Continuous spectrum of an integral operator I understand that the continuous spectrum of an operator are the $\lambda's$ such that $(\lambda-T)$ is injective but ran$(\lambda-T)$ isn't dense in the image. But i can't properly calculate it for a given example. For example consider the operator $T:C[0,1]\to C[0,1]$ such that $$Tf(x)=xf(x)+\int_0^xf(\xi)\:d\xi$$
I was able to determine the point spectrum of this operator, but what shall one do in order to find its continuous spectrum?
Thank you so much in advance.
 A: $$
   Tf = xf(x)+\int_0^xf(u)du, \;\; f\in C[0,1]. 
$$
The resolvent operator is $(T-\lambda I)^{-1}$, assuming the operator inverse exists and is bounded on $C[0,1]$. $g=(T-\lambda I)^{-1}f$ is a function such that
$$
           (x-\lambda)g(x)+\int_0^xg(u)du = f(x) \\
       (x-\lambda)\frac{d}{dx}\int_0^xg(u)du+\int_0^xg(u)du=f(x) \\
     \frac{d}{dx}\left[(x-\lambda)\int_0^xg(u)du\right]=f(x) \\
     (x-\lambda)\int_0^xg(u)du=\int_0^xf(u)du \\
      \int_0^xg(u)du = \frac{1}{x-\lambda}\int_0^xf(u)du \\
        g(x)=-\frac{1}{(x-\lambda)^2}\int_0^xf(u)du+\frac{1}{x-\lambda}f(x)
$$
Therefore,
$$
    (T-\lambda I)^{-1}f=-\frac{1}{(x-\lambda)^2}\int_0^xf(u)du+\frac{1}{x-\lambda}f(x)
$$
This is well-defined for a given $\lambda$ if, for all $f\in C[0,1]$, the right side of the above is in $C[0,1]$. So $\sigma(T)=[0,1]$.
A: The continuous spectrum is empty.
Fix $\lambda\in\sigma(T)$. From DisintegratingByParts' answer we know that $\lambda\in[0,1]$. We also know that
$$\tag1
(T-\lambda I)f(x)=\frac d{dx}\Big[(x-\lambda)\int_0^xf\Big].
$$
So if $g\in\operatorname{ran}(T-\lambda I)$ then
$$
g(x)=\frac d{dx}\Big[(x-\lambda)\int_0^xf\Big].
$$
We can write this as
$$
\int_0^x g=(x-\lambda)\,\int_0^x f.
$$
In particular, this forces
$$\tag2
\int_0^\lambda g=0.$$
Considering the real part,
$$\tag3
\int_0^\lambda \operatorname{Re}g=0.
$$
This implies that there exists $s\in[0,1]$ with $\operatorname{Re}g(s)=0$. Then
$$
\|1-g\|_\infty\geq|1-g(s)|\geq|1-\operatorname{Re}g(s)|=|1-0|=1.
$$
So $\operatorname{ran}(T-\lambda I)$ is not dense, which implies that $\lambda\not\in\sigma_c(T)$. As $\sigma_p(T)=\emptyset$, this shows that
$$
\sigma(T)=\sigma_r(T)=[0,1]. 
$$

Edit: $T-\lambda I$ is injective.
Let $\lambda\in[0,1]$. Suppose that $(T-\lambda I)f=0$. By $(1)$ above this means that there exists a constant $c$ such that
$$
(x-\lambda)\int_0^xf=c.
$$
Taking $x=\lambda$ we see that $c=0$. So, for any $x\ne\lambda$, we have that $\int_0^xf=0$. Taking the derivative we get that $f(x)=0$ with the possible exception of $x=\lambda$, but we get $f(\lambda)=0$ by continuity. So $T-\lambda I$ is injective.
