How to find the spectral radius of a compact operator $T:C[0,1]\to C[0,1]$ given by $Tf(x)= \int_{0}^{1-x} f(y)\,dy$ I proved yet that $||T||\leq 1$ and the eigen values are $\frac{1}{\frac{\pi}{2}+2n\pi}$.
hence my operator is compact i know that $\sigma(T)\subset [-1,1]$
The other side, i know that spectral radius is given by $r(T)=\max \{{|\lambda}|,\lambda\in\sigma(T)\}$ or equal to $\lim ||T^n||^{\frac{1}{n}}$ but i don't see how is the face of $T^n$, and for me $\max \{ \frac{1}{\frac{\pi}{2}+2n\pi}\}$ is $\frac{2\pi}{5}$ because $n\in\mathbb N$ right? but should be $0$ because $n\in \mathbb Z$ so $\min \frac{\pi}{2}+2n\pi $ goes to $-\infty$ right?
Perhaps somebody can help me please, or give me a hint.
Thank you
Hint: Non-zero points in the spectrum of a compact operator are all eigen values. Suppose $Tf=\lambda f$ and $\lambda \neq 0$. Then $\int_0^{1-x} f(y)dy =\lambda f$. Use induction to show that $f$ is a smooth function. Differentiate to get $-f(1-x)=\lambda f'(x)$ and differentiate again to get $f'(1-x)=\lambda f''(x)$. This gives $\lambda f'(1-x)=\lambda^{2} f''(x)$. Hence, $\lambda^{2} f''(x)=-f(x)$. Solve this to get all eigen values and eigen vectors.