Given the operator $T:L^2([0,1])\to L^2([0,1])$ defined by $$ T(f)(x)=\int_0^x \frac{f(t)}{1+t^2}dt $$ I have to find the spectrum of T, $\sigma(T)$.
Since I have already proved that T is compact, by the spectral theorem for compact operators I have that
- $0\in\sigma(T)$
- $\sigma(T)\setminus\{0\}=EV(T)\setminus\{0\}$, where $EV(T)$ is the set of the eigenvalues of T
So to solve the problem I have to find the eigenvalues of T, i.e. those $\lambda\in\mathbb{C}\setminus\{0\}$ such that the eigenvalues equation $$ Tf=\lambda f $$ admits a non-trivial solution $f\ne 0$. Writing explicitely T the equation becomes $$ \int_0^x \frac{f(t)}{1+t^2}dt=\lambda f(x)\quad\implies\quad f(0)=0 $$ Now since $\frac{f(t)}{\lambda(1+t^2)}\in L^1([0,1])$ and $f(0)=0$ I can write $$ f(x)=\int_0^x \frac{f(t)}{\lambda(1+t^2)}dt=f(0)+\int_0^x \frac{f(t)}{\lambda(1+t^2)}dt $$ and conclude that $f\in AC([0,1])$ where $AC([0,1])$ is the set of absolutely continuous functions from $[0,1]$ into $\mathbb{R}$. If what is written in bold is correct the exercise is solved since by the fact that f is absolutely continuos we have that f is differentiable almost everywhere and $$ f'(x) = \frac{f(x)}{\lambda(1+x^2)}. $$ So the eigenvalues of T are those $\lambda\ne 0$ such that the Cauchy problem $$ f'(x)-\frac{1}{\lambda(1+x^2)}f(x)=0\quad \mathrm{with}\quad f(0)=0 $$ admits a non-trivial solution $f\ne 0$. Then it's very easy to prove that there is no $\lambda\ne 0$ that fullfill that request and so we can conclude that $$ \sigma(T)=\{0\} $$ As I said I am not sure of the argument written in bold which is foundamental to go on, since the absolute continuity of $f$ is necessary to gain differentiability of $f$.
If $f, g \in L^2[0,1]$ then $fg \in L^1$ by Cauchy-Schwarz, and $h(x) = \int_0^x f(t) g(t)\; dt$ is continuous, since as $y \to x$, $h(y) - h(x) = \int_x^y f(t) g(t)\; dt \to 0$ by Dominated Convergence. If $f$ and $g$ are continuous then $h$ is $C^1$, etc.