Eigenvectors and eigenvalues in functional space

Question

Vector space $C_0$ consists of every function that is continuous in [0,1]
Tensor $A$ on the space $C$ is defined as $Af=\frac{1}{t}\int_0^t$$f(\varphi)d\varphi$ where $f(t)$ is a vector on space $C_0$.
When eigenvalue of the tensor $A$ is any real number between $\lambda \in(0,1]$,
how can show that the corresponding eigenvector $z(t)$ is
$z(t)=t^{\frac{1}{\lambda}-1}$
[$Af=\lambda f$ so $f=\lambda\frac{d}{dt}(tf)$] Any help would be much appreciated!!

Answer 1

Consider $f=\lambda\frac{d}{dt}(tf)$ here we have:

$f=\lambda(f+tf')\Rightarrow f(1-\lambda)-t\lambda f'=0$

This is a cauchy-Euler equation, so we try $f=t^n$ for some $n$ to be found.

Here $f'=n\lambda t^{n-1}$, so we have:

$(1-\lambda-n\lambda)z^n=0\Rightarrow n=1/\lambda-1$

Thus $f=t^{1/\lambda-1}$.

Answer 2

$$A\left(t^{\frac1\lambda-1}\right)=Az(t):=\frac1t\int\limits_0^t x^{\frac1\lambda-1}dx=\left.\frac1t\lambda x^\frac1\lambda\right|_0^t=\frac\lambda tt^{\frac1\lambda}=\lambda t^{\frac1\lambda -1}=\lambda z(t)\;\ldots$$