Is the Fractional integral operator well-defined?

Question

How to prove the fractional integral operator $J_{\alpha}:L^p(\Bbb R^+)\rightarrow L^p(\Bbb R^+)$ (of order $\alpha>0$) which is defined for each $f\in L^p(\Bbb R^+)$ by

$$J_{\alpha}f(x):={1\over \Gamma(\alpha)}\int_0^x(x-y)^{\alpha-1}f(y)\,dy~~~~~x\in\Bbb R^+$$ is well-defined?

Answer 1

$$ \begin{align} & \int_{a}^{b}\left|\int_{a}^{x}(x-y)^{\alpha-1}f(y)\,dy\right|^{2}dx \\ & \le \int_{a}^{b}\left[\int_{a}^{x}(x-y)^{\alpha/2-1/2}\{(x-y)^{\alpha/2-1/2}|f(y)|\}\,dy\right]^{2}dx \\ & \le \int_{a}^{b}\left[\int_{a}^{x}(x-y)^{\alpha-1}dy\int_{a}^{x}(x-y)^{\alpha-1}|f(y)|^{2}\,dy\right]dx \\ & = \frac{1}{\alpha}\int_{a}^{b}(x-a)^{\alpha}\int_{a}^{x}(x-y)^{\alpha-1}|f(y)|^{2}\,dy\,dx \\ & \le \frac{1}{\alpha}(b-a)^{\alpha}\int_{a}^{b}\int_{a}^{x}(x-y)^{\alpha-1}|f(y)|^{2}\,dy\,dx \\ & = \frac{1}{\alpha}(b-a)^{\alpha}\int_{a}^{b}\int_{y}^{b}(x-y)^{\alpha-1}|f(y)|^{2}dx\,dy \\ & = \frac{1}{\alpha^{2}}(b-a)^{\alpha}\int_{a}^{b}(b-y)^{\alpha}|f(y)|^{2}dy \\ & \le \frac{1}{\alpha^{2}}(b-a)^{2\alpha}\int_{a}^{b}|f(y)|^{2}dy \end{align} $$

Answer 2

So, for example, you would question $f(t) = (1-t)^{-49/100}$ which is in $L^2[0,1]$, but $$ J_{1/4}f(1) = +\infty $$ So you have to show the integral exists for almost all $x$, or in some weak sense, and the resulting function is in $L^2$. My guess is that you should consult a text on "singular integrals".