Find the eigenvalues of the operator $(Ax)(t) = \frac{1}{t^\alpha} \int_{0}^{t} s^{\alpha - 1} x(s) \, ds$ , $Re\alpha>\frac{1}{p}$, in $L^p[0,1]$ space.
I tried $(A-\lambda I)x(t)=\frac{1}{t^\alpha} \int_{0}^{t} s^{\alpha - 1} x(s) \, ds-\lambda x(t)$
$\int_{0}^{t} s^{\alpha - 1} x(s) \, ds=\frac{1}{t^\alpha}(\alpha(t^\alpha x(t)-\int_{0}^{t} s^{\alpha - 1} x'(s) \, ds)$
Where $\int_{0}^{t} s^{\alpha - 1} x'(s) \, ds) = x(t)t^{\alpha-1}-\int_{0}^{t}(\alpha-1) s^{\alpha - 2} x(s) \, ds)$
Please explain how to solve this problem
You just need to multiply the equation $(A-\lambda I)x(t) = 0$ with $t^\alpha$ tp get \begin{align} \int_0^t s^{\alpha-1}x(s) ds = \lambda t^\alpha x(t). \end{align} Now differentiate to get \begin{align} t^{\alpha-1}x(t)= \lambda \alpha t^{\alpha-1} x(t) + \lambda t^{\alpha} x'(t). \end{align} This yields \begin{align} x'(t) = \frac{t^{\alpha-1}x(t) -\lambda \alpha t^{\alpha-1} x(t)}{ \lambda t^{\alpha}} = \bigg(\frac{1}{\lambda} -\alpha \bigg)\frac{1}{t}x(t). \end{align} The solution to this is $x(t) = c t^{\frac{1}{\lambda} -\alpha}\in L^p([0,1])$. Thus, all $\lambda \in \mathbb{C}\setminus \{0\}$ are eigenvalues. It is straightforward to see that $\lambda = 0$ isnt an eigenvalue, because $Ax=0$ implies directly, that $x\equiv 0$ a.e. .