Prove boundedness of operator on $L^p$

Question

Let $\alpha \in (0,1)$, $p\in(1,\infty)$ and $T >0$

The operator is defined as $I^\alpha f(t) = \frac{1}{\Gamma(\alpha)}\int_{0}^{t} (t-r)^{\alpha-1}f(r)dr$. I want to prove that $I^\alpha \in B(L^p(0,T))$.

Let $H = L^p(0,T)$.

$||I^\alpha f||^p_{H}= \frac{1}{\Gamma^p(\alpha)} \int_{0}^{T} |\int_{0}^{t} s^{\alpha-1}f(t-s)ds|^p dt$.

Well, the $p$-th power is the problem and the inner divergent integral as $\alpha-1 <0$. I don't know how to proceed and estimate it properly.

Answer 1

I managed to solve this.

Let $H=L^p(0,T)$.

Note that $||I^\alpha f||^p_{H} = ||g*f||^p_{H}$, where $g(t)=t^{\alpha-1}$.

Now it's sufficient to use Young's inequality for convolutions:

For $p\in(1,\infty)$, $g\in L^1(0,T)$ and $f\in L^p(0,T)$ the inequality below holds:

$||g*f||_{H} \le ||g||_{L^1(0,T)} ||f||_H$.