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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.

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    $\begingroup$ You are almost correct, $\phi(t)$ has to be $\lambda$ when $f(t) \neq 0$. $\endgroup$ – user99914 Dec 3 '14 at 13:16
  • $\begingroup$ Okay, thanks. And how about the other spectrums? $\endgroup$ – Epsilondelta Dec 3 '14 at 14:09
  • $\begingroup$ If $\phi-\lambda$ is non-zero on $[0,2]$, then $\lambda\in\rho(T)$. If $\phi-\lambda$ is zero at some point of $[0,2]$, then the range of $T-\lambda I$ is not dense, which means $\lambda\in\sigma_{p}(T)$ or $\lambda\in\sigma_{r}(T)$; the first case occurs iff the zero set of $\phi-\lambda$ is such that a non-zero $f \in C[0,2]$ vanishes on the zero set of $\phi-\lambda$. $\endgroup$ – DisintegratingByParts Dec 3 '14 at 19:10
  • $\begingroup$ Thanks for the answer! Two questions: why the range of $I-\lambda$ is not dense, if $\phi-\lambda$ is zero at some point? And the other question: Should not you say at the end, that f has to vanish in the complement of the zero set of $\phi-\lambda$ ? $\endgroup$ – Epsilondelta Dec 4 '14 at 8:24
  • $\begingroup$ Minor note: the plural of spectrum is spectra. $\endgroup$ – Mariano Suárez-Álvarez Feb 9 '15 at 0:24
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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 $\rho(T)=\{\lambda\in\mathbb{C}\,:\,\lambda\neq\phi(t)\text{ for all }t\in[0,2]\}$.

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  • $\begingroup$ Wow. I understood most of the things. I will read it some more times, but one question: In the second part (continuous spectrum) you showed that $(\lambda I-T) $ does not have dense range. But the continuous spectrum is defined as the set of all $\lambda \in \mathbb C$ such that $T-\lambda$ is injective and $(T-\lambda)$ has dense range. So where is the mistake? $\endgroup$ – Epsilondelta Feb 8 '15 at 17:06
  • $\begingroup$ Right, I mixed up continuous and residual spectrum, haha. I will edit, thanks. $\endgroup$ – Jason Feb 9 '15 at 0:20

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