Operator $T : C[0,1] \rightarrow C[0,1]$ is defined by: $ (Tx)(t) = (1+t^2) x(t) , \forall x \in C [0,1]$. We assume that norm on $C[0,1] $ is standard supremum norm. I have to find $\rho(T), \sigma_{p}(T), \sigma_{c}(T) $and $\sigma_{r}(T) $.
The range of function $ 1+ t^2$, where $t \in [0,1]$ is $[1,2]$, so if $ \lambda \notin [1,2]$, the inverse $ (T - \lambda I ) ^{-1} $ exists and is bounded. It is defined on whole space $C[0,1]$, so the range of $ T - \lambda I $ is dense in $C[0,1] $. This means that the resolvent set $ \mathbb{C} \setminus [1,2] \subset \rho(T) $.
Now, if $\lambda \in [1,2] $, there is unique number $t_{0} $ which gives $ 1+ t_{0} ^2 - \lambda = 0$. If we assume that $ x \in C [0,1]$ is a vector such that $ (T - \lambda I ) x(t)= (1+ t_{0} ^2 - \lambda) x(t)= 0, \forall t \in [0,1] $, it follows from continuity of $x$ that $x = 0$. So, $T - \lambda I$ is injective.
I proved that $\operatorname{Range}(T - \lambda I) $ is not dense in $C[0,1] $ because all functions that are in range satisfy this: $ y (t_{0})= 0$, so $\sigma_{c}(T) = \emptyset $.
I know that $\sigma_{p}$ is the set of eigenvalues, so $\lambda \in \sigma_{p}$ if $(T-\lambda I) ^ {-1}$ doesn't exist. We have already proved that $T - \lambda I$ is injective, but I wonder for which values of $\lambda$ is it non-surjective? (I am also confused if we need surjectivity at all when we consider eigenvalues). If I take arbitrary $ y \in \operatorname{Range}(T - \lambda I )$, then $x(t) = (T - \lambda I ) ^{-1} (y) = \frac{y(t)}{1+ t ^2 - \lambda }, t\neq t_{0} $. What is then $x(t_{0})$ and is $x(t)$ continuous on $[0,1] $ at all?
Here is the solution of the task: $\sigma_{p} (T)=\sigma_{c} (T)= \emptyset$, $\sigma_{r}(T)= [1,2]$. Please help, at least give me some hint (I have recently started studying the spectral theory and this task would help me understand some basics better).
Since $\sigma(T) = \{1+t^2: t\in[0,1]\}=[1,2],$ therefore $\rho(T)= \mathbb{C} \setminus [1,2].$
Suppose $\lambda \in \sigma_p(T),$ then there exists $0\neq x \in C[0,1]$ such that $Tx=\lambda x.$ Since $x \neq 0,$ there exists $t_0 \in [0,1]$ such that $x(t_0)\neq 0.$ Moreover $x$ is continuous, hence there exists a neighborhood $U$ of $t_0$ in $[0,1]$ such that $x(t) \neq 0$ for all $t \in U.$ It follows that $$\lambda = 1+t^2 \; \forall\; t \in U$$which is absurd and hence $\sigma_p(T) = \emptyset.$
You have already shown $\sigma_c(T)=\emptyset$ and so $\sigma_r(T)=\sigma(T)=[1,2].$