_{\Bbb T} \left|\frac{2r\sin t}{1-2r\cos t + r^2} \right|\, dt\\ &\approx |x-y|\lim_{r\to1^-} (\log(r+1) - \log(r-1)) = \infty \end{align*}$$
It seems we need more careful estimation since directly using the Lipschitzness of $f$ in the above inequalities does not produce the required bound. In view of this result, though, would it be enough to show that the integral above is finite and $\widetilde f$ is Lipschitz-continuous?
Thanks for any help!