Define the operator $T$ on $L^2[0,1]$ by $(Tf)(x) = \int^x_0 f(y) dy + x f(x)$. What is the residual spectrum of $T$ and is the residual spectrum the whole spectrum?
My attempt: Solving the eigenvalue problem: $\int^x_0 f(y) dy + x f(x) = \lambda f(x)$ by first differentiate to get $2 f(x) = f'(x) (\lambda - x)$. Away from $\lambda$ then $f(x) = \frac{C}{|\lambda - x|^2}$. Since $\lambda f(0) = 0$, if $\lambda \neq 0$ then $f(0) = 0$. But this gives $f(x) \equiv 0$, so not possible. On the other hand, if $\lambda = 0$, then $f(x) = \frac{C}{x^2}$ for $x \in (0,1]$. However, this is also not possible since $\int^x_0 f(y)dy$ is undefined. Thus the point spectrum of $T$ is empty.
$T$ is obviously bounded, so is $(T - \lambda)$. By the Bounded Inverse Theorem, $(T-\lambda)^{-1}$ is bounded, so the continuous spectrum is empty.
For the residual spectrum, first find the point spectrum of the adjoint and then apply the relation $\sigma_{res}(T) \subset \overline{\sigma_{pt}(T^*)}$, where the bar denotes complex conjugate. Exchanging the order of integration, the adjoint $T^*$ is defined by $(T^*f)(x) = \int^1_x f(y)dy + x f(x)$. Similar to the previous part, $\sigma_{pt}(T^*) = \{1\}$ since $1$ is the only possible eigenvalue. Since all linear bounded operator on complex Hilbert space has non-empty spectrum, $\sigma_{res}(T)$ must equal $\overline{\sigma_{pt}(T^*)} = \{1\}$.
Questions:
1/ Could someone please help me check this? I am very unsure of my work since I am quite new to operator theory.
2/ Can someone please give me a different example that has the whole spectrum being the residual spectrum aside from the left-shift operator?
3/ How should one generally approach these problems?
4/ Is it true that if $T$ is defined on $L^2(0,1)$ instead, then the residual spectrum is the entire spectrum?
Thank you so much.