Suppose $f$ is a continuous function on $[0,\infty)$. For $\alpha > 0$, define $$I_\alpha[f](x)=\frac{1}{\Gamma(\alpha)}\int_0^x (x-t)^{\alpha-1} f(t)dt$$. Show that the derivate of $I_{\alpha+1}[f]$ is $I_\alpha[f]$ and $I_\alpha[I_\beta[f]]$ = $I_{\alpha+\beta}[f]$ for all $\alpha, \beta > 0$.
I just have no idea where to start on this problem.
On
Another way is to note that your operator $I_{\alpha}$ is the convolution on $[0,\infty)$ of $t^{-1+\alpha}/\Gamma(\alpha)$ and $f$. Applying the Laplace transform $\mathscr{L}$ gives $$ \mathscr{L}\{I_{\alpha}f\}=s^{\alpha}\mathscr{L}\{f\}, $$ $I_{\alpha}$ is a fractional derivative operator. In particular, $$ I_{\beta}I_{\alpha}f = I_{\alpha+\beta}f. $$ This also explains why $\frac{d}{dx}I_{\alpha}f = I_{\alpha+1}f$.
\begin{align} I_\alpha[I_\beta[f]](x)&=\frac{1}{\Gamma(\alpha)}\int_0^x(x-y)^{\alpha-1}I_\beta[f](y)\,dy\\&=\frac{1}{\Gamma(\alpha)\Gamma(\beta)}\int_0^x\color{blue}{\int_0^y}(x-y)^{\alpha-1}\color{blue}{(y-z)^{\beta-1}f(z)\,dz}\,dy\\\color{gray}{[\text{switch the integrations}]}&=\frac{1}{\Gamma(\alpha)\Gamma(\beta)}\int_0^x\color{blue}{\int_z^x(x-y)^{\alpha-1}(y-z)^{\beta-1}}f(z)\,\color{blue}{dy}\,dz\\\color{gray}{[\text{substitute }y=z+(x-z)t]}&=\frac{1}{\Gamma(\alpha)\Gamma(\beta)}\int_0^x\color{blue}{\int_0^1}(x-z)^{\alpha+\beta-1}\color{blue}{(1-t)^{\alpha-1}t^{\beta-1}}f(z)\,\color{blue}{dt}\,dz\\\color{gray}{[\text{apply the beta integral}]}&=\frac{1}{\Gamma(\alpha+\beta)}\int_0^x(x-z)^{\alpha+\beta-1}f(z)\,dz=I_{\alpha+\beta}[f](x). \end{align}