Sheet4 A1

 
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 $⇒σ(T)=[0,1]$. $σ_r(T)=[0,1),σ_c(T)=∅$.