Continuous spectrum of an integral operator

Question

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.

Answer 1

$$ Tf = xf(x)+\int_0^xf(u)du, \;\; f\in C[0,1]. $$ 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 $\sigma(T)=[0,1]$.

Answer 2

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 $$ \sigma(T)=\sigma_r(T)=[0,1]. $$

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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.