Let $f(\theta, t)$ be a Holder continuous function for every $t$ on the interval $\theta \in (\alpha,\beta)$. It is known that the application of a singular operator to this function results in another Holder continuous function, e.g. with the Hilbert kernel, $$ P.V. \int_{\alpha}^{\beta} \cot \left(\frac{\theta-\phi}{2}\right) f(\phi, t) d\phi $$ is also Holder continuous when $\theta \in (\alpha, \beta)$.
Now consider the convolution of this integral with an operator $\mu$ defined as $$ J = \int_{-\infty}^{t} \mu(t-\tau) \int_{\alpha}^{\beta} \cot \left(\frac{\theta-\phi}{2}\right) f(\phi, \tau) \; d\phi \; d\tau $$
where $\mu$ is a sufficiently smooth, non-singular kernel. Is $J$ holder continuous?
It seems plausible to me - changing the order of the integration (only one singular integral), it must be possible to write this as
$$ J = P.V. \int_{\alpha}^{\beta} \cot \left(\frac{\theta-\phi}{2}\right) \int_{-\infty}^{t} \mu(t-\tau) f(\phi, \tau) \; d\tau \; d\phi \\ = P.V. \int_{\alpha}^{\beta} \cot \left(\frac{\theta-\phi}{2}\right) F(\phi,\tau) d\phi $$ which is Holder continuous if the inner convolution results in a Holder continuous $F$.
How would one go about proving this? Also, what is the most general class of kernel functions $\mu$ for which J is Holder continuous? (I mean general in the sense of the least restrictions on smoothness etc.)
Thank you.