How to Find the Spectrum of an Integral Operator I need to find the spectrum of an operator $T: C([0,1]) \to C([0,1])$ defined by $(Tf)(t) = \int_0^t f(x) dx$. I know that the spectrum is the set of all values $\lambda$ such that $\lambda I - T$ is not invertible, but I'm not sure how to go about finding those values when integrals are involved. I've found several other posts about finding the spectrum of an integral operator here, but they all seem to involve functions of two variables, and that's throwing me off.
Any advice would be appreciated.
 A: The resolvent $(T-\lambda I)^{-1}$ is obtained by solving for $g$ in the following:
$$
                      g = (T-\lambda I)^{-1}f \\
                  (T-\lambda I)g= f \\
                \int_0^x g(t)dt-\lambda g(x)=f(x) \\
             \int_0^xg(t)dt-\lambda\frac{d}{dx}\int_0^x g(t)dt=f(x) \\
        \frac{d}{dx}\int_0^xg(t)dt-\frac{1}{\lambda}\int_0^x g(t)dt=-\frac{1}{\lambda}f(x) \\
       \frac{d}{dx}\left(e^{-x/\lambda}\int_0^xg(t)dt\right)=-\frac{1}{\lambda}e^{-x/\lambda}f(x) \\
       e^{-x/\lambda}\int_0^x g(t)dt = -\frac{1}{\lambda}\int_0^x e^{-u/\lambda} f(u)du \\
       \int_0^xg(t)dt=-\frac{1}{\lambda}\int_0^x e^{(x-u)/\lambda}f(u)du \\
        g(x)=-\frac{1}{\lambda}f(x)-\frac{1}{\lambda^2}\int_0^x  e^{(x-u)/\lambda}f(u)du
$$
So,
$$
   (T-\lambda I)^{-1}f = -\frac{1}{\lambda}f(x)-\frac{1}{\lambda^2}\int_0^x  e^{(x-u)/\lambda}f(u)du
$$
and the spectrum is $\sigma(T)=\{0\}$.
A: Another approach is to use the spectral radius formula $r(T) = \lim_{n\to\infty} \|T^n\|^{1/n}$.  By induction, you show that
$$ T^n f(t) = \int_0^t \frac{(t-x)^{n-1}}{(n-1)!} f(x) \, dx .$$
So $\|T^n\| = 1/n!$, and $\|T^n\|^{1/n} \to 0$.
A: The spectrum is $\{0\}$. To prove let $c \neq 0$ and this consider the equation $Tf-\frac  1 c f=g$ where $g \in C[0,1]$. If $h=Tf$ this becomes  $h-\frac  1c h'=g$ or $h'-ch=-cg$. Hence $(e^{-cx}h(x))'=-ce^{-cx} g(x)$. Integrating we get $h(x)=e^{cx} [C-\int_0^{x}e^{-ct} g(t) dt]$ for some constant $C$. Since $h(0)=0$ we get $C=0$. Next $f =h'=(-e^{cx} \int_0^{x}e^{-ct} g(t) dt)'$.  Now it is quite easy to see that $\|f\| \leq M \|g\|$ for a constant $M$ independent of $g$.   We have proved that for every $g$ there is a unique solution $f$ for $Tf-\frac  1 c f=g$ and that that map $g \to f$ is continuous. This means that $T-\frac  1 c I$ has  a bounded inverse whenever $c \neq 0$. 
