Resolvent set and spectrum of operator $T : C[-1,3] \rightarrow C[-1,3]$, where $Tx(t) = tx(t)$

Question

I would like a hint (maybe some steps) for the following exercise:

Consider the operator $T : C[-1,3] \rightarrow C[-1,3]$, $Tx(t) = tx(t)$, where $C$ is equipped with the supremum norm. Determine the sets $\rho(T)$, $\sigma_p(T)$ (point spectrum), $\sigma_c(T)$ (continous spectrum), and $\sigma_r(T)$ (residual spectrum).

Note: it is not homework, just studying for an exam and I feel like this exercise would help me understand the concepts on a concrete problem.

Answer 1

Here there are some steps, which are hints if you don't read the proof.

Claim 1: $\mathbb{C}\setminus[-1,3]\subset\rho(T)$.

Proof:

Let $\lambda\in\mathbb{C}\setminus[-1,3]$. Note that \begin{align} (\lambda-T)x=0\quad&\Longrightarrow\quad\lambda x(t)-tx(t)=0,\quad\forall\ t\in[-1,3]\\ &\Longrightarrow\quad x(t)=0, \quad\forall \ t\in[-1,3]\\ &\Longrightarrow\quad x=0 \end{align} This shows that $\lambda-T:C[-1,3]\to C[-1,3]$ is injective. Now, given $y\in C[-1,3]$ define $x:[-1,3]\to\mathbb{C}$ by \begin{align}x(t)=\frac{y(t)}{\lambda-t}.\end{align} Then, $x\in C[-1,3]$ and \begin{align}\lambda x(t)-tx(t)=y(t),\quad\forall\ t\in [-1,3].\end{align} In other words, $(\lambda-T)x=y$. This shows that $\lambda-T:C[-1,3]\to C[-1,3]$ is surjective. So, $(\lambda-T)^{-1}:C[-1,3]\to C[-1,3]$ exists. Also, \begin{align}\|(\lambda-T)^{-1}y\|_C=\sup_{t\in[-1,3]}\|\tfrac{y(t)}{\lambda-t}\|\leq\frac{1}{d}\|y\|_C,\quad\forall \ y\in C[-1,3]\end{align} where $d=\text{dist}(\lambda,[-1,3])$. This shows that $(\lambda-T)^{-1}$ is bounded. So, $\lambda\in \rho(T)$.

Claim 2: $[-1,3]\subset\sigma(T)$.

Proof:

Take $\lambda\in [-1,3]$. If $y\in \text{Im}(\lambda-T)$ then there exists $x\in C[-1,3]$ such that $(\lambda-T)x=y$ and thus \begin{align}y(\lambda)=\lim_{t\to\lambda}y(t)=\lambda x(\lambda)-\lambda x(\lambda)=0.\end{align} This shows that $y(\lambda)=0$ for all $y\in\text{Im}(\lambda-T)$ which implies that $\text{Im}(\lambda-T)$ is not dense in $C[-1,3]$ (because non-zero constant functions belongs to $C[-1,3]$). So, $\lambda\in\sigma(T)$ (otherwise $\text{Im}(\lambda-T)=D((\lambda-T)^{-1})$ would be dense in $C[-1,3]$).

Claim 3: $\rho(T)=\mathbb{C}\setminus[-1,3]$ and $\sigma(T)=[-1,3]$.

Proof:

Follows from claims 1 and 2.

Claim 4: $\sigma_p(T)=\varnothing$.

Proof:

If there exists $\lambda\in\sigma_p(T)$, then there exists a non-zero function $x\in C[-1,3]$ such that \begin{align} (\lambda -t)x(t)=0,\quad\forall \ t\in[-1,3].\tag{1}\end{align} As $x$ is continuous and non-zero, there are $t_0\in[-1,3]$ and $\delta>0$ such that \begin{align} x(t)\neq 0,\quad\forall\ t\in[t_0-\delta,t_0+\delta].\tag{2}\end{align} From $(1)$ and $(2)$, \begin{align}t=\lambda,\quad\forall \ t\in [t_0-\delta,t_0+\delta]\end{align} which is a contradiction. So, there is no element in $\sigma_p(T)$.

Claim 5: $\sigma_c(T)=\varnothing$.

Proof:

Assume that $(\lambda-T)^{-1}$ exists. As in the proof of Claim 2, $D((\lambda-T)^{-1})=\text{Im}(\lambda-T)$ is not dense in $C[-1,3]$.

Claim 6: $\sigma_r(T)=[-1,3]$.

Proof:

Follows from claims 4 and 5.