Let $X=C([0,2], ℂ)$ with the sup norm, and let
\[
g(t)= \begin{cases}t & \text { if } 0 ⩽ t ⩽ 1 \\ 1 & \text { if } 1Solution. $‖Tf‖=\sup|g||f|≤\sup|g|\sup|f|=\sup|f|=‖f‖⇒‖T‖≤1$.
$‖T1‖=1=‖1‖⇒‖T‖=1$.
[$⇒r_σ(T)≤‖T‖=1.$]
Now $σ_p(T)=?$
$Tf=λf$ i.e. $(g-λ)f=0$
i.e. $(g(t)-λ)⋅f(t)=0∀t$
If there is no $t$ such that $g(t)=λ$, then $f≡0$ and $λ$ isn't an eigenvalue.
If $∃t_0$ such that $g(t_0)=λ⇒λ∈[0,1]$
Case 1. $λ<1$, $t_0$ is unique
$⇒f(t)=0∀t≠t_0$
$f$ is continuous$⇒f≡0$
$λ∉σ_p(T)$.
Case 2. $λ=1$, $g(t)=1$ if $t≥1$; $g(t)≠1$ if $t<1$.
$⇒Tf=λf$ iff $f=0$ in $[0,1]$
e.g. $f(x)=(x-1)χ_[1,2]$.
∴$σ_p(T)=\{1\}$.
Now $σ_{ap}(T)=?$
$λ∈σ_{ap}(T)$ if $∃‖f_n‖=1,(λI-T)f_n→0$
i.e. $\sup|f_n|=1,\sup|λ-g||f_n|→0$
If $λ∉$Range of $g=[0,1]$.
$\min|λ-g|>a>0$
$a\sup|f_n|≤\sup|λ-g||f_n|→0$, contradicting $‖f_n‖=1$.
$⇒λ∉σ_{ap}(T)$.
If $0≤λ<1$
Pick an interval $J_n$ of size $\frac1n$ close to $t_0$
$‖f_n‖≤1$ in $t_0$
$f_n=0$ outside $J_n$, $f_n(t_0)=1$
$⇒λ∈σ_{ap}(T)$
$σ_{ap}(T)=[0,1]$
$σ(T)⊆\overline{D(0,1)}$
$σ(T)$ closed
$σ(T)⊇σ_{ap}(T)=[0,1]$
Claim. if $λ∉[0,1]$, then $λ∈ρ(T)$.
$(λI-T)f=h$
$(λ-g)f=h$
$f=\frac{h}{λ-g}$ continuous as $λ-g$ is continuous and nowhere 0
$⇒σ(T)=[0,1]$.
$σ_r(T)=[0,1),σ_c(T)=∅$.