Let $p \in [1, \infty], \alpha > 0, T > 0$ and define
$$I^{\alpha} f(t) - \frac{1}{\Gamma (\alpha)} \int_0^t(t-\tau)^{\alpha - 1}f(\tau)d\tau,$$
where $\Gamma$ is Gamma function. Show that $I^{\alpha} \ in B(L^p(0,T)).$
My idea was:
$|I^{\alpha} f(t)| = |\frac{1}{\Gamma (\alpha)}||\int_0^t(t-\tau)^{\alpha - 1}f(\tau)d\tau| \leq |\frac{1}{\Gamma (\alpha)}|\int_0^t|(t-\tau)^{\alpha - 1}||f(\tau)|d\tau \leq |\frac{1}{\Gamma (\alpha)}|\int_0^T|(t-\tau)^{\alpha - 1}||f(\tau)|d\tau \leq |\frac{1}{\Gamma (\alpha)}|(\int_0^T|(t-\tau)^{\alpha - 1}|^qd\tau)^{\frac{1}{q}}(\int_0^T|f(\tau)|^pd\tau)^{\frac{1}{p}}$,
where last inequality is Holders inequality. We get norm of $f$ multiplied by something, however I don't have an idea how to show that $|\frac{1}{\Gamma (\alpha)}|(\int_0^T|(t-\tau)^{\alpha - 1}|^qd\tau)^{\frac{1}{q}}$ is a finite value. Any clues for that or tips for other estimation would be really welcome.