TITLE: Spectrum of Indefinite Integral Operators
QUESTION [29 upvotes]: I've considered the following spectral problems for a  long time, I did not kow how to tackle them. Maybe they needs some skills with inequalities.
For the first, suppose $T:L^{2}[0,1]\rightarrow L^{2}[0,1]$ is defined by
$$Tf(x)=\int_{0}^{x} \! f(t) \, dt$$
How can I calculate:

the radius of the spectrum of $T$?
$T^{*}T$?
the norm of $T$?

I guess $r(T)$ should be $0$. but I did know how to prove it. My idea is use Fourier to tackle it, however it does not seem to work.
The other problem may be very similar to this one. Let $T:C[0,1]\rightarrow C[0,1]$ be  defined by
$$Tf(x)=\int_{0}^{1-x}f(t)dt$$
It is obvious that $T$ is compact and I guess its radius of spectrum is zero, but I do not know how to prove it.
Any references and advice will be much appreciated.

REPLY [3 votes]: thanks PZZ's answer for the first problem. Now I know why I did realize why I am failed in the second problem.
$$Tf(x)=\int_{0}^{1-x}f(t)dt$$
It is trival that this operator is is linaer compact operator according to Arzela-Ascoli theorem, so we get $\sigma(T)/{0} \subset \sigma_{p}(T)$, use simple computation we can get that $0\notin \sigma_{p}(T)$, for C[0,1] is a infinite dimensional space, we can get $0\in \sigma(T)$. $\forall \lambda \in \sigma_{p}(T)$, and $\lambda \neq 0$, we can get that 
$$\int_{0}^{1-x}f(t)dt=\lambda f(x)$$
which implies that $f \in C^{\infty}[0,1]$, then we can get
$$-f(1-x)=\lambda f'(x)$$
notice that 
$$f(1)=0$$
further we get 
$$\lambda^{2}f''(x)=-f(x)$$, solve this second order of differential equation, we can calulated every $\lambda_{n}=\frac{2\pi}{2n+1}, \qquad n\in Z$.