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.
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 $\rho(T)=\{\lambda\in\mathbb{C}\,:\,\lambda\neq\phi(t)\text{ for all }t\in[0,2]\}$.