Computation the different spectrums of an operator Let $X:=C^0([0,2],\mathbb C)$,$\phi\in X$ and $T\in L(X)$ defined as:
$$(Tf)(t):=\phi(t)f(t),t\in [0,2]$$
Compute: $\sigma_p(T),\sigma_c(T),\sigma_r(T),\sigma(T)$ and $\rho(T)$
I am quite new to this subject and I have no idea how to compute them. But I just need to compute the first 3 values, since $\sigma(T)=\sigma_p(T) \cup \sigma_c(T) \cup \sigma_r(T)$ (disjoint union) and $\rho(T)=\mathbb C$\ $\sigma(T)$
I want to start with the first one:
By definition $\sigma_p(T)=\{\lambda\in \mathbb C: T-\lambda$ is not injective }, so I need to find all $\lambda$ such that :
$(T-\lambda)f(t)=\phi(t)f(t)-\lambda f(t)$ has a non-trivial solution (in terms of $f(t)$)
Isn't that only possible if $\phi(t)$ is constant in $[0,2]$?
I am thankful for any kinds of tips for the first and of course for the other computations, also for useful references, tipps and tricks for the computation of the spectrum. 
 A: We know $\lambda\in\sigma_p(T)$ if and only if there exists $f\neq0$ such that $Tf=\lambda f$. Observe that given such an $f$, the set $\{t\in[0,2]\,:\,f(t)\ne2\}$ is open and non-empty, and so on this set we must have $\phi=\lambda$. Conversely, if there is an open set $U\subseteq[0,2]$ on which $\phi=\lambda$ then we may find a continuous function $f:[0,2]\rightarrow\mathbb{C}$ such that $f=0$ on $[0,2]\setminus U$. Hence we have found
$$\sigma_p(T)=\{\lambda\in\mathbb{C}\,:\,[\phi=\lambda]\ \text{has non-empty interior}\}$$
Next we look at the residual spectrum. Suppose $\lambda\in\mathbb{C}\setminus\sigma_p(T)$ is such that $\lambda-\phi$ has a zero in $[0,2]$. I claim $\lambda\in\sigma_r(T)$. Indeed, if $\lambda-\phi(t_0)=0$, note that for all $f\in X$ we have $(\lambda I-T)f(t_0)=0$, so letting $g(t)=1$ for all $t\in[0,2]$ we have $\|g-(\lambda I-T)f\|_\infty\ge1$ for all $f\in X$. Since $g\in X$, this shows $\lambda I-T$ does not have dense range, so $\lambda\in\sigma_r(T)$. Conversely, if $\lambda-\phi$ is never zero then for all $f\in X$, $g:=\frac{f}{\lambda-\phi}$ is continuous, so $f=(\lambda I-T)g$ and hence $\lambda I-T$ has dense range. This shows
$$\sigma_r(T)=\{\lambda\in\mathbb{C}\,:\,[\phi=\lambda]\ \text{is non-empty with empty interior}\}$$
Finally we look at $\sigma_c(T)$. However note that if $\lambda\in\mathbb{C}\setminus(\sigma_p(T)\cup\sigma_r(T))$ then $\lambda-\phi$ is never zero on $[0,2]$, so as noted above if $f\in X$ then $(\lambda I-T)g=f$ where $g=\frac1{\lambda-\phi}f$. Moreover, $\frac1{\lambda-\phi}(\lambda I-T)f=f$ for any $f\in X$. Hence $\lambda I-T$ is invertible with inverse $(\lambda I-T)^{-1}f=\frac1{\lambda-\phi}f$. We know $(\lambda I-T)^{-1}$ is bounded by the bounded inverse theorem, but you can also show it directly. Since $\lambda-\phi$ is never zero, there exists $\delta>0$ such that $|\lambda-\phi|\ge\delta$ by compactness of $[0,2]$. This implies $\|(\lambda I-T)^{-1}f\|_\infty\le\frac1\delta\|f\|_\infty$ for all $f\in X$. Hence we have shown
$$\sigma_c(T)=\emptyset$$
You can finish off your problem by noting $\sigma(T)=\{\lambda\in\mathbb{C}\,:\,\lambda=\phi(t)\text{ for some }t\in[0,2]\}$ and $\rho(T)=\{\lambda\in\mathbb{C}\,:\,\lambda\neq\phi(t)\text{ for all }t\in[0,2]\}$.
